Math Class 8.pdf

MATHEMATICS
Textbook for Class VIII
2021–22

First Edition
January 2008 Magha 1929
Reprinted
January 2009 Pausa 1930
November 2010 Kartika 1932
December 2015 Agrahayna 1937
December 2016 Pausa 1938
December 2017 Pausa 1939
August 2019 Bhadrapada 1941
PD 1107T RSP
© National Council of Educational
Research and Training, 2008
` 65.00
ALL RIGHTS RESERVED
q No part of this publication may be reproduced, stored in a retrieval system
or transmitted, in any form or by any means, electronic, mechanical,
photocopying, recording or otherwise without the prior permission of the
publisher.
q This book is sold subject to the condition that it shall not, by way of trade, be
lent, re-sold, hired out or otherwise disposed of without the publisher’s
consent, in any form of binding or cover other than that in which it is published.
q The correct price of this publication is the price printed on this page, Any
revised price indicated by a rubber stamp or by a sticker or by any other
means is incorrect and should be unacceptable.
OFFICES OF THE PUBLICATION
DIVISION, NCERT
NCERT Campus
Sri Aurobindo Marg
New Delhi 110 016 Phone : 011-26562708
108, 100 Feet Road
Hosdakere Halli Extension
Banashankari III Stage
Bengaluru 560 085 Phone : 080-26725740
Navjivan Trust Building
P.O. Navjivan
Ahmedabad 380 014 Phone : 079-27541446
CWC Campus
Opp. Dhankal Bus Stop
Panihati
Kolkata 700 114 Phone : 033-25530454
CWC Complex
Maligaon
Guwahati 781 021 Phone : 0361-2674869
Publication Team
Head, Publication : Anup Kumar Rajput
Division
Chief Editor : Shveta Uppal
Chief Production : Arun Chitkara
Officer
Chief Business Manager : Vipin Dewan
(In charge)
Editor : Bijnan Sutar
Production Assistant : Om Prakash
Cover Layout Design
Shweta Rao Digital Expressions
Illustrations
Prashant Soni
Printed on 80 GSM paper with NCERT
watermark
Published at the Publication Division by the
Secretary, National Council of Educational
Research and Training, Sri Aurobindo Marg,
New Delhi 110 016 and printed at Palak
Printers 6, Mohkampur Industrial Area,
Phase-II, Meerut
ISBN 978-81-7450-814-0
0852 – MATHEMATICS
Textbook for Class VIII
2021–22

Foreword
TheNationalCurriculumFramework,2005,recommendsthatchildren’slifeatschoolmustbelinked
to their life outside the school. This principle marks a departure from the legacy of bookish learning
whichcontinuestoshapeoursystemandcausesagapbetweentheschool,homeandcommunity.The
syllabiandtextbooksdevelopedonthebasisofNCFsignifyanattempttoimplementthisbasicidea.
They also attempt to discourage rote learning and the maintenance of sharp boundaries between
different subject areas.We hope these measures will take us significantly further in the direction of a
child-centredsystemofeducationoutlinedintheNationalPolicyonEducation(1986).
The success of this effort depends on the steps that school principals and teachers will take to
encouragechildrentoreflectontheirownlearningandtopursueimaginativeactivitiesandquestions.We
mustrecognisethat,givenspace,timeandfreedom,childrengeneratenewknowledgebyengagingwith
theinformationpassedontothembyadults.Treatingtheprescribedtextbookasthesolebasisofexamination
isoneofthekeyreasonswhyotherresourcesandsitesoflearningareignored.Inculcatingcreativityand
initiativeispossibleifweperceiveandtreatchildrenasparticipantsinlearning,notasreceiversofafixed
bodyofknowledge.
Theseaimsimplyconsiderablechangeinschoolroutinesandmodeoffunctioning.Flexibilityin
thedailytime-tableisasnecessaryasrigourinimplementingtheannualcalendarsothattherequired
numberofteachingdaysareactuallydevotedtoteaching.Themethodsusedforteachingandevaluation
will also determine how effective this textbook proves for making children’s life at school a happy
experience, rather than a source of stress or boredom. Syllabus designers have tried to address the
problemofcurricularburdenbyrestructuringandreorientingknowledgeatdifferentstageswithgreater
considerationforchildpsychologyandthetimeavailableforteaching.Thetextbookattemptstoenhance
thisendeavourbygivinghigherpriorityandspacetoopportunitiesforcontemplationandwondering,
discussioninsmallgroups,andactivitiesrequiringhands-onexperience.
NCERTappreciates the hard work done by the textbook development committee responsible
for this book. We wish to thank the Chairperson of the advisory group in science and mathematics,
Professor J.V. Narlikar and the ChiefAdvisor for this book, Dr H.K. Dewan for guiding the work of
this committee. Several teachers contributed to the development of this textbook; we are grateful to
their principals for making this possible.We are indebted to the institutions and organisations which
havegenerouslypermittedustodrawupontheirresources,materialandpersonnel.Asanorganisation
committed to systemic reform and continuous improvement in the quality of its products, NCERT
welcomescommentsandsuggestionswhichwillenableustoundertakefurtherrevisionandrefinement.
Director
NationalCouncilofEducational
ResearchandTraining
NewDelhi
30 November 2007
2021–22

Preface
Thisisthefinalbookoftheupperprimaryseries.Ithasbeenaninterestingjourneytodefinemathematics
learninginadifferentway.Theattempthasbeentoretainthenatureofmathematics,engagewiththe
question why learn mathematics while making an attempt to create materials that would address the
interestofthelearnersatthisstageandprovidesufficientandapproachablechallengetothem.There
havebeenmanyviewsonthepurposeofschoolmathematics.Theserangefromthefullyutilitarianto
theentirelyaestheticperceptions.Boththeseendupnotengagingwiththeconceptsandenrichingthe
apparatusavailabletothelearnerforparticipatinginlife.TheNCFemphasisestheneedfordeveloping
the ability to mathematise ideas and perhaps experiences as well.An ability to explore the ideas and
frameworkgivenbymathematicsinthestruggletofindaricherlifeandamoremeaningfulrelationship
withtheworldaround.
Thisisnoteveneasytocomprehend,farmoredifficulttooperationalise.ButNCFaddstothisan
evenmoredifficultgoal.Thetaskistoinvolveeveryoneofthatagegroupintheclassroomoroutside
indoingmathematics.Thisistheaimwehavebeenattemptingtomakeintheseries.
We have, therefore, provided space for children to engage in reflection, creating their own rules
anddefinitionsbasedonproblems/taskssolvedandfollowingtheirideaslogically.Theemphasisisnot
on remembering algorithms, doing complicated arithmetical problems or remembering proofs, but
understandinghowmathematicsworksandbeingabletoidentifythewayofmovingtowardssolving
problems.
Theimportantconcernforushasalsobeentoensurethatallstudentsatthisstagelearnmathematics
andbegintofeelconfidentinrelatingmathematics.Wehaveattemptedtohelpchildrenreadthebook
and to stop and reflect at each step where a new idea has been presented. In order to make the book
less formidable we have included illustrations and diagrams. These combined with the text help the
child comprehend the idea. Throughout the series and also therefore in this book we have tried to
avoidtheuseoftechnicalwordsandcomplexformulations.Wehaveleftmanythingsforthestudentto
describe and write in her own words.
Wehavemadeanattempttousechildfriendlylanguage.Toattractattentiontosomepointsblurbs
havebeenused.Theattempthasbeentoreducetheweightoflongexplanationsbyusingtheseandthe
diagrams.Theillustrationsandfillersalsoattempttobreakthemonotonyandprovidecontexts.
Class VIII is the bridge to Class IX where children will deal with more formal mathematics. The
attemptherehasbeentointroducesomeideasinawaythatismovingtowardsbecomingformal.The
tasksincludedexpectgeneralisationfromthegradualuseofsuchlanguagebythechild.
The team that developed this textbook consisted teachers with experience and appreciation of
childrenlearningmathematics.Thisteamalsoincludedpeoplewithexperienceofresearchinmathematics
teaching-learningandanexperienceofproducingmaterialsforchildren.Thefeedbackonthetextbooks
for Classes VI and VII was kept in mind while developing this textbook. This process of
development also included discussions with teachers during review workshop on the manuscript.
2021–22

vi
In the end, I would like to express the grateful thanks of our team to Professor Krishna Kumar,
Director, NCERT, Professor G. Ravindra, Joint Director, NCERT and Professor Hukum Singh,
Head,DESM,forgivingusanopportunitytoworkonthistaskwithfreedomandwithfullsupport.I
am also grateful to Professor J.V. Narlikar, Chairperson of the Advisory Group in Science and
Mathematicsforhissuggestions.IamalsogratefulforthesupportoftheteammembersfromNCERT,
Professor S.K. Singh Gautam, Dr V.P. Singh and in particular Dr Ashutosh K. Wazalwar who
coordinated this work and made arrangements possible. In the end I must thank the Publication
DepartmentofNCERTforitssupportandadviceandthosefromVidyaBhawanwhohelpedproduce
the book.
It need not be said but I cannot help mentioning that all the authors worked as a team and we
accepted ideas and advice from each other. We stretched ourselves to the fullest and hope that we
have done some justice to the challenge posed before us.
The process of developing materials is, however, a continuous one and we would hope to
make this book better. Suggestions and comments on the book are most welcome.
H.K.DEWAN
Chief Advisor
TextbookDevelopmentCommittee
2021–22

A Note for the Teacher
Thisisthethirdandthelastbookofthisseries.Itisacontinuationoftheprocessesinitiatedtohelpthe
learners in abstraction of ideas and principles of mathematics. Our students to be able to deal with
mathematical ideas and use them need to have the logical foundations to abstract and use postulates
andconstructnewformulations.ThemainpointsreflectedintheNCF-2005suggestrelatingmathematics
todevelopmentofwiderabilitiesinchildren,movingawayfromcomplexcalculationsandalgorithm
followingtounderstandingandconstructingaframeworkofunderstanding.Asyouknow,mathematical
ideasdonotdevelopbytellingthem.Theyalsodonotreachchildrenbymerelygivingexplanations.
Children need their own framework of concepts and a classroom where they are discussing ideas,
lookingforsolutionstoproblems,settingnewproblemsandfindingtheirownwaysofsolvingproblems
andtheirowndefinitions.
As we have said before, it is important to help children to learn to read the textbook and other
booksrelatedtomathematicswithunderstanding.Thereadingofmaterialsisclearlyrequiredtohelp
thechildlearnfurthermathematics.InClassVIIIpleasetakestockofwherethestudentshavereached
and give them more opportunities to read texts that use language with symbols and have brevity and
terseness with no redundancy. For this if you can, please get them to read other texts as well.You
couldalsohavethemrelatethephysicstheylearnandtheequationstheycomeacrossinchemistryto
theideastheyhavelearntinmathematics.Thesecross-disciplinaryreferenceswouldhelpthemdevelop
aframeworkandpurposeformathematics.Theyneedtobeabletoreconstructlogicalargumentsand
appreciatetheneedforkeepingcertainfactorsandconstraintswhiletheyrelatethemtootherareasas
well.ClassVIIIchildrenneedtohaveopportunityforallthis.
As we have already emphasised, mathematics at the Upper Primary Stage has to be close to the
experienceandenvironmentofthechildandbeabstractatthesametime.Fromthecomfortofcontext
and/ormodelslinkedtotheirexperiencetheyneedtomovetowardsworkingwithideas.Learningto
abstracthelpsformulateandunderstandarguments.Thecapacitytoseeinterrelationsamongconcepts
helpsusdealwithideasinothersubjectsaswell.Italsohelpsusunderstandandmakebetterpatterns,
maps,appreciateareaandvolumeandseesimilaritiesbetweenshapesandsizes.Whilethisisregarding
therelationshipofotherfieldsofknowledgetomathematics,itsmeaninginlifeandourenvironment
needs to be re-emphasised.
Children should be able to identify the principles to be used in contextual situations, for solving
problemssiftthroughandchoosetherelevantinformationasthefirstimportantstep.Oncestudentsdo
thattheyneedtobeabletofindthewaytousetheknowledgetheyhaveandreachwheretheproblem
requiresthemtogo. Theyneedtoidentifyanddefineaproblem,selectordesignpossiblesolutionsand
reviseorredesignthesteps,ifrequired.Astheygofurthertherewouldbemoretoofthistobedone.In
ClassVIIIwehavetogetthemtobeconsciousofthestepstheyfollow. Helpingchildrentodevelopthe
abilitytoconstructappropriatemodelsbybreakinguptheproblemsandevolvingtheirownstrategies
andanalysisofproblemsisextremelyimportant.Thisisintheplaceofgivingthemprescriptivealgorithms.
2021–22

viii
Cooperativelearning,learningthroughconversations,desireandcapacitytolearnfromeachother
andtherecognitionthatconversationisnotnoiseandconsultationnotcheatingisanimportantpartof
change in attitude for you as a teacher and for the students as well. They should be asked to make
presentations as a group with the inclusion of examples from the contexts of their own experiences.
Theyshouldbeencouragedtoreadthebookingroupsandformulateandexpresswhattheyunderstand
fromit.Theassessmentpatternhastorecogniseandappreciatethisandtheclassroomgroupsshould
besuchthatallchildrenenjoybeingwitheachotherandarecontributingtothelearningofthegroup.
Asyouwouldhaveseendifferentgroupsusedifferentstrategies.Someofthesearenotasefficientas
others as they reflect the modeling done and reflect the thinking used.All these are appropriate and
need to be analysed with children. The exposure to a variety of strategies deepens the mathematical
understanding. Each group moves from where it is and needs to be given an opportunity for that.
For conciseness we present the key ideas of mathematics learning that we would like you to
rememberinyourclassroom.
1. Enquirytounderstandisoneofthenaturalwaysbywhichstudentsacquireandconstructknowledge.
Theprocesscanusegenerationofobservationstoacquireknowledge.Studentsneedtodealwith
differentformsofquestioningandchallenginginvestigations-explorative,open-ended,contextual
andevenerrordetectionfromgeometry,arithmeticandgeneralisingittoalgebraicrelationsetc.
2. Children need to learn to provide and follow logical arguments, find loopholes in the arguments
presented and understand the requirement of a proof. By now children have entered the formal
stage.Theyneedtobeencouragedtoexercisecreativityandimaginationandtocommunicatetheir
mathematicalreasoningbothverballyandinwriting.
3. Themathematicsclassroomshouldrelatelanguagetolearningofmathematics.Childrenshouldtalk
about their ideas using their experiences and language. They should be encouraged to use their
ownwordsandlanguagebutalsograduallyshifttoformallanguageanduseofsymbols.
4. Thenumbersystemhasbeentakentothelevelofgeneralisationofrationalnumbersandtheirproperties
anddevelopingaframeworkthatincludesallprevioussystemsassub-setsofthegeneralisedrational
numbers.Generalisationsaretobepresentedinmathematicallanguageandchildrenhavetoseethat
algebraanditslanguagehelpsusexpressalotoftextinsmallsymbolicforms.
5. As before children should be required to set and solve a lot of problems. We hope that as the
nature of the problems set up by them becomes varied and more complex, they would become
confidentoftheideastheyaredealingwith.
6. ClassVIIIbookhasattemptedtobringtogetherthedifferentaspectsofmathematicsandemphasise
thecommonality.Unitarymethod,Ratioandproportion,Interestanddividendsareallpartofone
common logical framework. The idea of variable and equations is needed wherever we need to
findanunknownquantityinanybranchofmathematics.
We hope that the book will help children learn to enjoy mathematics and be confident in the
conceptsintroduced.Wewanttorecommendthecreationofopportunityforthinkingindividuallyand
collectively.
We look forward to your comments and suggestions regarding the book and hope that you will
send interesting exercises, activities and tasks that you develop during the course of teaching, to be
includedinthefutureeditions.Thiscanonlyhappenifyouwouldfindtimetolistencarefullytochildren
andidentifygapsandontheotherhandalsofindtheplaceswheretheycanbegivenspacetoarticulate
theirideasandverbalisetheirthoughts.
2021–22

Textbook Development Committee
CHAIRPERSON, ADVISORY GROUP IN SCIENCE AND MATHEMATICS
J.V. Narlikar, Emeritus Professor,Chairman,Advisory Committee, Inter University Centre for
Astronomy and Astrophysics(IUCCA),Ganeshkhind,PuneUniversity,Pune
CHIEF ADVISOR
H.K. Dewan,Vidya Bhawan Society, Udaipur, Rajasthan
CHIEF COORDINATOR
Hukum Singh, Professor and Head, DESM, NCERT, New Delhi
MEMBERS
AnjaliGupte,Teacher,VidyaBhawanPublicSchool,Udaipur, Rajasthan
AvantikaDam,TGT,CIEExperimentalBasicSchool,DepartmentofEducation,Delhi
B.C. Basti, Senior Lecturer, Regional Institute of Education, Mysore, Karnataka
H.C. Pradhan, Professor, Homi Bhabha Centre for Science Education, TIFR, Mumbai
Maharashtra
K.A.S.S.V.KameshwarRao,Lecturer, Regional Institute of Education, Shyamala Hills
Bhopal (M.P.)
Mahendra Shankar, Lecturer (S.G.) (Retd.), NCERT, New Delhi
Meena Shrimali,Teacher,Vidya Bhawan Senior Secondary School, Udaipur, Rajasthan
P. Bhaskar Kumar,PGT, Jawahar Navodaya Vidyalaya, Lepakshi, Distt.Anantpur (A.P.)
R.Athmaraman, Mathematics Education Consultant,TI Matric Higher Secondary School and
AMTI,Chennai,TamilNadu
RamAvtar, Professor (Retd.), NCERT, New Delhi
ShaileshShirali,RishiValleySchool,RishiValley,Madanapalle(A.P.)
S.K.S. Gautam, Professor, DEME, NCERT, New Delhi
ShradhaAgarwal, Principal, Florets International School, Panki, Kanpur (U.P.)
Srijata Das, Senior Lecturer in Mathematics, SCERT, New Delhi
V.P. Singh, Reader, DESM, NCERT, New Delhi
MEMBER-COORDINATOR
Ashutosh K.Wazalwar, Professor, DESM, NCERT, New Delhi
2021–22

ACKNOWLEDGEMENTS
The Council gratefully acknowledges the valuable contributions of the following participants of the
TextbookReviewWorkshop:ShriPradeepBhardwaj,TGT(Mathematics)BalSthaliPublicSecondary
School, Kirari, Nangloi, New Delhi; Shri Sankar Misra, Teacher in Mathematics, Demonstration
Multipurpose School, Regional Institute of Education, Bhubaneswar (Orissa); Shri Manohar
M. Dhok, Supervisor, M.P. Deo Smruti Lokanchi Shala, Nagpur (Maharashtra); Shri Manjit Singh
Jangra, Maths teacher, Government Senior Secondary School, Sector-4/7, Gurgoan (Haryana);
Dr.RajendraKumarPooniwala,U.D.T.,GovernmentSubhashExcellenceSchool,Burhanpur(M.P.);
ShriK.Balaji,TGT(Mathematics),KendriyaVidyalayaNo.1,Tirupati(A.P.);Ms.MalaMani,Amity
InternationalSchool,Sector-44,Noida;Ms.OmlataSingh,TGT(Mathematics),PresentationConvent
Senior Secondary School, Delhi; Ms. Manju Dutta,Army Public School, Dhaula Kuan, New Delhi;
Ms.NirupamaSahni,TGT(Mathematics),ShriMahaveerDigambarJainSeniorSecondarySchool,
Jaipur(Rajasthan);ShriNageshShankarMone,HeadMaster,KantilalPurshottamDasShahPrashala,
Vishrambag,Sangli(Maharashtra);ShriAnilBhaskarJoshi,Seniorteacher(Mathematics),Manutai
Kanya Shala, Tilak Road, Akola (Maharashtra); Dr. Sushma Jairath, Reader, DWS, NCERT,
New Delhi; Shri Ishwar Chandra, Lecturer (S.G.) (Retd.) NCERT, New Delhi.
TheCouncilisgratefulforthesuggestions/commentsgivenbythefollowingparticipantsduringthe
workshopofthemathematicsTextbookDevelopmentCommittee–ShriSanjayBoliaandShriDeepak
MantrifromVidyaBhawanBasicSchool,Udaipur;ShriInderMohanSinghChhabra,VidyaBhawan
EducationalResourceCentre,Udaipur.
TheCouncilacknowledgesthecomments/suggestionsgivenbyDr.R.P.Maurya,Reader,DESM,
NCERT, New Delhi; Dr. Sanjay Mudgal, Lecturer, DESM, NCERT, New Delhi; Dr.T.P. Sharma,
Lecturer, DESM, NCERT, New Delhi for the improvement of the book.
The Council acknowledges the support and facilities provided by Vidya Bhawan Society and
its staff, Udaipur, for conducting workshops of the development committee at Udaipur and to the
Director,CentreforScienceEducationandCommunication(CSEC),DelhiUniversityforproviding
libraryhelp.
TheCouncilacknowledgestheacademicandadministrativesupportofProfessorHukumSingh,
Head, DESM, NCERT.
The Council also acknowledges the efforts of Sajjad Haider Ansari, Rakesh Kumar,
NeelamWalecha,DTPOperators;KanwarSingh, CopyEditor;AbhimanuMohanty,ProofReader,
Deepak Kapoor, Computer Station Incharge, DESM, NCERTfor technical assistance,APC Office
and theAdministrativeStaff, DESM, NCERTand the Publication Department of the NCERT.
2021–22

Contents
Foreword iii
Preface v
Chapter 1 Rational Numbers 1
Chapter 2 Linear Equations in One Variable 21
Chapter 3 Understanding Quadrilaterals 37
Chapter 4 Practical Geometry 57
Chapter 5 Data Handling 69
Chapter 6 Squares and Square Roots 89
Chapter 7 Cubes and Cube Roots 109
Chapter 8 Comparing Quantities 117
Chapter 9 Algebraic Expressions and Identities 137
Chapter 10 Visualising Solid Shapes 153
Chapter 11 Mensuration 169
Chapter 12 Exponents and Powers 193
Chapter 13 Direct and Inverse Proportions 201
Chapter 14 Factorisation 217
Chapter 15 Introduction to Graphs 231
Chapter 16 Playing with Numbers 249
Answers 261
Just for Fun 275
2021–22

Constitution of India
Fundamental Duties
Part IV A (Article 51 A)
It shall be the duty of every citizen of India —
(a) to abide by the Constitution and respect its ideals and institutions, the
National Flag and the National Anthem;
(b) to cherish and follow the noble ideals which inspired our national struggle
for freedom;
(c) to uphold and protect the sovereignty, unity and integrity of India;
(d) to defend the country and render national service when called upon to
do so;
(e) to promote harmony and the spirit of common brotherhood amongst all
the people of India transcending religious, linguistic and regional or
sectional diversities; to renounce practices derogatory to the dignity of
women;
(f) to value and preserve the rich heritage of our composite culture;
(g) to protect and improve the natural environment including forests, lakes,
rivers, wildlife and to have compassion for living creatures;
(h) to develop the scientific temper, humanism and the spirit of inquiry and
reform;
(i) to safeguard public property and to abjure violence;
(j) to strive towards excellence in all spheres of individual and collective
activity so that the nation constantly rises to higher levels of endeavour
and achievement;
*(k) who is a parent or guardian, to provide opportunities for education to
his child or, as the case may be, ward between the age of six and
fourteen years.
Note: The Article 51A containing Fundamental Duties was inserted by the Constitution
(42nd Amendment) Act, 1976 (with effect from 3 January 1977).
*(k) was inserted by the Constitution (86th Amendment) Act, 2002 (with effect from
1 April 2010).
2021–22

RATIONAL NUMBERS 1
1.1 Introduction
InMathematics,wefrequentlycomeacrosssimpleequationstobesolved.Forexample,
theequation x + 2 = 13 (1)
is solved when x =11, because this value of x satisfies the given equation.The solution
11 is a natural number. On the other hand, for the equation
x + 5 = 5 (2)
the solution gives the whole number 0 (zero). If we consider only natural numbers,
equation (2) cannot be solved.To solve equations like (2), we added the number zero to
thecollectionofnaturalnumbersandobtainedthewholenumbers.Evenwholenumbers
willnotbesufficienttosolveequationsoftype
x + 18 = 5 (3)
Do you see ‘why’? We require the number –13 which is not a whole number.This
led us to think of integers, (positive and negative). Note that the positive integers
correspondtonaturalnumbers.Onemaythinkthatwehaveenoughnumberstosolveall
simpleequationswiththeavailablelistofintegers.Nowconsidertheequations
2x = 3 (4)
5x + 7 = 0 (5)
forwhichwecannotfindasolutionfromtheintegers.(Checkthis)
Weneedthenumbers
3
2
tosolveequation(4)and
7
5
−
tosolve
equation(5).Thisleadsustothecollectionofrationalnumbers.
We have already seen basic operations on rational
numbers.Wenowtrytoexploresomepropertiesofoperations
on the different types of numbers seen so far.
Rational Numbers
CHAPTER
1
2021–22

2 MATHEMATICS
1.2 Properties of Rational Numbers
1.2.1 Closure
(i) Wholenumbers
Let us revisit the closure property for all the operations on whole numbers in brief.
Operation Numbers Remarks
Addition 0 + 5 = 5, a whole number Whole numbers are closed
4 + 7 = ... . Is it a whole number? underaddition.
In general, a + b is a whole
number for any two whole
numbers a and b.
Subtraction 5 – 7 = – 2, which is not a Whole numbers are not closed
wholenumber. undersubtraction.
Multiplication 0 × 3 = 0, a whole number Whole numbers are closed
3×7=....Isitawholenumber? undermultiplication.
In general, if a and b are any two
whole numbers, their product ab
isawholenumber.
Division 5 ÷ 8 =
5
8
, which is not a
wholenumber.
Check for closure property under all the four operations for natural numbers.
(ii) Integers
Let us now recall the operations under which integers are closed.
Addition – 6 + 5 = – 1, an integer Integers are closed under
Is – 7 + (–5) an integer? addition.
Is 8 + 5 an integer?
In general, a + b is an integer
for any two integers a and b.
Subtraction 7 – 5 = 2, an integer Integers are closed under
Is 5 – 7 an integer? subtraction.
– 6 – 8 = – 14, an integer
Whole numbers are not closed
underdivision.
2021–22

RATIONAL NUMBERS 3
– 6 – (– 8) = 2, an integer
Is 8 – (– 6) an integer?
Ingeneral,foranytwointegers
a and b, a – b is again an integer.
Check if b – a is also an integer.
Multiplication 5 × 8 = 40, an integer Integers are closed under
Is – 5 × 8 an integer? multiplication.
– 5 × (– 8) = 40, an integer
Ingeneral,foranytwointegers
a and b, a × b is also an integer.
Division 5 ÷ 8 =
5
8
, which is not Integers are not closed
aninteger.
underdivision.
You have seen that whole numbers are closed under addition and multiplication but
notundersubtractionanddivision.However,integersareclosedunderaddition,subtraction
andmultiplicationbutnotunderdivision.
(iii) Rational numbers
Recallthatanumberwhichcanbewrittenintheform
p
q ,wherepandqareintegers
and q ≠ 0 is called a rational number. For example,
2
3
− ,
6
7
,
9
5
−
are all rational
numbers. Since the numbers 0, –2, 4 can be written in the form
p
q , they are also
rationalnumbers.(Checkit!)
(a) You know how to add two rational numbers. Let us add a few pairs.
3 ( 5)
8 7
−
+ =
21 ( 40) 19
56 56
+ − −
= (arationalnumber)
3 ( 4)
8 5
− −
+ =
15 ( 32)
...
40
− + −
= Isitarationalnumber?
4 6
7 11
+ = ... Isitarationalnumber?
We find that sum of two rational numbers is again a rational number. Check it
for a few more pairs of rational numbers.
We say that rational numbers are closed under addition. That is, for any
two rational numbers a and b, a + b is also a rational number.
(b) Willthedifferenceoftworationalnumbersbeagainarationalnumber?
We have,
5 2
7 3
−
− =
5 3 – 2 7 29
21 21
− × × −
= (arationalnumber)
2021–22

4 MATHEMATICS
TRY THESE
5 4
8 5
− =
25 32
40
−
= ... Isitarationalnumber?
3
7
8
5
−
−





 = ... Isitarationalnumber?
Trythisforsomemorepairsof rationalnumbers.Wefindthatrationalnumbers
are closed under subtraction. That is, for any two rational numbers a and
b, a – b is also a rational number.
(c) Let us now see the product of two rational numbers.
2 4
3 5
−
× =
8 3 2 6
;
15 7 5 35
−
× = (both the products are rational numbers)
4 6
5 11
−
− × = ... Isitarationalnumber?
Takesomemorepairsofrationalnumbersandcheckthattheirproductisagain
arationalnumber.
We say that rational numbers are closed under multiplication. That
is, for any two rational numbers a and b, a × b is also a rational
number.
(d) We note that
5 2 25
3 5 6
− −
÷ = (arationalnumber)
2 5
...
7 3
÷ = . Is it a rational number?
3 2
...
8 9
− −
÷ = . Is it a rational number?
Canyousaythatrationalnumbersareclosedunderdivision?
We find that for any rational number a, a ÷ 0 is not defined.
So rational numbers are not closed under division.
However,ifweexcludezerothenthecollectionof,allotherrationalnumbersis
closedunderdivision.
Fillintheblanksinthefollowingtable.
Numbers Closed under
addition subtraction multiplication division
Rationalnumbers Yes Yes ... No
Integers ... Yes ... No
Wholenumbers ... ... Yes ...
Naturalnumbers ... No ... ...
2021–22

RATIONAL NUMBERS 5
1.2.2 Commutativity
(i) Wholenumbers
Recall the commutativity of different operations for whole numbers by filling the
followingtable.
Addition 0 + 7 = 7 + 0 = 7 Additioniscommutative.
2 + 3 = ... + ... = ....
For any two whole
numbers a and b,
a + b = b + a
Subtraction ......... Subtractionisnotcommutative.
Multiplication ......... Multiplicationiscommutative.
Division ......... Divisionisnotcommutative.
Checkwhetherthecommutativityoftheoperationsholdfornaturalnumbersalso.
(ii) Integers
Fill in the following table and check the commutativity of different operations for
integers:
Addition ......... Additioniscommutative.
Subtraction Is 5 – (–3) = – 3 – 5? Subtractionisnotcommutative.
Multiplication ......... Multiplicationiscommutative.
Division ......... Divisionisnotcommutative.
(a) Addition
You know how to add two rational numbers. Let us add a few pairs here.
2 5 1 5 2 1
and
3 7 21 7 3 21
− −
 
+ = + =
 
 
So,
2 5 5 2
3 7 7 3
− −
 
+ = + 
 
Also,
−
+
−






6
5
8
3
= ... and
Is
−
+
−





 =
−





 +
−






6
5
8
3
6
5
8
3
?
2021–22

6 MATHEMATICS
TRY THESE
Is
3 1 1 3
8 7 7 8
− −
 
+ = + 
 
?
You find that two rational numbers can be added in any order. We say that
addition is commutative for rational numbers. That is, for any two rational
numbers a and b, a + b = b + a.
(b) Subtraction
Is
2 5 5 2
3 4 4 3
− = − ?
Is
1 3 3 1
2 5 5 2
− = − ?
Youwillfindthatsubtractionisnotcommutativeforrationalnumbers.
Note thatsubtractionisnotcommutativeforintegersandintegersarealsorational
numbers.So,subtractionwillnotbecommutativeforrationalnumberstoo.
(c) Multiplication
We have,
−
× =
−
= ×
−






7
3
6
5
42
15
6
5
7
3
Is
−
×
−





 =
−
×
−






8
9
4
7
4
7
8
9
?
Check for some more such products.
You will find that multiplication is commutative for rational numbers.
In general, a × b = b × a for any two rational numbers a and b.
(d) Division
Is
5 3 3 5
?
4 7 7 4
− −
 
÷ = ÷  
 
You will find that expressions on both sides are not equal.
Sodivisionisnotcommutativeforrationalnumbers.
Completethefollowingtable:
Numbers Commutative for
Rationalnumbers Yes ... ... ...
Integers ... No ... ...
Wholenumbers ... ... Yes ...
Naturalnumbers ... ... ... No
2021–22

RATIONAL NUMBERS 7
1.2.3 Associativity
(i) Wholenumbers
Recalltheassociativityofthefouroperationsforwholenumbersthroughthistable:
Addition ......... Additionisassociative
Subtraction ......... Subtractionisnotassociative
Multiplication Is 7 × (2 × 5) = (7 × 2) × 5? Multiplicationisassociative
Is 4 × (6 × 0) = (4 × 6) × 0?
For any three whole
numbers a, b and c
a × (b × c) = (a × b) × c
Division ......... Divisionisnotassociative
Fillinthistableandverifytheremarksgiveninthelastcolumn.
Checkforyourselftheassociativityofdifferentoperationsfornaturalnumbers.
(ii) Integers
Associativityofthefouroperationsforintegerscanbeseenfromthistable
Addition Is (–2) + [3 + (– 4)] Additionisassociative
= [(–2) + 3)] + (– 4)?
Is (– 6) + [(– 4) + (–5)]
= [(– 6) +(– 4)] + (–5)?
For any three integers a, b and c
a + (b + c) = (a + b) + c
Subtraction Is 5 – (7 – 3) = (5 – 7) – 3? Subtractionisnotassociative
Multiplication Is 5 × [(–7) × (– 8) Multiplicationisassociative
= [5 × (–7)] × (– 8)?
Is (– 4) × [(– 8) × (–5)]
= [(– 4) × (– 8)] × (–5)?
For any three integers a, b and c
a × (b × c) = (a × b) × c
Division Is [(–10) ÷ 2] ÷ (–5) Divisionisnotassociative
= (–10) ÷ [2 ÷ (– 5)]?
2021–22

8 MATHEMATICS
(a) Addition
We have
2 3 5 2 7 27 9
3 5 6 3 30 30 10
−  −  − − − −
   
+ + = + = =
   
 
   
 
2 3 5 1 5 27 9
3 5 6 15 6 30 10
− − − − − −
     
+ + = + = =
   
 
     
So,
2 3 5 2 3 5
3 5 6 3 5 6
−  −  − −
     
+ + = + +
   
   
     
 
Find
−
+ +
−












−
+





 +
−






1
2
3
7
4
3
1
2
3
7
4
3
and .Arethetwosumsequal?
Take some more rational numbers, add them as above and see if the two sums
are equal. We find that addition is associative for rational numbers. That
is, for any three rational numbers a, b and c, a + (b + c) = (a + b) + c.
(b) Subtraction
You already know that subtraction is not associative for integers, then what
aboutrationalnumbers.
Is
−
−
−
−





 = −
−











 −
2
3
4
5
1
2
2
3
4
5
1
2
?
Checkforyourself.
Subtraction is not associative for rational numbers.
(c) Multiplication
Letuschecktheassociativityformultiplication.
7 5 2 7 10 70 35
3 4 9 3 36 108 54
− − − −
 
× × = × = =
 
 
−
×





 × =
7
3
5
4
2
9
...
Wefindthat
7 5 2 7 5 2
3 4 9 3 4 9
− −
   
× × = × ×
   
   
Is
2 6 4 2 6 4
?
3 7 5 3 7 5
− −
   
× × = × ×
   
   
Take some more rational numbers and check for yourself.
We observe that multiplication is associative for rational numbers. That is
for any three rational numbers a, b and c, a × (b × c) = (a × b) × c.
2021–22

RATIONAL NUMBERS 9
TRY THESE
(d) Division
Recallthatdivisionisnotassociativeforintegers,thenwhataboutrationalnumbers?
Let us see if
1 1 2 1 1 2
2 3 5 2 3 5
−  − 
   
÷ ÷ = ÷ ÷
 
 
 
   
 
We have, LHS =
1 1 2
2 3 5
−
 
÷ ÷
 
 
=
1
2
1
3
5
2
÷
−
×





 (reciprocal of
2
5
is
5
2
)
=
1 5
2 6
 
÷ −
 
 
= ...
RHS =
1 1 2
2 3 5
 − 
 
÷ ÷
 
 
 
 
=
1 3 2
2 1 5
−
 
× ÷
 
 
=
3 2
2 5
−
÷ = ...
Is LHS = RHS? Check for yourself. You will find that division is
not associative for rational numbers.
Completethefollowingtable:
Numbers Associative for
Rationalnumbers ... ... ... No
Integers ... ... Yes ...
Wholenumbers Yes ... ... ...
Naturalnumbers ... No ... ...
Example 1: Find
3 6 8 5
7 11 21 22
− −
     
+ + +
     
     
Solution:
3 6 8 5
7 11 21 22
− −
     
+ + +
     
     
=
198
462
252
462
176
462
105
462
+
−





 +
−





 +





 (Note that 462 is the LCM of
7, 11, 21 and 22)
=
198 252 176 105
462
− − +
=
125
462
−
2021–22

10 MATHEMATICS
We can also solve it as.
3 6 8 5
7 11 21 22
− −
   
+ + +
   
   
=
3 8 6 5
7 21 11 22
 −  −
   
+ + +
 
   
   
 
(byusingcommutativityandassociativity)
=
9 8
21
12 5
22
+ −





 +
− +






( )
(LCM of 7 and 21 is 21; LCM of 11 and 22 is 22)
=
1 7
21 22
−
 
+ 
 
=
22 147 125
462 462
− −
=
Doyouthinkthepropertiesofcommutativityandassociativitymadethecalculationseasier?
Example 2: Find
−
× × ×
−






4
5
3
7
15
16
14
9
Solution: We have
−
× × ×
−






4
5
3
7
15
16
14
9
= −
×
×





 ×
× −
×






4 3
5 7
15 14
16 9
( )
=
−
×
−





 =
− × −
×
=
12
35
35
24
12 35
35 24
1
2
( )
We can also do it as.
−
× × ×
−






4
5
3
7
15
16
14
9
=
−
×





 × ×
−












4
5
15
16
3
7
14
9
(Usingcommutativityandassociativity)
=
3 2
4 3
− −
 
× 
 
=
1
2
1.2.4 The role of zero (0)
Lookatthefollowing.
2 + 0 = 0 + 2 = 2 (Addition of 0 to a whole number)
– 5 + 0 = ... + ... = – 5 (Addition of 0 to an integer)
2
7
−
+ ... = 0 +
2
7
−
 
 
 
=
2
7
−
(Addition of 0 to a rational number)
2021–22

RATIONAL NUMBERS 11
THINK, DISCUSS AND WRITE
You have done such additions earlier also. Do a few more such additions.
Whatdoyouobserve?Youwillfindthatwhenyouadd0toawholenumber,thesum
isagainthatwholenumber.Thishappensforintegersandrationalnumbersalso.
Ingeneral, a + 0 = 0 + a = a, where a is a whole number
b + 0 = 0 + b = b, where b is an integer
c + 0 = 0 + c = c, wherec is a rational number
Zero is called the identity for the addition of rational numbers. It is the additive
identity for integers and whole numbers as well.
1.2.5 The role of 1
We have,
5 × 1 = 5 = 1 × 5 (Multiplicationof1withawholenumber)
2
7
−
× 1 = ... × ... =
2
7
−
3
8
× ... = 1 ×
3
8
=
3
8
What do you find?
Youwillfindthatwhenyoumultiplyanyrationalnumberwith1,yougetbackthesame
rationalnumberastheproduct.Checkthisforafewmorerationalnumbers.Youwillfind
that, a × 1 = 1 × a = a for any rational number a.
We say that 1 is the multiplicative identity for rational numbers.
Is1themultiplicativeidentityforintegers?Forwholenumbers?
If a property holds for rational numbers, will it also hold for integers? For whole
numbers?Whichwill?Whichwillnot?
1.2.6 Negative of a number
Whilestudyingintegersyouhavecomeacrossnegativesofintegers.Whatisthenegative
of 1? It is – 1 because 1 + (– 1) = (–1) + 1 = 0
So, what will be the negative of (–1)? It will be 1.
Also, 2 + (–2) = (–2) + 2 = 0, so we say 2 is the negative or additive inverse of
–2 and vice-versa. In general, for an integer a, we have, a + (– a) = (– a) + a = 0; so, a
is the negative of – a and – a is the negative of a.
For the rational number
2
3
, we have,
2
3
2
3
+ −





 =
2 ( 2)
0
3
+ −
=
2021–22

12 MATHEMATICS
Also, −





 +
2
3
2
3
= 0 (How?)
Similarly,
8
...
9
−
+ = ... +
−





 =
8
9
0
... +
−






11
7
=
−





 + =
11
7
0
...
Ingeneral,forarationalnumber
a
b
,wehave,
a
b
a
b
a
b
a
b
+ −





 = −





 + = 0 .Wesay
that
a
b
− is the additive inverse of
a
b
and
a
b
is the additive inverse of −






a
b
.
1.2.7 Reciprocal
By which rational number would you multiply
8
21
, to get the product 1? Obviously by
21 8 21
, since 1
8 21 8
× = .
Similarly,
5
7
−
must be multiplied by
7
5
−
so as to get the product 1.
We say that
21
8
is the reciprocal of
8
21
and
7
5
−
is the reciprocal of
5
7
−
.
Can you say what is the reciprocal of 0 (zero)?
Istherearationalnumberwhichwhenmultipliedby0gives1? Thus,zerohasnoreciprocal.
We say that a rational number
c
d
is called the reciprocal or multiplicative inverse of
another non-zero rational number
a
b
if 1
a c
b d
× = .
1.2.8 Distributivity of multiplication over addition for rational
numbers
To understand this, consider the rational numbers
3 2
,
4 3
−
and
5
6
−
.
−
× +
−












3
4
2
3
5
6
=
−
×
+ −






3
4
4 5
6
( ) ( )
=
−
×
−






3
4
1
6
=
3 1
24 8
=
Also
3 2
4 3
−
× =
3 2 6 1
4 3 12 2
− × − −
= =
×
2021–22

RATIONAL NUMBERS 13
TRY THESE
And
3 5
4 6
− −
× =
5
8
Therefore
−
×





 +
−
×
−






3
4
2
3
3
4
5
6
=
1 5 1
2 8 8
−
+ =
Thus,
−
× +
−






3
4
2
3
5
6
=
−
×





 +
−
×
−






3
4
2
3
3
4
5
6
Find using distributivity. (i)
7
5
3
12
7
5
5
12
×
−












+ ×






(ii)
9
16
4
12
9
16
3
9
×






+ ×
−






Example 3: Writetheadditiveinverseofthefollowing:
(i)
7
19
−
(ii)
21
112
Solution:
(i)
7
19
is the additive inverse of
7
19
−
because
7
19
−
+
7
19
=
7 7 0
19 19
− +
= = 0
(ii) The additive inverse of
21
112
is
21
112
−
(Check!)
Example 4: Verify that – (– x) is the same as x for
(i) x =
13
17
(ii)
21
31
x
−
=
Solution: (i) We have, x =
13
17
The additive inverse of x =
13
17
is – x =
13
17
−
since
13
17
13
17
0
+
−





 = .
The same equality
13
17
13
17
0
+
−





 = , shows that the additive inverse of
13
17
−
is
13
17
or −
−






13
17
=
13
17
, i.e., – (– x) = x.
(ii) Additive inverse of
21
31
x
−
= is – x =
21
31
since
21 21
0
31 31
−
+ = .
The same equality
21 21
0
31 31
−
+ = , shows that the additive inverse of
21
31
is
21
31
−
,
i.e., – (– x) = x.
Distributivity of Multi-
plication over Addition
and Subtraction.
For all rational numbers a, b
and c,
a (b + c) = ab + ac
a (b – c) = ab – ac
When you use distributivity, you
split a product as a sum or
difference of two products.
2021–22

14 MATHEMATICS
Example 5: Find
2 3 1 3 3
5 7 14 7 5
−
× − − ×
Solution:
2 3 1 3 3
5 7 14 7 5
−
× − − × =
2 3 3 3 1
5 7 7 5 14
−
× − × − (bycommutativity)
=
2
5
3
7
3
7
3
5
1
14
×
−
+
−





 × −
=
−
+





 −
3
7
2
5
3
5
1
14
(bydistributivity)
=
3 1
1
7 14
−
× − =
6 1 1
14 2
− − −
=
EXERCISE 1.1
1. Usingappropriatepropertiesfind.
(i)
2 3 5 3 1
3 5 2 5 6
− × + − × (ii)
2
5
3
7
1
6
3
2
1
14
2
5
× −





 − × + ×
2. Writetheadditiveinverseofeachofthefollowing.
(i)
2
8
(ii)
5
9
−
(iii)
6
5
−
−
(iv)
2
9
−
(v)
19
6
−
3. Verify that – (– x) = x for.
(i) x =
11
15
(ii)
13
17
x = −
4. Findthemultiplicativeinverseofthefollowing.
(i) – 13 (ii)
13
19
−
(iii)
1
5
(iv)
5 3
8 7
− −
×
(v) – 1
2
5
−
× (vi) – 1
5. Namethepropertyundermultiplicationusedineachofthefollowing.
(i)
4 4 4
1 1
5 5 5
− −
× = × = − (ii)
13 2 2 13
17 7 7 17
− − −
− × = ×
(iii)
19 29
1
29 19
−
× =
−
6. Multiply
6
13
by the reciprocal of
7
16
−
.
7. Tell what property allows you to compute
1
3
6
4
3
1
3
6
4
3
× ×





 ×





 ×
as .
8. Is
8
9
themultiplicativeinverseof
1
1
8
− ?Whyorwhynot?
9. Is0.3themultiplicativeinverseof
1
3
3
? Why or why not?
2021–22

RATIONAL NUMBERS 15
10. Write.
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) Therationalnumberthatisequaltoitsnegative.
11. Fillintheblanks.
(i) Zero has ________ reciprocal.
(ii) The numbers ________ and ________ are their own reciprocals
(iii) The reciprocal of – 5 is ________.
(iv) Reciprocal of
1
x
, where x ≠ 0 is ________.
(v) The product of two rational numbers is always a _______.
(vi) The reciprocal of a positive rational number is ________.
1.3 Representation of Rational Numbers on the
Number Line
Youhavelearnttorepresentnaturalnumbers,wholenumbers,integers
andrationalnumbersonanumberline.Letusrevisethem.
Naturalnumbers
(i)
Wholenumbers
(ii)
Integers
(iii)
Rational numbers
(iv)
(v)
The point on the number line (iv) which is half way between 0 and 1 has been
labelled
1
2
.Also, the first of the equally spaced points that divides the distance between
0 and 1 into three equal parts can be labelled
1
3
, as on number line (v). How would you
labelthesecondofthesedivisionpointsonnumberline(v)?
The line extends
indefinitely only to the
right side of 1.
The line extends indefinitely
to the right, but from 0.
There are no numbers to the
left of 0.
The line extends
indefinitely on both sides.
Do you see any numbers
between –1, 0; 0, 1 etc.?
The line extends indefinitely
on both sides. But you can
now see numbers between
–1, 0; 0, 1 etc.
2021–22

16 MATHEMATICS
The point to be labelled is twice as far from and to the right of 0 as the point
labelled
1
3
.Soitistwotimes
1
3
,i.e.,
2
3
.Youcancontinuetolabelequally-spacedpointson
the number line in the same way. In this continuation, the next marking is 1. You can
seethat1isthesameas
3
3
.
Then comes
4 5 6
, ,
3 3 3
(or 2),
7
3
and so on as shown on the number line (vi)
(vi)
Similarly, to represent
1
8
, the number line may be divided into eight equal parts as
shown:
We use the number
1
8
to name the first point of this division. The second point of
division will be labelled
2
8
, the third point
3
8
, and so on as shown on number
line(vii)
(vii)
Anyrationalnumbercanberepresentedonthenumberlineinthisway.Inarational
number,thenumeralbelowthebar,i.e.,thedenominator,tellsthenumberofequal
parts into which the first unit has been divided. The numeral above the bar i.e., the
numerator, tells ‘how many’of these parts are considered. So, a rational number
such as
4
9
means four of nine equal parts on the right of 0 (number line viii) and
for
7
4
−
, we make 7 markings of distance
1
4
each on the left of zero and starting
from0.Theseventhmarkingis
7
4
−
[numberline(ix)].
(viii)
(ix)
2021–22

RATIONAL NUMBERS 17
TRY THESE
Writetherationalnumberforeachpointlabelledwithaletter.
(i)
(ii)
1.4 Rational Numbers between Two Rational Numbers
Can you tell the natural numbers between 1 and 5? They are 2, 3 and 4.
How many natural numbers are there between 7 and 9? There is one and it is 8.
How many natural numbers are there between 10 and 11? Obviously none.
List the integers that lie between –5 and 4. They are – 4, – 3, –2, –1, 0, 1, 2, 3.
How many integers are there between –1 and 1?
How many integers are there between –9 and –10?
You will find a definite number of natural numbers (integers) between two natural
numbers(integers).
How many rational numbers are there between
3
10
and
7
10
?
You may have thought that they are only
4 5
,
10 10
and
6
10
.
But you can also write
3
10
as
30
100
and
7
10
as
70
100
. Now the numbers,
31 32 33
, ,
100 100 100
68 69
, ... ,
100 100
, are all between
3
10
and
7
10
. The number of these rational numbers is 39.
Also
3
10
can be expressed as
3000
10000
and
7
10
as
7000
10000
. Now, we see that the
rational numbers
3001 3002 6998 6999
, ,..., ,
10000 10000 10000 10000
are between
3
10
and
7
10
. These
are 3999 numbers in all.
In this way, we can go on inserting more and more rational numbers between
3
10
and
7
10
.Sounlikenaturalnumbersandintegers,thenumberofrationalnumbersbetween
tworationalnumbersisnotdefinite.Hereisonemoreexample.
How many rational numbers are there between
1
10
−
and
3
10
?
Obviously
0 1 2
, ,
10 10 10
arerationalnumbersbetweenthegivennumbers.
2021–22

18 MATHEMATICS
If we write
1
10
−
as
10000
100000
−
and
3
10
as
30000
100000
, we get the rational numbers
9999 9998
, ,...,
100000 100000
− − 29998
100000
−
,
29999
100000
, between
1
10
−
and
3
10
.
You will find that you get countless rational numbers between any two given
rational numbers.
Example 6: Write any 3 rational numbers between –2 and 0.
Solution: –2 can be written as
20
10
−
and 0 as
0
10
.
Thus we have
19 18 17 16 15 1
, , , , , ...,
10 10 10 10 10 10
− − − − − −
between –2 and 0.
You can take any three of these.
Example 7: Find any ten rational numbers between
5
6
−
and
5
8
.
Solution:Wefirstconvert
5
6
−
and
5
8
torationalnumberswiththesamedenominators.
5 4 20
6 4 24
− × −
=
×
and
5 3 15
8 3 24
×
=
×
Thus we have
19 18 17 14
, , ,...,
24 24 24 24
− − −
as the rational numbers between
20
24
−
and
15
24
.
You can take any ten of these.
Another Method
Letusfindrationalnumbersbetween1and2.Oneofthemis 1.5or
1
1
2
or
3
2
.Thisisthe
mean of 1 and 2.You have studied mean in ClassVII.
We find that between any two given numbers, we need not necessarily get an
integer but there will always lie a rational number.
We can use the idea of mean also to find rational numbers between any two given
rationalnumbers.
Example 8: Find a rational number between
1
4
and
1
2
.
Solution:We find the mean of the given rational numbers.
1
4
1
2
2
+





 ÷ =
1 2
4
2
3
4
1
2
3
8
+





 ÷ = × =
3
8
lies between
1
4
and
1
2
.
This can be seen on the number line also.
2021–22

RATIONAL NUMBERS 19
We find the mid point ofAB which is C, represented by
1
4
1
2
2
+





 ÷ =
3
8
.
We find that
1 3 1
4 8 2
< < .
If a and b are two rational numbers, then
2
a b
+
is a rational number between a and
b such that a <
2
a b
+
< b.
This again shows that there are countless number of rational numbers between any
twogivenrationalnumbers.
Example 9: Find three rational numbers between
1
4
and
1
2
.
Solution: We find the mean of the given rational numbers.
As given in the above example, the mean is
3
8
and
1 3 1
4 8 2
< < .
Wenowfindanotherrationalnumberbetween
1 3
and
4 8
.Forthis,weagainfindthemean
of
1 3
and
4 8
. That is,
1
4
3
8
2
+





 ÷ =
5 1 5
8 2 16
× =
1 5 3 1
4 16 8 2
< < <
Now find the mean of
3 1
and
8 2
. We have,
3
8
1
2
2
+





 ÷ =
7 1
8 2
× =
7
16
Thus we get
1 5 3 7 1
4 16 8 16 2
< < < < .
Thus,
5 3 7
, ,
16 8 16
are the three rational numbers between
1 1
and
4 2
.
Thiscanclearlybeshownonthenumberlineasfollows:
In the same way we can obtain as many rational numbers as we want between two
givenrationalnumbers.Youhavenoticedthattherearecountlessrationalnumbersbetween
anytwogivenrationalnumbers.
2021–22

20 MATHEMATICS
EXERCISE 1.2
1. Represent these numbers on the number line. (i)
7
4
(ii)
5
6
−
2. Represent
2 5 9
, ,
11 11 11
− − −
on the number line.
3. Writefiverationalnumberswhicharesmallerthan2.
4. Find ten rational numbers between
2 1
and
5 2
−
.
5. Findfiverationalnumbersbetween.
(i)
2
3
and
4
5
(ii)
3
2
−
and
5
3
(iii)
1
4
and
1
2
6. Writefiverationalnumbersgreaterthan–2.
7. Find ten rational numbers between
3
5
and
3
4
.
WHAT HAVE WE DISCUSSED?
1. Rationalnumbersareclosed undertheoperationsofaddition,subtractionandmultiplication.
2. Theoperationsadditionandmultiplicationare
(i) commutativeforrationalnumbers.
(ii) associative for rational numbers.
3. The rational number 0 is theadditive identity for rational numbers.
4. Therationalnumber1isthe multiplicativeidentityforrationalnumbers.
5. The additive inverse of the rational number
a
b
is
a
b
− and vice-versa.
6. The reciprocal or multiplicative inverse of the rational number
a
b
is
c
d
if 1
a c
b d
× = .
7. Distributivity of rational numbers: For all rational numbers a,b and c,
a(b + c) = ab + ac and a(b – c) = ab – ac
8. Rational numbers can be represented on a number line.
9. Betweenanytwogivenrationalnumberstherearecountlessrationalnumbers.Theideaofmean
helpsustofindrationalnumbersbetweentworationalnumbers.
2021–22

LINEAR EQUATIONS IN ONE VARIABLE 21
2.1 Introduction
Intheearlierclasses,youhavecomeacrossseveralalgebraicexpressionsandequations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x2
+ 1, y + y2
Some examples of equations are: 5x = 25, 2x – 3 = 9,
5 37
2 , 6 10 2
2 2
y z
+ = + = −
Youwouldrememberthatequationsusetheequality(=)sign;itismissinginexpressions.
Ofthesegivenexpressions,manyhavemorethanonevariable.Forexample,2xy+5
has two variables.We however, restrict to expressions with only one variable when we
formequations.Moreover,theexpressionsweusetoformequationsarelinear.Thismeans
that the highest power of the variable appearing in the expression is 1.
Thesearelinearexpressions:
2x, 2x + 1, 3y – 7, 12 – 5z,
5
( – 4) 10
4
x +
Thesearenotlinearexpressions:
x2
+ 1, y + y2
, 1 + z + z2
+ z3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Letusbrieflyrevisewhatweknow:
(a) An algebraic equation is an equality
involving variables. It has an equality sign.
Theexpressionontheleftoftheequalitysign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
Linear Equations in
One Variable
CHAPTER
2
2x – 3 = 7
2x – 3 = LHS
7 = RHS
2021–22

22 MATHEMATICS
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certainvaluesofthevariable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
Weassumethatthetwosidesoftheequationarebalanced.
We perform the same mathematical operations on both
sidesoftheequation,sothatthebalanceisnotdisturbed.
Afewsuchstepsgivethesolution.
2.2 Solving Equations which have Linear Expressions
on one Side and Numbers on the other Side
Letusrecallthetechniqueofsolvingequationswithsomeexamples.Observethesolutions;
theycanbeanyrationalnumber.
Example 1: Find the solution of 2x – 3 = 7
Solution:
Step 1 Add 3 to both sides.
2x – 3 + 3 = 7 + 3 (The balance is not disturbed)
or 2x = 10
Step 2 Next divide both sides by 2.
2
2
x
=
10
2
or x = 5 (requiredsolution)
Example 2: Solve 2y + 9 = 4
Solution: Transposing 9 to RHS
2y = 4 – 9
or 2y = – 5
Dividingbothsidesby2, y =
5
2
−
(solution)
To check the answer: LHS = 2
−






5
2
+ 9 = – 5 + 9 = 4 = RHS (as required)
Do you notice that the solution
−






5
2
is a rational number? In Class VII, the equations
we solved did not have such solutions.
x = 5 is the solution of the equation
2x – 3 = 7. For x = 5,
LHS = 2 × 5 – 3 = 7 = RHS
On the other hand x = 10 is not a solution of the
equation. For x = 10, LHS = 2 × 10 –3 = 17.
This is not equal to the RHS
2021–22

Example 3: Solve
5
3 2
x
+ =
3
2
−
Solution: Transposing
5
2
to the RHS, we get
3
x
=
3 5 8
2 2 2
−
− = −
or
3
x
= – 4
Multiplybothsidesby3, x = – 4 × 3
or x = – 12 (solution)
Check: LHS =
12 5 5 8 5 3
4
3 2 2 2 2
− + −
− + = − + = = = RHS (as required)
Do you now see that the coefficient of a variable in an equation need not be an integer?
Example 4: Solve
15
4
– 7x = 9
Solution: We have
15
4
– 7x = 9
or – 7x = 9 –
15
4
(transposing
15
4
to R H S)
or – 7x =
21
4
or x =
21
4 ( 7)
× −
(dividing both sides by – 7)
or x =
3 7
4 7
×
−
×
or x =
3
4
− (solution)
Check: LHS =
15
4
7
3
4
−
−





 =
15 21 36
9
4 4 4
+ = = = RHS (as required)
EXERCISE 2.1
Solvethefollowingequations.
1. x – 2 = 7 2. y + 3 = 10 3. 6 = z + 2
4.
3 17
7 7
x
+ = 5. 6x = 12 6. 10
5
t
=
7.
2
18
3
x
= 8. 1.6 =
1.5
y
9. 7x – 9 = 16
2021–22

24 MATHEMATICS
10. 14y – 8 = 13 11. 17 + 6p = 9 12.
7
1
3 15
x
+ =
2.3 Some Applications
Webeginwithasimpleexample.
Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the
numbers?
We have a puzzle here.We do not know either of the two numbers, and we have to
findthem.Wearegiventwoconditions.
(i) One of the numbers is 10 more than the other.
(ii) Theirsumis74.
We already know from ClassVII how to proceed. If the smaller number is taken to
be x, the larger number is 10 more than x, i.e.,x + 10. The other condition says that
the sum of these two numbers x and x + 10 is 74.
This means that x + (x + 10) = 74.
or 2x + 10 = 74
Transposing 10 to RHS, 2x = 74 – 10
or 2x = 64
Dividingbothsidesby2, x = 32. This is one number.
Theothernumberis x + 10 = 32 + 10 = 42
The desired numbers are 32 and 42. (Their sum is indeed 74 as given and also one
number is 10 more than the other.)
We shall now consider several examples to show how useful this method is.
Example 5: What should be added to twice the rational number
7
3
−
to get
3
7
?
Solution: Twicetherationalnumber
7
3
−
is 2
7
3
14
3
×
−





 =
−
.Supposexaddedtothis
numbergives
3
7
;i.e.,
x +
−






14
3
=
3
7
or
14
3
x − =
3
7
or x =
3 14
7 3
+ (transposing
14
3
to RHS)
=
(3 3) (14 7)
21
× + ×
=
9 98 107
21 21
+
= .
2021–22

Thus
107
21
should be added to 2
7
3
×
−





 to give
3
7
.
Example 6: The perimeter of a rectangle is 13 cm and its width is
3
2
4
cm. Find its
length.
Solution: Assume the length of the rectangle to be x cm.
The perimeter of the rectangle = 2 × (length + width)
= 2 × (x +
3
2
4
)
= 2
11
4
x +






The perimeter is given to be 13 cm. Therefore,
2
11
4
x +





 = 13
or
11
4
x + =
13
2
(dividingbothsidesby2)
or x =
13 11
2 4
−
=
26 11 15 3
3
4 4 4 4
− = =
Thelengthoftherectangleis
3
3
4
cm.
Example 7: The present age of Sahil’s mother is three times the present age of Sahil.
After 5 years their ages will add to 66 years. Find their present ages.
Solution: Let Sahil’s present age be x years.
It is given that this sum is 66 years.
Therefore, 4x + 10 = 66
This equation determines Sahil’s present age which is x years. To solve the equation,
We could also choose Sahil’s age
5 years later to be x and proceed.
Why don’t you try it that way?
Sahil Mother Sum
Present age x 3x
Age 5 years later x + 5 3x + 5 4x + 10
2021–22

26 MATHEMATICS
we transpose 10 to RHS,
4x = 66 – 10
or 4x = 56
or x =
56
4
= 14 (solution)
Thus,Sahil’spresentageis14yearsandhismother’sageis42years.(Youmayeasily
check that 5 years from now the sum of their ages will be 66 years.)
Example 8: Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If
he has in all a sum of ` 77, how many coins of each denomination does he have?
Solution: Let the number of five-rupee coins that Bansi has be x. Then the number of
two-rupee coins he has is 3 times x or 3x.
TheamountBansihas:
(i) from 5 rupee coins, ` 5 × x = ` 5x
(ii) from 2 rupee coins, ` 2 × 3x = ` 6x
Hence the total money he has = ` 11x
But this is given to be ` 77; therefore,
11x = 77
or x =
77
11
= 7
Thus, numberoffive-rupeecoins= x = 7
and number of two-rupee coins = 3x = 21 (solution)
(You can check that the total money with Bansi is ` 77.)
Example 9: The sum of three consecutive multiples of 11 is 363. Find these
multiples.
Solution: If x is a multiple of 11, the next multiple is x + 11. The next to this is
x + 11 + 11 or x + 22. So we can take three consecutive multiples of 11 as x, x + 11 and
x + 22.
Itisgiventhatthesumoftheseconsecutive
multiples of 11 is 363. This will give the
followingequation:
x + (x + 11) + (x + 22) = 363
or x + x + 11 + x + 22 = 363
or 3x + 33 = 363
or 3x = 363 – 33
or 3x = 330
Rs 5
Rs 2
Alternatively, we may think of the multiple
of 11 immediately before x. This is (x – 11).
Therefore, we may take three consecutive
multiples of 11 as x – 11, x, x + 11.
In this case we arrive at the equation
(x – 11) + x + (x + 11) = 363
or 3x = 363
2021–22

or x =
330
3
= 110
Hence,thethreeconsecutivemultiples
are 110, 121, 132 (answer).
We can see that we can adopt different ways to find a solution for the problem.
Example 10: The difference between two whole numbers is 66. The ratio of the two
numbers is 2 : 5. What are the two numbers?
Solution: Since the ratio of the two numbers is 2 : 5, we may take one number to be
2x and the other to be 5x. (Note that 2x : 5x is same as 2 : 5.)
The difference between the two numbers is (5x – 2x). It is given that the difference
is 66. Therefore,
5x – 2x = 66
or 3x = 66
or x = 22
Since the numbers are 2x and 5x, they are 2 × 22 or 44 and 5 × 22 or 110, respectively.
The difference between the two numbers is 110 – 44 = 66 as desired.
Example 11: Deveshi has a total of ` 590 as currency notes in the denominations of
` 50, ` 20 and ` 10. The ratio of the number of ` 50 notes and ` 20 notes is 3:5. If she has
a total of 25 notes, how many notes of each denomination she has?
Solution: Let the number of ` 50 notes and ` 20 notes be 3x and 5x, respectively.
But she has 25 notes in total.
Therefore, the number of ` 10 notes = 25 – (3x + 5x) = 25 – 8x
Theamountshehas
from ` 50 notes : 3x × 50 = ` 150x
from ` 20 notes : 5x × 20 = ` 100x
from ` 10 notes : (25 – 8x) × 10 = ` (250 – 80x)
Hence the total money she has =150x + 100x + (250 – 80x) = ` (170x + 250)
But she has ` 590. Therefore, 170x + 250 = 590
or 170x = 590 – 250 = 340
or x =
340
170
= 2
The number of ` 50 notes she has = 3x
= 3 × 2 = 6
The number of ` 20 notes she has = 5x = 5 × 2 = 10
The number of ` 10 notes she has = 25 – 8x
= 25 – (8 × 2) = 25 – 16 = 9
or x =
363
3
= 121. Therefore,
x = 121, x – 11 = 110, x + 11 = 132
Hence, the three consecutive multiples are
110, 121, 132.
2021–22

28 MATHEMATICS
EXERCISE 2.2
1. If you subtract
1
2
fromanumberandmultiplytheresultby
1
2
, you get
1
8
. What is
thenumber?
2. Theperimeterofarectangularswimmingpoolis154m.Itslengthis2mmorethan
twice its breadth. What are the length and the breadth of the pool?
3. Thebaseofanisoscelestriangleis
4
cm
3
.Theperimeterofthetriangleis
2
4 cm
15
.
Whatisthelengthofeitheroftheremainingequalsides?
4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
6. Three consecutive integers add up to 51. What are these integers?
7. Thesumofthreeconsecutivemultiplesof8is888.Findthemultiples.
8. Threeconsecutiveintegersaresuchthatwhentheyaretakeninincreasingorderand
multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
9. The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their
ages will be 56 years. What are their present ages?
10. Thenumberofboysandgirlsinaclassareintheratio7:5.Thenumberofboysis8
more than the number of girls. What is the total class strength?
11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years
older than Baichung. The sum of the ages of all the three is 135 years. What is the
age of each one of them?
12. FifteenyearsfromnowRavi’sagewillbefourtimeshispresentage.WhatisRavi’s
present age?
13. Arationalnumberissuchthatwhenyoumultiplyitby
5
2
andadd
2
3
totheproduct,
you get
7
12
− . What is the number?
14. Lakshmiisacashierinabank.Shehascurrencynotesofdenominations
` 100, ` 50 and ` 10, respectively. The ratio of the number of these
notes is 2:3:5. The total cash with Lakshmi is ` 4,00,000. How many
notes of each denomination does she have?
15. I have a total of ` 300 in coins of denomination ` 1, ` 2 and ` 5. The
numberof`2coinsis3timesthenumberof` 5coins.Thetotalnumberof
coinsis160. Howmanycoinsofeachdenominationarewithme?
16. The organisers of an essay competition decide that a winner in the
competitiongetsaprizeof ` 100andaparticipantwhodoesnotwingets
a prize of ` 25. The total prize money distributed is ` 3,000. Find the
numberofwinners,ifthetotalnumberofparticipantsis63.
2021–22

2.4 Solving Equations having the Variable on
both Sides
An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7,
the two expressions are 2x – 3 and 7. In most examples that we have come across so
far, the RHS is just a number. But this need not always be so; both sides could have
expressions with variables. For example, the equation 2x – 3 = x + 2 has expressions
with a variable on both sides; the expression on the LHS is (2x – 3) and the expression
on the RHS is (x + 2).
• Wenowdiscusshowtosolvesuchequationswhichhaveexpressionswiththevariable
on both sides.
Example 12: Solve 2x – 3 = x + 2
Solution: We have
2x = x + 2 + 3
or 2x = x + 5
or 2x – x = x + 5 – x (subtracting x from both sides)
or x = 5 (solution)
Here we subtracted from both sides of the equation, not a number (constant), but a
term involving the variable.We can do this as variables are also numbers.Also, note that
subtracting x from both sides amounts to transposing x to LHS.
Example 13: Solve 5x +
7 3
14
2 2
x
= −
Solution: Multiply both sides of the equation by 2.We get
2 5
7
2
× +






x = 2
3
2
14
× −






x
(2 × 5x) + 2
7
2
×





 = 2
3
2
2 14
×





 − ×
x ( )
or 10x + 7 = 3x – 28
or 10x – 3x + 7 = – 28 (transposing 3x to LHS)
or 7x + 7 = – 28
or 7x = – 28 – 7
or 7x = – 35
or x =
35
7
−
or x = – 5 (solution)
2021–22

30 MATHEMATICS
EXERCISE 2.3
Solvethefollowingequationsandcheckyourresults.
1. 3x = 2x + 18 2. 5t – 3 = 3t – 5 3. 5x + 9 = 5 + 3x
4. 4z + 3 = 6 + 2z 5. 2x – 1 = 14 – x 6. 8x + 4 = 3 (x – 1) + 7
7. x =
4
5
(x + 10) 8.
2
3
x
+ 1 =
7
3
15
x
+ 9. 2y +
5
3
=
26
3
y
−
10. 3m = 5 m –
8
5
2.5 Some More Applications
Example 14:Thedigitsofatwo-digitnumberdifferby3. Ifthedigitsareinterchanged,
and the resulting number is added to the original number, we get 143. What can be the
originalnumber?
Solution: Take, for example, a two-digit number, say, 56. It can be written as
56 = (10 × 5) + 6.
If the digits in 56 are interchanged, we get 65, which can be written as (10 × 6 ) + 5.
Let us take the twodigitnumbersuchthatthedigit in the units place is b. The digit
in the tens place differs from b by 3. Let us take it as b + 3. So the two-digit number
is 10 (b + 3) + b = 10b + 30 + b = 11b + 30.
Withinterchangeofdigits,theresultingtwo-digitnumberwillbe
10b + (b + 3) = 11b + 3
If we add these two two-digit numbers, their sum is
(11b + 30) + (11b + 3) = 11b + 11b + 30 + 3 = 22b + 33
It is given that the sum is 143. Therefore, 22b + 33 = 143
or 22b = 143 – 33
or 22b = 110
or b =
110
22
or b = 5
The units digit is 5 and therefore the tens digit is 5 + 3
which is 8. The number is 85.
Check: Oninterchangeofdigitsthenumberwegetis
58. The sum of 85 and 58 is 143 as given.
The statement of the
example is valid for both 58
and 85 and both are correct
answers.
Could we take the tens
place digit to be
(b – 3)? Try it and see
what solution you get.
Remember, this is the solution
when we choose the tens digits to
be 3 more than the unit’s digits.
What happens if we take the tens
digit to be (b – 3)?
2021–22

Example 15: Arjun is twice as old as Shriya. Five years ago his age was three times
Shriya’s age. Find their present ages.
Solution: Let us take Shriya’s present age to be x years.
ThenArjun’s present age would be 2x years.
Shriya’s age five years ago was (x – 5) years.
Arjun’s age five years ago was (2x – 5) years.
It is given thatArjun’s age five years ago was three times Shriya’sage.
Thus, 2x – 5 = 3(x – 5)
or 2x – 5 = 3x – 15
or 15 – 5 = 3x – 2x
or 10 = x
So, Shriya’s present age = x = 10 years.
Therefore,Arjun’s present age = 2x = 2 × 10 = 20 years.
EXERCISE 2.4
1. Aminathinksofanumberandsubtracts
5
2
fromit.Shemultipliestheresultby8.The
resultnowobtainedis3timesthesamenumbershethoughtof.Whatisthenumber?
2. Apositive number is 5 times another number. If 21 is added to both the numbers,
then one of the new numbers becomes twice the other new number. What are the
numbers?
3. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is
foundthattheresultingnewnumber isgreaterthantheoriginalnumberby27.What
isthetwo-digitnumber?
4. One of the two digits of a two digit number is three times the other digit. If you
interchange the digits of this two-digit number and add the resulting number to the
originalnumber,youget88. Whatistheoriginalnumber?
5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five
years from now will be one third of his mother’s present age. What are their
present ages?
6. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The
length and breadth of the plot are in the ratio 11:4.At the rate `100 per metre it will
cost the village panchayat ` 75000 to fence the plot. What are the dimensions of
theplot?
7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that
costs him ` 50 per metre and trouser material that costs him ` 90 per metre.
2021–22

32 MATHEMATICS
Forevery3metersoftheshirtmaterialhebuys2metres
of the trouser material. He sells the materials at 12%
and10%profitrespectively. Histotalsaleis` 36,600.
Howmuchtrousermaterialdidhebuy?
8. Halfofaherdofdeeraregrazinginthefieldandthree
fourthsoftheremainingareplayingnearby. Therest9
are drinking water from the pond. Find the number of
deer in the herd.
9. Agrandfatheristentimesolderthanhisgranddaughter.
He is also 54 years older than her. Find their present ages.
10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s
age. Find their present ages.
2.6 Reducing Equations to Simpler Form
Example 16: Solve
6 1 3
1
3 6
x x
+ −
+ =
Solution: Multiplyingbothsidesoftheequationby6,
6 (6 1)
6 1
3
x +
+ × =
6( 3)
6
x −
or 2 (6x + 1) + 6 = x – 3
or 12x + 2 + 6 = x – 3 (opening the brackets )
or 12x + 8 = x – 3
or 12x – x + 8 = – 3
or 11x + 8 = – 3
or 11x = –3 – 8
or 11x = –11
or x = – 1 (requiredsolution)
Check: LHS =
6( 1) 1 6 1
1 1
3 3
− + − +
+ = + =
5 3 5 3 2
3 3 3 3
− − + −
+ = =
RHS =
( 1) 3 4 2
6 6 3
− − − −
= =
LHS = RHS. (as required)
Example 17: Solve 5x – 2 (2x – 7) = 2 (3x – 1) +
7
2
Solution: Let us open the brackets,
LHS = 5x – 4x + 14 = x + 14
Why 6? Because it is the
smallest multiple (or LCM)
of the given denominators.
2021–22

Did you observe how we
simplified the form of the given
equation? Here, we had to
multiply both sides of the
equation by the LCM of the
denominators of the terms in the
expressions of the equation.
RHS = 6x – 2 +
7
2
=
4 7 3
6 6
2 2 2
x x
− + = +
The equation is x + 14 = 6x +
3
2
or 14 = 6x – x +
3
2
or 14 = 5x +
3
2
or 14 –
3
2
= 5x (transposing
3
2
)
or
28 3
2
−
= 5x
or
25
2
= 5x
or x =
25 1 5 5 5
2 5 2 5 2
×
× = =
×
Therefore, required solution is x =
5
2
.
Check: LHS =
=
25 25 25
2(5 7) 2( 2) 4
2 2 2
− − = − − = + =
25 8 33
2 2
+
=
RHS =
=
26 7 33
2 2
+
= = LHS. (as required)
EXERCISE 2.5
Solvethefollowinglinearequations.
1.
1 1
2 5 3 4
x x
− = + 2.
3 5
21
2 4 6
n n n
− + = 3.
8 17 5
7
3 6 2
x x
x + − = −
Note, in this example we
brought the equation to a
simpler form by opening
brackets and combining like
terms on both sides of the
equation.
2021–22

34 MATHEMATICS
4.
5 3
3 5
x x
− −
= 5.
3 2 2 3 2
4 3 3
t t
t
− +
− = − 6.
1 2
1
2 3
m m
m
− −
− = −
Simplifyandsolvethefollowinglinearequations.
7. 3(t – 3) = 5(2t + 1) 8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0
9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
10. 0.25(4f – 3) = 0.05(10f – 9)
2.7 Equations Reducible to the Linear Form
Example 18: Solve
1 3
2 3 8
x
x
+
=
+
Solution:Observe that the equation is not a linear equation, since the expression on its
LHS is not linear. But we can put it into the form of a linear equation. We multiply both
sides of the equation by (2x + 3),
x
x
x
+
+





 × +
1
2 3
2 3
( ) =
3
(2 3)
8
x
× +
Notice that (2x + 3) gets cancelled on the LHS We have then,
x + 1 =
3 (2 3)
8
x +
We have now a linear equation which we know how to solve.
Multiplyingbothsidesby8
8 (x + 1) = 3 (2x + 3)
or 8x + 8 = 6x + 9
or 8x = 6x + 9 – 8
or 8x = 6x + 1
or 8x – 6x = 1
or 2x = 1
or x =
1
2
The solution is x =
1
2
.
Check : Numerator of LHS =
1
2
+ 1 =
1 2 3
2 2
+
=
Denominator of LHS = 2x + 3 =
1
2
2
× + 3 = 1 + 3 = 4
This step can be
directly obtained by
‘cross-multiplication’
Note that
2x + 3 ≠ 0 (Why?)
2021–22

LHS = numerator ÷ denominator =
3
4
2
÷ =
3 1 3
2 4 8
× =
LHS = RHS.
Example 19:Present ages ofAnu and Raj are in the ratio 4:5. Eight years from now
the ratio of their ages will be 5:6. Find their present ages.
Solution: Let the present ages ofAnu and Raj be 4x years and 5x years respectively.
After eight years.Anu’s age = (4x + 8) years;
After eight years, Raj’s age = (5x + 8) years.
Therefore, the ratio of their ages after eight years =
4 8
5 8
x
x
+
+
This is given to be 5 : 6
Therefore,
4 8
5 8
x
x
+
+
=
5
6
Cross-multiplicationgives 6 (4x + 8) = 5 (5x + 8)
or 24x + 48 = 25x + 40
or 24x + 48 – 40 = 25x
or 24x + 8 = 25x
or 8 = 25x – 24x
or 8 = x
Therefore, Anu’s present age = 4x = 4 × 8 = 32 years
Raj’s present age = 5x = 5 × 8 = 40 years
EXERCISE 2.6
Solvethefollowingequations.
1.
8 3
2
3
x
x
−
= 2.
9
15
7 6
x
x
=
−
3.
4
15 9
z
z
=
+
4.
3 4 2
2 – 6 5
y
y
+ −
= 5.
7 4 4
2 3
y
y
+ −
=
+
6. The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of
their ages will be 3:4. Find their present ages.
7. The denominator of a rational number is greater than its numerator by 8. If the
numerator is increased by 17 and the denominator is decreased by 1, the number
obtained is
3
2
. Find the rational number.
2021–22

36 MATHEMATICS
1. Analgebraicequationisanequalityinvolvingvariables.Itsaysthatthevalueoftheexpressionon
one side of the equality sign is equal to the value of the expression on the other side.
2. The equations we study in Classes VI, VII and VIII are linear equations in one variable. In such
equations,theexpressionswhichformtheequationcontainonlyonevariable.Further,theequations
are linear, i.e., the highest power of the variable appearing in the equation is 1.
3. Alinearequationmayhaveforitssolutionanyrationalnumber.
4. An equation may have linear expressions on both sides. Equations that we studied in Classes VI
and VII had just a number on one side of the equation.
5. Just as numbers, variables can, also, be transposed from one side of the equation to the other.
6. Occasionally,theexpressionsformingequationshavetobesimplifiedbeforewecansolvethem
byusualmethods.Someequationsmaynotevenbelineartobeginwith,buttheycanbebrought
toalinearformbymultiplyingbothsidesoftheequationbyasuitableexpression.
7. Theutilityoflinearequationsisintheirdiverseapplications;differentproblemsonnumbers,ages,
perimeters, combination of currency notes, and so on can be solved using linear equations.
2021–22

UNDERSTANDING QUADRILATERALS 37
3.1 Introduction
You know that the paper is a model for a plane surface. When you join a number of
points without lifting a pencil from the paper (and without retracing any portion of the
drawing other than single points), you get a plane curve.
Trytorecalldifferentvarietiesofcurvesyouhaveseenintheearlierclasses.
Matchthefollowing:(Caution!Afiguremaymatchtomorethanonetype).
Figure Type
(1) (a) Simpleclosedcurve
(2) (b) Aclosedcurvethatisnotsimple
(3) (c) Simple curve that is not closed
(4) (d) Not a simple curve
Compare your matchings with those of your friends. Do they agree?
3.2 Polygons
A simple closed curve made up of only line segments is called apolygon.
Curves that are polygons Curves that are not polygons
Understanding
Quadrilaterals
CHAPTER
3
2021–22

38 MATHEMATICS
Try to give a few more examples and non-examples for a polygon.
Drawaroughfigureofapolygonandidentifyitssidesandvertices.
3.2.1 Classification of polygons
We classify polygons according to the number of sides (or vertices) they have.
Number of sides Classification Sample figure
or vertices
3 Triangle
4 Quadrilateral
5 Pentagon
6 Hexagon
7 Heptagon
8 Octagon
9 Nonagon
10 Decagon

n n-gon
3.2.2 Diagonals
Adiagonalisalinesegmentconnectingtwonon-consecutiveverticesofapolygon(Fig3.1).
Fig 3.1
2021–22

Can you name the diagonals in each of the above figures? (Fig 3.1)
Is PQ a diagonal? What about LN ?
You already know what we mean by interiorand exterior of a closed curve (Fig 3.2).
Interior Exterior
Theinteriorhasaboundary.Doestheexteriorhaveaboundary?Discusswithyourfriends.
3.2.3 Convex and concave polygons
Here are some convex polygons and some concave polygons. (Fig 3.3)
Fig 3.2
Fig 3.3
Convex polygons Concave polygons
Canyoufindhowthesetypesofpolygonsdifferfromoneanother?Polygonsthatare
convexhavenoportionsoftheirdiagonalsintheirexteriorsoranylinesegmentjoiningany
twodifferentpoints,intheinteriorofthepolygon,lieswhollyintheinteriorofit.Isthistrue
withconcavepolygons?Studythefiguresgiven.Thentrytodescribeinyourownwords
what we mean by a convex polygon and what we mean by a concave polygon. Give two
roughsketchesofeachkind.
Inourworkinthisclass,wewillbedealingwithconvexpolygonsonly.
3.2.4 Regular and irregular polygons
Aregularpolygonisboth‘equiangular’and‘equilateral’.Forexample,asquarehassidesof
equal length and angles of equal measure. Hence it is a regular polygon.Arectangle is
equiangularbutnotequilateral.Isarectanglearegularpolygon?Isanequilateraltrianglea
regularpolygon?Why?
2021–22

40 MATHEMATICS
Regular polygons Polygons that are not regular
DO THIS
[Note: Use of or indicates segments of equal length].
Inthepreviousclasses,haveyoucomeacrossanyquadrilateralthatisequilateralbutnot
equiangular?Recallthequadrilateralshapesyousawinearlierclasses–Rectangle,Square,
Rhombusetc.
Isthereatrianglethatisequilateralbutnotequiangular?
3.2.5 Angle sum property
Do you remember the angle-sum property of a triangle? The sum of the measures of the
three angles of a triangle is 180°. Recall the methods by which we tried to visualise this
fact. We now extend these ideas to a quadrilateral.
1. Takeanyquadrilateral,sayABCD(Fig3.4).Divide
itintotwotriangles,bydrawingadiagonal.Youget
six angles 1, 2, 3, 4, 5 and 6.
Usetheangle-sumpropertyofatriangleandargue
how the sum of the measures of ∠A, ∠B, ∠C and
∠D amounts to 180° + 180° = 360°.
2. Take four congruent card-board copies of any quadrilateralABCD, with angles
as shown [Fig 3.5 (i)].Arrange the copies as shown in the figure, where angles
∠1, ∠2, ∠3, ∠4 meet at a point [Fig 3.5 (ii)].
Fig 3.4
What can you say about the sum of the angles ∠1, ∠2, ∠3 and ∠4?
[Note: We denote the angles by ∠1, ∠2, ∠3, etc., and their respective measures
by m∠1, m∠2, m∠3, etc.]
The sum of the measures of the four angles of a quadrilateral is___________.
You may arrive at this result in several other ways also.
Fig 3.5
(i)
(ii)
For doing this you may
have to turn and match
appropriate corners so
that they fit.
2021–22

3. As before consider quadrilateralABCD (Fig 3.6). Let Pbe any
pointinitsinterior.JoinPtoverticesA,B,CandD. Inthefigure,
consider ∆PAB. From this we see x = 180° – m∠2 – m∠3;
similarly from ∆PBC, y = 180° – m∠4 – m∠5, from ∆PCD,
z = 180º – m∠6 – m∠7 and from ∆PDA, w = 180º – m∠8
– m∠1. Use this to find the total measure m∠1 + m∠2 + ...
+ m∠8, does it help you to arrive at the result? Remember
∠x + ∠y + ∠z + ∠w = 360°.
4. These quadrilaterals were convex. What would happen if the
quadrilateralisnotconvex?ConsiderquadrilateralABCD.Splitit
intotwotrianglesandfindthesumoftheinteriorangles(Fig3.7).
EXERCISE 3.1
1. Givenherearesomefigures.
(1) (2) (3) (4)
(5) (6) (7) (8)
Classifyeachofthemonthebasisofthefollowing.
(a) Simplecurve (b) Simpleclosedcurve (c) Polygon
(d) Convexpolygon (e) Concavepolygon
2. Howmanydiagonalsdoeseachofthefollowinghave?
(a) Aconvexquadrilateral (b) Aregularhexagon (c) Atriangle
3. Whatisthesumofthemeasuresoftheanglesofaconvexquadrilateral?Willthisproperty
holdifthequadrilateralisnotconvex?(Makeanon-convexquadrilateralandtry!)
4. Examine the table. (Each figure is divided into triangles and the sum of the angles
deduced from that.)
Figure
Side 3 4 5 6
Angle sum 180º 2 × 180° 3 × 180° 4 × 180°
= (4 – 2) × 180° = (5 – 2) × 180° = (6 – 2) × 180°
Fig 3.6
Fig 3.7
2021–22

42 MATHEMATICS
What can you say about the angle sum of a convex polygon with number of sides?
(a) 7 (b) 8 (c) 10 (d) n
5. Whatisaregularpolygon?
State the name of a regular polygon of
(i) 3 sides (ii) 4 sides (iii) 6 sides
6. Findtheanglemeasurexinthefollowingfigures.
(a) (b)
(c) (d)
7.
(a) Find x + y + z (b) Find x + y + z + w
3.3 Sum of the Measures of the Exterior Angles of a
Polygon
On many occasions a knowledge of exterior angles may throw light on the nature of
interioranglesandsides.
2021–22

DO THIS
Fig 3.8
TRY THESE
Draw a polygon on the floor, using a piece of chalk.
(In the figure, a pentagonABCDE is shown) (Fig 3.8).
We want to know the total measure of angles, i.e,
m∠1 + m∠2 + m∠3 + m∠4 + m∠5. Start at A. Walk
along AB . On reaching B, you need to turn through an
angle of m∠1, to walk along BC . When you reach at C,
you need to turn through an angle of m∠2 to walk along
CD.Youcontinuetomoveinthismanner,untilyoureturn
tosideAB.Youwouldhaveinfactmadeonecompleteturn.
Therefore, m∠1 + m∠2 + m∠3 + m∠4 + m∠5 = 360°
This is true whatever be the number of sides of the polygon.
Therefore, the sum of the measures of the external angles of any polygon is 360°.
Example 1: Find measure x in Fig 3.9.
Solution: x + 90° + 50° + 110° = 360° (Why?)
x + 250° = 360°
x = 110°
Take a regular hexagon Fig 3.10.
1. What is the sum of the measures of its exterior angles x, y, z, p, q, r?
2. Is x = y = z = p = q = r? Why?
3. What is the measure of each?
(i) exteriorangle (ii) interiorangle
4. Repeat this activity for the cases of
(i) aregularoctagon (ii) a regular 20-gon
Example 2: Find the number of sides of a regular polygon whose each exterior angle
has a measure of 45°.
Solution:Total measure of all exterior angles = 360°
Measure of each exterior angle = 45°
Therefore, the number of exterior angles =
360
45
= 8
The polygon has 8 sides.
Fig 3.9
Fig 3.10
2021–22

44 MATHEMATICS
EXERCISE 3.2
1. Findxinthefollowingfigures.
(a) (b)
2. Find the measure of each exterior angle of a regular polygon of
(i) 9 sides (ii) 15 sides
3. Howmanysidesdoesaregularpolygonhaveifthemeasureofanexteriorangleis24°?
4. How many sides does a regular polygon have if each of its interior angles
is 165°?
5. (a) Isitpossibletohavearegularpolygonwithmeasureofeachexteriorangleas22°?
(b) Can it be an interior angle of a regular polygon? Why?
6. (a) Whatistheminimuminterioranglepossibleforaregularpolygon?Why?
(b) Whatisthemaximumexterioranglepossibleforaregularpolygon?
3.4 Kinds of Quadrilaterals
Based on the nature of the sides or angles of a quadrilateral, it gets special names.
3.4.1 Trapezium
Trapeziumisaquadrilateralwithapairofparallelsides.
These are trapeziums These are not trapeziums
Studytheabovefiguresanddiscusswithyourfriendswhysomeofthemaretrapeziums
while some are not. (Note: The arrow marks indicate parallel lines).
1. Takeidenticalcut-outsofcongruenttrianglesofsides3cm,4cm,5cm.Arrange
them as shown (Fig 3.11).
Fig 3.11
DO THIS
2021–22

DO THIS
You get a trapezium. (Check it!) Which are the parallel sides here? Should the
non-parallelsidesbeequal?
Youcangettwomoretrapeziumsusingthesamesetoftriangles.Findthemoutand
discusstheirshapes.
2. Takefourset-squaresfromyour and yourfriend’s instrumentboxes.Usedifferent
numbersofthemtoplaceside-by-sideandobtaindifferenttrapeziums.
If the non-parallel sides of a trapezium are of equal length, we call it an isosceles
trapezium.Didyougetanisocelestrapeziuminanyofyourinvestigationsgivenabove?
3.4.2 Kite
Kite is a special type of a quadrilateral. The sides with the same markings in each figure
are equal. For exampleAB =AD and BC = CD.
Fig 3.12
Fig 3.13
Show that
∆ABC and
∆ADC are
congruent .
What do we
infer from
this?
These are kites These are not kites
Study these figures and try to describe what a kite is. Observe that
(i) A kite has 4 sides (It is a quadrilateral).
(ii) There are exactly two distinct consecutive pairs of sides of equal length.
Check whether a square is a kite.
Take a thick white sheet.
Fold the paper once.
DrawtwolinesegmentsofdifferentlengthsasshowninFig3.12.
Cut along the line segments and open up.
You have the shape of a kite (Fig 3.13).
Hasthekiteanylinesymmetry?
Fold both the diagonals of the kite. Use the set-square to check if they cut at
rightangles.Arethediagonalsequalinlength?
Verify(bypaper-foldingormeasurement)ifthediagonalsbisecteachother.
By folding an angle of the kite on its opposite, check for angles of equal measure.
Observethediagonalfolds;dotheyindicateanydiagonalbeingananglebisector?
Share your findings with others and list them.Asummary of these results are
givenelsewhereinthechapterforyourreference.
2021–22

46 MATHEMATICS
3.4.3 Parallelogram
Aparallelogram is a quadrilateral.As the name suggests, it has something to do with
parallellines.
AB CD

AB ED

BC FE

Study these figures and try to describe in your own words what we mean by a
parallelogram.Shareyourobservationswithyourfriends.
Check whether a rectangle is also a parallelogram.
Take two different rectangular cardboard strips of different widths (Fig 3.14).
Strip 1 Strip 2
Placeonestriphorizontallyanddrawlinesalong
itsedgeasdrawninthefigure(Fig3.15).
Nowplacetheotherstripinaslantpositionover
thelinesdrawnandusethistodrawtwomorelines
as shown (Fig 3.16).
Thesefourlinesencloseaquadrilateral.Thisismadeupoftwopairsofparallellines
(Fig 3.17).
DO THIS
Fig 3.14
Fig 3.15
Fig 3.16 Fig 3.17
AB DC

AD BC

LM ON

LO MN

QP SR

QS PR

These are parallelograms These are not parallelograms
2021–22

DO THIS
Itisaparallelogram.
A parallelogram is a quadrilateral whose opposite sides are parallel.
3.4.4 Elements of a parallelogram
Therearefoursidesandfouranglesinaparallelogram.Someoftheseare
equal.Therearesometermsassociatedwiththeseelementsthatyouneed
toremember.
GivenaparallelogramABCD(Fig3.18).
AB and DC, are opposite sides. AD and BC form another pair of opposite sides.
∠A and ∠C are a pair of opposite angles; another pair of opposite angles would be
∠B and ∠D.
AB and BC are adjacent sides. This means, one of the sides starts where the other
ends.Are BC and CD adjacent sides too? Try to find two more pairs of adjacent sides.
∠A and ∠B are adjacent angles. They are at the ends of the same side. ∠B and ∠C
are also adjacent. Identify other pairs of adjacent angles of the parallelogram.
Take cut-outs of two identical parallelograms, sayABCD andA′B′C′D′ (Fig 3.19).
Here AB is same as A B
′ ′ except for the name. Similarly the other corresponding
sides are equal too.
Place A B
′ ′ over DC.Dotheycoincide?Whatcanyounowsayaboutthelengths
AB and DC?
Similarlyexaminethelengths AD and BC .Whatdoyoufind?
You may also arrive at this result by measuring AB and DC.
Property: The opposite sides of a parallelogram are of equal length.
Take two identical set squares with angles 30° – 60° – 90°
andplacethemadjacentlytoformaparallelogramasshown
inFig3.20.Doesthishelpyoutoverifytheaboveproperty?
Youcanfurtherstrengthenthisidea
throughalogicalargumentalso.
Consideraparallelogram
ABCD (Fig 3.21). Draw
any one diagonal, say AC.
Fig 3.19
TRY THESE
Fig 3.20
Fig 3.21
Fig 3.18
2021–22

48 MATHEMATICS
Fig 3.22
DO THIS
TRY THESE
Lookingattheangles,
∠1 = ∠2 and ∠3 = ∠4 (Why?)
Since in triangles ABC and ADC, ∠1 = ∠2, ∠3 = ∠4
and AC is common, so, byASAcongruency condition,
∆ ABC ≅ ∆ CDA (How isASAused here?)
Thisgives AB = DC and BC =AD.
Example 3: Find the perimeter of the parallelogram PQRS (Fig 3.22).
Solution: In a parallelogram, the opposite sides have same length.
Therefore, PQ = SR = 12 cm and QR = PS = 7 cm
So, Perimeter= PQ + QR + RS + SP
= 12 cm + 7 cm + 12 cm + 7 cm = 38 cm
3.4.5 Angles of a parallelogram
We studied a property of parallelograms concerning the (opposite) sides.What can we
say about the angles?
LetABCD be a parallelogram (Fig 3.23). Copy it on
a tracing sheet. Name this copy asA′B′C′D′. Place
A′B′C′D′ on ABCD. Pin them together at the point
wherethediagonalsmeet.Rotatethetransparentsheet
by180°.Theparallelogramsstillconcide;butyounow
findA′lyingexactlyonCandvice-versa;similarlyB′
lies on D and vice-versa.
Does this tell you anything about the measures of the anglesAand C? Examine the
same for angles B and D. State your findings.
Property: The opposite angles of a parallelogram are of equal measure.
Take two identical 30° – 60° – 90° set-squares and form a parallelogram as before.
Does the figure obtained help you to confirm the above property?
Youcanfurtherjustifythisideathroughlogicalarguments.
If AC and BD are the diagonals of the
parallelogram,(Fig3.24)youfindthat
∠1 =∠2 and ∠3 = ∠4 (Why?)
Fig 3.23
Fig 3.24
2021–22

Fig 3.26
Studying ∆ABC and ∆ADC (Fig 3.25) separately, will help you to see that byASA
congruencycondition,
∆ ABC ≅ ∆ CDA (How?)
Fig 3.25
This shows that ∠B and ∠D have same measure. In the same way you can get
m∠A = m ∠C.
Alternatively, ∠1 = ∠2 and ∠3 = ∠4, we have, m∠A= ∠1+∠4 = ∠2+∠C m∠C
Example 4: In Fig 3.26, BEST is a parallelogram. Find the values x, y and z.
Solution: S is opposite to B.
So, x = 100° (opposite angles property)
y = 100° (measure of angle corresponding to ∠x)
z = 80° (since ∠y, ∠z is a linear pair)
We now turn our attention to adjacent angles of a parallelogram.
In parallelogramABCD,(Fig3.27).
∠A and ∠D are supplementary since
DC AB
and with transversal DA , these
two angles are interior opposite.
∠Aand∠B are also supplementary.Canyou
say‘why’?
AD BC
and BA is a transversal, making ∠A and ∠B interior opposite.
Identifytwomorepairsofsupplementaryanglesfromthefigure.
Property: The adjacent angles in a parallelogram are supplementary.
Example 5: InaparallelogramRING,(Fig3.28)ifm∠R=70°,findalltheotherangles.
Solution: Given m∠R = 70°
Then m∠N = 70°
because ∠R and ∠N are opposite angles of a parallelogram.
Since ∠R and ∠I are supplementary,
m∠I = 180° – 70° = 110°
Also, m∠G = 110° since ∠G is opposite to ∠I
Thus, m∠R = m∠N = 70° and m∠I = m∠G = 110°
Fig 3.27
Fig 3.28
2021–22

50 MATHEMATICS
DO THIS
After showing m∠R = m∠N = 70°, can you find m∠I and m∠G by any other
method?
3.4.6 Diagonals of a parallelogram
Thediagonalsofaparallelogram,ingeneral,arenotofequallength.
(Didyoucheckthisinyourearlieractivity?)However,thediagonals
ofaparallelogramhaveaninterestingproperty.
Take a cut-out of a parallelogram, say,
ABCD (Fig 3.29). Let its diagonals AC and DB meet at O.
Find the mid point of AC by a fold, placing C onA. Is the
mid-point same as O?
Doesthisshowthatdiagonal DB bisectsthediagonal AC atthepointO?Discussit
withyourfriends.Repeattheactivitytofindwherethemidpointof DB couldlie.
Property: The diagonals of a parallelogram bisect each other (at the point of their
intersection, of course!)
To argue and justify this property is not very
difficult. From Fig 3.30, applyingASAcriterion, it
is easy to see that
∆ AOB ≅ ∆ COD (How is ASAused here?)
Thisgives AO = CO and BO = DO
Example 6: In Fig 3.31 HELP is a parallelogram. (Lengths are in cms). Given that
OE = 4 and HL is 5 more than PE? Find OH.
Solution : If OE = 4 then OP also is 4 (Why?)
So PE = 8, (Why?)
Therefore HL= 8 + 5 = 13
Hence OH =
1
13
2
× = 6.5 (cms)
EXERCISE 3.3
1. Given a parallelogramABCD. Complete each
statementalongwiththedefinitionorpropertyused.
(i) AD = ...... (ii) ∠ DCB = ......
(iii) OC = ...... (iv) m ∠DAB + m ∠CDA = ......
Fig 3.31
Fig 3.29
Fig 3.30
2021–22

2. Considerthefollowingparallelograms.Findthevaluesoftheunknownsx,y,z.
(i) (ii)
30
(iii) (iv) (v)
3. Can a quadrilateralABCD be a parallelogram if
(i) ∠D + ∠B = 180°? (ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles
ofequalmeasure.
5. Themeasuresof twoadjacentanglesofaparallelogramareintheratio3:2.Findthemeasureofeach
oftheanglesoftheparallelogram.
6. Twoadjacentanglesofaparallelogramhaveequalmeasure.Findthe
measureofeachoftheanglesoftheparallelogram.
7. TheadjacentfigureHOPEisaparallelogram.Findtheanglemeasures
x, y and z. State the properties you use to find them.
8. The following figures GUNS and RUNS are parallelograms.
Find x andy. (Lengths are in cm)
9.
In the above figure both RISK and CLUE are parallelograms. Find the value of x.
(i) (ii)
2021–22

52 MATHEMATICS
DO THIS
10. Explainhowthisfigureisatrapezium.Whichofitstwosidesareparallel?(Fig3.32)
11. Find m∠C in Fig 3.33 if AB DC
.
12. Find the measure of ∠P and ∠S if SP RQ
in Fig 3.34.
(Ifyoufindm∠R,istheremorethanonemethodtofindm∠P?)
3.5 Some Special Parallelograms
3.5.1 Rhombus
We obtain a Rhombus (which, you will see, is a parallelogram) as a special case of kite
(whichisnotaaparallelogram).
Recall the paper-cut kite you made earlier.
Kite-cut Rhombus-cut
When you cut alongABC and opened up, you got a kite. Here lengthsAB and
BCweredifferent.IfyoudrawAB=BC,thenthekiteyouobtainiscalledarhombus.
Fig 3.33
Fig 3.34
Fig 3.32
Note that the sides of rhombus are all of same
length;thisisnotthecasewiththekite.
A rhombus is a quadrilateral with sides of equal
length.
Sincetheoppositesidesofarhombushavethesame
length,itisalsoaparallelogram.So,arhombushasall
the properties of a parallelogram and also that of a
kite.Trytolistthemout.Youcanthenverifyyourlist
withthechecklistsummarisedinthebookelsewhere. Kite Rhombus
2021–22

DO THIS
The most useful property of a rhombus is that of its diagonals.
Property: The diagonals of a rhombus are perpendicular bisectors of one another.
Take a copy of rhombus. By paper-folding verify if the point of intersection is the
mid-pointofeachdiagonal.Youmayalsocheckiftheyintersectatrightangles,using
the corner of a set-square.
Hereisanoutlinejustifyingthispropertyusinglogicalsteps.
ABCD is a rhombus (Fig 3.35). Therefore it is a parallelogram too.
Since diagonals bisect each other, OA = OC and OB = OD.
We have to show that m∠AOD = m∠COD = 90°
It can be seen that by SSS congruency criterion
∆AOD ≅ ∆ COD
Therefore, m ∠AOD = m ∠COD
Since ∠AOD and ∠COD are a linear pair,
m ∠AOD = m ∠COD = 90°
Example 7:
RICE is a rhombus (Fig 3.36). Find x, y,z.Justify your findings.
Solution:
x = OE y = OR z = sideoftherhombus
= OI (diagonals bisect) = OC (diagonals bisect) = 13 (all sides are equal )
= 5 = 12
3.5.2 A rectangle
Arectangleisaparallelogramwithequalangles(Fig3.37).
Whatisthefullmeaningofthisdefinition?Discusswithyourfriends.
If the rectangle is to be equiangular, what could be
the measure of each angle?
Let the measure of each angle be x°.
Then 4x° = 360° (Why)?
Therefore, x° = 90°
Thuseachangleofarectangleisarightangle.
So,arectangleisaparallelograminwhicheveryangleisarightangle.
Beingaparallelogram,therectanglehasoppositesidesofequallengthanditsdiagonals
bisect each other.
Fig 3.35
Since AO = CO (Why?)
AD = CD (Why?)
OD = OD
Fig 3.36
Fig 3.37
2021–22

54 MATHEMATICS
Inaparallelogram,thediagonalscanbeofdifferentlengths.(Checkthis);butsurprisingly
therectangle(beingaspecialcase)hasdiagonalsofequallength.
Property: The diagonals of a rectangle are of equal length.
This is easy to justify. IfABCD is a rectangle (Fig 3.38), then looking at triangles
ABC andABD separately [(Fig 3.39) and (Fig 3.40) respectively], we have
∆ ABC ≅ ∆ ABD
Thisisbecause AB = AB (Common)
BC = AD (Why?)
m ∠A = m ∠B = 90° (Why?)
ThecongruencyfollowsbySAScriterion.
Thus AC = BD
and inarectanglethediagonals,besidesbeingequalinlengthbisecteachother(Why?)
Example 8: RENT is a rectangle (Fig 3.41). Its diagonals meet at O. Find x, if
OR = 2x + 4 and OT = 3x + 1.
Solution: OT is half of the diagonal TE ,
OR is half of the diagonal RN .
Diagonalsareequalhere.(Why?)
So, their halves are also equal.
Therefore 3x + 1 = 2x + 4
or x = 3
3.5.3 A square
A square is a rectangle with equal sides.
This means a square has all the
propertiesofarectanglewithanadditional
requirementthatallthesideshaveequal
length.
The square, like the rectangle, has
diagonalsofequallength.
Inarectangle,thereisnorequirement
for the diagonals to be perpendicular to
one another, (Check this).
Fig 3.40
Fig 3.39
Fig 3.38
Fig 3.41
BELT is a square, BE = EL = LT = TB
∠B, ∠E, ∠L, ∠T are right angles.
BL = ET and BL ET
⊥ .
OB = OL and OE = OT.
2021–22

DO THIS
In a square the diagonals.
(i) bisect one another (squarebeingaparallelogram)
(ii) areofequallength (square being a rectangle) and
(iii) are perpendicular to one another.
Hence,wegetthefollowingproperty.
Property: The diagonals of a square are perpendicular bisectors of each other.
Take a square sheet, say PQRS (Fig 3.42).
Fold along both the diagonals.Are their mid-points the same?
Check if the angle at O is 90° by using a set-square.
This verifies the property stated above.
Wecanjustifythisalsobyarguinglogically:
ABCD is a square whose diagonals meet at O (Fig 3.43).
OA = OC (Since the square is a parallelogram)
By SSS congruency condition, we now see that
∆ AOD ≅ ∆ COD (How?)
Therefore, m∠AOD = m∠COD
Theseanglesbeingalinearpair,eachisrightangle.
EXERCISE 3.4
1. State whetherTrue or False.
(a) Allrectanglesaresquares (e) Allkitesarerhombuses.
(b) Allrhombusesareparallelograms (f) Allrhombusesarekites.
(c) Allsquaresarerhombusesandalsorectangles (g) Allparallelogramsaretrapeziums.
(d) Allsquaresarenotparallelograms. (h) Allsquaresaretrapeziums.
2. Identifyallthequadrilateralsthathave.
(a) foursidesofequallength (b) fourrightangles
3. Explainhowasquareis.
(i) aquadrilateral (ii) aparallelogram (iii) arhombus (iv) arectangle
4. Namethequadrilateralswhosediagonals.
(i) bisect each other (ii) are perpendicular bisectors of each other (iii) are equal
5. Explainwhyarectangleisaconvexquadrilateral.
6. ABCisaright-angledtriangleandOisthemidpointoftheside
oppositetotherightangle.ExplainwhyOisequidistantfromA,
B and C. (The dotted lines are drawn additionally to help you).
Fig 3.42
Fig 3.43
2021–22

56 MATHEMATICS
1. Amasonhasmadeaconcreteslab.Heneedsittoberectangular.Inwhatdifferent
ways can he make sure that it is rectangular?
2. A square was defined as a rectangle with all sides equal. Can we define it as
rhombuswithequalangles?Explorethisidea.
3. Canatrapeziumhaveallanglesequal?Canithaveallsidesequal?Explain.
Quadrilateral Properties
(1) Opposite sides are equal.
(2) Opposite angles are equal.
(3) Diagonalsbisectoneanother.
(1) Allthepropertiesofaparallelogram.
(2) Diagonals are perpendicular to each other.
(1) Allthepropertiesofaparallelogram.
(2) Eachoftheanglesisarightangle.
(3) Diagonalsareequal.
Allthepropertiesofaparallelogram,
rhombus and a rectangle.
(1) Thediagonalsareperpendicular
tooneanother
(2) One of the diagonals bisects the other.
(3) In the figure m∠B = m∠D but
m∠A ≠ m∠C.
Parallelogram:
A quadrilateral
with each pair of
opposite sides
parallel.
Rhombus:
A parallelogram with sides
ofequallength.
Rectangle:
A parallelogram
witharightangle.
Square: A rectangle
with sides of equal
length.
Kite:Aquadrilateral
withexactlytwopairs
of equal consecutive
sides
2021–22

PRACTICAL GEOMETRY 57
DO THIS
4.1 Introduction
You have learnt how to draw triangles in Class VII. We require three measurements
(of sides and angles) to draw a unique triangle.
Since three measurements were enough to draw a triangle, a natural question arises
whetherfourmeasurementswouldbesufficienttodrawauniquefoursidedclosedfigure,
namely,aquadrilateral.
Takeapairofsticksofequallengths,say
10 cm. Take another pair of sticks of
equal lengths, say, 8 cm. Hinge them up
suitablytogetarectangleoflength10cm
and breadth 8 cm.
Thisrectanglehasbeencreatedwith
the4availablemeasurements.
Now just push along the breadth of
the rectangle. Is the new shape obtained,
still a rectangle (Fig 4.2)? Observe
that the rectangle has now become
a parallelogram. Have you altered the
lengths of the sticks? No! The
measurements ofsidesremain the same.
Give another push to the newly
obtained shape in a different direction;
what do you get? You again get a
parallelogram,whichisaltogetherdifferent
(Fig 4.3), yet the four measurements
remainthesame.
Thisshowsthat4measurementsofaquadrilateralcannotdetermineituniquely.
Can5measurementsdetermineaquadrilateraluniquely?Letusgobacktotheactivity!
Practical Geometry
CHAPTER
4
Fig 4.1
Fig 4.2
Fig 4.3
2021–22

58 MATHEMATICS
Youhaveconstructedarectanglewith
twostickseachoflength10cmandother
two sticks each of length 8 cm. Now
introduce another stick of length equal to
BD and tie it along BD (Fig 4.4). If you
push the breadth now, does the shape
change?No!Itcannot,withoutmakingthe
figure open. The introduction of the fifth
stickhasfixedtherectangleuniquely,i.e.,
there is no other quadrilateral (with the
givenlengthsofsides)possiblenow.
Thus,weobservethatfivemeasurementscandetermineaquadrilateraluniquely.
Butwillanyfivemeasurements(ofsidesandangles)besufficienttodrawaunique
quadrilateral?
Arshad has five measurements of a quadrilateralABCD. These are AB = 5 cm,
∠A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique
quadrilateral?Givereasonsforyouranswer.
4.2 Constructing a Quadrilateral
We shall learn how to construct a unique quadrilateral given the following
measurements:
• When four sides and one diagonal are given.
• When two diagonals and three sides are given.
• When two adjacent sides and three angles are given.
• When three sides and two included angles are given.
• When other special properties are known.
Let us take up these constructions one-by-one.
4.2.1 When the lengths of four sides and a diagonal are given
Weshallexplainthisconstructionthroughanexample.
Example 1: Construct a quadrilateral PQRS
where PQ = 4 cm,QR = 6 cm, RS = 5 cm,
PS = 5.5 cm and PR = 7 cm.
Solution: [A rough sketch will help us in
visualisingthequadrilateral.Wedrawthisfirstand
mark the measurements.] (Fig 4.5)
Fig 4.4
Fig 4.5
2021–22

Step 1 From the rough sketch, it is easy to see that ∆PQR
canbeconstructedusingSSSconstructioncondition.
Draw ∆PQR (Fig 4.6).
Step 2 Now, we have to locate the fourth point S.This ‘S’
would be on the side opposite to Q with reference to
PR. For that, we have two measurements.
S is 5.5 cm away from P. So, with Pas centre, draw
an arc of radius 5.5 cm. (The point S is somewhere
on this arc!) (Fig 4.7).
Step 3 S is 5 cm away from R. So with R as centre, draw an arc of radius 5 cm (The
point S is somewhere on this arc also!) (Fig 4.8).
Fig 4.6
Fig 4.7
Fig 4.8
2021–22

60 MATHEMATICS
Step 4 S should lie on both the arcs drawn.
Soitisthepointofintersectionofthe
twoarcs.MarkSandcompletePQRS.
PQRS is the required quadrilateral
(Fig4.9).
(i) We saw that 5 measurements of a quadrilateral can determine a quadrilateral
uniquely.Doyouthinkanyfivemeasurementsofthequadrilateralcandothis?
(ii) Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and
AS = 6.5 cm? Why?
(iii) Can you draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm? Why?
(iv) Astudent attempted to draw a quadrilateral PLAYwhere PL= 3 cm, LA= 4 cm,
AY = 4.5 cm, PY = 2 cm and LY = 6 cm, but could not draw it. What is
the reason?
[Hint:Discussitusingaroughsketch].
EXERCISE 4.1
1. Constructthefollowingquadrilaterals.
(i) QuadrilateralABCD. (ii) QuadrilateralJUMP
AB = 4.5 cm JU = 3.5 cm
BC = 5.5 cm UM = 4 cm
CD = 4 cm MP = 5 cm
AD = 6 cm PJ = 4.5 cm
AC = 7 cm PU = 6.5 cm
(iii) ParallelogramMORE (iv) RhombusBEST
OR = 6 cm BE = 4.5 cm
RE = 4.5 cm ET = 6 cm
EO = 7.5 cm
Fig 4.9
2021–22

Fig 4.12
4.2.2 When two diagonals and three sides are given
Whenfoursidesandadiagonalweregiven,wefirstdrewatrianglewiththeavailabledata
and then tried to locate the fourth point. The same technique is used here.
Example 2: Construct a quadrilateralABCD, given that BC = 4.5 cm,AD = 5.5 cm,
CD = 5 cm the diagonalAC = 5.5 cm and diagonal BD = 7 cm.
Solution:
HereistheroughsketchofthequadrilateralABCD
(Fig 4.10). Studying this sketch, we can easily see
that it is possible to draw ∆ACD first (How?).
Step 1 Draw ∆ACD using SSS
construction(Fig4.11).
(WenowneedtofindBatadistance
of 4.5 cm from C and 7 cm from D).
Step 2 With D as centre, draw an arc of radius 7 cm. (B is somewhere
on this arc) (Fig 4.12).
Step 3 With C as centre, draw an arc of
radius 4.5 cm (B is somewhere on
thisarcalso) (Fig 4.13).
Fig 4.13
Fig 4.11
Fig 4.10
2021–22

62 MATHEMATICS
Step 4 Since B lies on both the arcs, B is
the point intersection of the two
arcs.MarkBandcompleteABCD.
ABCDistherequiredquadrilateral
(Fig 4.14).
Fig 4.14
1. In the above example, can we draw the quadrilateral by drawing ∆ABD first and
thenfindthefourthpointC?
2. CanyouconstructaquadrilateralPQRSwithPQ=3cm,RS=3cm,PS=7.5cm,
PR = 8 cm and SQ = 4 cm? Justify your answer.
EXERCISE 4.2
(i) quadrilateralLIFT (ii) QuadrilateralGOLD
LI = 4 cm OL = 7.5 cm
IF = 3 cm GL = 6 cm
TL = 2.5 cm GD = 6 cm
LF = 4.5 cm LD = 5 cm
IT = 4 cm OD = 10 cm
(iii) RhombusBEND
BN = 5.6 cm
DE = 6.5 cm
4.2.3 When two adjacent sides and three angles are known
As before, we start with constructing a triangle and then look for the fourth point to
completethequadrilateral.
Example 3: Construct a quadrilateral MIST where MI = 3.5 cm, IS = 6.5 cm,
∠M = 75°, ∠I = 105° and ∠S = 120°.
2021–22

Solution:
Here is a rough sketch that would help us in deciding our steps of
construction.We give only hints for various steps (Fig 4.15).
Step 1 How do you locate the points? What choice do you make for the base and what
is the first step? (Fig 4.16)
Fig 4.16
Step 2 Make ∠ISY = 120° at S (Fig 4.17).
Fig 4.17
Fig 4.15
2021–22

64 MATHEMATICS
Step 3 Make ∠IMZ = 75° at M. (where will SYand MZ meet?) Mark that point asT.
We get the required quadrilateral MIST(Fig 4.18).
Fig 4.18
1. Can you construct the above quadrilateral MIST if we have 100° at M instead
of 75°?
2. Can you construct the quadrilateral PLAN if PL = 6 cm, LA = 9.5 cm, ∠P = 75°,
∠L =150° and ∠A = 140°? (Hint: Recall angle-sum property).
3. Inaparallelogram,thelengthsofadjacentsidesareknown.Dowestillneedmeasures
of the angles to construct as in the example above?
EXERCISE 4.3
(i) QuadrilateralMORE (ii) QuadrilateralPLAN
MO = 6 cm PL = 4 cm
OR = 4.5 cm LA = 6.5 cm
∠M = 60° ∠P = 90°
∠O = 105° ∠A= 110°
∠R = 105° ∠N = 85°
(iii) ParallelogramHEAR (iv) Rectangle OKAY
HE = 5 cm OK = 7 cm
EA = 6 cm KA = 5 cm
∠R = 85°
2021–22

4.2.4 When three sides and two included angles are given
Under this type, when you draw a rough sketch, note carefully the “included” angles
inparticular.
Example 4: Construct a quadrilateral ABCD, where
AB = 4 cm, BC = 5 cm, CD = 6.5 cm and ∠B = 105° and
∠C = 80°.
Solution:
Wedrawaroughsketch,asusual,togetanideaofhowwecan
start off. Then we can devise a plan to locate the four points
(Fig 4.19). Fig 4.19
Fig 4.20
Step 2 ThefourthpointDisonCYwhichisinclinedat80°toBC.Somake∠BCY=80°
atConBC(Fig4.21).
Step 1 Start with taking BC = 5 cm on B. Draw an angle of 105° along BX. LocateA
4 cm away on this. We now have B, C andA(Fig 4.20).
Fig 4.21
2021–22

66 MATHEMATICS
Step 3 D is at a distance of 6.5 cm on CY. With
C as centre, draw an arc of length 6.5 cm.
It cuts CY at D (Fig 4.22).
1. In the above example, we first drew BC. Instead, what could have been be the
otherstartingpoints?
2. We used some five measurements to draw quadrilaterals so far. Can there be
differentsetsoffivemeasurements(otherthanseensofar)todrawaquadrilateral?
Thefollowingproblemsmayhelpyouinansweringthequestion.
(i) QuadrilateralABCD withAB = 5 cm, BC = 5.5 cm, CD = 4 cm,AD = 6 cm
and ∠B = 80°.
(ii) Quadrilateral PQRS with PQ = 4.5 cm, ∠P = 70°, ∠Q = 100°, ∠R = 80°
and ∠S = 110°.
Constructafewmoreexamplesofyourowntofindsufficiency/insufficiencyofthe
data for construction of a quadrilateral.
Fig 4.22
Step 4 CompletethequadrilateralABCD.ABCDistherequiredquadrilateral(Fig4.23).
Fig 4.23
2021–22

TRY THESE
EXERCISE 4.4
(i) QuadrilateralDEAR (ii) QuadrilateralTRUE
DE = 4 cm TR = 3.5 cm
EA = 5 cm RU = 3 cm
AR = 4.5 cm UE = 4 cm
∠E = 60° ∠R = 75°
∠A = 90° ∠U = 120°
4.3 Some Special Cases
To draw a quadrilateral, we used 5 measurements in our work. Is there any quadrilateral
whichcanbedrawnwithlessnumberofavailablemeasurements?Thefollowingexamples
examinesuchspecialcases.
Example 5: Draw a square of side 4.5 cm.
Solution: Initiallyitappearsthatonlyonemeasurementhasbeengiven.Actually
wehavemanymoredetailswithus,becausethefigureisaspecialquadrilateral,
namelyasquare.Wenowknowthateachofitsanglesisarightangle. (Seethe
roughfigure)(Fig4.24)
Thisenablesustodraw∆ABCusingSAScondition.ThenDcanbeeasily
located.Try yourself now to draw the square with the given measurements.
Example 6: Is it possible to construct a rhombusABCD whereAC = 6 cm
and BD = 7 cm? Justify your answer.
Solution: Only two (diagonal) measurements of the rhombus are given. However,
since it is a rhombus, we can find more help from its properties.
Thediagonalsofarhombusareperpendicularbisectors
of one another.
So,firstdrawAC=7cmandthenconstructitsperpendicularbisector.
Let them meet at 0. Cut off 3 cm lengths on either side of the drawn
bisector.You now get B and D.
Draw the rhombus now, based on the method described above
(Fig 4.25).
1. HowwillyouconstructarectanglePQRSifyouknow
onlythelengthsPQandQR?
2. Construct the kite EASYifAY= 8 cm, EY= 4 cm
and SY = 6 cm (Fig 4.26). Which properties of the
kite did you use in the process?
Fig 4.24
Fig 4.25
Fig 4.26
2021–22

68 MATHEMATICS
EXERCISE 4.5
Drawthefollowing.
1. The square READ with RE = 5.1 cm.
2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
3. A rectangle with adjacent sides of lengths 5 cm and 4 cm.
4. Aparallelogram OKAYwhere OK = 5.5 cm and KA= 4.2 cm. Is it unique?
1. Fivemeasurementscandetermineaquadrilateraluniquely.
2. Aquadrilateralcanbeconstructeduniquelyifthelengthsofitsfoursidesandadiagonalisgiven.
3. A quadrilateral can be constructed uniquely if its two diagonals and three sides are known.
4. A quadrilateral can be constructed uniquely if its two adjacent sides and three angles are known.
5. Aquadrilateralcanbeconstructeduniquelyifitsthreesidesandtwoincludedanglesaregiven.
2021–22

DATA HANDLING 69
5.1 Looking for Information
Inyourday-to-daylife,youmighthavecomeacrossinformation,suchas:
(a) Runs made by a batsman in the last 10 test matches.
(b) Number of wickets taken by a bowler in the last 10 ODIs.
(c) Marks scored by the students of your class in the Mathematics unit test.
(d) Number of story books read by each of your friends etc.
Theinformationcollectedinallsuchcasesiscalleddata.Dataisusuallycollectedin
thecontextofasituationthatwewanttostudy.Forexample,ateachermayliketoknow
theaverageheightofstudentsinherclass.Tofindthis,shewillwritetheheightsofallthe
students in her class, organise the data in a systematic manner and then interpret it
accordingly.
Sometimes,dataisrepresentedgraphically togiveaclearideaofwhatitrepresents.
Do you remember the different types of graphs which we have learnt in earlier classes?
1. APictograph:Pictorialrepresentationofdatausingsymbols.
Data Handling
CHAPTER
5
= 100 cars ← One symbol stands for 100 cars
July = 250 denotes
1
2
of 100
August = 300
September = ?
(i) How many cars were produced in the month of July?
(ii) Inwhichmonthweremaximumnumberofcarsproduced?
2021–22

70 MATHEMATICS
2. Abar graph:Adisplay of information using bars of uniform width, their heights
beingproportionaltotherespectivevalues.
Bar heights give the
quantity for each
category.
Bars are of equal width
with equal gaps in
between.
(i) Whatistheinformationgivenbythebargraph?
(ii) Inwhichyearistheincreaseinthenumberofstudentsmaximum?
(iii) Inwhichyearisthenumberofstudentsmaximum?
(iv) State whether true or false:
‘The number of students during 2005-06 is twice that of 2003-04.’
3. Double Bar Graph:Abar graph showing two sets of data simultaneously. It is
useful for the comparison of the data.
(i) Whatistheinformationgivenbythedoublebargraph?
(ii) In which subject has the performance improved the most?
(iii) In which subject has the performance deteriorated?
(iv) In which subject is the performance at par?
2021–22

DATA HANDLING 71
If we change the position of any of the bars of a bar graph, would it change the
informationbeingconveyed?Why?
1. Month July August September October November December
Numberof 1000 1500 1500 2000 2500 1500
watches sold
2. Children who prefer School A School B School C
Walking 40 55 15
Cycling 45 25 35
3. Percentage wins in ODI by 8 top cricket teams.
Teams From Champions Last 10
Trophy to World Cup-06 ODI in 07
SouthAfrica 75% 78%
Australia 61% 40%
SriLanka 54% 38%
New Zealand 47% 50%
England 46% 50%
Pakistan 45% 44%
West Indies 44% 30%
India 43% 56%
TRY THESE
Draw an appropriate graph to represent the given information.
5.2 Organising Data
Usually,dataavailabletousisinanunorganisedformcalledrawdata.Todrawmeaningful
inferences,weneedtoorganisethedatasystematically.Forexample,agroupofstudents
was asked for their favourite subject. The results were as listed below:
Art,Mathematics,Science,English,Mathematics,Art,English,Mathematics,English,
Art,Science,Art,Science,Science,Mathematics,Art,English, Art,Science,Mathematics,
Science,Art.
Which is the most liked subject and the one least liked?
2021–22

72 MATHEMATICS
TRY THESE
It is not easy to answer the question looking at the choices written haphazardly.We
arrange the data inTable 5.1 using tally marks.
Table 5.1
Subject Tally Marks Number of Students
Art | | | | | | 7
Mathematics | | | | 5
Science | | | | | 6
English | | | | 4
The number of tallies before each subject gives the number of students who like that
particularsubject.
This is known as the frequency of that subject.
Frequency gives the number of times that a particular entry occurs.
FromTable 5.1, FrequencyofstudentswholikeEnglishis4
Frequency of students who like Mathematics is 5
The table made is known as frequency distributiontable as it gives the number
of times an entry occurs.
1. A group of students were asked to say which animal they would like most to have
as a pet. The results are given below:
dog, cat, cat, fish, cat, rabbit, dog, cat, rabbit, dog, cat, dog, dog, dog, cat, cow,
fish, rabbit, dog, cat, dog, cat, cat, dog, rabbit, cat, fish, dog.
Make a frequency distribution table for the same.
5.3 Grouping Data
Thedataregardingchoiceofsubjectsshowedtheoccurrenceofeachoftheentriesseveral
times. For example,Art is liked by 7 students, Mathematics is liked by 5 students and so
on (Table 5.1). This information can be displayed graphically using a pictograph or a
bargraph.Sometimes,however,wehavetodealwithalargedata.Forexample,consider
the following marks (out of 50) obtained in Mathematics by 60 students of Class VIII:
21, 10, 30, 22, 33, 5, 37, 12, 25, 42, 15, 39, 26, 32, 18, 27, 28, 19, 29, 35, 31, 24,
36, 18, 20, 38, 22, 44, 16, 24, 10, 27, 39, 28, 49, 29, 32, 23, 31, 21, 34, 22, 23, 36, 24,
36, 33, 47, 48, 50, 39, 20, 7, 16, 36, 45, 47, 30, 22, 17.
If we make a frequency distribution table for each observation, then the table would
be too long, so, for convenience, we make groups of observations say, 0-10, 10-20 and
so on, and obtain a frequency distribution of the number of observations falling in each
2021–22

DATA HANDLING 73
group. Thus, the frequency distribution table for the above data can be.
Table 5.2
Groups Tally Marks Frequency
0-10 | | 2
10-20 | | | | | | | | 10
20-30 | | | | | | | | | | | | | | | | | 21
30-40 | | | | | | | | | | | | | | | | 19
40-50 | | | | | | 7
50-60 | 1
Total 60
Datapresentedinthismannerissaidtobegroupedandthedistributionobtainediscalled
groupedfrequency distribution. It helps us to draw meaningful inferences like –
(1) Most of the students have scored between 20 and 40.
(2) Eight students have scored more than 40 marks out of 50 and so on.
Each of the groups 0-10, 10-20, 20-30, etc., is called a Class Interval (or briefly
a class).
Observe that 10 occurs in both the classes, i.e., 0-10 as well as 10-20. Similarly, 20
occursinclasses10-20and20-30.Butit is not possible that anobservation(say10or20)
canbelongsimultaneouslytotwoclasses.Toavoidthis,weadopttheconventionthatthe
commonobservationwillbelongtothehigherclass,i.e., 10 belongs to the class interval
10-20 (and not to 0-10). Similarly, 20 belongs to 20-30 (and not to 10-20). In the class
interval, 10-20, 10 is called the lower class limit and 20 is called the upper class limit.
Similarly,intheclassinterval20-30,20isthelowerclasslimitand30istheupperclasslimit.
Observethatthedifferencebetweentheupperclasslimitandlowerclasslimitforeachofthe
classintervals0-10,10-20,20-30etc., isequal,(10inthiscase).Thisdifferencebetween
theupperclasslimitandlowerclasslimitiscalled thewidthorsizeoftheclassinterval.
TRY THESE
1. Study the following frequency distribution table and answer the questions
given below.
Frequency Distribution of Daily Income of 550 workers of a factory
Table 5.3
Class Interval Frequency
(Daily Income in `) (Number of workers)
100-125 45
125-150 25
2021–22

74 MATHEMATICS
150-175 55
175-200 125
200-225 140
225-250 55
250-275 35
275-300 50
300-325 20
Total 550
(i) Whatisthesizeoftheclassintervals?
(ii) Whichclasshasthehighestfrequency?
(iii) Whichclasshasthelowestfrequency?
(iv) What is the upper limit of the class interval 250-275?
(v) Which two classes have the same frequency?
2. Constructafrequencydistributiontableforthedataonweights(inkg)of20students
of a class using intervals 30-35, 35-40 and so on.
40, 38, 33, 48, 60, 53, 31, 46, 34, 36, 49, 41, 55, 49, 65, 42, 44, 47, 38, 39.
5.3.1 Bars with a difference
Let us again consider the grouped frequency distribution of the marks obtained by 60
students in Mathematics test. (Table 5.4)
Table 5.4
Class Interval Frequency
0-10 2
10-20 10
20-30 21
30-40 19
40-50 7
50-60 1
Total 60
This is displayed graphically as in the
adjoininggraph(Fig5.1).
Isthisgraphinanywaydifferentfromthe
bargraphswhichyouhavedrawninClassVII?
Observe that, here we have represented the
groups of observations (i.e., class intervals) Fig 5.1
2021–22

DATA HANDLING 75
TRY THESE
on the horizontal axis. The height of the bars show the frequency of the class-interval.
Also, there is no gap between the bars as there is no gap between the class-intervals.
Thegraphicalrepresentationofdatainthismanneriscalledahistogram.
Thefollowinggraphisanotherhistogram(Fig5.2).
Fromthebarsofthishistogram,wecananswerthefollowingquestions:
(i) How many teachers are of age 45 years or more but less than 50 years?
(ii) How many teachers are of age less than 35 years?
1. Observe the histogram (Fig 5.3) and answer the questions given below.
Fig 5.3
(i) Whatinformationisbeinggivenbythehistogram?
(ii) Whichgroupcontainsmaximumgirls?
Fig 5.2
2021–22

76 MATHEMATICS
(iii) How many girls have a height of 145 cms and more?
(iv) If we divide the girls into the following three categories, how many would
there be in each?
150 cm and more — GroupA
140 cm to less than 150 cm — Group B
Less than 140 cm — Group C
EXERCISE 5.1
1. For which of these would you use a histogram to show the data?
(a) The number of letters for different areas in a postman’s bag.
(b) Theheightofcompetitorsinanathleticsmeet.
(c) The number of cassettes produced by 5 companies.
(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m. at a
station.
Give reasons for each.
2. The shoppers who come to a departmental store are marked as: man (M), woman
(W),boy(B)orgirl(G).Thefollowinglistgivestheshopperswhocameduringthe
firsthourinthemorning:
W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W
W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W
Makeafrequencydistributiontableusingtallymarks.Drawabargraphtoillustrateit.
3. The weekly wages (in `) of 30 workers in a factory are.
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845,
804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Using tally marks make a frequency table with intervals as 800–810, 810–820 and
so on.
4. Draw a histogram for the frequency table made for the data in Question 3, and
answerthefollowingquestions.
(i) Whichgrouphasthemaximumnumberofworkers?
(ii) How many workers earn ` 850 and more?
(iii) How many workers earn less than ` 850?
5. Thenumberofhoursforwhichstudentsofaparticularclasswatchedtelevisionduring
holidaysisshownthroughthegivengraph.
Answerthefollowing.
(i) ForhowmanyhoursdidthemaximumnumberofstudentswatchTV?
(ii) How many students watched TV for less than 4 hours?
2021–22

DATA HANDLING 77
(iii) How many students spent more than 5 hours in watching TV?
5.4 Circle Graph or Pie Chart
Have you ever come across data represented in circular form as shown (Fig 5.4)?
The time spent by a child during a day Age groups of people in a town
(i) (ii)
These are called circle graphs. A circle graph shows the relationship between a
whole and its parts. Here, the whole circle is divided into sectors. The size of each sector
isproportionaltotheactivityorinformationitrepresents.
For example, in the above graph, the proportion of the sector for hours spent in sleeping
=
number of sleeping hours
whole day
=
8 hours 1
24 hours 3
=
So, this sector is drawn as
1
rd
3
part of the circle. Similarly, the proportion of the sector
for hours spent in school =
number of school hours
whole day
=
6 hours 1
24 hours 4
=
Fig 5.4
2021–22

78 MATHEMATICS
Sothissectorisdrawn
1
th
4
of thecircle.Similarly,thesizeofothersectorscanbefound.
Addupthefractionsforalltheactivities.Doyougetthetotalasone?
A circle graph is also called a pie chart.
TRY THESE
Fig 5.5
1. Eachofthefollowingpiecharts(Fig5.5)givesyouadifferentpieceofinformationaboutyourclass.
Findthefractionofthecirclerepresentingeachoftheseinformation.
(i) (ii) (iii)
2. Answer the following questions based on the pie chart
given (Fig 5.6 ).
(i) Whichtypeofprogrammesareviewedthemost?
(ii) Which two types of programmes have number of
viewersequaltothosewatchingsportschannels?
Viewers watching different types
of channels on T.V.
5.4.1 Drawing pie charts
The favourite flavours of ice-creams for
studentsofaschoolisgiveninpercentages
asfollows.
Flavours Percentage of students
Preferring the flavours
Chocolate 50%
Vanilla 25%
Otherflavours 25%
Let us represent this data in a pie chart.
Thetotalangleatthecentreofacircleis360°.Thecentralangleofthesectorswillbe
Fig 5.6
2021–22

DATA HANDLING 79
a fraction of 360°. We make a table to find the central angle of the sectors (Table 5.5).
Table 5.5
Flavours Students in per cent In fractions Fraction of 360°
preferring the flavours
Chocolate 50%
50 1
100 2
=
1
2
of 360° = 180°
Vanilla 25%
25 1
100 4
=
1
4
of 360° = 90°
Otherflavours 25%
25 1
100 4
=
1
4
of 360° = 90°
Fig 5.7
1. Drawacirclewithanyconvenientradius.
Mark its centre (O) and a radius (OA).
2. The angle of the sector for chocolate is 180°.
Use the protractor to draw ∠AOB = 180°.
3. Continuemarkingtheremainingsectors.
Example 1: Adjoining pie chart (Fig 5.7) gives the expenditure (in percentage)
onvariousitemsandsavingsofafamilyduringamonth.
(i) Onwhichitem,theexpenditurewasmaximum?
(ii) Expenditure on which item is equal to the total
savingsofthefamily?
(iii) Ifthemonthlysavingsofthefamilyis`3000,what
isthemonthlyexpenditureonclothes?
Solution:
(i) Expenditureismaximumonfood.
(ii) Expenditure on Education of children is the same
(i.e., 15%) as the savings of the family.
2021–22

80 MATHEMATICS
(iii) 15% represents ` 3000
Therefore, 10% represents `
3000
10
15
× = ` 2000
Example 2: On a particular day, the sales (in rupees) of different items of a baker’s
shop are given below.
ordinary bread : 320
fruitbread : 80
cakes and pastries : 160 Draw a pie chart for this data.
biscuits : 120
others : 40
Total : 720
Solution: We find the central angle of each sector. Here the total sale = ` 720. We
thushavethistable.
Item Sales (in `) In Fraction Central Angle
OrdinaryBread 320
320 4
720 9
=
4
360 160
9
× ° = °
Biscuits 120
120 1
720 6
=
1
360 60
6
× ° = °
Cakes and pastries 160
160 2
720 9
=
2
360 80
9
× ° = °
FruitBread 80
80 1
720 9
=
1
360 40
9
× ° = °
Others 40
40 1
720 18
=
1
360 20
18
× ° = °
Now, we make the pie chart (Fig 5.8):
Fig 5.8
2021–22

DATA HANDLING 81
TRY THESE
Draw a pie chart of the data given below.
The time spent by a child during a day.
Sleep — 8 hours
School — 6 hours
Home work — 4 hours
Play — 4 hours
Others — 2 hours
Which form of graph would be appropriate to display the following data.
1. Production of food grains of a state.
Year 2001 2002 2003 2004 2005 2006
Production 60 50 70 55 80 85
(in lakh tons)
2. Choice of food for a group of people.
Favourite food Number of people
NorthIndian 30
SouthIndian 40
Chinese 25
Others 25
Total 120
3. The daily income of a group of a factory workers.
Daily Income Number of workers
(in Rupees) (in a factory)
75-100 45
100-125 35
125-150 55
150-175 30
175-200 50
200-225 125
225-250 140
Total 480
2021–22

82 MATHEMATICS
EXERCISE 5.2
1. Asurveywasmadetofindthetypeofmusic
thatacertaingroupofyoungpeoplelikedin
acity.Adjoiningpiechartshowsthefindings
ofthissurvey.
Fromthispiechartanswerthefollowing:
(i) If20peoplelikedclassicalmusic,how
manyyoungpeopleweresurveyed?
(ii) Which type of music is liked by the
maximumnumberofpeople?
(iii) If a cassette company were to make
1000 CD’s, how many of each type
wouldtheymake?
2. A group of 360 people were asked to vote
for their favourite season from the three
seasonsrainy,winterandsummer.
(i) Which season got the most votes?
(ii) Find the central angle of each sector.
(iii) Draw a pie chart to show this
information.
3. Draw a pie chart showing the following information. The table shows the colours
preferred by a group of people.
Colours Number of people
Blue 18
Green 9
Red 6
Yellow 3
Total 36
4. The adjoining pie chart gives the marks scored in an examination by a student in
Hindi,English,Mathematics,Social ScienceandScience.Ifthetotalmarksobtained
by the students were 540, answer the following questions.
(i) In which subject did the student score 105
marks?
(Hint:for540marks,thecentralangle=360°.
So, for 105 marks, what is the central angle?)
(ii) How many more marks were obtained by the
studentinMathematicsthaninHindi?
(iii) Examine whether the sum of the marks
obtained in Social Science and Mathematics
ismorethanthatinScienceandHindi.
(Hint:Juststudythecentralangles).
Find the proportion of each sector. For example,
Blue is
18 1
36 2
= ; Green is
9 1
36 4
= and so on. Use
this to find the corresponding angles.
Season No. of votes
Summer 90
Rainy 120
Winter 150
2021–22

DATA HANDLING 83
TRY THESE
5. The number of students in a hostel, speaking different languages is given below.
Display the data in a pie chart.
Language Hindi English Marathi Tamil Bengali Total
Number 40 12 9 7 4 72
of students
5.5 Chance and Probability
Sometimesithappensthatduringrainyseason,youcarryaraincoateveryday
anditdoesnotrainformanydays.However,bychance,onedayyouforget to
take the raincoat and it rains heavily on that day.
Sometimes it so happens that a student prepares 4 chapters out of 5, very well
foratest.Butamajorquestionisaskedfromthechapterthatsheleftunprepared.
Everyone knows that a particular train runs in time but the day you reach
wellintimeitislate!
You face a lot of situations such as these where you take a chance and it
doesnotgothewayyouwantitto.Canyougivesomemoreexamples?These
areexampleswherethechancesofacertainthinghappeningornothappening
are not equal. The chances of the train being in time or being late are not the
same. When you buy a ticket which is wait listed, you do take a chance.You
hopethatitmightgetconfirmedbythetimeyoutravel.
We however, consider here certain experiments whose results have an equal chance
ofoccurring.
5.5.1 Getting a result
You might have seen that before a cricket match starts, captains of the two teams go out
to toss a coin to decide which team will bat first.
What are the possible results you get when a coin is tossed? Of course, Head or Tail.
Imagine that you are the captain of one team and your friend is the captain of the other
team.You toss a coin and ask your friend to make the call. Can you control the result of
the toss? Can you get a head if you want one? Or a tail if you want that? No, that is not
possible. Such an experiment is called a random experiment. Head orTail are the two
outcomesofthisexperiment.
1. If you try to start a scooter, what are the possible outcomes?
2. When a die is thrown, what are the six possible outcomes?
Oh!
my
raincoat.
2021–22

84 MATHEMATICS
3. When you spin the wheel shown, what are the possible outcomes? (Fig 5.9)
Listthem.
(Outcome here means the sector at which the pointer stops).
4. Youhaveabagwithfiveidenticalballsofdifferentcoloursandyouaretopullout
(draw) a ball without looking at it; list the outcomes you would
get (Fig 5.10).
In throwing a die:
• Does the first player have a greater chance of getting a six?
• Would the player who played after him have a lesser chance of getting a six?
• Suppose the second player got a six. Does it mean that the third player would not
have a chance of getting a six?
5.5.2 Equally likely outcomes:
Acoinistossedseveraltimesandthenumberoftimeswegetheadortailisnoted.Letus
look at the result sheet where we keep on increasing the tosses:
Fig 5.10
Fig 5.9
Number of tosses Tally marks (H) Number of heads Tally mark (T) Number of tails
50 | | | | | | | | | | | | 27 | | | | | | | | | | | | 23
| | | | | | | | | | | | | | | | |
60 | | | | | | | | | | | | 28 | | | | | | | | | | | | 32
| | | | | | | | | | | | | | | | | | | | | | | | |
70 ... 33 ... 37
80 ... 38 ... 42
90 ... 44 ... 46
100 ... 48 ... 52
2021–22

DATA HANDLING 85
Observe that as you increase the number of tosses more and more, the number of
heads and the number of tails come closer and closer to each other.
This could also be done with a die, when tossed a large number of times. Number of
each of the six outcomes become almost equal to each other.
In such cases, we may say that the different outcomes of the experiment are equally
likely.This means that each of the outcomes has the same chance of occurring.
5.5.3 Linking chances to probability
Consider the experiment of tossing a coin once. What are the outcomes? There are only
twooutcomes–HeadorTail.Boththeoutcomesareequallylikely.Likelihoodofgetting
a head is one out of two outcomes, i.e.,
1
2
. In other words, we say that the probability of
getting a head=
1
2
.Whatistheprobabilityofgettingatail?
Now take the example of throwing a die marked with 1, 2, 3, 4, 5, 6 on its faces (one
number on one face). If you throw it once, what are the outcomes?
The outcomes are: 1, 2, 3, 4, 5, 6. Thus, there are six equally likely outcomes.
Whatistheprobabilityofgettingtheoutcome‘2’?
Itis
Whatistheprobabilityofgettingthenumber5?Whatistheprobabilityofgettingthe
number 7? What is the probability of getting a number 1 through 6?
5.5.4 Outcomes as events
Each outcome of an experiment or a collection of outcomes make an event.
For example in the experiment of tossing a coin, getting a Head is an event and getting a
Tailisalsoanevent.
In case of throwing a die, getting each of the outcomes 1, 2, 3, 4, 5 or 6 is an event.
← Number of outcomes giving 2
← Numberofequallylikelyoutcomes.
1
6
2021–22

86 MATHEMATICS
TRY THESE
Is getting an even number an event? Since an even number could be 2, 4 or 6, getting an
evennumberisalsoanevent.Whatwillbetheprobabilityofgettinganevennumber?
Itis
Example 3: Abag has 4 red balls and 2 yellow balls. (The balls are identical in all
respects other than colour).Aball is drawn from the bag without looking into the bag.
What is probability of getting a red ball? Is it more or less than getting a yellow ball?
Solution: There are in all (4 + 2 =) 6 outcomes of the event. Getting a red ball
consists of 4 outcomes. (Why?)
Therefore, the probability of getting a red ball is
4
6
=
2
3
. In the same way the probability
ofgettingayellowball=
2 1
6 3
= (Why?).Therefore,theprobabilityofgettingaredballis
morethanthatofgettingayellowball.
Suppose you spin the wheel
1. (i) Listthenumberofoutcomesofgettingagreensector
and not getting a green sector on this wheel
(Fig 5.11).
(ii) Find the probability of getting a green sector.
(iii) Find the probability of not getting a green sector.
5.5.5 Chance and probability related to real life
We talked about the chance that it rains just on the day when we do not carry a rain coat.
What could you say about the chance in terms of probability? Could it be one in 10
daysduringarainyseason?Theprobabilitythatitrainsisthen
1
10
.Theprobabilitythatit
doesnotrain=
9
10
.(Assumingrainingornotrainingonadayareequallylikely)
Theuseofprobabilityismadeinvariouscasesinreallife.
1. Tofindcharacteristicsofalargegroupbyusingasmall
part of the group.
For example, during elections ‘an exit poll’ is taken.
Thisinvolvesaskingthepeoplewhomtheyhavevoted
for, when they come out after voting at the centres
which are chosen off hand and distributed over the
wholearea.Thisgivesanideaofchanceofwinningof
each candidate and predictions are made based on it
accordingly.
← Number of outcomes that make the event
← Totalnumberofoutcomesoftheexperiment.
3
6
Fig 5.11
2021–22

DATA HANDLING 87
2. Metrological Department predicts weather by observing trends from the data over
many years in the past.
EXERCISE 5.3
1. List the outcomes you can see in these experiments.
(a) Spinningawheel (b) Tossingtwocoinstogether
2. When a die is thrown, list the outcomes of an event of getting
(i) (a) a prime number (b) not a prime number.
(ii) (a) a number greater than 5 (b) a number not greater than 5.
3. Findthe.
(a) Probability of the pointer stopping on D in (Question 1-(a))?
(b) Probabilityofgettinganacefromawellshuffleddeckof52playingcards?
(c) Probability of getting a red apple.(See figure below)
4. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in
aboxandmixedwell.Oneslipischosenfromtheboxwithoutlookingintoit.What
istheprobabilityof.
(i) gettinganumber6?
(ii) gettinganumberlessthan6?
(iii) gettinganumbergreaterthan6?
(iv) gettinga1-digitnumber?
5. If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector,
whatistheprobabilityofgettingagreensector?Whatistheprobabilityofgettinga
non blue sector?
6. FindtheprobabilitiesoftheeventsgiveninQuestion2.
1. Datamostlyavailabletousinanunorganisedformiscalledrawdata.
2. Inordertodrawmeaningfulinferencesfromanydata,weneedtoorganisethedatasystematically.
2021–22

88 MATHEMATICS
3. Frequencygives the number of times that a particular entry occurs.
4. Rawdatacanbe‘grouped’andpresentedsystematicallythrough‘groupedfrequencydistribution’.
5. Grouped data can be presented using histogram.Histogram is a type of bar diagram, where the
classintervalsareshownonthehorizontalaxisandtheheightsofthebarsshowthefrequencyof
the class interval.Also, there is no gap between the bars as there is no gap between the class
intervals.
6. Data can also presented using circle graph or pie chart.Acircle graph shows the relationship
between a whole and its part.
7. There are certain experiments whose outcomes have an equal chance of occurring.
8. A random experiment is one whose outcome cannot be predicted exactly in advance.
9. Outcomes of an experiment are equally likely if each has the same chance of occurring.
10. Probabilityofanevent=
Number of outcomes that make an event
Total number of outcomes of the experiment
,whentheoutcomes
areequallylikely.
11. One or more outcomes of an experiment make an event.
12. Chances and probability are related to real life.
2021–22

SQUARES AND SQUARE ROOTS 89
6.1 Introduction
You know that the area of a square = side × side (where ‘side’ means ‘the length of
aside’).Studythefollowingtable.
Side of a square (in cm) Area of the square (in cm2
)
1 1 × 1 = 1 = 12
2 2 × 2 = 4 = 22
3 3 × 3 = 9 = 32
5 5 × 5 = 25 = 52
8 8 × 8 = 64 = 82
a a × a = a2
What is special about the numbers 4, 9, 25, 64 and other such numbers?
Since, 4 can be expressed as 2 × 2 = 22
, 9 can be expressed as 3 × 3 = 32
, all such
numbers can be expressed as the product of the number with itself.
Such numbers like 1, 4, 9, 16, 25, ... are known as square numbers.
In general, if a natural number m can be expressed as n2
, where n is also a natural
number, then m is a square number. Is 32 a square number?
We know that 52
= 25 and 62
= 36. If 32 is a square number, it must be the square of
a natural number between 5 and 6. But there is no natural number between 5 and 6.
Therefore 32 is not a square number.
Considerthefollowingnumbersandtheirsquares.
Number Square
1 1 × 1 = 1
2 2 × 2 = 4
Squares and Square
Roots
CHAPTER
6
2021–22

90 MATHEMATICS
TRY THESE
3 3 × 3 = 9
4 4 × 4 = 16
5 5 × 5 = 25
6 -----------
7 -----------
8 -----------
9 -----------
10 -----------
From the above table, can we enlist the square numbers between 1 and 100?Are
there any natural square numbers upto 100 left out?
You will find that the rest of the numbers are not square numbers.
Thenumbers1,4,9,16...aresquarenumbers.Thesenumbersarealsocalledperfect
squares.
1. Find the perfect square numbers between (i) 30 and 40 (ii) 50 and 60
6.2 Properties of Square Numbers
Following table shows the squares of numbers from 1 to 20.
Number Square Number Square
1 1 11 121
2 4 12 144
3 9 13 169
4 16 14 196
5 25 15 225
6 36 16 256
7 49 17 289
8 64 18 324
9 81 19 361
10 100 20 400
Study the square numbers in the above table. What are the ending digits (that is, digits in
the units place) of the square numbers?All these numbers end with 0, 1, 4, 5, 6 or 9 at
units place. None of these end with 2, 3, 7 or 8 at unit’s place.
Can we say that if a number ends in 0, 1, 4, 5, 6 or 9, then it must be a square
number?Thinkaboutit.
1. Canwesaywhetherthefollowingnumbersareperfectsquares?Howdoweknow?
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222
(v) 1069 (vi) 2061
TRY THESE
Can you
complete it?
2021–22

Writefivenumberswhichyoucandecidebylookingattheirunitsdigitthattheyare
not square numbers.
2. Write five numbers which you cannot decide just by looking at their units digit
(or units place) whether they are square numbers or not.
• Studythefollowingtableofsomenumbersandtheirsquaresandobservetheone’s
place in both.
Table 1
Number Square Number Square Number Square
1 1 11 121 21 441
2 4 12 144 22 484
3 9 13 169 23 529
4 16 14 196 24 576
5 25 15 225 25 625
6 36 16 256 30 900
7 49 17 289 35 1225
8 64 18 324 40 1600
9 81 19 361 45 2025
10 100 20 400 50 2500
Thefollowingsquarenumbersendwithdigit1.
Square Number
1 1
81 9
121 11
361 19
441 21
Write the next two square numbers which end in 1 and their corresponding numbers.
You will see that if a number has 1 or 9 in the units place, then it’s square ends in 1.
• Let us consider square numbers ending in 6.
Square Number
16 4
36 6
196 14
256 16
TRY THESE
Which of 1232
, 772
, 822
,
1612
, 1092
would end with
digit1?
TRY THESE
Whichofthefollowingnumberswouldhavedigit
6 at unit place.
(i) 192
(ii) 242
(iii) 262
(iv) 362
(v) 342
2021–22

92 MATHEMATICS
TRY THESE
TRY THESE
We can see that when a square number ends in 6, the number whose square it is, will
have either 4 or 6 in unit’s place.
Can you find more such rules by observing the numbers and their squares (Table 1)?
Whatwillbethe“one’sdigit”inthesquareofthefollowingnumbers?
(i) 1234 (ii) 26387 (iii) 52698 (iv) 99880
(v) 21222 (vi) 9106
• Considerthefollowingnumbersandtheirsquares.
102
= 100
202
= 400
802
= 6400
1002
= 10000
2002
= 40000
7002
= 490000
9002
= 810000
If a number contains 3 zeros at the end, how many zeros will its square have ?
What do you notice about the number of zeros at the end of the number and the
number of zeros at the end of its square?
Can we say that square numbers can only have even number of zeros at the end?
• See Table 1 with numbers and their squares.
Whatcanyousayaboutthesquaresofevennumbersandsquaresofoddnumbers?
1. The square of which of the following numbers would be an odd number/an even
number?Why?
(i) 727 (ii) 158 (iii) 269 (iv) 1980
2. Whatwillbethenumberofzerosinthesquareofthefollowingnumbers?
(i) 60 (ii) 400
6.3 Some More Interesting Patterns
1. Addingtriangularnumbers.
Doyouremembertriangularnumbers(numberswhosedotpatternscanbearranged
astriangles)?
*
* * *
* ** * **
* ** *** * ***
* ** *** **** * ****
1 3 6 10 15
But we have
four zeros
But we have
two zeros
We have
one zero
We have
two zeros
2021–22

Ifwecombinetwoconsecutivetriangularnumbers,wegetasquarenumber,like
1 + 3 = 4 3 + 6 = 9 6 + 10 = 16
= 22
= 32
= 42
2. Numbers between square numbers
Let us now see if we can find some interesting pattern between two consecutive
squarenumbers.
1 (= 12
)
2, 3, 4 (= 22
)
5, 6, 7, 8, 9 (= 32
)
10, 11, 12, 13, 14, 15, 16 (= 42
)
17, 18, 19, 20, 21, 22, 23, 24, 25 (= 52
)
Between 12
(=1) and 22
(= 4) there are two (i.e., 2 × 1) non square numbers 2, 3.
Between 22
(= 4) and 32
(= 9) there are four (i.e., 2 × 2) non square numbers 5, 6, 7, 8.
Now, 32
= 9, 42
= 16
Therefore, 42
– 32
= 16 – 9 = 7
Between 9(=32
) and 16(= 42
) the numbers are 10, 11, 12, 13, 14, 15 that is, six
non-square numbers which is 1 less than the difference of two squares.
We have 42
= 16 and 52
= 25
Therefore, 52
– 42
= 9
Between16(=42
)and25(=52
)the numbersare17,18,... ,24 that is, eight non square
numbers which is 1 less than the difference of two squares.
Consider 72
and 62
. Can you say how many numbers are there between 62
and 72
?
If we think of any natural number n and (n + 1), then,
(n + 1)2
– n2
= (n2
+ 2n + 1) – n2
= 2n + 1.
We find that between n2
and (n + 1)2
there are 2n numbers which is 1 less than the
difference of two squares.
Thus, in general we can say that there are 2n non perfect square numbers between
the squares of the numbers n and (n + 1). Check for n = 5, n = 6 etc., and verify.
Two non square numbers
between the two square
numbers 1 (=12
) and 4(=22
).
4 non square numbers
between the two square
numbers 4(=22
) and 9(32
).
8 non square
numbers between
the two square
numbers 16(= 42
)
and 25(=52
).
6 non square numbers between
the two square numbers 9(=32
)
and 16(= 42
).
2021–22

94 MATHEMATICS
TRY THESE
1. How many natural numbers lie between 92
and 102
? Between 112
and 122
?
2. Howmanynonsquarenumbersliebetweenthefollowingpairsofnumbers
(i) 1002
and 1012
(ii) 902
and 912
(iii) 10002
and 10012
3. Addingoddnumbers
Considerthefollowing
1 [one odd number] = 1 = 12
1 + 3 [sum of first two odd numbers] = 4 = 22
1 + 3 + 5 [sum of first three odd numbers] = 9 = 32
1 + 3 + 5 + 7 [... ] = 16 = 42
1 + 3 + 5 + 7 + 9 [... ] = 25 = 52
1 + 3 + 5 + 7 + 9 + 11 [... ] = 36 = 62
So we can say that the sum of first n odd natural numbers is n2
.
Looking at it in a different way, we can say: ‘If the number is a square number, it has
to be the sum of successive odd numbers starting from 1.
Consider those numbers which are not perfect squares, say 2, 3, 5, 6, ... . Can you
express these numbers as a sum of successive odd natural numbers beginning from 1?
Youwillfindthatthesenumberscannotbeexpressedinthisform.
Consider the number 25. Successively subtract 1, 3, 5, 7, 9, ... from it
(i) 25 – 1 = 24 (ii) 24 – 3 = 21 (iii) 21 – 5 = 16 (iv) 16 – 7 = 9
(v) 9 – 9 = 0
This means, 25 = 1 + 3 + 5 + 7 + 9.Also, 25 is a perfect square.
Now consider another number 38, and again do as above.
(i) 38 – 1 = 37 (ii) 37 – 3 = 34 (iii) 34 – 5 = 29 (iv) 29– 7 = 22
(v) 22 – 9 = 13 (vi) 13 – 11 = 2 (vii) 2 – 13 = – 11
This shows that we are not able to express 38 as the
sumofconsecutiveoddnumbersstartingwith1.Also,38is
not a perfect square.
So we can also say that if a natural number cannot be
expressed as a sum of successive odd natural numbers
starting with 1, then it is not a perfect square.
We can use this result to find whether a number is a perfect
square or not.
4. A sum of consecutive natural numbers
32
= 9 = 4 + 5
52
= 25 = 12 + 13
72
= 49 = 24 + 25
TRY THESE
Find whether each of the following
numbers is a perfect square or not?
(i) 121 (ii) 55 (iii) 81
(iv) 49 (v) 69
First Number
=
2
3 1
2
−
Second Number
=
2
3 1
2
+
2021–22

TRY THESE
Vow! we can express the
square of any odd number as
the sum of two consecutive
positive integers.
92
= 81 = 40 + 41
112
= 121 = 60 + 61
152
= 225 = 112 + 113
1. Expressthefollowingasthesumoftwoconsecutiveintegers.
(i) 212
(ii) 132
(iii) 112
(iv) 192
2. Doyouthinkthereverseisalsotrue,i.e.,isthesumofanytwoconsecutivepositive
integers is perfect square of a number? Give example to support your answer.
5. Product of two consecutive even or odd natural numbers
11 × 13 = 143 = 122
– 1
Also 11 × 13 = (12 – 1) × (12 + 1)
Therefore, 11 × 13 = (12 – 1) × (12 + 1) = 122
– 1
Similarly, 13 × 15 = (14 – 1) × (14 + 1) = 142
– 1
29 × 31 = (30 – 1) × (30 + 1) = 302
– 1
44 × 46 = (45 – 1) × (45 + 1) = 452
– 1
So in general we can say that (a + 1) × (a – 1) = a2
– 1.
6. Some more patterns in square numbers
TRY THESE
Writethesquare,makinguseoftheabove
pattern.
(i) 1111112
(ii) 11111112
TRY THESE
Can you find the square of the following
numbersusingtheabovepattern?
(i) 66666672
(ii) 666666672
Observe the squares of numbers; 1, 11, 111 ... etc. They give a beautiful pattern:
12
= 1
112
= 1 2 1
1112
= 1 2 3 2 1
11112
= 1 2 3 4 3 2 1
111112
= 1 2 3 4 5 4 3 2 1
111111112
= 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
Anotherinterestingpattern.
72
= 49
672
= 4489
6672
= 444889
66672
= 44448889
666672
= 4444488889
6666672
= 444444888889
Thefunisinbeingabletofindoutwhythishappens.May
beitwouldbeinterestingforyoutoexploreandthinkabout
suchquestionseveniftheanswerscomesomeyearslater.
2021–22

96 MATHEMATICS
EXERCISE 6.1
1. Whatwillbetheunitdigitofthesquaresofthefollowingnumbers?
(i) 81 (ii) 272 (iii) 799 (iv) 3853
(v) 1234 (vi) 26387 (vii) 52698 (viii) 99880
(ix) 12796 (x) 55555
2. Thefollowingnumbersareobviouslynotperfectsquares.Givereason.
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222
(v) 64000 (vi) 89722 (vii) 222000 (viii) 505050
3. Thesquaresofwhichofthefollowingwouldbeoddnumbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
4. Observethefollowingpatternandfindthemissingdigits.
112
= 121
1012
= 10201
10012
= 1002001
1000012
= 1 ......... 2 ......... 1
100000012
= ...........................
5. Observethefollowingpatternandsupplythemissingnumbers.
112
= 1 2 1
1012
= 1 0 2 0 1
101012
= 102030201
10101012
= ...........................
............2
= 10203040504030201
6. Usingthegivenpattern,findthemissingnumbers.
12
+ 22
+ 22
= 32
22
+ 32
+ 62
= 72
32
+ 42
+ 122
= 132
42
+ 52
+ _2
= 212
52
+ _2
+ 302
= 312
62
+ 72
+ _2
= __2
7. Withoutadding,findthesum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
9. Howmanynumbersliebetweensquaresofthefollowingnumbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
To find pattern
Third number is related to first and second
number. How?
Fourth number is related to third number.
How?
2021–22

TRY THESE
6.4 Finding the Square of a Number
Squares of small numbers like 3, 4, 5, 6, 7, ... etc. are easy to find. But can we find the
square of 23 so quickly?
The answer is not so easy and we may need to multiply 23 by 23.
Thereisawaytofindthiswithouthavingtomultiply23×23.
We know 23 = 20 + 3
Therefore 232
= (20 + 3)2
= 20(20 + 3) + 3(20 + 3)
= 202
+ 20 × 3 + 3 × 20 + 32
= 400 + 60 + 60 + 9 = 529
Example 1: Findthesquareofthefollowingnumberswithoutactualmultiplication.
(i) 39 (ii) 42
Solution: (i) 392
= (30 + 9)2
= 30(30 + 9) + 9(30 + 9)
= 302
+ 30 × 9 + 9 × 30 + 92
= 900 + 270 + 270 + 81 = 1521
(ii) 422
= (40 + 2)2
= 40(40 + 2) + 2(40 + 2)
= 402
+ 40 × 2 + 2 × 40 + 22
= 1600 + 80 + 80 + 4 = 1764
6.4.1 Other patterns in squares
Considerthefollowingpattern:
252
= 625 = (2 × 3) hundreds + 25
352
= 1225 = (3 × 4) hundreds + 25
752
= 5625 = (7 × 8) hundreds + 25
1252
= 15625 = (12 × 13) hundreds + 25
Now can you find the square of 95?
Findthesquaresofthefollowingnumberscontaining5inunit’splace.
(i) 15 (ii) 95 (iii) 105 (iv) 205
6.4.2 Pythagorean triplets
32
+ 42
= 9 + 16 = 25 = 52
The collection of numbers 3, 4 and 5 is known as Pythagorean triplet. 6, 8, 10 is
alsoaPythagoreantriplet,since
62
+ 82
= 36 + 64 = 100 = 102
Again,observethat
52
+ 122
= 25 + 144 = 169 = 132
. The numbers 5, 12, 13 form another such triplet.
Consider a number with unit digit 5, i.e., a5
(a5)2
= (10a + 5)2
= 10a(10a + 5) + 5(10a + 5)
= 100a2
+ 50a + 50a + 25
= 100a(a + 1) + 25
= a(a + 1) hundred + 25
2021–22

98 MATHEMATICS
Canyoufindmoresuchtriplets?
For any natural number m 1, we have (2m)2
+ (m2
– 1)2
= (m2
+ 1)2
. So, 2m,
m2
– 1 and m2
+ 1 forms a Pythagorean triplet.
TrytofindsomemorePythagoreantripletsusingthisform.
Example 2: Write a Pythagorean triplet whose smallest member is 8.
Solution: We can get Pythagorean triplets by using general form 2m,m2
– 1, m2
+ 1.
Let us first take m2
– 1 = 8
So, m2
= 8 + 1 = 9
whichgives m = 3
Therefore, 2m = 6 and m2
+ 1 = 10
The triplet is thus 6, 8, 10. But 8 is not the smallest member of this.
So, let us try 2m = 8
then m = 4
We get m2
– 1 = 16 – 1 = 15
and m2
+ 1 = 16 + 1 = 17
The triplet is 8, 15, 17 with 8 as the smallest member.
Example 3: Find a Pythagorean triplet in which one member is 12.
Solution: If we take m2
– 1 = 12
Then, m2
= 12 + 1 = 13
Then the value ofmwill not be an integer.
So, we try to take m2
+ 1 = 12.Again m2
= 11 will not give an integer value for m.
So, let us take 2m = 12
then m = 6
Thus, m2
– 1 = 36 – 1 = 35 and m2
+ 1 = 36 + 1 = 37
Therefore, the required triplet is 12, 35, 37.
Note:AllPythagoreantripletsmaynotbeobtainedusingthisform.Forexampleanother
triplet 5, 12, 13 also has 12 as a member.
EXERCISE 6.2
1. Findthesquareofthefollowingnumbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93
(v) 71 (vi) 46
2. Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv) 18
6.5 Square Roots
Studythefollowingsituations.
(a) Area of a square is 144 cm2
. What could be the side of the square?
2021–22

We know that the area of a square = side2
If we assume the length of the side to be ‘a’, then 144 = a2
To find the length of side it is necessary to find a number whose square is 144.
(b) What is the length of a diagonal of a square of side 8 cm (Fig 6.1)?
Can we use Pythagoras theorem to solve this ?
We have, AB2
+ BC2
= AC2
i.e., 82
+ 82
= AC2
or 64 + 64 = AC2
or 128 = AC2
AgaintogetACweneedtothinkofanumberwhosesquareis128.
(c) In a right triangle the length of the hypotenuse and a side are
respectively 5 cm and 3 cm (Fig 6.2).
Can you find the third side?
Let x cm be the length of the third side.
UsingPythagorastheorem 52
= x2
+ 32
25 – 9 = x2
16 = x2
Again, to find x we need a number whose square is 16.
In all the above cases, we need to find a number whose square is known. Finding the
number with the known square is known as finding the square root.
6.5.1 Finding square roots
The inverse (opposite) operation of addition is subtraction and the inverse operation
of multiplication is division. Similarly, finding the square root is the inverse operation
of squaring.
We have, 12
= 1, therefore square root of 1 is 1
22
32
Fig 6.1
Fig 6.2
TRY THESE
(i) 112
= 121. What is the square root of 121?
(ii) 142
= 196. What is the square root of 196?
Since 92
= 81,
and (–9)2
= 81
We say that square
roots of 81 are 9 and –9.
(–1)2
= 1. Is –1, a square root of 1? (–2)2
= 4. Is –2, a square root of 4?
(–9)2
= 81. Is –9 a square root of 81?
From the above, you may say that there are two integral square roots of a perfect square
number. In this chapter, we shall take up only positive square root of a natural number.
Positive square root of a number is denoted by the symbol .
For example: 4 = 2 (not –2); 9 = 3 (not –3) etc.
2021–22

100 MATHEMATICS
Statement Inference Statement Inference
12
= 1 1 = 1 62
= 36 36 = 6
22
= 4 4 = 2 72
= 49 49 = 7
32
= 9 9 = 3 82
= 64 64 = 8
42
= 16 16 = 4 92
= 81 81 = 9
52
= 25 25 = 5 102
= 100 100 = 10
6.5.2 Finding square root through repeated subtraction
Doyourememberthatthesumofthefirstnoddnaturalnumbersisn2
?Thatis,everysquare
numbercanbeexpressedasasumofsuccessiveoddnaturalnumbersstartingfrom1.
Consider 81. Then,
(i) 81 – 1 = 80 (ii) 80 – 3 = 77 (iii) 77 – 5 = 72 (iv) 72 – 7 = 65
(v) 65 – 9 = 56 (vi) 56 – 11 = 45 (vii) 45 – 13 = 32 (viii) 32 – 15 = 17
(ix) 17 – 17 = 0
From 81 we have subtracted successive odd
numbers starting from 1 and obtained 0 at 9th
step.
Therefore 81 = 9.
Canyoufindthesquarerootof 729usingthismethod?
Yes,butitwillbetimeconsuming.Letustrytofinditin
asimplerway.
6.5.3 Finding square root through prime factorisation
Considertheprimefactorisationofthefollowingnumbersandtheirsquares.
Prime factorisation of a Number Prime factorisation of its Square
6 = 2 × 3 36 = 2 × 2 × 3 × 3
8 = 2 × 2 × 2 64 = 2 × 2 × 2 × 2 × 2 × 2
12 = 2 × 2 × 3 144 = 2 × 2 × 2 × 2 × 3 × 3
15 = 3 × 5 225 = 3 × 3 × 5 × 5
Howmanytimesdoes2occurintheprimefactorisationof6?Once.Howmanytimes
does2occurintheprimefactorisationof36?Twice.Similarly,observetheoccurrenceof
3 in 6 and 36 of 2 in 8 and 64 etc.
You will find that each prime factor in the prime factorisation of the
square of a number, occurs twice the number of times it occurs in the
primefactorisationofthenumberitself.Letususethistofindthesquare
root of a given square number, say 324.
We know that the prime factorisation of 324 is
324 = 2 × 2 × 3 × 3 × 3 × 3
TRY THESE
Byrepeatedsubtractionofoddnumbersstarting
from1,findwhetherthefollowingnumbersare
perfectsquaresornot?Ifthenumberisaperfect
squarethenfinditssquareroot.
(i) 121
(ii) 55
(iii) 36
(iv) 49
(v) 90
2 324
2 162
3 81
3 27
3 9
3
2021–22

Bypairingtheprimefactors,weget
324 = 2 × 2 × 3 × 3 × 3 × 3 = 22
× 32
× 32
= (2 × 3 × 3)2
So, 324 = 2 × 3 × 3 = 18
Similarlycanyoufindthesquarerootof 256?Primefactorisationof 256is
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Bypairingtheprimefactorsweget,
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2 × 2 × 2)2
Therefore, 256 = 2 × 2 × 2 × 2 = 16
Is 48 a perfect square?
We know 48 = 2 × 2 × 2 × 2 × 3
Since all the factors are not in pairs so 48 is not a perfect square.
Suppose we want to find the smallest multiple of 48 that is a perfect square, how
should we proceed? Making pairs of the prime factors of 48 we see that 3 is the only
factor that does not have a pair. So we need to multiply by 3 to complete the pair.
Hence 48 × 3 = 144 is a perfect square.
Can you tell by which number should we divide 48 to get a perfect square?
The factor 3 is not in pair, so if we divide 48 by 3 we get 48 ÷ 3 = 16 = 2 × 2 × 2 × 2
and this number 16 is a perfect square too.
Example 4: Find the square root of 6400.
Solution: Write 6400 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
Therefore 6400 = 2 × 2 × 2 × 2 × 5 = 80
Example 5: Is 90 a perfect square?
Solution: We have 90 = 2 × 3 × 3 × 5
The prime factors 2 and 5 do not occur in pairs. Therefore, 90 is not a perfect square.
That 90 is not a perfect square can also be seen from the fact that it has only one zero.
Example 6: Is 2352 a perfect square? If not, find the smallest multiple of 2352 which
is a perfect square. Find the square root of the new number.
Solution: We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If3getsapairthenthenumberwillbecomeperfectsquare.So,wemultiply2352by3toget,
2352 × 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair.Therefore, 2352 × 3 = 7056 is a perfect square.
Thus the required smallest multiple of 2352 is 7056 which is a perfect square.
And, 7056 = 2 × 2 × 3 × 7 = 84
Example 7: Find the smallest number by which 9408 must be divided so that the
quotient is a perfect square. Find the square root of the quotient.
2 256
2 128
2 64
2 32
2 16
2 8
2 4
2
2 90
3 45
3 15
5
2 2352
2 1176
2 588
2 294
3 147
7 49
7
2 6400
2 3200
2 1600
2 800
2 400
2 200
2 100
2 50
5 25
5
2021–22

102 MATHEMATICS
Solution: We have, 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7
If we divide 9408 by the factor 3, then
9408 ÷ 3 = 3136 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 which is a perfect square. (Why?)
Therefore,therequiredsmallestnumberis3.
And, 3136 = 2 × 2 × 2 × 7 = 56.
Example 8: Findthesmallestsquarenumberwhichisdivisiblebyeachofthenumbers
6, 9 and 15.
Solution: Thishastobedoneintwosteps.Firstfindthesmallestcommonmultipleand
then find the square number needed. The least number divisible by each one of 6, 9 and
15 is their LCM. The LCM of 6, 9 and 15 is 2 × 3 × 3 × 5 = 90.
Prime factorisation of 90 is 90 = 2 × 3 × 3 × 5.
We see that prime factors 2 and 5 are not in pairs. Therefore 90 is not a perfect
square.
In order to get a perfect square, each factor of 90 must be paired. So we need to
make pairs of 2 and 5. Therefore, 90 should be multiplied by 2 × 5, i.e., 10.
Hence, the required square number is 90 × 10 = 900.
EXERCISE 6.3
1. What could be the possible ‘one’s’digits of the square root of each of the following
numbers?
(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025
2. Withoutdoinganycalculation,findthenumberswhicharesurelynotperfectsquares.
(i) 153 (ii) 257 (iii) 408 (iv) 441
3. Find the square roots of 100 and 169 by the method of repeated subtraction.
4. FindthesquarerootsofthefollowingnumbersbythePrimeFactorisationMethod.
(i) 729 (ii) 400 (iii) 1764 (iv) 4096
(v) 7744 (vi) 9604 (vii) 5929 (viii) 9216
(ix) 529 (x) 8100
5. Foreachofthefollowingnumbers,findthesmallestwholenumberbywhichitshould
be multiplied so as to get a perfect square number.Also find the square root of the
square number so obtained.
(i) 252 (ii) 180 (iii) 1008 (iv) 2028
(v) 1458 (vi) 768
6. Foreachofthefollowingnumbers,findthesmallestwholenumberbywhichitshould
be divided so as to get a perfect square. Also find the square root of the square
number so obtained.
(i) 252 (ii) 2925 (iii) 396 (iv) 2645
(v) 2800 (vi) 1620
7. The students of ClassVIII of a school donated ` 2401 in all, for Prime Minister’s
NationalReliefFund.Eachstudentdonatedasmanyrupeesasthenumberofstudents
in the class. Find the number of students in the class.
2 6, 9, 15
3 3, 9, 15
3 1, 3, 5
5 1, 1, 5
1, 1, 1
2021–22

8. 2025 plants are to be planted in a garden in such a way that each row contains as
many plants as the number of rows. Find the number of rows and the number of
plants in each row.
9. Findthesmallestsquarenumberthatisdivisiblebyeachofthenumbers4,9and10.
10. Findthesmallestsquarenumberthatisdivisiblebyeachofthenumbers8,15and20.
6.5.4 Finding square root by division method
Whenthenumbersarelarge,eventhemethodoffindingsquarerootbyprimefactorisation
becomeslengthyanddifficult.ToovercomethisproblemweuseLongDivisionMethod.
For this we need to determine the number of digits in the square root.
Seethefollowingtable:
Number Square
10 100 whichisthesmallest3-digitperfectsquare
31 961 which is the greatest 3-digit perfect square
32 1024 whichisthesmallest4-digitperfectsquare
99 9801 which is the greatest 4-digit perfect square
So, what can we say about the number of digits in the square root if a perfect
square is a 3-digit or a 4-digit number? We can say that, if a perfect square is a
3-digit or a 4-digit number, then its square root will have 2-digits.
Can you tell the number of digits in the square root of a 5-digit or a 6-digit
perfect square?
The smallest 3-digit perfect square number is 100 which is the square of 10 and the
greatest 3-digit perfect square number is 961 which is the square of 31. The smallest
4-digitsquarenumberis1024whichisthesquareof32andthegreatest4-digitnumberis
9801 which is the square of 99.
Canwesaythatifaperfectsquareisofn-digits,thenitssquarerootwillhave
2
n
digits if n is even or
( 1)
2
n +
if n is odd?
Theuseofthenumberofdigitsinsquarerootofanumberisusefulinthefollowingmethod:
• Consider the following steps to find the square root of 529.
Can you estimate the number of digits in the square root of this number?
Step 1 Placeabarovereverypairofdigitsstartingfromthedigitatone’splace.Ifthe
number of digits in it is odd, then the left-most single digit too will have a bar.
Thus we have, 529 .
Step 2 Findthelargestnumberwhosesquareislessthanorequaltothenumberunderthe
extreme left bar(22
532
).Takethisnumberasthedivisorandthequotient
with the numberundertheextreme left barasthedividend(here 5).Divideand
gettheremainder(1inthiscase).
2
2 529
– 4
1
2021–22

104 MATHEMATICS
Step 3 Bring down the number under the next bar (i.e., 29 in this case) to the right of
the remainder. So the new dividend is 129.
Step 4 Double the quotient and enter it with a blank on its right.
Step 5 Guess a largest possible digit to fill the blank which will also become the new
digit in the quotient, such that when the new divisor is multiplied to the new
quotient the product is less than or equal to the dividend.
In this case 42 × 2 = 84.
As 43 × 3 = 129 so we choose the new digit as 3. Get the remainder.
Step 6 Since the remainder is 0 and no digits are left in the given number, therefore,
529 = 23.
• Now consider 4096
Step 1 Place a bar over every pair of digits starting from the one’s digit. ( 40 96).
Step 2 Findthelargestnumberwhosesquareislessthanorequaltothenumberunder
theleft-mostbar(62
4072
).Takethisnumberasthedivisorandthenumber
under the left-most bar as the dividend. Divide and get the remainder i.e., 4 in
this case.
Step 3 Bringdownthenumberunderthenextbar (i.e.,96)totherightoftheremainder.
The new dividend is 496.
Step 4 Double the quotient and enter it with a blank on its right.
Step 5 Guessalargestpossibledigittofilltheblankwhichalsobecomesthenewdigitinthe
quotientsuchthatwhenthenewdigitismultipliedtothenewquotienttheproductis
lessthanorequaltothedividend.Inthiscaseweseethat124×4=496.
So the new digit in the quotient is 4. Get the remainder.
Step 6 Since the remainder is 0 and no bar left, therefore, 4096 = 64.
Estimatingthenumber
We use bars to find the number of digits in the square root of a perfect square number.
529 = 23 and 4096 = 64
Inboththenumbers529and4096therearetwobarsandthenumberofdigitsintheir
square root is 2. Can you tell the number of digits in the square root of 14400?
By placing bars we get 14400. Since there are 3 bars, the square root will be of 3 digit.
2
2 529
– 4
129
6
6 4096
– 36
4
23
2 529
– 4
43 129
–129
0
2
2 529
– 4
4_ 129
6 4
6 4096
– 36
124 496
– 496
0
6
6 4096
– 36
496
6
6 4096
– 36
12_ 496
2021–22

TRY THESE
Without calculating square roots, find the number of digits in the square root of the
followingnumbers.
(i) 25600 (ii) 100000000 (iii) 36864
Example 9: Find the square root of : (i) 729 (ii)1296
Solution:
(i) (ii)
Example 10: Find the least number that must be subtracted from 5607 so as to get
a perfect square.Also find the square root of the perfect square.
Solution: Let us try to find 5607 by long division method. We get the
remainder 131. It shows that 742
is less than 5607 by 131.
This means if we subtract the remainder from the number, we get a perfect square.
Therefore, the required perfect square is 5607 – 131 = 5476. And, 5476 = 74.
Example 11: Find the greatest 4-digit number which is a perfect square.
Solution: Greatest number of 4-digits = 9999. We find 9999 by long division
method. The remainder is 198. This shows 992
is less than 9999 by 198.
This means if we subtract the remainder from the number, we get a perfect square.
Therefore, the required perfect square is 9999 – 198 = 9801.
And, 9801 = 99
Example 12: Find the least number that must be added to 1300 so as to get a
perfect square.Also find the square root of the perfect square.
Solution: We find 1300 by long division method.The remainder is 4.
This shows that 362
1300.
Next perfect square number is 372
= 1369.
Hence, the number to be added is 372
– 1300 = 1369 – 1300 = 69.
6.6 Square Roots of Decimals
Consider 17.64
Step 1 To find the square root of a decimal number we put bars on the integral part
(i.e.,17)ofthenumberintheusualmanner.Andplacebarsonthedecimalpart
Therefore 1296 36
=
Therefore 729 27
=
74
7 5607
– 49
144 707
–576
131
99
9 9999
– 81
189 1899
– 1701
198
36
3 1300
– 9
66 400
– 396
4
27
2 729
– 4
47 329
329
0
36
3 1296
– 9
66 396
396
0
2021–22

106 MATHEMATICS
(i.e.,64)oneverypairofdigitsbeginningwiththefirstdecimalplace.Proceed
as usual. We get 17.64.
Step 2 Now proceed in a similar manner.The left most bar is on 17 and 42
17 52
.
Take this number as the divisor and the number under the left-most bar as the
dividend, i.e., 17. Divide and get the remainder.
Step 3 Theremainderis1.Writethenumberunderthenextbar(i.e.,64)totherightof
this remainder, to get 164.
Step 4 Double the divisor and enter it with a blank on its right.
Since 64 is the decimal part so put a decimal point in the
quotient.
Step 5 We know 82 × 2 = 164, therefore, the new digit is 2.
Divideandgettheremainder.
Step 6 Since the remainder is 0 and no bar left, therefore 17.64 4.2
= .
Example 13: Find the square root of 12.25.
Solution:
Which way to move
Consideranumber176.341.Putbarsonbothintegralpartanddecimalpart.Inwhatway
is putting bars on decimal part different from integral part? Notice for 176 we start from
theunit’splaceclosetothedecimalandmovetowardsleft.Thefirstbarisover76andthe
second bar over 1. For .341, we start from the decimal and move towards right. First bar
is over 34 and for the second bar we put 0 after 1 and make .3410 .
Example 14: Area of a square plot is 2304 m2
. Find the side of the square plot.
Solution: Area of square plot = 2304 m2
Therefore, side of the square plot = 2304 m
Wefindthat, 2304 = 48
Thus, the side of the square plot is 48 m.
Example 15: There are 2401 students in a school. P.T. teacher wants them to stand
inrowsandcolumnssuchthatthenumberofrowsisequaltothenumberofcolumns.Find
the number of rows.
4
4 17.64
– 16
1
4.
4 17.64
– 16
82 164
4.2
4 17.64
–16
82 164
– 164
0
4
4 17.64
– 16
8_ 1 64
Therefore, 12.25 3.5
=
3.5
3 12.25
– 9
65 325
325
0
48
4 2304
–16
88 704
704
0
2021–22

TRY THESE
Solution: Let the number of rows be x
So, the number of columns = x
Therefore, number of students = x × x = x2
Thus, x2
= 2401 gives x = 2401 = 49
The number of rows = 49.
6.7 Estimating Square Root
Considerthefollowingsituations:
1. Deveshi has a square piece of cloth of area 125 cm2
. She wants to know whether
she can make a handkerchief of side 15 cm. If that is not possible she wants to
know what is the maximum length of the side of a handkerchief that can be made
from this piece.
2. Meena and Shobha played a game. One told a number and other gave its square
root. Meena started first. She said 25 and Shobha answered quickly as 5. Then
Shobhasaid81andMeenaanswered9.Itwenton,tillatonepointMeenagavethe
number 250. And Shobha could not answer. Then Meena asked Shobha if she
could atleast tell a number whose square is closer to 250.
In all such cases we need to estimate the square root.
We know that 100 250 400 and 100 = 10 and 400 = 20.
So 10 250 20
But still we are not very close to the square number.
We know that 152
= 225 and 162
= 256
Therefore, 15 250 16 and 256 is much closer to 250 than 225.
So, 250 is approximately 16.
Estimatethevalueofthefollowingtothenearestwholenumber.
(i) 80 (ii) 1000 (iii) 350 (iv) 500
EXERCISE 6.4
1. FindthesquarerootofeachofthefollowingnumbersbyDivisionmethod.
(i) 2304 (ii) 4489 (iii) 3481 (iv) 529
(v) 3249 (vi) 1369 (vii) 5776 (viii) 7921
(ix) 576 (x) 1024 (xi) 3136 (xii) 900
2. Findthenumberofdigitsinthesquarerootofeachofthefollowingnumbers(without
anycalculation).
(i) 64 (ii) 144 (iii) 4489 (iv) 27225
(v) 390625
49
4 2401
–16
89 801
801
0
2021–22

108 MATHEMATICS
3. Findthesquarerootofthefollowingdecimalnumbers.
(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25
(v) 31.36
4. Findtheleastnumberwhichmustbesubtractedfromeachofthefollowingnumbers
so as to get a perfect square. Also find the square root of the perfect square so
obtained.
(i) 402 (ii) 1989 (iii) 3250 (iv) 825
(v) 4000
5. Find the least number which must be added to each of the following numbers so as
to get a perfect square.Also find the square root of the perfect square so obtained.
(i) 525 (ii) 1750 (iii) 252 (iv) 1825
(v) 6412
6. Find the length of the side of a square whose area is 441 m2
.
7. In a right triangleABC, ∠B = 90°.
(a) IfAB = 6 cm, BC = 8 cm, findAC (b) IfAC=13cm,BC=5cm,findAB
8. A gardener has 1000 plants. He wants to plant these in such a way that the number
of rows and the number of columns remain same. Find the minimum number of
plants he needs more for this.
9. There are 500 children in a school. For a P.T. drill they have to stand in such a
mannerthatthenumberofrowsisequaltonumberofcolumns.Howmanychildren
wouldbeleftoutinthisarrangement.
1. If a natural number m can be expressed as n2
, where n is also a natural number, then m is a
squarenumber.
2. All square numbers end with 0, 1, 4, 5, 6 or 9 at units place.
3. Square numbers can only have even number of zeros at the end.
4. Square root is the inverse operation of square.
5. There are two integral square roots of a perfect square number.
Positive square root of a number is denoted by the symbol .
For example, 32
= 9 gives 9 3
=
2021–22

CUBES AND CUBE ROOTS 109
7.1 Introduction
This is a story about one of India’s great mathematical geniuses, S. Ramanujan. Once
anotherfamousmathematicianProf.G.H.Hardycametovisithiminataxiwhosenumber
was1729.WhiletalkingtoRamanujan,Hardydescribedthisnumber
“adullnumber”.Ramanujanquicklypointedoutthat1729wasindeed
interesting. He said it is the smallest number that can be expressed
as a sum of two cubes in two different ways:
1729 = 1728 + 1 = 123
+ 13
1729 = 1000 + 729 = 103
+ 93
1729 has since been known as the Hardy – Ramanujan Number,
even though this feature of 1729 was known more than 300 years
beforeRamanujan.
How did Ramanujan know this? Well, he loved numbers.All
throughhislife,heexperimentedwithnumbers.Heprobablyfound
numbers that were expressed as the sum of two squares and sum of
two cubes also.
There are many other interesting patterns of cubes. Let us learn about cubes, cube
roots and many other interesting facts related to them.
7.2 Cubes
Youknowthattheword‘cube’isusedingeometry.Acube is
asolidfigurewhichhasallitssidesequal.Howmanycubesof
side 1 cm will make a cube of side 2 cm?
How many cubes of side 1 cm will make a cube of side 3 cm?
Consider the numbers 1, 8, 27, ...
These are called perfect cubes or cube numbers. Can you say why
they are named so? Each of them is obtained when a number is multiplied by
takingitthreetimes.
Cubes and Cube Roots
CHAPTER
7
Hardy – Ramanujan
Number
1729 is the smallest Hardy–
Ramanujan Number. There
are an infinitely many such
numbers. Few are 4104
(2, 16; 9, 15), 13832 (18, 20;
2, 24), Check it with the
numbersgiveninthebrackets.
Figures which have
3-dimensions are known as
solid figures.
2021–22

110 MATHEMATICS
The numbers 729, 1000, 1728
are also perfect cubes.
We note that 1 = 1 × 1 × 1 = 13
; 8 = 2 × 2 × 2 = 23
; 27 = 3 × 3 × 3 = 33
.
Since 53
= 5 × 5 × 5 = 125, therefore 125 is a cube number.
Is9acubenumber?No,as9=3×3andthereisnonaturalnumberwhichmultiplied
by taking three times gives 9. We can see also that 2 × 2 × 2 = 8 and 3 × 3 × 3 = 27. This
shows that 9 is not a perfect cube.
The following are the cubes of numbers from 1 to 10.
Table 1
Number Cube
1 13
= 1
2 23
= 8
3 33
= 27
4 43
= 64
5 53
= ____
6 63
= ____
7 73
= ____
8 83
= ____
9 93
= ____
10 103
= ____
There are only ten perfect cubes from 1 to 1000. (Check this). How many perfect
cubes are there from 1 to 100?
Observe the cubes of even numbers.Are they all even?What can you say about the
cubes of odd numbers?
Following are the cubes of the numbers from 11 to 20.
Table 2
Number Cube
11 1331
12 1728
13 2197
14 2744
15 3375
16 4096
17 4913
18 5832
19 6859
20 8000
We are odd so are
our cubes
We are even, so
are our cubes
Complete it.
2021–22

TRY THESE
each prime factor
appears three times
in its cubes
TRY THESE
Consider a few numbers having 1 as the one’s digit (or unit’s). Find the cube of each
of them. What can you say about the one’s digit of the cube of a number having 1 as the
one’sdigit?
Similarly, explore the one’s digit of cubes of numbers ending in 2, 3, 4, ... , etc.
Findtheone’sdigitofthecubeofeachofthefollowingnumbers.
(i) 3331 (ii) 8888 (iii) 149 (iv) 1005
(v) 1024 (vi) 77 (vii) 5022 (viii) 53
7.2.1 Some interesting patterns
1. Adding consecutive odd numbers
Observe the following pattern of sums of odd numbers.
1 = 1 = 13
3 + 5 = 8 = 23
7 + 9 + 11 = 27 = 33
13 + 15 + 17 + 19 = 64 = 43
21 + 23 + 25 + 27 + 29 = 125 = 53
Is it not interesting? How many consecutive odd numbers will be needed to obtain
the sum as 103
?
Expressthefollowingnumbersasthesumofoddnumbersusingtheabovepattern?
(a) 63
(b) 83
(c) 73
Considerthefollowingpattern.
23
– 13
= 1 + 2 × 1 × 3
33
– 23
= 1 + 3 × 2 × 3
43
– 33
= 1 + 4 × 3 × 3
Usingtheabovepattern,findthevalueofthefollowing.
(i) 73
– 63
(ii) 123
– 113
(iii) 203
– 193
(iv) 513
– 503
2. Cubes and their prime factors
Considerthefollowingprimefactorisationofthenumbersandtheircubes.
Prime factorisation Prime factorisation
of a number of its cube
4 = 2 × 2 43
= 64 = 2 × 2 × 2 × 2 × 2 × 2 = 23
× 23
6 = 2 × 3 63
= 216 = 2 × 2 × 2 × 3 × 3 × 3 = 23
× 33
15 = 3 × 5 153
= 3375 = 3 × 3 × 3 × 5 × 5 × 5 = 33
× 53
12 = 2 × 2 × 3 123
= 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 23
× 23
× 33
2021–22

112 MATHEMATICS
TRY THESE
Observe that each prime factor of a number appears
threetimesintheprimefactorisationofitscube.
Intheprimefactorisationofanynumber,ifeachfactor
appears three times, then, is the number a perfect cube?
Think about it. Is 216 a perfect cube?
By prime factorisation, 216 = 2 × 2 × 2 × 3 × 3 × 3
Each factor appears 3 times. 216 = 23
× 33
= (2 × 3)3
= 63
which is a perfect cube!
Is 729 a perfect cube? 729 = 3 × 3 × 3 × 3 × 3 × 3
Yes, 729 is a perfect cube.
Now let us check for 500.
Prime factorisation of 500 is 2 × 2 × 5 × 5 × 5.
So, 500 is not a perfect cube.
Example 1: Is 243 a perfect cube?
Solution: 243 = 3 × 3 × 3 × 3 × 3
Intheabovefactorisation3×3remainsaftergroupingthe3’sintriplets.Therefore,243is
not a perfect cube.
Whichofthefollowingareperfectcubes?
1. 400 2. 3375 3. 8000 4. 15625
5. 9000 6. 6859 7. 2025 8. 10648
7.2.2 Smallest multiple that is a perfect cube
Raj made a cuboid of plasticine. Length, breadth and height of the cuboid are 15 cm,
30 cm, 15 cm respectively.
Anuaskshowmanysuchcuboidswillsheneedtomakeaperfectcube?Canyoutell?
Raj said, Volume of cuboid is 15 × 30 × 15 = 3 × 5 × 2 × 3 × 5 × 3 × 5
= 2 × 3 × 3 × 3 × 5 × 5 × 5
Since there is only one 2 in the prime factorisation. So we need 2 × 2, i.e., 4 to make
it a perfect cube. Therefore, we need 4 such cuboids to make a cube.
Example 2: Is 392 a perfect cube? If not, find the smallest natural number by which
392 must be multiplied so that the product is a perfect cube.
Solution: 392 = 2 × 2 × 2 × 7 × 7
The prime factor 7 does not appear in a group of three. Therefore, 392 is not a perfect
cube. To make its a cube, we need one more 7. In that case
392 × 7 = 2 × 2 × 2 × 7 × 7 × 7 = 2744 which is a perfect cube.
factors can be
grouped in triples
There are three
5’s in the product but
only two 2’s.
Do you remember that
am
× bm
= (a × b)m
2 216
2 108
2 54
3 27
3 9
3 3
1
2021–22

Hencethesmallestnaturalnumberbywhich392shouldbemultipliedtomakeaperfect
cube is 7.
Example 3: Is 53240 a perfect cube? If not, then by which smallest natural number
should 53240 be divided so that the quotient is a perfect cube?
Solution: 53240 = 2 × 2 × 2 × 11 × 11 × 11 × 5
The prime factor 5 does not appear in a group of three. So, 53240 is not a perfect cube.
Inthefactorisation5appearsonlyonetime.Ifwedividethenumberby5,thentheprime
factorisationofthequotientwillnotcontain5.
So, 53240 ÷ 5 = 2 × 2 × 2 × 11 × 11 × 11
Hence the smallest number by which 53240 should be divided to make it a perfect
cube is 5.
The perfect cube in that case is = 10648.
Example 4: Is 1188 a perfect cube? If not, by which smallest natural number should
1188 be divided so that the quotient is a perfect cube?
Solution: 1188 = 2 × 2 × 3 × 3 × 3 × 11
The primes 2 and 11 do not appear in groups of three. So, 1188 is not a perfect cube. In
the factorisation of 1188 the prime 2 appears only two times and the prime 11 appears
once. So, if we divide 1188 by 2 × 2 × 11 = 44, then the prime factorisation of the
quotientwillnotcontain2and11.
Hence the smallest natural number by which 1188 should be divided to make it a
perfect cube is 44.
And the resulting perfect cube is 1188 ÷ 44 = 27 (=33
).
Example 5: Is 68600 a perfect cube? If not, find the smallest number by which 68600
must be multiplied to get a perfect cube.
Solution: We have, 68600 = 2 × 2 × 2 × 5 × 5 × 7 × 7 × 7. In this factorisation, we
find that there is no triplet of 5.
So, 68600 is not a perfect cube. To make it a perfect cube we multiply it by 5.
Thus, 68600 × 5 = 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7
= 343000, which is a perfect cube.
Observe that 343 is a perfect cube. From Example 5 we know that 343000 is also
perfect cube.
Check which of the following are perfect cubes. (i) 2700 (ii) 16000 (iii) 64000
(iv) 900 (v) 125000 (vi) 36000 (vii) 21600 (viii) 10,000 (ix) 27000000 (x) 1000.
What pattern do you observe in these perfect cubes?
2021–22

114 MATHEMATICS
EXERCISE 7.1
1. Whichofthefollowingnumbersarenotperfectcubes?
(i) 216 (ii) 128 (iii) 1000 (iv) 100
(v) 46656
2. Findthesmallestnumberbywhicheachofthefollowingnumbersmustbemultiplied
to obtain a perfect cube.
(i) 243 (ii) 256 (iii) 72 (iv) 675
(v) 100
3. Findthesmallestnumberbywhicheachofthefollowingnumbersmustbedividedto
obtain a perfect cube.
(i) 81 (ii) 128 (iii) 135 (iv) 192
(v) 704
4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such
cuboids will he need to form a cube?
7.3 Cube Roots
Ifthevolumeofacubeis125cm3
,whatwouldbethelengthofitsside?Togetthelength
of the side of the cube, we need to know a number whose cube is 125.
Finding the square root, as you know, is the inverse operation of squaring. Similarly,
findingthecuberootistheinverseoperationoffindingcube.
We know that 23
= 8; so we say that the cube root of 8 is 2.
We write 3
8 = 2. The symbol 3 denotes ‘cube-root.’
Considerthefollowing:
Statement Inference Statement Inference
13
= 1 3
1 = 1 63
= 216 3
216 = 6
23
= 8 3
8 = 3 3
2 = 2 73
= 343 3
343 = 7
33
= 27 3
27 = 3 3
3 = 3 83
= 512 3
512 = 8
43
= 64 3
64 = 4 93
= 729 3
729 = 9
53
= 125 3
125 = 5 103
= 1000 3
1000 = 10
7.3.1 Cube root through prime factorisation method
Consider 3375.We find its cube root by prime factorisation:
3375 = 3 × 3 × 3 × 5 × 5 × 5 = 33
× 53
= (3 × 5)3
Therefore, cube root of 3375 = 3
3375 = 3 × 5 = 15
Similarly,tofind 3
74088 ,wehave,
2021–22

74088 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 7 = 23
× 33
× 73
= (2 × 3 × 7)3
Therefore, 3
74088 = 2 × 3 × 7 = 42
Example 6: Find the cube root of 8000.
Solution: Prime factorisation of 8000 is 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5
So, 3
8000 = 2 × 2 × 5 = 20
Example 7: Find the cube root of 13824 by prime factorisation method.
Solution:
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 23
× 23
× 23
× 33
.
Therefore, 3
13824 = 2 × 2 × 2 × 3 = 24
State true or false: for any integer m, m2
m3
. Why?
7.3.2 Cube root of a cube number
If you know that the given number is a cube number then following method can be used.
Step 1 Take any cube number say 857375 and start making groups of three digits
startingfromtherightmostdigitofthenumber.
857
second group
↓
375
first group
↓
We can estimate the cube root of a given cube number through a step by
step process.
We get 375 and 857 as two groups of three digits each.
Step 2 First group, i.e., 375 will give you the one’s (or unit’s) digit of the required
cube root.
The number 375 ends with 5. We know that 5 comes at the unit’s place of a
number only when it’s cube root ends in 5.
So, we get 5 at the unit’s place of the cube root.
Step 3 Now take another group, i.e., 857.
We know that 93
= 729 and 103
= 1000.Also, 729 857 1000. We take
the one’s place, of the smaller number 729 as the ten’s place of the required
cube root. So, we get 3
857375 95
= .
Example 8: Find the cube root of 17576 through estimation.
Solution: The given number is 17576.
Step 1 Formgroupsofthreestartingfromtherightmostdigitof17576.
2021–22

116 MATHEMATICS
17 576. In this case one group i.e., 576 has three digits whereas 17 has only
twodigits.
Step 2 Take 576.
The digit 6 is at its one’s place.
We take the one’s place of the required cube root as 6.
Step 3 Take the other group, i.e., 17.
Cube of 2 is 8 and cube of 3 is 27. 17 lies between 8 and 27.
The smaller number among 2 and 3 is 2.
The one’s place of 2 is 2 itself. Take 2 as ten’s place of the cube root of
17576.
Thus, 3
17576 26
= (Check it!)
EXERCISE 7.2
1. Findthecuberootofeachofthefollowingnumbersbyprimefactorisationmethod.
(i) 64 (ii) 512 (iii) 10648 (iv) 27000
(v) 15625 (vi) 13824 (vii) 110592 (viii) 46656
(ix) 175616 (x) 91125
2. State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) Thecubeofasingledigitnumbermaybeasingledigitnumber.
3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what
is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
1. Numbers like 1729, 4104, 13832, are known as Hardy – Ramanujan Numbers. They can be
expressed as sum of two cubes in two different ways.
2. Numbersobtainedwhenanumberismultipliedbyitselfthreetimesareknownascubenumbers.
For example 1, 8, 27, ... etc.
3. If in the prime factorisation of any number each factor appears three times, then the number is a
perfect cube.
4. The symbol 3 denotes cube root. For example 3
27 3
= .
2021–22

COMPARING QUANTITIES 117
8.1 Recalling Ratios and Percentages
We know, ratio means comparing two quantities.
Abasket has two types of fruits, say, 20 apples and 5 oranges.
Then, the ratio of the number of oranges to the number of apples = 5 : 20.
The comparison can be done by using fractions as,
5
20
=
1
4
The number of oranges is
1
4
th the number of apples. In terms of ratio, this is
1 : 4, read as, “1 is to 4”
Number of apples to number of oranges =
20 4
5 1
= which means, the number of apples
is 4 times the number of oranges. This comparison can also be done using percentages.
There are 5 oranges out of 25 fruits.
So percentage of oranges is
5 4 20
20%
25 4 100
× = = OR
[Denominator made 100].
Since contains only apples and oranges,
So, percentage of apples + percentage of oranges = 100
or percentage of apples + 20 = 100
or percentage of apples = 100 – 20 = 80
Thus the basket has 20% oranges and 80% apples.
Example 1: Apicnic is being planned in a school for Class VII. Girls are 60% of the
total number of students and are 18 in number.
Thepicnicsiteis55kmfromtheschoolandthetransportcompanyischargingattherate
of ` 12 per km. The total cost of refreshments will be ` 4280.
Comparing Quantities
CHAPTER
8
By unitary method:
Out of 25 fruits, number of oranges are 5.
So out of 100 fruits, number of oranges
=
5
100
25
× = 20.
OR
2021–22

118 MATHEMATICS
Canyoutell.
1. The ratio of the number of girls to the number of boys in the class?
2. The cost per head if two teachers are also going with the class?
3. If their first stop is at a place 22 km from the school, what per cent of the total
distance of 55 km is this? What per cent of the distance is left to be covered?
Solution:
1. To find the ratio of girls to boys.
AshimaandJohncameupwiththefollowinganswers.
They needed to know the number of boys and also the total number of students.
Ashima did this John used the unitary method
Let the total number of students There are 60 girls out of 100 students.
be x. 60% of x is girls. There is one girl out of
100
60
students.
Therefore, 60% of x = 18 So, 18 girls are out of how many students?
60
100
x
× = 18 OR Number of students =
100
18
60
×
or, x =
18 100
60
×
= 30 = 30
Number of students = 30.
So, the number of boys = 30 – 18 = 12.
Hence, ratio of the number of girls to the number of boys is 18 : 12 or
18
12
=
3
2
.
3
2
is written as 3 : 2 and read as 3 is to 2.
2. To find the cost per person.
Transportationcharge = Distance both ways × Rate
= ` (55 × 2) × 12
= ` 110 × 12 = ` 1320
Total expenses = Refreshmentcharge
+Transportationcharge
= ` 4280 + ` 1320
= ` 5600
Total number of persons =18 girls + 12 boys + 2 teachers
= 32 persons
Ashima and John then used unitary method to find the cost per head.
For 32 persons, amount spent would be ` 5600.
The amount spent for 1 person = `
5600
32
= ` 175.
3. The distance of the place where first stop was made = 22 km.
2021–22

Tofindthepercentageofdistance:
Ashima used this method: John used the unitary method:
22 22 100
40%
55 55 100
= × = Out of 55 km, 22 km are travelled.
OR Out of 1 km,
22
55
km are travelled.
Out of 100 km,
22
55
× 100 km are travelled.
That is 40% of the total distance is travelled.
TRY THESE
She is multiplying
100
the ratio by =1
100
and converting to
percentage.
Bothcameoutwiththesameanswerthatthedistancefromtheirschooloftheplacewhere
they stopped at was 40% of the total distance they had to travel.
Therefore, the percent distance left to be travelled = 100% – 40% = 60%.
Inaprimaryschool,theparentswereaskedaboutthenumberofhourstheyspendperday
inhelpingtheirchildrentodohomework.Therewere90parentswhohelpedfor
1
2
hour
to
1
1
2
hours.Thedistributionofparentsaccordingtothetimeforwhich,
they saidtheyhelpedisgivenintheadjoiningfigure;20%helpedfor
morethan
1
1
2
hoursperday;
30% helped for
1
2
hour to
1
1
2
hours; 50% did not help at all.
Usingthis,answerthefollowing:
(i) How many parents were surveyed?
(ii) How many said that they did not help?
(iii) How many said that they helped for more than
1
1
2
hours?
EXERCISE 8.1
1. Findtheratioofthefollowing.
(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km (c) 50 paise to ` 5
2. Convertthefollowingratiostopercentages.
(a) 3 : 4 (b) 2 : 3
3. 72% of 25 students are interested in mathematics. How many are not interested
inmathematics?
4. A football team won 10 matches out of the total number of matches they played. If
their win percentage was 40, then how many matches did they play in all?
5. IfChamelihad` 600leftafterspending75%ofhermoney,howmuchdidshehave
inthebeginning?
2021–22

120 MATHEMATICS
6. If 60% people in a city like cricket, 30% like football and the remaining like other
games, then what per cent of the people like other games? If the total number of
people is 50 lakh, find the exact number who like each type of game.
8.2 Finding the Increase or Decrease Per cent
Weoftencomeacrosssuchinformationinourdailylifeas.
(i) 25% off on marked prices (ii) 10% hike in the price of petrol
Let us consider a few such examples.
Example 2: The price of a scooter was ` 34,000 last year. It has increased by 20%
this year.What is the price now?
Solution:
OR
Amita said that she would first find
the increase in the price, which is 20% of
` 34,000, and then find the new price.
20% of ` 34000 = `
20
34000
100
×
= ` 6800
New price = Old price + Increase
= ` 34,000 + ` 6,800
= ` 40,800
Similarly, a percentage decrease in price would imply finding the actual decrease
followedbyitssubtractionthefromoriginalprice.
Suppose in order to increase its sale, the price of scooter was decreased by 5%.
Then let us find the price of scooter.
Price of scooter = ` 34000
Reduction = 5% of ` 34000
= `
5
34000
100
× = ` 1700
New price = Old price – Reduction
= ` 34000 – ` 1700 = ` 32300
We will also use this in the next section of the chapter.
8.3 Finding Discounts
Discount is a reduction given on the Marked Price
(MP) of the article.
Thisisgenerallygiventoattractcustomerstobuy
goodsortopromotesalesofthegoods.Youcanfind
the discount by subtracting its sale price from its
marked price.
So, Discount = Marked price – Sale price
Sunita used the unitary method.
20% increase means,
` 100 increased to ` 120.
So, ` 34,000 will increase to?
Increased price = `
120
34000
100
×
= ` 40,800
2021–22

TRY THESE
Example 3: An item marked at ` 840 is sold for ` 714. What is the discount and
discount%?
Solution: Discount= Marked Price – Sale Price
= ` 840 – ` 714
= ` 126
Since discount is on marked price, we will have to use marked price as the base.
On marked price of ` 840, the discount is ` 126.
On MP of ` 100, how much will the discount be?
Discount=
126
100%
840
× = 15%
You can also find discount when discount % is given.
Example 4: The list price of a frock is ` 220.
Adiscountof20%isannouncedonsales.Whatistheamount
of discount on it and its sale price.
Solution: Markedpriceissameasthelistprice.
20% discount means that on ` 100 (MP), the discount is ` 20.
By unitary method, on `1 the discount will be `
20
100
.
On ` 220, discount = `
20
220
100
× = ` 44
The sale price = (` 220 – ` 44) or ` 176
Rehana found the sale price like this —
A discount of 20% means for a MP of ` 100, discount is ` 20. Hence the sale price is
` 80. Using unitary method, when MP is ` 100, sale price is ` 80;
When MP is ` 1, sale price is `
80
100
.
Hence when MP is ` 220, sale price = `
80
220
100
× = ` 176.
1. A shop gives 20% discount. What would the sale price of each of these be?
(a) A dress marked at ` 120 (b) A pair of shoes marked at ` 750
(c) A bag marked at ` 250
2. A table marked at ` 15,000 is available for ` 14,400. Find the discount given and
the discount per cent.
3. Analmirahissoldat` 5,225afterallowingadiscountof5%.Finditsmarkedprice.
Even though the
discount was not
found, I could find
the sale price
directly.
2021–22

122 MATHEMATICS
8.3.1 Estimation in percentages
Your bill in a shop is ` 577.80 and the shopkeeper gives a discount of 15%. How would
you estimate the amount to be paid?
(i) Round off the bill to the nearest tens of ` 577.80, i.e., to ` 580.
(ii) Find 10% of this, i.e., `
10
580 58
100
× = ` .
(iii) Take half of this, i.e.,
1
58 29
2
× = ` .
(iv) Add the amounts in (ii) and (iii) to get ` 87.
Youcouldthereforereduceyourbillamountby` 87orbyabout` 85,whichwillbe
` 495 approximately.
1. Tryestimating20%ofthesamebillamount. 2.Try finding 15% of ` 375.
8.4 Prices Related to Buying and Selling (Profit and Loss)
For the school fair (mela) I am going to put a stall of lucky dips. I will charge ` 10 for one
lucky dip but I will buy items which are worth ` 5.
So you are making a profit of 100%.
No, I will spend ` 3 on paper to wrap the gift and tape. So my expenditure is ` 8.
This gives me a profit of ` 2, which is,
2
100% 25%
8
× = only.
Sometimes when an article is bought, some additional expenses are made while buying or
before selling it. These expenses have to be included in the cost price.
These expenses are sometimes referred to as overhead charges. These may include
expenses like amount spent on repairs, labour charges, transportation etc.
8.4.1 Finding cost price/selling price, profit %/loss%
Example 5: Sohan bought a second hand refrigerator for ` 2,500, thenspent` 500on
itsrepairs and solditfor` 3,300. Find his loss or gain per cent.
Solution: CostPrice(CP)= ` 2500+ ` 500(overheadexpensesareaddedtogiveCP)
= ` 3000
Sale Price (SP) = ` 3300
As SP CP, he made a profit = ` 3300 – ` 3000 = ` 300
His profit on ` 3,000, is ` 300. How much would be his profit on ` 100?
Profit
300 30
100% % 10%
3000 3
= × = = P% =
P
100
CP
×
2021–22

TRY THESE
TRY THESE
1. Find selling price (SP) if a profit of 5% is made on
(a) a cycle of ` 700 with ` 50 as overhead charges.
(b) a lawn mower bought at ` 1150 with ` 50 as transportation charges.
(c) a fan bought for ` 560 and expenses of ` 40 made on its repairs.
Example 6: A shopkeeper purchased 200 bulbs for ` 10 each. However 5 bulbs
were fused and had to be thrown away.The remaining were sold at ` 12 each. Find the
gain or loss %.
Solution: Cost price of 200 bulbs = ` 200 × 10 = ` 2000
5 bulbs were fused. Hence, number of bulbs left = 200 – 5 = 195
These were sold at ` 12 each.
The SP of 195 bulbs = ` 195 × 12 = ` 2340
He obviously made a profit (as SP CP).
Profit = ` 2340 – ` 2000 = ` 340
On ` 2000, the profit is ` 340. How much profit is made on ` 100? Profit
=
340
100%
2000
× = 17%.
Example 7: Meenu bought two fans for ` 1200 each. She sold one at
a loss of 5% and the other at a profit of 10%. Find the selling price of
each.Also find out the total profit or loss.
Solution: Overall CP of each fan = ` 1200. One is sold at a loss of 5%.
This means if CP is ` 100, SP is ` 95.
Therefore, when CP is ` 1200, then SP = `
95
1200
100
× = ` 1140
Also second fan is sold at a profit of 10%.
It means, if CPis ` 100, SPis ` 110.
Therefore, when CP is ` 1200, then SP = `
110
1200
100
× = ` 1320
Was there an overall loss or gain?
We need to find the combined CP and SP to say
whether there was an overall profit or loss.
Total CP = ` 1200 + ` 1200 = ` 2400
Total SP = ` 1140 + ` 1320 = ` 2460
Since total SP total CP, a profit of ` (2460 – 2400) or ` 60 has been made.
1. A shopkeeper bought two TV sets at ` 10,000 each. He sold one at a profit 10%
and the other at a loss of 10%. Find whether he made an overall profit or loss.
CP is ` 10
SP is ` 12
2021–22

124 MATHEMATICS
8.5 Sales Tax/Value Added Tax/Goods and Services Tax
Theteachershowedtheclassabillinwhichthefollowingheadswerewritten.
Bill No. Date
Menu
S.No. Item Quantity Rate Amount
Billamount
+ ST (5%)
Total
Sales tax (ST) is charged by the government on the sale of an item. It is collected by the
shopkeeper from the customer and given to the government. This is, therefore, always on
the selling price of an item and is added to the value of the bill. There is another type of tax
which is included in the prices known as Value Added Tax (VAT).
From July 1, 2017, Government of India introduced GST which stands for Goods and
Services Tax which is levied on supply of goods or services or both.
Example 8: (Finding Sales Tax) The cost of a pair of
roller skates at a shop was ` 450. The sales tax charged was
5%.Findthebillamount.
Solution: On ` 100, the tax paid was ` 5.
On ` 450, the tax paid would be = `
5
450
100
×
= ` 22.50
Bill amount = Cost of item + Sales tax = ` 450 + ` 22.50 = ` 472.50.
Example 9: (Value Added Tax (VAT)) Waheeda bought an air cooler for ` 3300
including a tax of 10%. Find the price of the air cooler beforeVATwas added.
Solution: The price includes the VAT, i.e., the value added tax. Thus, a 10% VAT
means if the price withoutVATis ` 100 then price includingVATis ` 110.
Now, when price includingVATis ` 110, original price is ` 100.
Hencewhenpriceincludingtaxis` 3300,theoriginalprice= `
100
3300 3000.
110
× = `
Example 10:Salim bought an article for ` 784 which included GST of 12% . What is
the price of the article before GST was added?
Solution: Let original price of the article be ` 100. GST = 12%.
Price after GST is included = ` (100+12) = ` 112
When the selling price is ` 112 then original price = ` 100.
When the selling price is ` 784, then original price = `
100
784
12
× = ` 700
2021–22

1. Two times a number is a 100% increase in the number. If we take half the number
what would be the decrease in per cent?
2. By what per cent is ` 2,000 less than ` 2,400? Is it the same as the per cent by
which ` 2,400 is more than ` 2,000?
EXERCISE 8.2
1. Aman got a 10% increase in his salary. If his new salary is ` 1,54,000, find his
originalsalary.
2. On Sunday 845 people went to the Zoo. On Monday only 169 people went. What
is the per cent decrease in the people visiting the Zoo on Monday?
3. A shopkeeper buys 80 articles for ` 2,400 and sells them for a profit of
16%. Find the selling price of one article.
4. The cost of an article was ` 15,500. ` 450 were spent on its repairs. If it is
sold for a profit of 15%, find the selling price of the article.
5. A VCR and TV were bought for ` 8,000 each. The shopkeeper made a
loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss
percent on the whole transaction.
6. During a sale, a shop offered a discount of 10% on the
marked prices of all the items. What would a customer
have to pay for a pair of jeans marked at ` 1450 and
two shirts marked at ` 850 each?
7. A milkman sold two of his buffaloes for ` 20,000 each.
On one he made a gain of 5% and on the other a loss of
10%. Find his overall gain or loss. (Hint: Find
CP of each)
8. The price of a TV is ` 13,000. The sales tax charged on
it is at the rate of 12%. Find the amount that Vinod will
have to pay if he buys it.
9. Arun bought a pair of skates at a sale where the discount given was 20%. If the
amount he pays is ` 1,600, find the marked price.
10. I purchased a hair-dryer for ` 5,400 including 8% VAT. Find the price before VAT
was added.
11. An article was purchased for ` 1239 including GST of 18%. Find the price of the
article before GST was added?
8.6 Compound Interest
You might have come across statements like “one year interest for FD (fixed deposit) in
the bank @ 9% per annum” or ‘Savings account with interest @ 5% per annum’.
2021–22

126 MATHEMATICS
TRY THESE
Interestistheextramoneypaidbyinstitutionslikebanksorpostofficesonmoney
deposited(kept)withthem.Interestisalsopaidbypeoplewhentheyborrowmoney.
We already know how to calculate Simple Interest.
Example 10: Asumof` 10,000isborrowedatarateofinterest15%perannumfor2
years.Findthesimpleinterestonthissumandtheamounttobepaidattheendof2years.
Solution: On ` 100, interest charged for 1 year is ` 15.
So, on ` 10,000, interest charged =
15
10000
100
× = ` 1500
Interest for 2 years = ` 1500 × 2 = ` 3000
Amount to be paid at the end of 2 years = Principal+Interest
= ` 10000 + ` 3000 = ` 13000
Find interest and amount to be paid on ` 15000 at 5% per annum after 2 years.
My father has kept some money in the post office for 3 years. Every year the money
increases as more than the previous year.
We have some money in the bank. Every year some interest is added to it, which is
shown in the passbook. This interest is not the same, each year it increases.
Normally,theinterestpaidorchargedisneversimple.Theinterestiscalculatedonthe
amount of the previous year.This is known as interest compounded or Compound
Interest (C.I.).
Let us take an example and find the interest year by year. Each year our sum or
principalchanges.
Calculating Compound Interest
A sum of ` 20,000 is borrowed by Heena for 2 years at an interest of 8% compounded
annually. Find the Compound Interest (C.I.) and the amount she has to pay at the end of
2 years.
Aslamaskedtheteacherwhetherthismeansthattheyshouldfindtheinterestyearby
year.The teacher said ‘yes’, and asked him to use the following steps :
1. Find the Simple Interest (S.I.) for one year.
Let the principal for the first year be P1
. Here, P1
= ` 20,000
SI1
= SI at 8% p.a. for 1st year = `
20000 8
100
×
= ` 1600
2. Thenfindtheamountwhichwillbepaidorreceived.Thisbecomesprincipalforthe
nextyear.
Amount at the end of 1st year = P1
+ SI1
= ` 20000 + ` 1600
= ` 21600 = P2
(Principal for 2nd year)
2021–22

3. Againfindtheinterestonthissumforanotheryear.
SI2
= SI at 8% p.a.for 2nd year = `
21600 8
100
×
= ` 1728
4. Find the amount which has to be paid or received at the end of second year.
Amount at the end of 2nd year = P2
+ SI2
= ` 21600 + ` 1728
= ` 23328
Totalinterestgiven= ` 1600 + ` 1728
= ` 3328
Reeta asked whether the amount would be different for simple interest. The teacher
told her to find the interest for two years and see for herself.
SI for 2 years = `
20000 8 2
100
× ×
= ` 3200
Reeta said that when compound interest was used Heena would pay ` 128 more.
Let us look at the difference between simple interest and compound interest.We start
with ` 100.Try completing the chart.
Under Under
Simple Interest Compound Interest
First year Principal ` 100.00 ` 100.00
Interest at 10% ` 10.00 ` 10.00
Year-endamount ` 110.00 ` 110.00
Second year Principal ` 100.00 ` 110.00
Interest at 10% ` 10.00 ` 11.00
Year-endamount `(110 + 10) = ` 120 ` 121.00
Third year Principal ` 100.00 ` 121.00
Interest at 10% ` 10.00 ` 12.10
Year-endamount `(120 + 10) = ` 130 ` 133.10
Note that in 3 years,
Interest earned by Simple Interest = ` (130 – 100) = ` 30, whereas,
Interest earned by Compound Interest = ` (133.10 – 100) = ` 33.10
NotealsothatthePrincipalremainsthesameunderSimpleInterest,whileitchanges
year after year under compound interest.
Which
means you
pay interest
on the
interest
accumulated
till then!
2021–22

128 MATHEMATICS
8.7 Deducing a Formula for Compound Interest
Zubedaaskedherteacher,‘Is thereaneasierwaytofindcompoundinterest?’Theteacher
said‘Thereisa shorterwayof findingcompoundinterest.Letustrytofindit.’
Suppose P1
is the sum on which interest is compounded annually at a rate of R%
perannum.
Let P1
= ` 5000 and R = 5. Then by the steps mentioned above
1. SI1
= `
5000 5 1
100
× ×
or SI1
= `
1
P R 1
100
× ×
so, A1
= ` 5000 +
5000 5 1
100
× ×
or A1
= P1
+ SI1
= 1
1
P R
P
100
+
= ` 5000 1
5
100
+





 = P2
= P
R
P
1 2
1
100
+





 =
2. SI2
= ` 5000 1
5
100
5 1
100
+





 ×
×
or SI2
=
2
P R 1
100
× ×
= `
5000 5
100
1
100
×
+






5
= P
R R
1 1
100 100
+





 ×
=
P R R
1
100
1
100
+






A2
= `
5 5000 5 5
5000 1 1
100 100 100
×
   
+ + +
   
   
` A2
= P2
+ SI2
= ` 5000 1
5
100
1
5
100
+





 +





 = P
R
P
R R
1 1
1
100 100
1
100
+





 + +






= ` 5000 1
5
100
2
+





 = P3
= P
R R
1 1
100
1
100
+





 +






= P
R
P
1
2
3
1
100
+





 =
Proceeding in this way the amount at the end of n years will be
An
= P
R
1 1
100
+






n
Or, we can say A = P
R
1
100
+






n
2021–22

So, Zubeda said, but using this we get only the formula for the amount to be paid at
the end of n years, and not the formula for compound interest.
Aruna at once said that we know CI =A– P, so we can easily find the compound
interest too.
Example 11: Find CI on ` 12600 for 2 years at 10% per annum compounded
annually.
Solution: We have,A= P 1
100
+






R
n
, where Principal (P) = ` 12600, Rate (R) = 10,
Number of years (n) = 2
= ` 12600 1
10
100
2
+





 = ` 12600
11
10
2






= `
11 11
12600
10 10
× × = ` 15246
CI = A – P = ` 15246 – ` 12600 = ` 2646
8.8 Rate Compounded Annually or Half Yearly
(Semi Annually)
You may want to know why ‘compounded
annually’ was mentioned after ‘rate’. Does it
meananything?
It does, because we can also have interest
ratescompoundedhalfyearlyorquarterly.Let
us see what happens to ` 100 over a period of
oneyearifaninterestiscompoundedannually
orhalfyearly.
TRY THESE
1. Find CI on a sum of ` 8000 for
2 years at 5% per annum
compoundedannually.
P = ` 100 at 10% per P = ` 100 at 10% per annum
annumcompoundedannually compoundedhalfyearly
The time period taken is 1 year The time period is 6 months or
1
2
year
I = `
100 10 1
Rs 10
100
× ×
= I = `
1
100 10
2
5
100
× ×
= `
A = ` 100 + ` 10 A = ` 100 + ` 5 = ` 105
= ` 110 Now for next 6 months the P = ` 105
So, I = `
1
105 10
2
100
× ×
= ` 5.25
and A = ` 105 + ` 5.25 = ` 110.25
Rate
becomes
half
Time period and rate when interest not compounded
annually
The time period after which the interest is added each
time to form a new principal is called the conversion
period. When the interest is compounded half yearly,
there are two conversion periods in a year each after 6
months. In such situations, the half yearly rate will be
half of the annual rate. What will happen if interest is
compounded quarterly? In this case, there are 4
conversion periods in a year and the quarterly rate will
be one-fourth of the annual rate.
2021–22

130 MATHEMATICS
TRY THESE
Do you see that, if interest is compounded half yearly, we compute the interest two
times. So time period becomes twice and rate is taken half.
Find the time period and rate for each .
1. Asum taken for
1
1
2
years at 8% per annum is compounded half yearly.
2. Asum taken for 2 years at 4% per annum compounded half yearly.
A sum is taken for one year at 16% p.a. If interest is compounded after every three
months,howmanytimeswillinterestbechargedinoneyear?
Example 12: What amount is to be repaid on a loan of ` 12000 for 1
1
2
years at 10%
perannumcompoundedhalfyearly.
Solution:
Principal for first 6 months = `
`
`
`
` 12,000 Principal for first 6 months = `
`
`
`
` 12,000
There are 3 half years in
1
1
2
years. Time = 6 months =
6 1
year year
12 2
=
Therefore, compounding has to be done 3 times. Rate = 10%
Rate of interest = halfof10% I = `
1
12000 10
2
100
× ×
= ` 600
= 5%halfyearly A = P + I = ` 12000 + ` 600
A = P
R
1
100
+






n
= `12600. It is principal for next 6 months.
= ` 12000 1
100
3
+






5
I = `
1
12600 10
2
100
× ×
= ` 630
= `
21 21 21
12000
20 20 20
× × × Principal for third period = ` 12600 + ` 630
= ` 13,891.50 = ` 13,230.
I = `
1
13230 10
2
100
× ×
= ` 661.50
A = P + I = ` 13230 + ` 661.50
= ` 13,891.50
2021–22

TRY THESE
Find the amount to be paid
1. At the end of 2 years on ` 2,400 at 5% per annum compounded annually.
2. At the end of 1 year on ` 1,800 at 8% per annum compounded quarterly.
Example 13: Find CI paid when a sum of ` 10,000 is invested for 1 year and
3monthsat8
1
2
%perannumcompoundedannually.
Solution: Mayuri first converted the time in years.
1 year 3 months =
3
1
12
year =
1
1
4
years
Mayuritriedputtingthevaluesintheknownformulaandcameupwith:
A = ` 10000 1
17
200
1
1
4
+






Nowshewasstuck.Sheaskedherteacherhowwouldshefindapowerwhichisfractional?
Theteacherthengaveherahint:
Find the amount for the whole part, i.e., 1 year in this case. Then use this as principal
to get simple interest for
1
4
year more. Thus,
A = ` 10000 1
17
200
+






= ` 10000 ×
217
200
= ` 10,850
Now this would act as principal for the next
1
4
year. We find the SI on ` 10,850
for
1
4
year.
SI = `
1
10850 17
4
100 2
× ×
×
= `
10850 1 17
800
× ×
= ` 230.56
2021–22

132 MATHEMATICS
Interest for first year = ` 10850 – ` 10000 = ` 850
And, interest for the next
1
4
year = ` 230.56
Therefore, total compound Interest = 850 + 230.56 = ` 1080.56.
8.9 Applications of Compound Interest Formula
TherearesomesituationswherewecouldusetheformulaforcalculationofamountinCI.
Here are a few.
(i) Increase (or decrease) in population.
(ii) The growth of a bacteria if the rate of growth is known.
(iii) The value of an item, if its price increases or decreases in the intermediate years.
Example 14: The population of a city was 20,000 in the year 1997. It increased at
the rate of 5% p.a. Find the population at the end of the year 2000.
Solution: There is 5% increase in population every year, so every new year has new
population.Thus,wecansayitisincreasingincompoundedform.
Populationinthebeginningof1998=20000 (wetreatthisastheprincipalforthe1st year)
Increase at 5% =
5
20000 1000
100
× =
Population in 1999 = 20000 + 1000 = 21000
Increase at 5% =
5
21000 1050
100
× =
Population in 2000 = 21000 + 1050
= 22050
Increase at 5% =
5
22050
100
×
= 1102.5
At the end of 2000 the population = 22050 + 1102.5 = 23152.5
or, Population at the end of 2000 = 20000 1
5
100
3
+






=
21 21 21
20000
20 20 20
× × ×
= 23152.5
So, the estimated population = 23153.
Treat as
the Principal
for the
2nd year.
Treat as
the Principal
for the
3rd year.
2021–22

TRY THESE
Aruna asked what is to be done if there is a decrease. The teacher then considered
thefollowingexample.
Example 15: ATV was bought at a price of ` 21,000.After one year the value of the
TVwasdepreciatedby5%(Depreciationmeansreductionofvalueduetouseandageof
the item). Find the value of the TVafter one year.
Solution:
Principal= ` 21,000
Reduction= 5% of ` 21000 per year
= `
21000 5 1
100
× ×
= ` 1050
value at the end of 1 year = ` 21000 – ` 1050 = ` 19,950
Alternately,We may directly get this as follows:
value at the end of 1 year = ` 21000 1
5
100
−






= ` 21000 ×
19
20
= ` 19,950
1. Amachinery worth ` 10,500 depreciated by 5%. Find its value after one year.
2. Find the population of a city after 2 years, which is at present 12 lakh, if the rate
of increase is 4%.
EXERCISE 8.3
1. Calculatetheamountandcompoundintereston
(a) ` 10,800 for 3 years at 12
1
2
% per annum compounded annually.
(b) ` 18,000 for 2
1
2
years at 10% per annum compounded annually.
(c) ` 62,500 for 1
1
2
years at 8% per annum compounded half yearly.
(d) ` 8,000 for 1 year at 9% per annum compounded half yearly.
(YoucouldusetheyearbyyearcalculationusingSIformulatoverify).
(e) ` 10,000 for 1 year at 8% per annum compounded half yearly.
2. Kamala borrowed ` 26,400 from a Bank to buy a scooter at a rate of 15% p.a.
compoundedyearly.Whatamountwillshepayattheendof2yearsand4monthsto
cleartheloan?
(Hint:FindAfor2yearswithinterestiscompoundedyearlyandthenfindSIonthe
2nd year amount for
4
12
years).
2021–22

134 MATHEMATICS
3. Fabina borrows ` 12,500 at 12% per annum for 3 years at simple interest and
Radha borrows the same amount for the same time period at 10% per annum,
compounded annually.Who pays more interest and by how much?
4. I borrowed ` 12,000 from Jamshed at 6% per annum simple interest for 2 years.
Had I borrowed this sum at 6% per annum compound interest, what extra amount
would I have to pay?
5. Vasudevan invested ` 60,000 at an interest rate of 12% per annum compounded
halfyearly.Whatamountwouldheget
(i) after6months?
(ii) after 1 year?
6. Arif took a loan of ` 80,000 from a bank. If the rate of interest is 10% per annum,
findthedifferenceinamountshewouldbepayingafter
1
1
2
yearsiftheinterestis
(i) compoundedannually.
(ii) compoundedhalfyearly.
7. Maria invested ` 8,000 in a business. She would be paid interest at 5% per annum
compoundedannually.Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
8. Find the amount and the compound interest on ` 10,000 for
1
1
2
years at 10% per
annum, compounded half yearly.Would this interest be more than the interest he
wouldgetifitwascompoundedannually?
9. FindtheamountwhichRamwillgeton`4096,ifhegaveitfor18monthsat
1
12 %
2
perannum,interestbeingcompoundedhalfyearly.
10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) findthepopulationin2001.
(ii) what would be its population in 2005?
11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the
rateof2.5%perhour.Findthebacteriaattheendof2hoursifthecountwasinitially
5, 06,000.
12. A scooter was bought at ` 42,000. Its value
depreciated at the rate of 8% per annum.
Find its value after one year.
2021–22

1. Discount is a reduction given on marked price.
Discount = Marked Price – Sale Price.
2. Discountcanbecalculatedwhendiscountpercentageisgiven.
Discount = Discount % of Marked Price
3. Additional expenses made after buying an article are included in the cost price and are known
as overhead expenses.
CP = Buying price + Overhead expenses
4. Sales tax is charged on the sale of an item by the government and is added to the BillAmount.
Salestax=Tax%ofBillAmount
5. GSTstands for Goods and Services Tax and is levied on supply of goods or services or both.
6. Compoundinterestistheinterestcalculatedonthepreviousyear’samount(A=P+I)
7. (i) Amountwheninterestiscompoundedannually
= P
R
1
100
+






n
; P is principal, R is rate of interest, n is time period
(ii) Amountwheninterestiscompoundedhalfyearly
= P
R
2
1
200
+






n
R
is half yearly rate and
2 = number of ’half-years’
2
n





2021–22

136 MATHEMATICS
NOTES
2021–22

ALGEBRAIC EXPRESSIONS AND IDENTITIES 137
9.1 What are Expressions?
In earlier classes, we have already become familiar with what algebraic expressions
(orsimplyexpressions)are.Examplesofexpressionsare:
x + 3, 2y – 5, 3x2
, 4xy + 7 etc.
You can form many more expressions.As you know expressions are formed from
variablesandconstants.Theexpression2y–5isformedfromthevariableyandconstants
2 and 5. The expression 4xy + 7 is formed from variablesx and y and constants 4 and 7.
We know that, the value of y in the expression, 2y – 5, may be anything. It can be
2, 5, –3, 0,
5 7
, –
2 3
etc.; actually countless different values. The value of an expression
changes with the value chosen for the variables it contains. Thus as ytakes on different
values, the value of 2y – 5 goes on changing. When y = 2, 2y – 5 = 2(2) – 5 = –1; when
y = 0, 2y – 5 = 2 × 0 –5 = –5, etc. Find the value of the expression 2y – 5 for the other
givenvaluesofy.
Number line and an expression:
Considertheexpressionx+5.LetussaythevariablexhasapositionXonthenumberline;
Xmaybeanywhereonthenumberline,butitisdefinitethatthevalueofx+5isgivenby
apointP,5unitstotherightof X.Similarly,thevalueofx–4willbe4unitstotheleftof
X and so on.
What about the position of 4x and 4x + 5?
Thepositionof4xwillbepointC;thedistanceofCfromtheoriginwillbefourtimes
thedistanceofXfromtheorigin.ThepositionDof4x+5willbe5unitstotherightofC.
Algebraic Expressions
and Identities
CHAPTER
9
2021–22

138 MATHEMATICS
TRY THESE
TRY THESE
1. Give five examples of expressions containing one variable and five examples of
expressionscontainingtwovariables.
2. Show on the number line x, x – 4, 2x + 1, 3x – 2.
9.2 Terms, Factors and Coefficients
Take the expression 4x + 5.This expression is made up of two terms, 4x and 5. Terms
are added to form expressions. Terms themselves can be formed as the product of
factors. The term 4xis the product of its factors 4 and x. The term 5
is made up of just one factor, i.e., 5.
The expression 7xy – 5x has two terms 7xy and –5x. The term
7xy is a product of factors 7, x and y. The numerical factor of a term
iscalleditsnumericalcoefficientorsimplycoefficient.Thecoefficient
in the term 7xy is 7 and the coefficient in the term –5x is –5.
9.3 Monomials, Binomials and Polynomials
Expressionthatcontainsonlyonetermiscalledamonomial.Expressionthatcontainstwo
termsis called abinomial.Anexpressioncontainingthreeterms isatrinomialandsoon.
In general, an expression containing, one or more terms with non-zero coefficient (with
variableshavingnonnegativeintegersasexponents)iscalledapolynomial.Apolynomial
maycontainanynumberofterms, oneormorethanone.
Examplesofmonomials: 4x2
, 3xy, –7z, 5xy2
, 10y, –9, 82mnp, etc.
Examplesof binomials: a + b, 4l + 5m, a + 4, 5 –3xy, z2
– 4y2
, etc.
Examplesoftrinomials: a + b + c, 2x + 3y – 5, x2
y – xy2
+ y2
, etc.
Examplesofpolynomials: a + b + c + d, 3xy, 7xyz – 10, 2x + 3y + 7z, etc.
1. Classifythefollowingpolynomialsasmonomials,binomials,trinomials.
– z + 5, x + y + z, y + z + 100, ab – ac, 17
2. Construct
(a) 3binomialswithonlyx as a variable;
(b) 3 binomials with x and yas variables;
(c) 3 monomials with xand y as variables;
(d) 2polynomialswith4ormoreterms.
9.4 Like and Unlike Terms
Lookatthefollowingexpressions:
7x, 14x, –13x, 5x2
, 7y, 7xy, –9y2
, –9x2
, –5yx
Liketermsfromtheseare:
(i) 7x, 14x, –13x are like terms.
(ii) 5x2
and –9x2
are like terms.
TRY THESE
Identify the coefficient of each
termintheexpression
x2
y2
– 10x2
y + 5xy2
– 20.
2021–22

TRY THESE
(iii) 7xy and –5yx are like terms.
Why are 7x and 7y not like?
Why are 7x and 7xy not like?
Why are 7x and 5x2
not like?
Writetwotermswhicharelike
(i) 7xy (ii) 4mn2
(iii) 2l
9.5 Addition and Subtraction of Algebraic Expressions
In the earlier classes, we have also learnt how to add and subtract algebraic expressions.
For example, to add 7x2
– 4x + 5 and 9x – 10, we do
7x2
– 4x + 5
+ 9x – 10
7x2
+ 5x – 5
Observe how we do the addition. We write each expression to be added in a separate
row. While doing so we write like terms one below the other, and add them, as shown.
Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some
moreexamples.
Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.
Solution: Writing the three expressions in separate rows, with like terms one below
the other, we have
7xy + 5yz –3zx
+ 4yz + 9zx – 4y
+ –2xy – 3zx + 5x (Note xz is same as zx)
5xy + 9yz +3zx + 5x – 4y
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y
inthesecondexpressionand5xinthethirdexpression,arecarriedoverastheyare,since
they have no like terms in the other expressions.
Example 2: Subtract 5x2
– 4y2
+ 6y – 3 from 7x2
– 4xy + 8y2
+ 5x – 3y.
Solution:
7x2
– 4xy + 8y2
+ 5x – 3y
5x2
– 4y2
+ 6y – 3
(–) (+) (–) (+)
2x2
– 4xy + 12y2
+ 5x – 9y + 3
2021–22

140 MATHEMATICS
Note that subtraction of a number is the same as addition of its additive inverse.
Thus subtracting–3isthesameasadding+3.Similarly,subtracting6yisthesameas
adding–6y; subtracting – 4y2
is the same as adding 4y2
and so on. The signs in the
third row written below each term in the second row help us in knowing which
operation has to be performed.
EXERCISE 9.1
1. Identifytheterms,theircoefficientsforeachofthefollowingexpressions.
(i) 5xyz2
– 3zy (ii) 1 + x + x2
(iii) 4x2
y2
– 4x2
y2
z2
+ z2
(iv) 3 – pq + qr – rp (v)
2 2
x y
xy
+ − (vi) 0.3a – 0.6ab + 0.5b
2. Classify the following polynomials as monomials, binomials, trinomials. Which
polynomialsdonotfitinanyofthesethreecategories?
x + y, 1000, x + x2
+ x3
+ x4
, 7 + y + 5x, 2y – 3y2
, 2y – 3y2
+ 4y3
, 5x – 4y + 3xy,
4z – 15z2
, ab + bc + cd + da, pqr, p2
q + pq2
, 2p + 2q
3. Addthefollowing.
(i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2
q2
– 3pq + 4, 5 + 7pq – 3p2
q2
(iv) l2
+ m2
, m2
+ n2
, n2
+ l2
,
2lm + 2mn + 2nl
4. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p2
q – 3pq + 5pq2
– 8p + 7q – 10 from
18 – 3p – 11q + 5pq – 2pq2
+ 5p2
q
9.6 Multiplication of Algebraic Expressions: Introduction
(i) Look at the following patterns of dots.
Pattern of dots Total number of dots
4 × 9
5 × 7
2021–22

m × n
(m + 2) × (n + 3)
(ii) Canyounowthinkofsimilarothersituationsinwhich
twoalgebraicexpressionshavetobemultiplied?
Ameena gets up. She says, “We can think of area of
a rectangle.” The area of a rectangle is l × b, where l
is the length, and b is breadth. If the length of the
rectangle is increased by 5 units, i.e., (l + 5) and
breadth is decreased by 3 units , i.e., (b – 3) units,
the area of the new rectangle will be (l + 5) × (b – 3).
(iii) Can you think about volume? (The volume of a
rectangular box is given by the product of its length,
breadthandheight).
(iv) Saritapointsoutthatwhenwebuythings,wehaveto
carryoutmultiplication.Forexample,if
price of bananas per dozen = ` p
and for the school picnic bananas needed = z dozens,
then we have to pay = ` p × z
Suppose, the price per dozen was less by ` 2 and the bananas needed were less by
4 dozens.
Then, price of bananas per dozen = ` (p – 2)
and bananas needed = (z – 4) dozens,
Therefore, we would have to pay = ` (p – 2) × (z – 4)
To find the area of a rectangle, we
have to multiply algebraic
expressions like l × b or
(l + 5) × (b – 3).
Here the number of rows
is increased by
2, i.e., m + 2 and number
of columns increased by
3, i.e., n + 3.
To find the number of
dots we have to multiply
the expression for the
number of rows by the
expression for the
number of columns.
2021–22

142 MATHEMATICS
Notice that all the three
products of monomials, 3xy,
15xy, –15xy, are also
monomials.
TRY THESE
Canyouthinkoftwomoresuchsituations,wherewemayneedtomultiplyalgebraic
expressions?
[Hint: • Think of speed and time;
• Thinkofinteresttobepaid,theprincipalandtherateofsimpleinterest;etc.]
Inalltheaboveexamples,wehadtocarryoutmultiplicationoftwoormorequantities.If
the quantities are given by algebraic expressions, we need to find their product. This
means that we should know how to obtain this product. Let us do this systematically.To
beginwithweshalllookatthemultiplicationoftwomonomials.
9.7 Multiplying a Monomial by a Monomial
9.7.1 Multiplying two monomials
Webeginwith
4 × x = x + x + x + x = 4x as seen earlier.
Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x
Now, observe the following products.
(i) x × 3y = x × 3 × y = 3 × x × y = 3xy
(ii) 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y = 15xy
(iii) 5x × (–3y) = 5 × x × (–3) × y
= 5 × (–3) × x × y = –15xy
Note that 5 × 4 = 20
i.e., coefficient of product = coefficient of
first monomial × coefficient of second
monomial;
and x × x2
= x3
i.e., algebraic factor of product
= algebraic factor of first monomial
× algebraic factor of second monomial.
Somemoreusefulexamplesfollow.
(iv) 5x × 4x2
= (5 × 4) × (x × x2
)
= 20 × x3
= 20x3
(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz)
= –20 × (x × x × yz) = –20x2
yz
Observe how we collect the powers of different variables
in the algebraic parts of the two monomials. While doing
so, we use the rules of exponents and powers.
9.7.2 Multiplying three or more monomials
Observethefollowingexamples.
(i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz
(ii) 4xy × 5x2
y2
× 6x3
y3
= (4xy × 5x2
y2
) × 6x3
y3
= 20x3
y3
× 6x3
y3
= 120x3
y3
× x3
y3
= 120 (x3
× x3
) × (y3
× y3
) = 120x6
× y6
= 120x6
y6
It is clear that we first multiply the first two monomials and then multiply the resulting
monomial by the third monomial. This method can be extended to the product of any
numberofmonomials.
2021–22

TRY THESE
Find 4x × 5y × 7z
First find 4x × 5y and multiply it by 7z;
or first find 5y × 7z and multiply it by 4x.
Is the result the same? What do you observe?
Doestheorderinwhichyoucarryoutthemultiplicationmatter?
Example 3: Complete the table for area of a rectangle with given length and breadth.
Solution: length breadth area
3x 5y 3x × 5y = 15xy
9y 4y2
..............
4ab 5bc ..............
2l2
m 3lm2
..............
Example 4: Find the volume of each rectangular box with given length, breadth
andheight.
length breadth height
(i) 2ax 3by 5cz
(ii) m2
n n2
p p2
m
(iii) 2q 4q2
8q3
Solution: Volume = length × breadth × height
Hence, for (i) volume= (2ax) × (3by) × (5cz)
= 2 × 3 × 5 × (ax) × (by) × (cz) = 30abcxyz
for (ii) volume= m2
n × n2
p × p2
m
= (m2
× m) × (n × n2
) × (p × p2
) = m3
n3
p3
for (iii) volume= 2q × 4q2
× 8q3
= 2 × 4 × 8 × q × q2
× q3
= 64q6
EXERCISE 9.2
1. Findtheproductofthefollowingpairsofmonomials.
(i) 4, 7p (ii) – 4p, 7p (iii) – 4p, 7pq (iv) 4p3
, – 3p
(v) 4p, 0
2. Findtheareasofrectangleswiththefollowingpairsofmonomialsastheirlengthsand
breadthsrespectively.
(p, q); (10m, 5n); (20x2
, 5y2
); (4x, 3x2
); (3mn, 4np)
We can find the product in other way also.
4xy × 5x2
y2
× 6x3
y3
= (4 × 5 × 6) × (x × x2
× x3
) × (y × y2
× y3
)
= 120 x6
y6
2021–22

144 MATHEMATICS
TRY THESE
3. Complete the table of products.
2x –5y 3x2
– 4xy 7x2
y –9x2
y2
2x 4x2 . . . . . . . . . . . . . . .
–5y . . . . . .
–15x2
y . . . . . . . . .
3x2 . . . . . . . . . . . . . . . . . .
– 4xy . . . . . . . . . . . . . . . . . .
7x2
y . . . . . . . . . . . . . . . . . .
–9x2
y2 . . . . . . . . . . . . . . . . . .
4. Obtainthevolumeofrectangularboxeswiththefollowinglength,breadthandheight
respectively.
(i) 5a, 3a2
, 7a4
(ii) 2p, 4q, 8r (iii) xy, 2x2
y, 2xy2
(iv) a, 2b, 3c
5. Obtain the product of
(i) xy, yz, zx (ii) a, – a2
, a3
(iii) 2, 4y, 8y2
, 16y3
(iv) a, 2b, 3c, 6abc (v) m, – mn, mnp
9.8 Multiplying a Monomial by a Polynomial
9.8.1 Multiplying a monomial by a binomial
Let us multiply the monomial 3x by the binomial 5y + 2, i.e., find 3x × (5y + 2) = ?
Recall that 3xand (5y + 2) represent numbers.Therefore, using the distributive law,
3x × (5y + 2) = (3x × 5y) + (3x × 2) = 15xy + 6x
We commonly use distributive law in our calculations. For example:
7 × 106 = 7 × (100 + 6)
= 7 × 100 + 7 × 6 (Here, we used distributive law)
= 700 + 42 = 742
7 × 38 = 7 × (40 – 2)
= 7 × 40 – 7 × 2 (Here, we used distributive law)
= 280 – 14 = 266
Similarly, (–3x) × (–5y + 2) = (–3x) × (–5y) + (–3x) × (2) = 15xy – 6x
and 5xy × (y2
+ 3) = (5xy × y2
) + (5xy × 3) = 5xy3
+ 15xy.
What about a binomial × monomial? For example, (5y + 2) × 3x = ?
We may use commutative law as : 7 × 3 = 3 × 7; or in general a × b = b × a
Similarly, (5y + 2) × 3x = 3x × (5y + 2) = 15xy + 6x as before.
Find the product (i) 2x (3x + 5xy) (ii) a2
(2ab – 5c)
Firstmonomial →
Secondmonomial↓
2021–22

9.8.2 Multiplying a monomial by a trinomial
Consider 3p × (4p2
+ 5p + 7).As in the earlier case, we use distributive law;
3p × (4p2
+ 5p + 7) = (3p × 4p2
) + (3p × 5p) + (3p × 7)
= 12p3
+ 15p2
+ 21p
Multiplyeachtermofthetrinomialbythemonomialandaddproducts.
Observe, by using the distributive law, we are able to carry out the
multiplication term by term.
Example 5: Simplify the expressions and evaluate them as directed:
(i) x (x – 3) + 2 for x = 1, (ii) 3y (2y – 7) – 3 (y – 4) – 63 for y = –2
Solution:
(i) x (x – 3) + 2 = x2
– 3x + 2
For x = 1, x2
– 3x + 2 = (1)2
– 3 (1) + 2
= 1 – 3 + 2 = 3 – 3 = 0
(ii) 3y (2y – 7) – 3 (y – 4) – 63 = 6y2
– 21y – 3y + 12 – 63
= 6y2
– 24y – 51
For y = –2, 6y2
– 24y – 51 = 6 (–2)2
– 24(–2) – 51
= 6 × 4 + 24 × 2 – 51
= 24 + 48 – 51 = 72 – 51 = 21
Example 6: Add
(i) 5m (3 – m) and 6m2
– 13m (ii) 4y (3y2
+ 5y – 7) and 2 (y3
– 4y2
+ 5)
Solution:
(i) First expression = 5m (3 – m) = (5m × 3) – (5m × m) = 15m – 5m2
Now adding the second expression to it,15m – 5m2
+ 6m2
– 13m = m2
+ 2m
(ii) The first expression = 4y (3y2
+ 5y – 7) = (4y × 3y2
) + (4y × 5y) + (4y × (–7))
= 12y3
+ 20y2
– 28y
The second expression = 2 (y3
– 4y2
+ 5) = 2y3
+ 2 × (– 4y2
) + 2 × 5
= 2y3
– 8y2
+ 10
Addingthetwoexpressions, 12y3
+ 20y2
– 28y
+ 2y3
– 8y2
+ 10
14y3
+ 12y2
– 28y + 10
Example 7: Subtract 3pq (p – q) from 2pq (p + q).
Solution: We have 3pq (p – q) = 3p2
q – 3pq2
and
2pq (p + q) = 2p2
q + 2pq2
Subtracting, 2p2
q + 2pq2
3p2
q – 3pq2
– +
– p2
q + 5pq2
TRY THESE
Findtheproduct:
(4p2
+ 5p + 7) × 3p
2021–22

146 MATHEMATICS
EXERCISE 9.3
1. Carryoutthemultiplicationoftheexpressionsineachofthe followingpairs.
(i) 4p, q + r (ii) ab, a – b (iii) a + b, 7a2
b2
(iv) a2
– 9, 4a
(v) pq + qr + rp, 0
2. Complete the table.
First expression Second expression Product
(i) a b + c + d . . .
(ii) x + y – 5 5xy . . .
(iii) p 6p2
– 7p + 5 . . .
(iv) 4p2
q2
p2
– q2 . . .
(v) a + b + c abc . . .
3. Find the product.
(i) (a2
) × (2a22
) × (4a26
) (ii)
2
3
9
10
2 2
xy x y





 ×
−






(iii) −





 ×






10
3
6
5
3 3
pq p q (iv) x × x2
× x3
× x4
4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =
1
2
.
(b) Simplify a (a2
+ a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1
(iii) a = – 1.
5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)
(b) Add: 2x (z – x – y) and 2y (z – y – x)
(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )
(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c) from 4c ( – a + b + c )
9.9 Multiplying a Polynomial by a Polynomial
9.9.1 Multiplying a binomial by a binomial
Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b). We do this
step-by-step,aswedidinearliercases,followingthedistributivelawofmultiplication,
(3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b)
= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)
= 6a2
+ 9ab + 8ba + 12b2
= 6a2
+ 17ab + 12b2
(Since ba = ab)
When we carry out term by term multiplication, we expect 2 × 2 = 4 terms to be
present. But two of these are like terms, which are combined, and hence we get 3 terms.
In multiplication of polynomials with polynomials, we should always look for like
terms, if any, and combine them.
Observe, every term in one
binomial multiplies every
term in the other binomial.
2021–22

Example 8: Multiply
(i) (x – 4) and (2x + 3) (ii) (x – y) and (3x + 5y)
Solution:
(i) (x – 4) × (2x + 3) = x × (2x + 3) – 4 × (2x + 3)
= (x × 2x) + (x × 3) – (4 × 2x) – (4 × 3) = 2x2
+ 3x – 8x – 12
= 2x2
– 5x – 12 (Adding like terms)
(ii) (x – y) × (3x + 5y)= x × (3x + 5y) – y × (3x + 5y)
= (x × 3x) + (x × 5y) – (y × 3x) – ( y × 5y)
= 3x2
+ 5xy – 3yx – 5y2
= 3x2
+ 2xy – 5y2
(Adding like terms)
Example 9: Multiply
(i) (a + 7) and (b – 5) (ii) (a2
+ 2b2
) and (5a – 3b)
Solution:
(i) (a + 7) × (b – 5) = a × (b – 5) + 7 × (b – 5)
= ab – 5a + 7b – 35
Notethattherearenoliketermsinvolvedinthismultiplication.
(ii) (a2
+ 2b2
) × (5a – 3b) = a2
(5a – 3b) + 2b2
× (5a – 3b)
= 5a3
– 3a2
b + 10ab2
– 6b3
9.9.2 Multiplying a binomial by a trinomial
Inthismultiplication,weshallhavetomultiplyeachofthethreetermsinthetrinomialby
each of the two terms in the binomial. We shall get in all 3 × 2 = 6 terms, which may
reduce to 5 or less, if the term by term multiplication results in like terms. Consider
( 7)
binomial
a +
×
2
( 3 5)
trinomial
a a
+ +

= a × (a2
+ 3a + 5) + 7 × (a2
+ 3a + 5)
= a3
+ 3a2
+ 5a + 7a2
+ 21a + 35
= a3
+ (3a2
+ 7a2
) + (5a + 21a) + 35
= a3
+ 10a2
+ 26a + 35 (Why are there only 4
termsinthefinalresult?)
Example 10: Simplify (a + b) (2a – 3b + c) – (2a – 3b) c.
Solution: We have
(a + b) (2a – 3b + c) = a (2a – 3b + c) + b (2a – 3b + c)
= 2a2
– 3ab + ac + 2ab – 3b2
+ bc
= 2a2
– ab – 3b2
+ bc + ac (Note, –3ab and 2ab
areliketerms)
and (2a – 3b) c = 2ac – 3bc
Therefore,
(a + b) (2a – 3b + c) – (2a – 3b) c = 2a2
– ab – 3b2
+ bc + ac – (2ac – 3bc)
= 2a2
– ab – 3b2
+ bc + ac – 2ac + 3bc
= 2a2
– ab – 3b2
+ (bc + 3bc) + (ac – 2ac)
= 2a2
– 3b2
– ab + 4bc – ac
[usingthedistributivelaw]
2021–22

148 MATHEMATICS
EXERCISE 9.4
1. Multiplythebinomials.
(i) (2x + 5) and (4x – 3) (ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2
) and (3pq – 2q2
)
(vi)
2. Find the product.
(i) (5 – 2x) (3 + x) (ii) (x + 7y) (7x – y)
(iii) (a2
+ b) (a + b2
) (iv) (p2
– q2
) (2p + q)
3. Simplify.
(i) (x2
– 5) (x + 5) + 25 (ii) (a2
+ 5) (b3
+ 3) + 5
(iii) (t + s2
) (t2
– s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(v) (x + y)(2x + y) + (x + 2y)(x – y) (vi) (x + y)(x2
– xy + y2
)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c)(a + b – c)
9.10 What is an Identity?
Considertheequality (a + 1) (a +2) = a2
+ 3a + 2
We shall evaluate both sides of this equality for some value of a, say a = 10.
For a = 10, LHS = (a + 1) (a + 2) = (10 + 1) (10 + 2) = 11 × 12 = 132
RHS = a2
+ 3a + 2 = 102
+ 3 × 10 + 2 = 100 + 30 + 2 = 132
Thus, the values of the two sides of the equality are equal for a = 10.
Let us now take a = –5
LHS = (a + 1) (a + 2) = (–5 + 1) (–5 + 2) = (– 4) × (–3) = 12
RHS = a2
+ 3a + 2 = (–5)2
+ 3 (–5) + 2
= 25 – 15 + 2 = 10 + 2 = 12
Thus, for a = –5, also LHS = RHS.
We shall find that for any value of a, LHS = RHS. Such an equality, true forevery
value of the variable in it, is called an identity. Thus,
(a + 1) (a + 2) = a2
+ 3a + 2 is an identity.
An equation is true for only certain values of the variable in it. It is not true for
all values of the variable. For example, consider the equation
a2
+ 3a + 2 = 132
It is true for a = 10, as seen above, but it is not true for a = –5 or for a = 0 etc.
Try it: Show that a2
+ 3a + 2 = 132 is not true for a = –5 and for a = 0.
9.11 Standard Identities
Weshallnowstudythreeidentitieswhichareveryusefulinourwork.Theseidentitiesare
obtainedbymultiplyingabinomialbyanotherbinomial.
2021–22

TRY THESE
TRY THESE
Let us first consider the product (a + b) (a + b) or (a + b)2
.
(a + b)2
= (a + b) (a + b)
= a(a + b) + b (a + b)
= a2
+ ab + ba + b2
= a2
+ 2ab + b2
(since ab = ba)
Thus (a + b)2
= a2
+ 2ab + b2
(I)
Clearly,thisisanidentity,sincetheexpressionontheRHSisobtainedfromtheLHSby
actualmultiplication.Onemayverifythatforanyvalueofaandanyvalueofb,thevaluesof
thetwosidesareequal.
• Next we consider (a – b)2
= (a – b) (a – b) = a (a – b) – b (a – b)
We have = a2
– ab – ba + b2
= a2
– 2ab + b2
or (a – b)2
= a2
– 2ab + b2
(II)
• Finally, consider (a + b) (a – b). We have (a + b) (a – b) = a (a – b) + b (a – b)
= a2
– ab + ba – b2
= a2
– b2
(since ab = ba)
or (a + b) (a – b) = a2
– b2
(III)
The identities (I), (II) and (III) are known as standard identities.
1. Put – b in place of b in Identity (I). Do you get Identity (II)?
• Weshallnowworkoutonemoreusefulidentity.
(x + a) (x + b) = x (x + b) + a (x + b)
= x2
+ bx + ax + ab
or (x + a) (x + b) = x2
+ (a + b) x + ab (IV)
1. Verify Identity (IV), for a = 2, b = 3, x = 5.
2. Consider, the special case of Identity (IV) with a = b, what do you get? Is it
related to Identity (I)?
3. Consider, the special case of Identity (IV) with a = – cand b = – c.What do you
get? Is it related to Identity (II)?
4. Consider the special case of Identity (IV) with b = – a. What do you get? Is it
related to Identity (III)?
We can see that Identity (IV) is the general form of the other three identities also.
9.12 Applying Identities
Weshallnowseehow,formanyproblemsonmultiplicationofbinomialexpressionsand
alsoofnumbers,useoftheidentitiesgivesasimplealternativemethodofsolvingthem.
2021–22

150 MATHEMATICS
Example 11: Using the Identity (I), find (i) (2x + 3y)2
(ii) 1032
Solution:
(i) (2x + 3y)2
= (2x)2
+ 2(2x) (3y) + (3y)2
[UsingtheIdentity(I)]
= 4x2
+ 12xy + 9y2
We may work out (2x + 3y)2
directly.
(2x + 3y)2
= (2x + 3y) (2x + 3y)
= (2x) (2x) + (2x) (3y) + (3y) (2x) + (3y) (3y)
= 4x2
+ 6xy + 6 yx + 9y2
(as xy = yx)
= 4x2
+ 12xy + 9y2
UsingIdentity(I)gaveusanalternativemethodofsquaring(2x+3y).Doyounoticethat
the Identity method required fewer steps than the above direct method?You will realise
the simplicity of this method even more if you try to square more complicated binomial
expressions than (2x + 3y).
(ii) (103)2
= (100 + 3)2
= 1002
+ 2 × 100 × 3 + 32
(UsingIdentityI)
= 10000 + 600 + 9 = 10609
Wemayalsodirectlymultiply103by103andgettheanswer.DoyouseethatIdentity(I)
has given us a less tedious method than the direct method of squaring 103?Try squaring
1013.Youwillfindinthiscase,themethodofusingidentitiesevenmoreattractivethanthe
directmultiplicationmethod.
Example 12: Using Identity (II), find (i) (4p – 3q)2
(ii) (4.9)2
Solution:
(i) (4p – 3q)2
=(4p)2
– 2 (4p) (3q) + (3q)2
[UsingtheIdentity(II)]
= 16p2
– 24pq + 9q2
Doyouagreethatforsquaring(4p–3q)2
themethodofidentitiesisquickerthanthe
directmethod?
(ii) (4.9)2
=(5.0 – 0.1)2
= (5.0)2
– 2 (5.0) (0.1) + (0.1)2
= 25.00 – 1.00 + 0.01 = 24.01
Isitnotthat,squaring4.9usingIdentity(II)ismuchlesstediousthansquaringitby
directmultiplication?
Example 13: Using Identity (III), find
(i)
3
2
2
3
3
2
2
3
m n m n
+





 −





 (ii) 9832
– 172
(iii) 194 × 206
Solution:
(i)
3
2
2
3
3
2
2
3
m n m n
+





 −





 =
3
2
2
3
2 2
m n





 −






=
2 2
9 4
4 9
m n
−
(ii) 9832
– 172
= (983 + 17) (983 – 17)
[Here a = 983, b =17, a2
– b2
= (a + b) (a – b)]
Therefore, 9832
– 172
= 1000 × 966 = 966000
Try doing this directly.
You will realise how easy
our method of using
Identity (III) is.
2021–22

(iii) 194 × 206 = (200 – 6) × (200 + 6) = 2002
– 62
= 40000 – 36 = 39964
Example 14: Use the Identity (x + a) (x + b) = x2
+ (a + b) x + ab to find the
following:
(i) 501 × 502 (ii) 95 × 103
Solution:
(i) 501 × 502 = (500 + 1) × (500 + 2) = 5002
+ (1 + 2) × 500 + 1 × 2
= 250000 + 1500 + 2 = 251502
(ii) 95 × 103 = (100 – 5) × (100 + 3) = 1002
+ (–5 + 3) × 100 + (–5) × 3
= 10000 – 200 – 15 = 9785
EXERCISE 9.5
1. Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a – 7) (2a – 7)
(iv) (3a –
1
2
) (3a –
1
2
) (v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a2
+ b2
) (– a2
+ b2
) (vii) (6x – 7) (6x + 7) (viii) (– a + c) (– a + c)
(ix)
x y x y
2
3
4 2
3
4
+





 +





 (x) (7a – 9b) (7a – 9b)
2. Use the identity (x + a) (x + b) = x2
+ (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7) (ii) (4x + 5) (4x + 1)
(iii) (4x – 5) (4x – 1) (iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y) (vi) (2a2
+ 9) (2a2
+ 5)
(vii) (xyz – 4) (xyz – 2)
3. Findthefollowingsquaresbyusingtheidentities.
(i) (b – 7)2
(ii) (xy + 3z)2
(iii) (6x2
– 5y)2
(iv)
2
3
3
2
2
m n
+





 (v) (0.4p – 0.5q)2
(vi) (2xy + 5y)2
4. Simplify.
(i) (a2
– b2
)2
(ii) (2x + 5)2
– (2x – 5)2
(iii) (7m – 8n)2
+ (7m + 8n)2
(iv) (4m + 5n)2
+ (5m + 4n)2
(v) (2.5p – 1.5q)2
– (1.5p – 2.5q)2
(vi) (ab + bc)2
– 2ab2
c (vii) (m2
– n2
m)2
+ 2m3
n2
5. Show that.
(i) (3x + 7)2
– 84x = (3x – 7)2
(ii) (9p – 5q)2
+ 180pq = (9p + 5q)2
(iii) + 2mn =
2 2
16 9
9 16
m n
+
(iv) (4pq + 3q)2
– (4pq – 3q)2
= 48pq2
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
2021–22

152 MATHEMATICS
6. Usingidentities,evaluate.
(i) 712
(ii) 992
(iii) 1022
(iv) 9982
(v) 5.22
(vi) 297 × 303 (vii) 78 × 82 (viii) 8.92
(ix) 10.5 × 9.5
7. Using a2
– b2
= (a + b) (a – b), find
(i) 512
– 492
(ii) (1.02)2
– (0.98)2
(iii) 1532
– 1472
(iv) 12.12
– 7.92
8. Using (x + a) (x + b) = x2
+ (a + b) x + ab, find
(i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8
1. Expressions are formed from variables and constants.
2. Terms are added to form expressions. Terms themselves are formed as product of factors.
3. Expressionsthatcontainexactlyone,twoandthreetermsarecalledmonomials,binomials and
trinomialsrespectively.Ingeneral,anyexpressioncontainingoneormoretermswithnon-zero
coefficients(andwithvariableshavingnon-negativeintegersasexponents)iscalledapolynomial.
4. Like terms are formed from the same variables and the powers of these variables are the same,
too. Coefficients of like terms need not be the same.
5. While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them;
thenhandletheunliketerms.
6. Therearenumberofsituationsinwhichweneedtomultiplyalgebraicexpressions:forexample,in
finding area of a rectangle, the sides of which are given as expressions.
7. Amonomialmultipliedbyamonomialalwaysgivesamonomial.
8. Whilemultiplyingapolynomialbyamonomial,wemultiplyeveryterminthepolynomialbythe
monomial.
9. Incarryingoutthemultiplicationofapolynomialbyabinomial(ortrinomial),wemultiplytermby
term,i.e.,everytermofthepolynomialismultipliedbyeveryterminthebinomial(ortrinomial).
Note that in such multiplication, we may get terms in the product which are like and have to be
combined.
10. Anidentityisanequality,whichistrueforallvaluesofthevariablesintheequality.
Ontheotherhand,anequationistrueonlyforcertainvaluesofitsvariables.Anequationisnotan
identity.
11. Thefollowingarethestandardidentities:
(a + b)2
= a2
+ 2ab + b2
(I)
(a – b)2
= a2
– 2ab + b2
(II)
(a + b) (a – b) = a2
– b2
(III)
12. Another useful identity is (x + a) (x + b) = x2
+ (a + b) x + ab (IV)
13. Theabovefouridentitiesareusefulincarryingoutsquaresandproductsofalgebraicexpressions.
They also allow easy alternative methods to calculate products of numbers and so on.
2021–22

VISUALISING SOLID SHAPES 153
DO THIS
10.1 Introduction
InClassVII,youhavelearntaboutplaneshapesandsolidshapes.Planeshapeshavetwo
measurementslikelengthandbreadthandthereforetheyarecalledtwo-dimensionalshapes
whereasasolidobjecthasthreemeasurementslikelength,breadth,heightordepth.Hence,
they are called three-dimensional shapes. Also, a solid object occupies some space.
Two-dimensionalandthree-dimensionalfigurescanalsobebrieflynamedas2-Dand3-
Dfigures.Youmayrecallthattriangle,rectangle,circleetc.,are2-Dfigureswhilecubes,
cylinders,cones,spheresetc.arethree-dimensionalfigures.
Matchthefollowing:(Firstoneisdoneforyou)
Shape Type of Shape Name of the shape
3-dimensional Sphere
2-Dimensional Cylinder
3-dimensional Square
2-dimensional Circle
Visualising Solid
Shapes
CHAPTER
10
2021–22

154 MATHEMATICS
DO THIS
Matchthefollowingpictures(objects)withtheirshapes:
3-dimensional Cuboid
3-dimensional Cube
2-dimensional Cone
3-dimensional Triangle
Notethatalltheaboveshapesaresingle.However,inourpracticallife,manyatimes,we
comeacrosscombinationsofdifferentshapes.Forexample,lookatthefollowingobjects.
A tent A tin Softy (ice-cream)
A cone surmounted A cylinderical shell A cone surmounted by a
on a cylinder hemisphere
A photoframe A bowl Tomb on a pillar
A rectangular path A hemispherical shell Cylinder surmounted
by a hemisphere
Picture (object) Shape
(i) An agricultural field Two rectangular cross paths inside a
rectangularpark.
2021–22

(ii) Agroove Acircularpatharoundacircularground.
(iii) Atoy Atriangularfieldadjoiningasquarefield.
(iv) Acircularpark Acone taken out of a cylinder.
(v) A cross path Ahemispheresurmountedonacone.
10.2 Views of 3D-Shapes
Youhavelearntthata3-dimensionalobjectcanlookdifferentlyfromdifferentpositionsso
they can be drawn from different perspectives. For example, a given hut can have the
followingviews.
A hut Front view Side view Top view
similarly,aglasscanhavethefollowingviews.
A glass Side view Top view
Whyisthetopviewoftheglassapairofconcentriccircles?Willthesideviewappeardifferentiftakenfrom
some other direction? Think about this! Now look at the different views of a brick.
T
op
Side
Front
2021–22

156 MATHEMATICS
DO THIS
A brick Front view Side view Top view
We can also get different views of figures made by joining cubes. For example.
Solid Side view Front view Top view
made of three cubes
Solid Top view Front view Side view
made of four cubes
Solid Side view Front view Top view
made of four cubes
Observedifferentthingsaroundyoufromdifferentpositions.Discusswithyourfriends
theirvariousviews.
Top
Side
Front
Side
Front
Top
Front Side
Top
Top
Side
Front
2021–22

EXERCISE 10.1
1. For each of the given solid, the two views are given. Match for each solid the
corresponding top and front views. The first one is done for you.
Object Side view Top view
(a) (i) (i)
A bottle
(b) (ii) (ii)
A weight
(c) (iii) (iii)
A flask
(d) (iv) (iv)
Cup and Saucer
(e) (v) (v)
Container
2021–22

158 MATHEMATICS
2. For each of the given solid, the three views are given. Identify for each solid the corresponding top,
frontandsideviews.
(a) Object (i) (ii) (iii)
An almirah
(b)
A Match box
(c)
ATelevision
(d)
A car
Side
Front
T
op
Side
Front
T
op
Side
Front
Top
Side
Front
Top
2021–22

3. Foreachgivensolid,identifythetopview,frontviewandsideview.
(a)
(i) (ii) (iii)
(b)
(i) (ii) (iii)
(c)
(i) (ii) (iii)
(d)
(i) (ii) (iii)
(e)
(i) (ii) (iii)
2021–22

160 MATHEMATICS
4. Draw the front view, side view and top view of the given objects.
(a) Amilitarytent (b) Atable
(c) Anut (d) Ahexagonalblock
(e) Adice (f) Asolid
10.3 Mapping Space Around Us
You have been dealing with maps since you were in primary,classes.InGeography,you
have been asked to locate a particular State, a particular river, a mountain etc., on a map.
In History, you might have been asked to locate a particular place where some event had
occured long back. You have traced routes of rivers, roads, railwaylines, traders and
manyothers.
Howdowereadmaps?Whatcanweconcludeandunderstandwhilereadingamap?
What information does a map have and what it does not have? Is it any different from a
picture?Inthissection,wewilltrytofindanswerstosomeofthesequestions.Lookatthe
mapofahousewhosepictureisgivenalongside(Fig10.1).
Fig 10.1
Top
Side
Front
Top
Side
Front
Top
Side
Front
Top
Side
Front
2021–22

Whatcanweconcludefromtheaboveillustration?Whenwedrawapicture,weattempt
torepresentrealityasitisseenwithallitsdetails,whereas,amapdepictsonlythelocationof
anobject,inrelationtootherobjects.Secondly,differentpersonscangivedescriptionsof
picturescompletelydifferentfromoneanother,dependinguponthepositionfromwhichthey
are looking at the house. But, this is not true in the case of a map. The map of the house
remainsthesameirrespectiveofthepositionoftheobserver.Inotherwords,perspective
is very important for drawing a picture but it is not relevant for a map.
Now, look at the map (Fig 10.2), which has been drawn by
seven year old Raghav, as the route from his house to his school:
From this map, can you tell –
(i) howfarisRaghav’sschoolfromhishouse?
(ii) would every circle in the map depict a round about?
(iii) whoseschoolisnearertothehouse,Raghav’sorhissister’s?
Itisverydifficulttoanswertheabovequestionsonthebasisof
thegivenmap.Canyoutellwhy?
The reason is that we do not know if the distances have been
drawn properly or whether the circles drawn are roundabouts or
representsomethingelse.
Now look at another map drawn by his sister,
ten year old Meena, to show the route from her
house to her school (Fig 10.3).
Thismapisdifferentfromtheearliermaps.Here,
Meena has used different symbols for different
landmarks. Secondly, longer line segments have
been drawn for longer distances and shorter line
segments have been drawn for shorter distances,
i.e., she has drawn the map to a scale.
Now,youcananswerthefollowingquestions:
• How far is Raghav’s school from his
residence?
• Whose school is nearer to the house, Raghav’s or Meena’s?
• Which are the important landmarks on the route?
Thuswerealisethat,useofcertainsymbolsandmentioningofdistanceshashelpedus
read the map easily.Observe that the distances shown on the map are proportional to the
actualdistancesontheground.Thisisdonebyconsideringaproperscale.Whiledrawing
(or reading) a map, one must know, to what scale it has to be drawn (or has been drawn),
i.e.,howmuchofactualdistanceisdenotedby1mmor1cminthemap.Thismeans,thatif
onedrawsamap,he/shehastodecidethat1cmofspaceinthatmapshowsacertainfixed
distanceofsay1kmor10km.Thisscalecanvaryfrommaptomapbutnotwithinamap.
Forinstance,lookatthemapofIndiaalongsidethemapofDelhi.
My school
Fig 10.2
Myhouse
My sister’s school
Fig 10.3
2021–22

162 MATHEMATICS
You will find that when the maps are drawn of same size, scales and the distances in
the two maps will vary.That is 1 cm of space in the map of Delhi will represent smaller
distances as compared to the distances in the map of India.
Thelargertheplaceandsmallerthesizeofthemapdrawn,thegreateristhedistance
represented by 1 cm.
Thus,wecansummarisethat:
1. Amapdepictsthelocationofaparticularobject/placeinrelationtootherobjects/places.
2. Symbols are used to depict the different objects/places.
3. There is no reference or perspective in map, i.e., objects that are closer to the
observer are shown to be of the same size as those that are farther away. For
example, lookatthefollowingillustration(Fig10.4).
Fig 10.4
4. Maps use a scale which is fixed for a particular map. It reduces the real distances
proportionately to distances on the paper.
DO THIS
Fig 10.5
1. Look at the following map of a city (Fig 10.5).
(a) Colour the map as follows: Blue-water, Red-fire station, Orange-Library,
Yellow-schools, Green-Parks, Pink-Community Centre, Purple-Hospital,
Brown-Cemetry.
2021–22

(b) Mark a Green ‘X’at the intersection of 2nd street and Danim street.ABlack
‘Y’wheretherivermeetsthethirdstreet.Ared‘Z’attheintersectionofmain
street and 1st street.
(c) In magenta colour, draw a short street route from the college to the lake.
2. Draw a map of the route from your house to your school showing important
landmarks.
EXERCISE 10.2
1. Look at the given map of a city.
Answerthefollowing.
(a) Colourthemapasfollows:Blue-water,red-firestation,orange-library,yellow
- schools, Green - park, Pink - College, Purple - Hospital, Brown - Cemetery.
(b) Mark a green ‘X’ at the intersection of Road ‘C’ and Nehru Road, Green ‘Y’
at the intersection of Gandhi Road and RoadA.
(c) In red, draw a short street route from Library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?
2. Drawamapofyourclassroomusingproperscaleandsymbolsfordifferentobjects.
3. Draw a map of your school compound using proper scale and symbols for various
featureslikeplaygroundmainbuilding,gardenetc.
4. Drawamapgivinginstructionstoyourfriendsothatshereachesyourhousewithout
anydifficulty.
10.4 Faces, Edges and Vertices
Lookatthefollowingsolids!
Riddle
I have no vertices.
I have no flat
faces. Who am I?
2021–22

164 MATHEMATICS
Each of these solids is made up of polygonal regions which are called its faces;
these faces meet at edges which are line segments; and the edges meet at vertices which
are points. Such solids are called polyhedrons.
These are polyhedrons These are not polyhedrons
How are the polyhedrons different from the non-polyhedrons? Study the figures
carefully.Youknowthreeothertypesofcommonsolids.
Convex polyhedrons: You will recall the concept of convex polygons. The idea of
convexpolyhedronissimilar.
These are convex polyhedrons These are not convex polyhedrons
Regular polyhedrons: Apolyhedron is said to be regular if its faces are made up of
regular polygons and the same number of faces meet at each vertex.
Sphere Cylinder
Cone
2021–22

DO THIS
This polyhedron is regular. This polyhedon is not regular.All the sides
Its faces are congruent, regular are congruent; but the vertices are not
polygons. Vertices are formed by the formed by the same number of faces.
same number of faces 3 faces meet at A but
4 faces meet at B.
Twoimportantmembersofpolyhedronfamilyaroundareprismsandpyramids.
These are prisms These are pyramids
We say that a prism is a polyhedron whose base and top are congruent polygons
and whose other faces, i.e., lateral faces are parallelograms in shape.
On the other hand, a pyramid is a polyhedron whose base is a polygon (of any
numberofsides)andwhoselateralfacesaretriangleswithacommonvertex.(Ifyoujoin
all the corners of a polygon to a point not in its plane, you get a model for pyramid).
A prism or a pyramid is named after its base. Thus a hexagonal prism has a hexagon
asitsbase;andatriangularpyramidhasatriangleasitsbase.What,then,isarectangular
prism?Whatisasquarepyramid?Clearlytheirbasesarerectangleandsquarerespectively.
Tabulatethenumberoffaces,edgesandverticesforthefollowingpolyhedrons:
(Here‘V’standsfornumberofvertices,‘F’standsfornumberoffacesand‘E’stands
for number of edges).
Solid F V E F+V E+2
Cuboid
Triangularpyramid
Triangularprism
Pyramidwithsquarebase
Prism with square base
2021–22

166 MATHEMATICS
What do you infer from the last two columns? In each case, do you find
F + V = E + 2, i.e., F + V – E = 2? This relationship is called Euler’s formula.
Infactthisformulaistrueforanypolyhedron.
What happens to F,Vand E if some parts are sliced off from a solid? (To start with,
you may take a plasticine cube, cut a corner off and investigate).
EXERCISE 10.3
1. Can a polyhedron have for its faces
(i) 3triangles? (ii) 4triangles?
(iii) a square and four triangles?
2. Isitpossibletohaveapolyhedronwithanygivennumberoffaces?(Hint:Thinkof
apyramid).
3. Whichareprismsamongthefollowing?
(i) (ii)
(iii) (iv)
A table weight A box
4. (i) Howareprismsandcylindersalike?
(ii) How are pyramids and cones alike?
5. Is a square prism same as a cube? Explain.
6. VerifyEuler’sformulaforthesesolids.
(i) (ii)
A nail Unsharpened pencil
2021–22

7. UsingEuler’sformulafindtheunknown.
Faces ? 5 20
Vertices 6 ? 12
Edges 12 9 ?
8. Can a polyhedron have 10 faces, 20 edges and 15 vertices?
1. Recognising 2D and 3D objects.
2. Recognisingdifferentshapesinnestedobjects.
3. 3Dobjectshavedifferentviewsfromdifferentpositions.
4. Amapisdifferentfromapicture.
5. A map depicts the location of a particular object/place in relation to other objects/places.
6. Symbols are used to depict the different objects/places.
7. There is no reference or perspective in a map.
8. Mapsinvolveascalewhichisfixedforaparticularmap.
9. For any polyhedron,
F + V – E = 2
where‘F’standsfornumberoffaces,VstandsfornumberofverticesandEstandsfornumberof
edges.ThisrelationshipiscalledEuler’sformula.
2021–22

168 MATHEMATICS
NOTES
2021–22

MENSURATION 169
11.1 Introduction
We have learnt that for a closed plane figure, the perimeter is the distance around its
boundary and its area is the region covered by it. We found the area and perimeter of
variousplanefiguressuchastriangles,rectangles,circlesetc.We have also learnt to find
the area of pathways or borders in rectangular shapes.
In this chapter, we will try to solve problems related to perimeter and area of other
planeclosedfigureslikequadrilaterals.
We will also learn about surface area and volume of solids such as cube, cuboid and
cylinder.
11.2 Let us Recall
Let us take an example to review our previous knowledge.
This is a figure of a rectangular park (Fig 11.1) whose length is 30 m and width is 20 m.
(i) Whatisthetotallengthofthefencesurroundingit?Tofindthelengthofthefencewe
need to find the perimeter of this park, which is 100 m.
(Check it)
(ii) How much land is occupied by the park? To find the
land occupied by this park we need to find the area of
this park which is 600 square meters (m2
) (How?).
(iii) There is a path of one metre width running inside along
the perimeter of the park that has to be cemented.
If 1 bag of cement is required to cement 4 m2
area, how
manybagsofcementwouldberequiredtoconstructthe
cementedpath?
We can say that the number of cement bags used =
area of the path
area cemented by 1 bag
.
Area of cemented path =Area of park –Area of park not cemented.
Path is 1 m wide, so the rectangular area not cemented is (30 – 2) × (20 – 2) m2
.
That is 28 × 18 m2
.
Hence number of cement bags used = ------------------
(iv) There are two rectangular flower beds of size 1.5 m × 2 m each in the park as
shown in the diagram (Fig11.1) and the rest has grass on it. Find the area covered
by grass.
Mensuration
CHAPTER
11
Fig 11.1
2021–22

170 MATHEMATICS
TRY THESE
Area of rectangular beds = ------------------
Area of park left after cementing the path = ------------------
Area covered by the grass = ------------------
We can find areas of geometrical shapes other than rectangles also if certain
measurementsaregiven tous.Trytorecallandmatchthefollowing:
Diagram Shape Area
rectangle a × a
square b × h
triangle πb2
parallelogram
1
2
b h
×
circle a × b
Can you write an expression for the perimeter of each of the above shapes?
49 cm2
77 cm2
98 cm2
(a) Matchthefollowingfigureswiththeirrespectiveareasinthebox.
(b) Write the perimeter of each shape.
2021–22

MENSURATION 171
EXERCISE 11.1
1. A square and a rectangular field with
measurementsasgiveninthefigurehavethesame
perimeter.Which field has a larger area?
2. Mrs. Kaushik has a square plot with the
measurementasshowninthefigure.Shewantsto
construct a house in the middle of the plot.Agarden is developed
aroundthehouse.Findthetotalcostofdevelopingagardenaround
the house at the rate of ` 55 per m2
.
3. Theshapeofagardenisrectangularinthemiddleandsemicircular
attheendsasshowninthediagram.Findtheareaandtheperimeter
of this garden [Length of rectangle is
20 – (3.5 + 3.5) metres].
4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the
correspondingheightis10cm.Howmanysuchtilesarerequiredtocoverafloorof
area1080m2
?(Ifrequiredyoucansplitthetilesinwhateverwayyouwanttofillup
the corners).
5. Anantismovingaroundafewfoodpiecesofdifferentshapesscatteredonthefloor.
For which food-piece would the ant have to take a longer round? Remember,
circumference of a circle can be obtained by using the expression c = 2πr, where r
is the radius of the circle.
(a) (b) (c)
(b)
(a)
11.3 Area of Trapezium
Nazma owns a plot near a main road
(Fig11.2).Unlikesomeotherrectangular
plots in her neighbourhood, the plot has
only one pair of parallel opposite sides.
So,itisnearlyatrapeziuminshape.Can
you find out its area?
Let us name the vertices of this plot as
showninFig11.3.
By drawing EC ||AB, we can divide it
into two parts, one of rectangular shape
andtheotheroftriangularshape,(which
isrightangledatC),asshowninFig11.3. (b = c + a = 30 m)
Fig 11.3
Fig 11.2
2021–22

172 MATHEMATICS
DO THIS
TRY THESE
Area of ∆ ECD =
1
2
h × c =
1
12 10
2
× × = 60 m2
.
Area of rectangleABCE = h × a = 12 × 20 = 240 m2
.
AreaoftrapeziumABDE=areaof∆ECD+AreaofrectangleABCE=60+240=300m2
.
We can write the area by combining the two areas and write the area of trapezium as
area ofABDE =
1
2
h × c + h × a = h
c
a
2
+






= h
c a
h
c a a
+





 =
+ +






2
2 2
=
( )
2
b a
h
+
=
(sum of parallelsides)
height
2
By substituting the values of h,b and a in this expression, we find
( )
2
b a
h
+
= 300 m2
.
1. Nazma’ssisteralsohasatrapeziumshapedplot.Divideitintothreepartsasshown
(Fig 11.4). Show that the area of trapezium WXYZ
( )
2
a b
h
+
= .
Fig 11.4
2. If h = 10 cm, c = 6 cm, b = 12 cm,
d = 4 cm, find the values of each of
its parts separetely and add to find
theareaWXYZ.Verifyitbyputting
the values of h, a and b in the
expression
( )
2
h a b
+
.
Fig 11.5
1. DrawanytrapeziumWXYZonapiece
of graph paper as shown in the figure
and cut it out (Fig 11.5).
2. Find the mid point of XY by folding
the side and name itA(Fig 11.6).
Fig 11.6
2021–22

MENSURATION 173
DO THIS
TRY THESE
3. CuttrapeziumWXYZintotwopiecesbycuttingalongZA.Place∆ZYAasshown
in Fig11.7, whereAYis placed onAX.
What is the length of the base of the larger
triangle? Write an expression for the area of
thistriangle(Fig11.7).
4. The area of this triangle and the area of the trapezium WXYZ are same(How?).
Gettheexpressionfortheareaoftrapeziumbyusingtheexpressionforthearea
oftriangle.
Sotofindtheareaofatrapeziumweneedtoknowthelengthoftheparallelsidesandthe
perpendicular distance between these two parallel sides. Half the product of the sum of
thelengthsofparallelsidesandtheperpendiculardistancebetweenthemgivestheareaof
trapezium.
Findtheareaofthefollowingtrapeziums(Fig11.8).
(i) (ii)
Fig 11.7
Fig 11.8
InClassVIIwelearnttodrawparallelogramsofequalareaswithdifferentperimeters.
Canitbedonefortrapezium?Checkifthefollowingtrapeziumsareofequalareasbut
havedifferentperimeters(Fig11.9).
Fig 11.9
2021–22

174 MATHEMATICS
TRY THESE
Weknowthatallcongruentfiguresareequalinarea.Canwesayfiguresequalinarea
need to be congruent too?Are these figures congruent?
Draw at least three trapeziums which have different areas but equal perimeters on a
squared sheet.
11.4 Area of a General Quadrilateral
Ageneralquadrilateralcanbesplitintotwotrianglesbydrawingoneofitsdiagonals.This
“triangulation”helpsustofindaformulaforanygeneralquadrilateral.StudytheFig11.10.
AreaofquadrilateralABCD
= (area of ∆ ABC) + (area of ∆ ADC)
= (
1
2
AC × h1
) + (
1
2
AC × h2
)
=
1
2
AC × ( h1
+ h2
)
=
1
2
d ( h1
+ h2
) where d denotes the length of diagonalAC.
Example 1: Find the area of quadrilateral PQRS shown in Fig 11.11.
Solution: In this case, d = 5.5 cm, h1
= 2.5cm, h2
= 1.5 cm,
Area =
1
2
d ( h1
+ h2
)
=
1
2
× 5.5 × (2.5 + 1.5) cm2
=
1
2
× 5.5 × 4 cm2
= 11 cm2
Weknowthatparallelogramisalsoaquadrilateral.Letus
alsosplitsuchaquadrilateralintotwotriangles,findtheir
areasandhencethatoftheparallelogram.Doesthisagree
withtheformulathatyouknowalready?(Fig11.12)
11.4.1 Area of special quadrilaterals
Wecanusethesamemethodofsplittingintotriangles(whichwecalled“triangulation”)to
findaformulafortheareaofarhombus.In Fig11.13 ABCDisarhombus.Therefore,its
diagonals are perpendicular bisectors of each other.
Area of rhombusABCD = (area of ∆ ACD) + (area of ∆ ABC)
Fig 11.11
Fig 11.10
Fig 11.12
2021–22

MENSURATION 175
TRY THESE
Fig 11.14
= (
1
2
× AC × OD) + (
1
2
× AC × OB) =
1
2
AC × (OD + OB)
=
1
2
AC × BD =
1
2
d1
× d2
where AC = d1
and BD = d2
In other words, area of a rhombus is half the product of its diagonals.
Example 2: Findtheareaofarhombuswhosediagonalsareoflengths10cmand8.2cm.
Solution: Areaoftherhombus=
1
2
d1
d2
where d1
,d2
are lengths of diagonals.
=
1
2
× 10 × 8.2 cm2
= 41 cm2
.
Aparallelogramisdividedintotwocongruenttrianglesbydrawingadiagonalacross
it.Canwedivideatrapeziumintotwocongruenttriangles?
Find the area
ofthese
quadrilaterals
(Fig 11.14).
11.5 Area of a Polygon
Wesplitaquadrilateralintotrianglesandfinditsarea.Similarmethodscanbeusedtofind
the area of a polygon. Observe the following for a pentagon: (Fig 11.15, 11.16)
By constructing one diagonalAD and two perpendiculars BF
and CG on it, pentagonABCDE is divided into four parts. So,
area of ABCDE = area of right angled ∆ AFB + area of
trapezium BFGC + area of right angled ∆ CGD + area of
∆ AED. (Identify the parallel sides of trapezium BFGC.)
By constructing two diagonalsAC andAD the
pentagon ABCDE is divided into three parts.
So, area ABCDE = area of ∆ ABC + area of
∆ ACD + area of ∆ AED.
Fig 11.16
Fig 11.15
(ii)
(iii)
Fig 11.13
(i)
2021–22

176 MATHEMATICS
TRY THESE
Fig 11.17
Fig 11.18
(i) Divide the following polygons (Fig 11.17) into parts (triangles and trapezium) to
find out its area.
FI is a diagonal of polygon EFGHI NQ is a diagonal of polygon MNOPQR
(ii) PolygonABCDEisdividedintopartsasshownbelow(Fig11.18).Finditsareaif
AD = 8 cm,AH = 6 cm,AG = 4 cm,AF = 3 cm and perpendiculars BF = 2 cm,
CH = 3 cm, EG = 2.5 cm.
Area of PolygonABCDE = area of ∆AFB + ....
Area of ∆ AFB =
1
2
× AF × BF =
1
2
× 3 × 2 = ....
Area of trapezium FBCH = FH ×
(BF CH)
2
+
= 3 ×
(2 3)
2
+
[FH = AH – AF]
Area of ∆CHD =
1
2
× HD× CH = ....; Area of ∆ADE =
1
2
× AD × GE = ....
So, the area of polygonABCDE = ....
(iii) Find the area of polygon MNOPQR (Fig 11.19) if
MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm,
MA = 2 cm
NA,OC,QDandRBareperpendiculars to
diagonalMP.
Example 1: The area of a trapezium shaped field is 480 m2
, the distance between two
parallel sides is 15 m and one of the parallel side is 20 m. Find the other parallel side.
Solution: One of the parallel sides of the trapezium is a = 20 m, let another parallel
side be b, height h = 15 m.
Thegivenareaoftrapezium= 480 m2
.
Area of a trapezium =
1
2
h (a + b)
So 480 =
1
2
× 15 × (20 + b) or
480 2
15
×
= 20 + b
or 64 = 20 + b or b = 44 m
Hence the other parallel side of the trapezium is 44 m.
Fig 11.19
2021–22

MENSURATION 177
Example 2: The area of a rhombus is 240 cm2
and one of the diagonals is 16 cm. Find
theotherdiagonal.
Solution: Let length of one diagonal d1
= 16 cm
and lengthoftheotherdiagonal= d2
Area of the rhombus =
1
2
d1
. d2
= 240
So, 2
1
16
2
d
⋅ = 240
Therefore, d2
= 30 cm
Hence the length of the second diagonal is 30 cm.
Example 3: There is a hexagon MNOPQR of side 5 cm (Fig 11.20). Aman and
Ridhimadivideditintwodifferentways(Fig11.21).
Find the area of this hexagon using both ways.
Fig 11.20
Fig 11.21
Solution: Aman’s method:
SinceitisahexagonsoNQdividesthehexagonintotwocongruenttrapeziums.Youcan
verifyitbypaperfolding(Fig11.22).
Now area of trapezium MNQR =
(11 5)
4
2
+
× = 2 × 16 = 32 cm2
.
So the area of hexagon MNOPQR = 2 × 32 = 64 cm2
.
Ridhima’smethod:
∆ MNO and ∆ RPQ are congruent triangles with altitude
3 cm (Fig 11.23).
You can verify this by cutting off these two triangles and
placingthemononeanother.
Area of ∆ MNO =
1
2
× 8 × 3 = 12 cm2
= Area of ∆ RPQ
Area of rectangle MOPR = 8 × 5 = 40 cm2
.
Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm2
.
EXERCISE 11.2
1. The shape of the top surface of a table is a trapezium. Find its area
if its parallel sides are 1 m and 1.2 m and perpendicular distance
between them is 0.8 m.
Fig 11.23
Fig 11.22
Aman’s method
Ridhima’s method
2021–22

178 MATHEMATICS
2. The area of a trapezium is 34 cm2
and the length of one of the parallel sides is
10 cm and its height is 4 cm. Find the length of the other parallel side.
3. Length of the fence of a trapezium shaped fieldABCD is 120 m. If
BC=48m,CD=17mandAD = 40 m, find the area of this field. Side
AB is perpendicular to the parallel sidesAD and BC.
4. The diagonal of a quadrilateral shaped field is 24 m
and the perpendiculars dropped on it from the
remaining opposite vertices are 8 m and 13 m. Find
the area of the field.
5. Thediagonalsofarhombusare7.5cmand12cm.Find
its area.
6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm.
Ifoneofitsdiagonalsis8cmlong,findthelengthoftheotherdiagonal.
7. Thefloorofabuildingconsistsof3000tileswhicharerhombusshapedandeachof
itsdiagonalsare45cmand30cminlength.Findthetotalcostofpolishingthefloor,
if the cost per m2
is ` 4.
8. Mohan wants to buy a trapezium shaped field.
Its side along the river is parallel to and twice
thesidealongtheroad.Iftheareaofthisfieldis
10500 m2
and the perpendicular distance
betweenthetwoparallelsidesis100m,findthe
lengthofthesidealongtheriver.
9. Topsurfaceofaraisedplatformisintheshapeofaregularoctagonasshownin
the figure. Find the area of the octagonal surface.
10. There is a pentagonal shaped park as shown in the figure.
ForfindingitsareaJyotiandKavitadivideditintwodifferentways.
Findtheareaofthisparkusingbothways.Canyousuggestsomeotherway
offindingitsarea?
11. Diagramoftheadjacentpictureframehasouterdimensions=24cm×28cm
and inner dimensions 16 cm × 20 cm. Find the area of each section of
the frame, if the width of each section is same.
11.6 Solid Shapes
In your earlier classes you have studied that two dimensional figures can be identified as
thefacesofthreedimensionalshapes.Observethesolidswhichwehavediscussedsofar
(Fig 11.24).
2021–22

MENSURATION 179
DO THIS
Observe that some shapes have two or more than two identical (congruent) faces.
Namethem.Whichsolidhasallcongruentfaces?
Soaps, toys, pastes, snacks etc. often come in the packing of cuboidal, cubical or
cylindrical boxes. Collect, such boxes (Fig 11.25).
Fig 11.25
Fig 11.24
All six faces are rectangular,
and opposites faces are
identical. So there are three
pairs of identical faces.
Cuboidal Box Cubical Box
All six faces
are squares
and identical.
One curved surface
and two circular
faces which are
identical.
Cylindrical Box
Now take one type of box at a time. Cut out all the faces it has. Observe the shape of
each face and find the number of faces of the box that are identical by placing them on
each other.Write down your observations.
2021–22

180 MATHEMATICS
Fig 11.26
(This is a right
circular cylinder)
Fig 11.27
(This is not a right
circular cylinder)
Didyounoticethefollowing:
The cylinder has congruent circular faces that are parallel
toeachother(Fig11.26).Observethatthelinesegmentjoining
the center of circular faces is perpendicular to the base. Such
cylinders are known as right circularcylinders.We are only
going to study this type of cylinders, though there are other
types of cylinders as well (Fig 11.27).
Why is it incorrect to call the solid shown here a cylinder?
11.7 Surface Area of Cube, Cuboid and Cylinder
Imran,MonicaandJaspalarepaintingacuboidal,cubicalandacylindricalboxrespectively
of same height (Fig 11.28).
Fig 11.28
They try to determine who has painted more area. Hari suggested that finding the
surface area of each box would help them find it out.
To find the total surface area, find the area of each face and then add.The surface
area of a solid is the sum of the areas of its faces.To clarify further, we take each shape
one by one.
11.7.1 Cuboid
Suppose you cut open a cuboidal box
and lay it flat (Fig 11.29). We can see
a net as shown below (Fig 11.30).
Write the dimension of each side.
You know that a cuboid has three
pairs of identical faces. What
expression can you use to find the
area of each face?
Find the total area of all the faces
of the box. We see that the total surface area of a cuboid is area I + area II + area III +
area IV +area V + area VI
= h × l + b × l + b × h + l × h + b × h + l × b
Fig 11.29 Fig 11.30
2021–22

MENSURATION 181
DO THIS
TRY THESE
So total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl)
where h, l and b are the height, length and width of the cuboid respectively.
Suppose the height, length and width of the box shown above are 20 cm, 15 cm and
10cmrespectively.
Then the total surface area = 2 (20 × 15 + 20 × 10 + 10 × 15)
= 2 ( 300 + 200 + 150) = 1300 m2
.
Findthetotalsurfaceareaofthefollowing
cuboids (Fig 11.31):
Fig 11.32
(ii)
Fig 11.31
•
•
•
•
• The side walls (the faces excluding the top and
bottom) make the lateral surface area of the
cuboid. For example, the total area of all the four
wallsofthecuboidalroominwhichyouaresitting
is the lateral surface area of this room (Fig 11.32).
Hence,thelateralsurfaceareaofacuboidisgiven
by 2(h × l + b × h) or 2h (l + b).
(i) Cover the lateral surface of a cuboidal duster (which your teacher uses in the
class room) using a strip of brown sheet of paper, such that it just fits around the
surface. Remove the paper. Measure the area of the paper. Is it the lateral surface
area of the duster?
(ii) Measurelength,widthandheightofyourclassroomandfind
(a) the total surface area of the room, ignoring the area of windows and doors.
(b) the lateral surface area of this room.
(c) the total area of the room which is to be white washed.
1. Can we say that the total surface area of cuboid =
lateral surface area + 2 × area of base?
2. If we interchange the lengths of the base and the height
of a cuboid (Fig 11.33(i)) to get another cuboid
(Fig11.33(ii)),willitslateralsurfaceareachange? (i)
Fig 11.33
2021–22

182 MATHEMATICS
TRY THESE
DO THIS
(i) (ii) (iii)
Fig 11.34
Fig 11.35
(i) (ii)
11.7.2 Cube
Draw the pattern shown on a squared paper and cut it out [Fig 11.34(i)]. (You
know that this pattern is a net of a cube. Fold it along the lines [Fig 11.34(ii)]and
tape the edges to form a cube [Fig 11.34(iii)].
(a) Whatisthelength,widthandheightofthecube?Observethatallthefacesofa
cube are square in shape. This makes length, height and width of a cube equal
(Fig11.35(i)).
(b) Write the area of each of the faces.Are they equal?
(c) Write the total surface area of this cube.
(d) If each side of the cube is l, what will be the area of each face? (Fig 11.35(ii)).
Can we say that the total surface area of a cube of side l is 6l2
?
Find the surface area of cubeAand lateral surface area of cube B (Fig 11.36).
Fig 11.36
2021–22

MENSURATION 183
Fig 11.38
(i) Two cubes each with side b are joined to form a cuboid (Fig 11.37). What is the
surface area of this cuboid? Is it 12b2
? Is the surface area of cuboid formed by
joining three such cubes, 18b2
? Why?
DO THIS
(ii) (iii) (iv)
(i)
Fig 11.39
(i) Take a cylindrical can or box and trace the base of the can on graph paper and cut
it [Fig 11.39(i)]. Take another graph paper in such a way that its width is equal to
the height of the can. Wrap the strip around the can such that it just fits around the
can (remove the excess paper) [Fig 11.39(ii)].
Tape the pieces [Fig 11.39(iii)]togethertoformacylinder[Fig11.39(iv)].What is
the shape of the paper that goes around the can?
(ii) Howwillyouarrange12cubesofequallengthtoforma
cuboid of smallest surface area?
(iii) Afterthesurfaceareaofacubeispainted,thecubeiscut
into 64 smaller cubes of same dimensions (Fig 11.38).
Howmanyhavenofacepainted?1facepainted?2faces
painted? 3 faces painted?
11.7.3 Cylinders
Most of the cylinders we observe are right circular cylinders. For example, a tin, round
pillars, tube lights, water pipes etc.
Fig 11.37
2021–22

184 MATHEMATICS
TRY THESE
Ofcourseitisrectangularinshape.Whenyoutapethepartsofthiscylindertogether,
thelengthoftherectangularstripisequaltothecircumferenceofthecircle.Record
the radius (r) of the circular base, length (l)andwidth(h) of the rectangular strip.
Is 2πr = length of the strip. Check if the area of rectangular strip is 2πrh. Count
how many square units of the squared paper are used to form the cylinder.
Check if this count is approximately equal to 2πr (r + h).
(ii) We can deduce the relation 2πr(r +h) as the surface area of a cylinder in another
way. Imagine cutting up a cylinder as shown below (Fig 11.40).
Fig 11.40
The lateral (or curved) surface area of a cylinder is 2πrh.
The total surface area of a cylinder = πr2
+ 2πrh + πr2
= 2πr2
+ 2πrh or 2πr (r + h)
Findtotalsurfaceareaofthefollowingcylinders(Fig11.41)
Fig 11.41
Note that lateral surface area of a cylinder is the circumference of base × height of
cylinder.Canwewritelateralsurfaceareaofacuboidasperimeterof base×height
ofcuboid?
Example 4: An aquarium is in the form of a cuboid whose external measures are
80 cm × 30 cm × 40 cm. The base, side faces and back face are to be covered with a
coloured paper. Find the area of the paper needed?
Solution: Thelengthoftheaquarium= l = 80 cm
Widthoftheaquarium= b = 30 cm
Note: We take π to be
22
7
unless otherwise stated.
2021–22

MENSURATION 185
Heightoftheaquarium= h = 40 cm
Area of the base = l × b = 80 × 30 = 2400 cm2
Area of the side face = b × h = 30 × 40 = 1200 cm2
Area of the back face = l × h = 80 × 40 = 3200 cm2
Required area = Area of the base + area of the back face
+ (2 × area of a side face)
= 2400 + 3200 + (2 × 1200) = 8000 cm2
Hence the area of the coloured paper required is 8000 cm2
.
Example 5: The internal measures of a cuboidal room are 12 m × 8 m × 4 m. Find the
total cost of whitewashing all four walls of a room, if the cost of white washing is` 5 per
m2
.Whatwillbethecostofwhitewashingiftheceilingoftheroomisalsowhitewashed.
Solution: Let the length of the room = l = 12 m
Width of the room = b = 8 m
Height of the room = h = 4 m
Area of the four walls of the room = Perimeter of the base × Height of the room
= 2 (l + b) × h = 2 (12 + 8) × 4
= 2 × 20 × 4 = 160 m2
.
Cost of white washing per m2
= ` 5
Hence the total cost of white washing four walls of the room = ` (160 × 5) = ` 800
Area of ceiling is 12 × 8 = 96 m2
Costofwhitewashingtheceiling= ` (96 × 5) = ` 480
So the total cost of white washing = ` (800 + 480) = ` 1280
Example 6: Inabuildingthereare24cylindricalpillars.Theradiusofeachpillar
is 28 cm and height is 4 m. Find the total cost of painting the curved surface area of
all pillars at the rate of ` 8 per m2
.
Solution: Radiusofcylindricalpillar,r= 28 cm = 0.28 m
height, h= 4 m
curved surface area of a cylinder = 2πrh
curved surface area of a pillar =
22
2 0.28 4
7
× × × = 7.04 m2
curved surface area of 24 such pillar = 7.04 × 24 = 168.96 m2
cost of painting an area of 1 m2
= ` 8
Therefore, cost of painting 1689.6 m2
= 168.96 × 8 = ` 1351.68
Example 7: Findtheheightofacylinderwhoseradiusis7cmandthe
total surface area is 968 cm2
.
Solution: Let height of the cylinder = h, radius = r= 7cm
Total surface area = 2πr (h + r)
2021–22

186 MATHEMATICS
i.e., 2 ×
22
7
× 7 × (7 + h) = 968
h = 15 cm
Hence, the height of the cylinder is 15 cm.
EXERCISE 11.3
1. There are two cuboidal boxes as
shownintheadjoiningfigure.Which
box requires the lesser amount of
materialtomake?
2. A suitcase with measures 80 cm ×
48 cm × 24 cm is to be covered with
atarpaulincloth.Howmanymetresoftarpaulinofwidth96cmisrequiredtocover
100 such suitcases?
3. Find the side of a cube whose surface area is
600 cm2
.
4. Rukhsarpaintedtheoutsideofthecabinetof
measure 1 m × 2 m × 1.5 m. How much
surface area did she cover if she painted all except the bottom of the cabinet.
5. Daniel is painting the walls and ceiling of a
cuboidalhallwithlength,breadthandheight
of 15 m, 10 m and 7 m respectively. From
each can of paint 100 m2
of area is painted.
Howmanycansofpaintwillsheneedtopaint
the room?
6. Describehowthetwofiguresattherightarealikeandhowtheyaredifferent.Which
box has larger lateral surface area?
7. Aclosedcylindricaltankofradius7mandheight3mis
madefromasheetofmetal.Howmuchsheetofmetalis
required?
8. Thelateralsurfaceareaofahollowcylinderis4224cm2
.
It is cut along its height and formed a rectangular sheet
ofwidth33cm.Findtheperimeterofrectangularsheet?
9. A road roller takes 750 complete revolutions to move
onceovertolevelaroad.Findtheareaoftheroadifthe
diameter of a road roller is 84 cm and length is 1 m.
10. A company packages its milk powder in cylindrical
containerwhosebasehasadiameterof14cmandheight
20 cm. Company places a label around the surface of
thecontainer(asshowninthefigure).Ifthelabelisplaced
2 cm from top and bottom, what is the area of the label.
2021–22

MENSURATION 187
11.8 Volume of Cube, Cuboid and Cylinder
Amount of space occupied by a three dimensional object is called its volume. Try to
comparethevolumeofobjectssurroundingyou.Forexample,volumeofaroomisgreater
thanthevolumeofanalmirahkeptinsideit.Similarly,volumeofyourpencilboxisgreater
than the volume of the pen and the eraser kept inside it.
Can you measure volume of either of these objects?
Remember, we use square units to find the area of a
region.Herewewillusecubicunitstofindthevolumeofa
solid, as cube is the most convenient solid shape (just as
square is the most convenient shape to measure area of a
region).
For finding the area we divide the region into square
units, similarly, to find the volume of a solid we need to
divideitintocubicalunits.
Observe that the volume of each of the adjoining solids is
8 cubic units (Fig 11.42 ).
We can say that the volume of a solid is measured by
countingthenumberofunitcubesitcontains.Cubicunitswhichwegenerallyusetomeasure
volumeare
1 cubic cm = 1 cm × 1 cm × 1 cm = 1 cm3
= 10 mm × 10 mm × 10 mm = ............... mm3
1 cubic m = 1 m × 1 m × 1 m = 1 m3
= ............................... cm3
1 cubic mm = 1 mm × 1 mm × 1 mm = 1 mm3
= 0.1 cm × 0.1 cm × 0.1 cm = ...................... cm3
We now findsomeexpressionstofindvolumeofacuboid,cubeandcylinder. Let us
takeeach solidonebyone.
11.8.1 Cuboid
Take36cubesofequalsize(i.e.,lengthofeachcubeissame).Arrangethemtoformacuboid.
Youcanarrangetheminmanyways.Observethefollowingtableandfillintheblanks.
Fig 11.42
cuboid length breadth height l × b × h = V
(i) 12 3 1 12 × 3 × 1 = 36
(ii) ... ... ... ...
2021–22

188 MATHEMATICS
TRY THESE
DO THIS
What do you observe?
Sincewehaveused36cubestoformthesecuboids,volumeofeachcuboid
is36cubicunits.Alsovolumeofeachcuboidisequaltotheproductoflength,
breadth and height of the cuboid. From the above example we can say volume of cuboid
= l × b × h. Since l × b is the area of its base we can also say that,
Volume of cuboid = area of the base × height
Takeasheetofpaper.Measureits
area.Pileupsuchsheetsofpaper
of same size to make a cuboid
(Fig11.43).Measuretheheightof
this pile. Find the volume of the
cuboid by finding the product of
theareaofthesheetandtheheight
ofthispileofsheets.
Thisactivityillustratestheidea
that volume of a solid can be deduced by this method also (if the base and top of the
solid are congruent and parallel to each other and its edges are perpendicular to the
base).Canyouthinkofsuchobjectswhosevolumecanbefoundbyusingthismethod?
Findthevolumeofthefollowingcuboids(Fig11.44).
(i)
Fig 11.43
Fig 11.44
(iii) ... ... ... ...
(iv) ... ... ... ...
2021–22

MENSURATION 189
DO THIS
TRY THESE
11.8.2 Cube
The cube is a special case of a cuboid, where l = b = h.
Hence, volume of cube = l × l × l = l 3
Findthevolumeofthefollowingcubes
(a) with a side 4 cm (b) with a side 1.5 m
Arrange 64 cubes of equal size in as many ways as you can to form a cuboid.
Find the surface area of each arrangement. Can solid shapes of same volume have
same surface area?
A company sells biscuits. For packing purpose they are using cuboidal boxes:
box A→3 cm × 8 cm × 20 cm, box B → 4 cm × 12 cm × 10 cm. What size of the box
willbeeconomicalforthecompany?Why?Canyousuggestanyothersize(dimensions)
whichhasthesamevolumebutismoreeconomicalthanthese?
11.8.3 Cylinder
We know that volume of a cuboid can be found by finding the
productofareaofbaseanditsheight.Canwefindthevolumeof
acylinderinthesameway?
Just like cuboid, cylinder has got a top and a base which are
congruent and parallel to each other. Its lateral surface is also
perpendicular to the base, just like cuboid.
So theVolume of a cuboid = area of base × height
= l × b × h = lbh
Volumeofcylinder= area of base × height
= πr2
× h = πr2
h
TRY THESE
Findthevolumeofthefollowingcylinders.
(i) (ii)
2021–22

190 MATHEMATICS
11.9 Volume and Capacity
There is not much difference between these two words.
(a) Volume refers to the amount of space occupied by an object.
(b) Capacity refers to the quantity that a container holds.
Note: If a water tin holds 100 cm3
of water then the capacity of the water tin is 100 cm3
.
Capacity is also measured in terms of litres. The relation between litre and cm3
is,
1 mL = 1 cm3
,1 L = 1000 cm3
. Thus, 1 m3
= 1000000 cm3
= 1000 L.
Example 8: Find the height of a cuboid whose volume is 275 cm3
and base area
is 25 cm2
.
Solution: Volume of a cuboid = Base area × Height
Hence height of the cuboid =
Volume of cuboid
Base area
=
275
25
= 11 cm
Height of the cuboid is 11 cm.
Example 9: Agodown is in the form of a cuboid of measures 60 m × 40 m × 30 m.
How many cuboidal boxes can be stored in it if the volume of one box is 0.8 m3
?
Solution: Volume of one box = 0.8 m3
Volumeofgodown= 60 × 40 × 30 = 72000 m3
Number of boxes that can be stored in the godown =
Volume of the godown
Volume of one box
=
60× 40× 30
0.8
= 90,000
Hence the number of cuboidal boxes that can be stored in the godown is 90,000.
Example 10: Arectangularpaperof width14cmisrolledalongitswidthandacylinder
of radius 20 cm isformed.Findthevolumeofthecylinder(Fig11.45). (Take
22
7
forπ)
Solution: Acylinder is formed by rolling a rectangle about its width. Hence the width
of the paper becomes height and radius of the cylinder is 20 cm.
Fig 11.45
Heightofthecylinder= h = 14 cm
Radius= r = 20 cm
2021–22

MENSURATION 191
Volumeofthecylinder= V = π r2
h
=
22
20 20 14
7
× × × = 17600 cm3
Hence, the volume of the cylinder is 17600 cm3
.
Example 11: Arectangularpieceofpaper11cm×4cmisfoldedwithoutoverlapping
to make a cylinder of height 4 cm. Find the volume of the cylinder.
Solution: Length of the paper becomes the perimeter of the base of the cylinder and
widthbecomesheight.
Let radius of the cylinder = r and height = h
Perimeter of the base of the cylinder = 2πr = 11
or
22
2
7
r
× × = 11
Therefore, r =
7
4
cm
Volumeofthecylinder= V = πr2
h
=
22 7 7
4
7 4 4
× × × cm3
= 38.5 cm3
.
Hence the volume of the cylinder is 38.5 cm3
.
EXERCISE 11.4
1. Given a cylindrical tank, in which situation will you find surface area and in
whichsituationvolume.
(a) Tofindhowmuchitcanhold.
(b) Number of cement bags required to plaster it.
(c) Tofindthenumberofsmallertanksthatcanbefilledwithwaterfromit.
2. DiameterofcylinderAis7cm,andtheheightis14cm.Diameterof
cylinderBis14cmandheightis7cm.Withoutdoinganycalculations
can you suggest whose volume is greater? Verify it by finding the
volumeof boththecylinders.Checkwhetherthecylinderwithgreater
volume also has greater surface area?
3. Findtheheightofacuboidwhosebaseareais180cm2
andvolume
is 900 cm3
?
4. Acuboidisofdimensions60cm×54cm×30cm.Howmanysmallcubeswithside
6 cm can be placed in the given cuboid?
5. Find the height of the cylinder whose volume is 1.54 m3
and diameter of the base is
140 cm ?
6. A milk tank is in the form of cylinder whose radius is 1.5 m and
lengthis7m.Findthequantityofmilkinlitresthatcanbestored
inthetank?
7. If each edge of a cube is doubled,
(i) howmanytimeswillitssurfaceareaincrease?
(ii) howmanytimeswillitsvolumeincrease?
A
B
2021–22

192 MATHEMATICS
8. Water is pouring into a cubiodal reservoir at the rate of 60 litres per
minute. If the volume of reservoir is 108 m3
, find the number of hours it
willtaketofillthereservoir.
1. Area of
(i) atrapezium=halfofthesumofthelengthsofparallelsides×perpendiculardistancebetween
them.
(ii) a rhombus = half the product of its diagonals.
2. Surface area of a solid is the sum of the areas of its faces.
3. Surface area of
a cuboid = 2(lb + bh + hl)
a cube = 6l2
a cylinder = 2πr(r + h)
4. Amount of region occupied by a solid is called itsvolume.
5. Volumeof
a cuboid = l × b × h
a cube = l3
a cylinder = πr2
h
6. (i) 1 cm3
= 1 mL
(ii) 1L = 1000 cm3
(iii) 1 m3
= 1000000 cm3
= 1000L
2021–22

EXPONENTS AND POWERS 193
12.1 Introduction
Do you know?
Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have
already learnt in earlier class how to write such large numbers more
conveniently using exponents, as, 5.97 × 1024
kg.
We read 1024
as 10 raised to the power 24.
We know 25
= 2 × 2 × 2 × 2 × 2
and 2m
= 2 × 2 × 2 × 2 × ... × 2 × 2 ... (m times)
Let us now find what is 2– 2
is equal to?
12.2 Powers with Negative Exponents
You know that, 102
= 10 × 10 = 100
101
= 10 =
100
10
100
= 1 =
10
10
10– 1
= ?
Continuing the above pattern we get,10– 1
=
1
10
Similarly 10– 2
= 2
1 1 1 1 1
10
10 10 10 100 10
÷ = × = =
10– 3
= 3
1 1 1 1 1
10
100 100 10 1000 10
÷ = × = =
What is 10– 10
equal to?
Exponents and Powers
CHAPTER
12
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
2021–22

194 MATHEMATICS
TRY THESE
TRY THESE
Nowconsiderthefollowing.
33
= 3 × 3 × 3 = 27
32
= 3 × 3 = 9 =
27
3
31
= 3 =
9
3
3° = 1 =
3
3
So looking at the above pattern, we say
3– 1
= 1 ÷ 3 =
1
3
3– 2
=
1
3
3
÷ =
1
3 3
×
= 2
1
3
3– 3
= 2
1
3
3
÷ = 2
1
3
×
1
3
= 3
1
3
Youcannowfindthevalueof 2– 2
inasimilarmanner.
We have, 10– 2
= 2
1
10
or 102
= 2
1
10−
10– 3
= 3
1
10
or 103
= 3
1
10−
3– 2
= 2
1
3
or 32
= 2
1
3− etc.
In general, we can say that for any non-zero integer a, a– m
=
1
m
a
, where m is a
positiveinteger.a–m
isthemultiplicativeinverseofam
.
Findthemultiplicativeinverseofthefollowing.
(i) 2– 4
(ii) 10– 5
(iii) 7– 2
(iv) 5– 3
(v) 10– 100
We learnt how to write numbers like 1425 in expanded form using exponents as
1 × 103
+ 4 × 102
+ 2 × 101
+ 5 × 10°.
Let us see how to express 1425.36 in expanded form in a similar way.
We have 1425.36 = 1 × 1000 + 4 × 100 + 2 × 10 + 5 × 1 +
3 6
10 100
+
= 1 × 103
+ 4 × 102
+ 2 × 10 + 5 × 1 + 3 × 10– 1
+ 6 × 10– 2
The previous number is
divided by the base 3.
Expandthefollowingnumbersusingexponents.
(i) 1025.63 (ii) 1256.249
10– 1
=
1
10
, 10– 2
= 2
1 1
100
10
=
2021–22

TRY THESE
12.3 Laws of Exponents
We have learnt that for any non-zero integer a, am
×an
= am + n
, where m and n are natural
numbers.Does this law also hold if the exponents are negative? Let us explore.
(i) We know that 2 – 3
= 3
1
2
and 2 – 2
= 2
1
2
Therefore, 3 2
2 2
− −
× = 3 2 3 2 3 2
1 1 1 1
2 2 2 2 2 +
× = = =
×
2 – 5
(ii) Take (–3)– 4
× (–3)–3
(–3)– 4
×(–3)–3
= 4 3
1 1
( 3) ( 3)
×
− −
= 4 3 4 3
1 1
( 3) ( 3) ( 3) +
=
− × − −
= (–3)–7
(iii) Now consider 5–2
× 54
5–2
× 54
=
4
4 4 2
2 2
1 5
5 5
5 5
−
× = = = 5(2)
(iv) Now consider (–5)– 4
× (–5)2
(–5)– 4
× (–5)2
=
2
2
4 4 4 2
1 ( 5) 1
( 5)
( 5) ( 5) ( 5) ( 5)−
−
× − = =
− − − × −
= 4 2
1
( 5) −
−
= (–5)– (2)
In general, we can say that for any non-zero integer a,
am
× an
= am + n
, where m and n are integers.
Simplifyandwriteinexponentialform.
(i) (–2)–3
× (–2)– 4
(ii) p3
× p–10
(iii) 32
× 3–5
× 36
Onthesamelinesyoucanverifythefollowinglawsofexponents,wherea andbarenon
zero integers and m, n are any integers.
(i)
m
m n
n
a
a
a
−
= (ii) (am
)n
= amn
(iii) am
× bm
= (ab)m
(iv)
a
b
a
b
m
m
m
=





 (v) a0
= 1
Let us solve some examples using the above Laws of Exponents.
1
m
m
a
a
−
= for any non-zero integer a.
In Class VII, you have learnt that for any
non-zero integer a,
m
m n
n
a
a
a
−
= , where
m and n are natural numbers and m n.
These laws you have studied
in Class VII for positive
exponents only.
–5 is the sum of two exponents – 3 and – 2
(– 4) + (–3) = – 7
(–2) + 4 = 2
(– 4) + 2 = –2
2021–22

196 MATHEMATICS
Example 1: Find the value of
(i) 2–3
(ii) 2
1
3−
Solution:
(i)
3
3
1 1
2
8
2
−
= = (ii)
2
2
1
3 3 3 9
3−
= = × =
Example 2: Simplify
(i) (– 4)5
× (– 4)–10
(ii) 25
÷ 2– 6
Solution:
(i) (– 4)5
× (– 4)–10
= (– 4) (5 – 10)
= (– 4)–5
= 5
1
( 4)
−
(am
× an
= am + n
,
1
m
m
a
a
−
= )
(ii) 25
÷ 2– 6
= 25 – (– 6)
= 211
(am
÷ an
= am – n
)
Example 3: Express 4– 3
as a power with the base 2.
Solution: We have, 4 = 2 × 2 = 22
Therefore, (4)– 3
= (2 × 2)– 3
= (22
)– 3
= 22 × (– 3)
= 2– 6
[(am
)n
= amn
]
Example 4: Simplify and write the answer in the exponential form.
(i) (25
÷ 28
)5
× 2– 5
(ii) (– 4)– 3
× (5)– 3
× (–5)– 3
(iii)
3
1
(3)
8
−
× (iv) ( )
− ×






3
5
3
4
4
Solution:
(i) (25
÷ 28
)5
× 2– 5
= (25 – 8
)5
× 2– 5
= (2– 3
)5
× 2– 5
= 2– 15 – 5
= 2–20
= 20
1
2
(ii) (– 4)– 3
× (5)– 3
× (–5)–3
= [(– 4) × 5 × (–5)]– 3
= [100]– 3
= 3
1
100
[using the law am
× bm
= (ab)m
, a–m
=
1
m
a
]
(iii) 3 3 3 3 3 3
3 3
1 1 1
(3) (3) 2 3 (2 3) 6
8 2 6
− − − − − −
× = × = × = × = =
(iv) ( )
− ×






3
5
3
4
4
=
4
4
4
5
( 1 3)
3
− × × = (–1)4
× 34
×
4
4
5
3
= (–1)4
× 54
= 54
[(–1)4
= 1]
Example 5: Find m so that (–3)m + 1
× (–3)5
= (–3)7
Solution: (–3)m + 1
× (–3)5
= (–3)7
(–3)m + 1+ 5
= (–3)7
(–3)m + 6
= (–3)7
Onboththesidespowershavethesamebasedifferentfrom1and–1,sotheirexponents
must be equal.
2021–22

Therefore, m + 6 = 7
or m = 7 – 6 = 1
Example 6: Find the value of
2
3
2






−
.
Solution:
2
3
2
3
3
2
9
4
2 2
2
2
2





 = = =
− −
−
Example 7: Simplify (i)
1
3
1
2
1
4
2 3 2





 −
















÷






− − −
(ii)
–7 –5
5 8
8 5
   
×
   
   
Solution:
(i)
1
3
1
2
1
4
2 3 2





 −
















÷






− − −
=
1
3
1
2
1
4
2
2
3
3
2
2
−
−
−
−
−
−
−






÷
=
3
1
2
1
4
1
9 8 16
1
16
2
2
3
3
2
2
−






÷ = − ÷ =
{ }
(ii)
5
8
8
5
7 5





 ×






− −
=
7 5 7 5
( 7) – ( 5) ( 5) ( 7)
7 5 5 7
5 8 5 8
5 8
8 5 5 8
− − − −
− − − − −
− − − −
× = × = ×
=
2
2 2
2
8 64
5 8
25
5
−
× = =
EXERCISE 12.1
1. Evaluate.
(i) 3–2
(ii) (– 4)– 2
(iii) 1
2
5






−
2. Simplifyandexpresstheresultinpowernotationwithpositiveexponent.
(i) (– 4)5
÷ (– 4)8
(ii)
1
23
2






(iii) ( )
− ×






3
5
3
4
4
(iv) (3– 7
÷ 3– 10
) × 3– 5
(v) 2– 3
× (–7)– 3
3. Findthevalueof.
(i) (3° + 4– 1
) × 22
(ii) (2– 1
× 4– 1
) ÷ 2– 2
(iii)
1
2
1
3
1
4
2 2 2





 +





 +






− − −
2
3
2
3
3
2
3
2
2 2
2
2
2
2





 = = =






− −
−
In general,
a
b
b
a
m m





 =






−
an
= 1 only if n = 0. This will work for any a.
For a = 1, 11
= 12
= 13
= 1– 2
= ... = 1 or (1)n
=
1forinfinitelymanyn.
For a = –1,
(–1)0
= (–1)2
= (–1)4
= (–1)–2
= ... = 1 or
(–1)p
= 1 for any even integer p.
2021–22

198 MATHEMATICS
(iv) (3– 1
+ 4– 1
+ 5– 1
)0
(v)
−
















−
2
3
2 2
4. Evaluate(i)
1 3
4
8 5
2
−
−
×
(ii) (5–1
× 2–1
) × 6–1
5. Find the value of m for which 5m
÷ 5– 3
= 55
.
6. Evaluate(i)
1
3
1
4
1 1 1





 −
















− − −
(ii)
7. Simplify.
(i)
4
3 8
25
( 0)
5 10
t
t
t
−
− −
×
≠
× ×
(ii)
5 5
7 5
3 10 125
5 6
− −
− −
× ×
×
12.4 Use of Exponents to Express Small Numbers in
Standard Form
Observethefollowingfacts.
1. The distance from the Earth to the Sun is 149,600,000,000 m.
2. The speed of light is 300,000,000 m/sec.
3. Thickness of Class VII Mathematics book is 20 mm.
4. The average diameter of a Red Blood Cell is 0.000007 mm.
5. The thickness of human hair is in the range of 0.005 cm to 0.01 cm.
6. The distance of moon from the Earth is 384, 467, 000 m (approx).
7. The size of a plant cell is 0.00001275 m.
8. Average radius of the Sun is 695000 km.
9. Mass of propellant in a space shuttle solid rocket booster is 503600 kg.
10. Thickness of a piece of paper is 0.0016 cm.
11. Diameter of a wire on a computer chip is 0.000003 m.
12. The height of Mount Everest is 8848 m.
Observe that there are few numbers which we can read like 2 cm, 8848 m,
6,95,000 km. There are some large
numberslike150,000,000,000mand
some very small numbers like
0.000007 m.
Identifyverylargeandverysmall
numbers from the above facts and
writethemintheadjacenttable:
We have learnt how to express
very large numbers in standard form
inthepreviousclass.
For example: 150,000,000,000 = 1.5 × 1011
Now, let us try to express 0.000007 m in standard form.
Very large numbers Very small numbers
150,000,000,000 m 0.000007 m
--------------- ---------------
--------------- ---------------
--------------- ---------------
--------------- ---------------
2021–22

TRY THESE
0.000007 =
7
1000000
= 6
7
10
= 7 × 10– 6
0.000007 m = 7 × 10– 6
m
Similarly,considerthethicknessofapieceofpaper
which is 0.0016 cm.
0.0016 =
4
4
16 1.6 10
1.6 10 10
10000 10
−
×
= = × ×
= 1.6 × 10– 3
Therefore, we can say thickness of paper is 1.6 × 10– 3
cm.
1. Writethefollowingnumbersinstandardform.
(i) 0.000000564 (ii) 0.0000021 (iii) 21600000 (iv) 15240000
2. Writeallthefactsgiveninthestandardform.
12.4.1 Comparing very large and very small numbers
The diameter of the Sun is 1.4 × 109
m and the diameter of the Earth is 1.2756 × 107
m.
Suppose you want to compare the diameter of the Earth, with the diameter of the Sun.
Diameter of the Sun = 1.4 × 109
m
Diameter of the earth = 1.2756 × 107
m
Therefore
9
7
1.4 10
1.2756 10
×
×
=
9–7
1.4 10
1.2756
×
=
1.4 100
1.2756
×
whichisapproximately100
So, the diameter of the Sun is about 100 times the diameter of the earth.
LetuscomparethesizeofaRedBloodcellwhichis0.000007mtothatofaplantcellwhich
is0.00001275m.
Size of Red Blood cell = 0.000007 m = 7 × 10– 6
m
Sizeofplantcell= 0.00001275 = 1.275 × 10– 5
m
Therefore,
6
5
7 10
1.275 10
−
−
×
×
=
6 (–5) –1
7 10 7 10
1.275 1.275
− −
× ×
= =
0.7 0.7 1
1.275 1.3 2
= = (approx.)
So a red blood cell is half of plant cell in size.
Mass of earth is 5.97 × 1024
kg and mass of moon is 7.35 × 1022
kg. What is the
totalmass?
Total mass = 5.97 × 1024
kg + 7.35 × 1022
kg.
= 5.97 × 100 × 1022
+ 7.35 × 1022
= 597 × 1022
+ 7.35 × 1022
= (597 + 7.35) × 1022
= 604.35 × 1022
kg.
The distance between Sun and Earth is 1.496 × 1011
m and the distance between
Earth and Moon is 3.84 × 108
m.
During solar eclipse moon comes in between Earth and Sun.
At that time what is the distance between Moon and Sun.
When we have to add numbers in
standard form, we convert them into
numbers with the same exponents.
2021–22

200 MATHEMATICS
Distance between Sun and Earth = 1.496 × 1011
m
Distance between Earth and Moon = 3.84 × 108
m
Distance between Sun and Moon = 1.496 × 1011
– 3.84 × 108
= 1.496 × 1000 × 108
– 3.84 × 108
= (1496 – 3.84) × 108
m = 1492.16 × 108
m
Example 8: Express the following numbers in standard form.
(i) 0.000035 (ii) 4050000
Solution: (i) 0.000035 = 3.5 × 10– 5
(ii) 4050000 = 4.05 × 106
Example 9: Expressthefollowingnumbersinusualform.
(i) 3.52 × 105
(ii) 7.54 × 10– 4
(iii) 3 × 10– 5
Solution:
(i) 3.52 × 105
= 3.52 × 100000 = 352000
(ii) 7.54 × 10– 4
= 4
7.54 7.54
10000
10
= = 0.000754
(iii) 3 × 10– 5
= 5
3 3
100000
10
= = 0.00003
EXERCISE 12.2
1. Expressthefollowingnumbersinstandardform.
(i) 0.0000000000085 (ii) 0.00000000000942
(iii) 6020000000000000 (iv) 0.00000000837
(v) 31860000000
2. Expressthefollowingnumbersinusualform.
(i) 3.02 × 10– 6
(ii) 4.5 × 104
(iii) 3 × 10– 8
(iv) 1.0001 × 109
(v) 5.8 × 1012
(vi) 3.61492 × 106
3. Expressthenumberappearinginthefollowingstatementsinstandardform.
(i) 1 micron is equal to
1
1000000
m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm
4. In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of
thickness 0.016 mm. What is the total thickness of the stack.
Again we need to convert
numbers in standard form into
a numbers with the same
exponents.
1. Numberswithnegativeexponentsobeythefollowinglawsofexponents.
(a) am
× an
= am+n
(b) am
÷ an
= am–n
(c) (am
)n
= amn
(d) am
× bm
= (ab)m
(e) a0
= 1 (f)
m
m
m
a a
b b
 
= 
 
2. Verysmallnumberscanbeexpressedinstandardformusingnegativeexponents.
2021–22

DIRECT AND INVERSE PROPORTIONS 201
13.1 Introduction
Mohan prepares tea for himself and his sister. He uses 300 mL of
water, 2 spoons of sugar, 1 spoon of tea leaves and 50 mL of milk.
How much quantity of each item will he need, if he has to make tea
for five persons?
If two students take 20 minutes to arrange chairs for an assembly,
then how much time would five students take to do the same job?
We come across many such situations in our day-to-day life, where we
need to see variation in one quantity bringing in variation in the other
quantity.
Forexample:
(i) If the number of articles purchased increases, the total cost also increases.
(ii) More the money deposited in a bank, more is the interest earned.
(iii) As the speed of a vehicle increases, the time taken to cover the same distance
decreases.
(iv) Foragivenjob,morethenumberofworkers,lesswillbethetimetakentocomplete
the work.
Observe that change in one quantity leads to change in the other quantity.
Write five more such situations where change in one quantity leads to change in
anotherquantity.
How do we find out the quantity of each item needed by Mohan? Or, the time five
students take to complete the job?
To answer such questions, we now study some concepts of variation.
13.2 Direct Proportion
If the cost of 1 kg of sugar is ` 36, then what would be the cost of 3 kg sugar? It is` 108.
Direct and Inverse
Proportions
CHAPTER
13
2021–22

202 MATHEMATICS
Similarly, we can find the cost of 5 kg or 8 kg of sugar. Study the following table.
Observe that as weight of sugar increases, cost also increases in such a manner that
theirratioremainsconstant.
Take one more example. Suppose a car uses 4 litres of petrol to travel a distance of
60km.Howfarwillittravelusing12litres?Theansweris180km.Howdidwecalculate
it?Sincepetrolconsumedinthesecondinstanceis12litres,i.e.,threetimesof4litres,the
distance travelled will also be three times of 60 km. In other words, when the petrol
consumption becomes three-fold, the distance travelled is also three fold the previous
one.Lettheconsumptionofpetrolbexlitresandthecorrespondingdistancetravelledbe
y km.Now, complete the following table:
Petrol in litres (x) 4 8 12 15 20 25
Distance in km (y) 60 ... 180 ... ... ...
Wefindthatasthevalueofxincreases,valueofyalsoincreasesinsuchawaythatthe
ratio
x
y
doesnotchange;itremainsconstant(sayk).Inthiscase,itis
1
15
(checkit!).
We say that x and y are in direct proportion, if =
x
k
y
or x = ky.
In this example,
4 12
60 180
= , where 4 and 12 are the quantities of petrol consumed in
litres (x) and 60 and 180 are the distances (y) in km. So when x and y are in direct
proportion,wecanwrite 1 2
1 2
x x
y y
= .[y1
,y2
arevaluesofycorrespondingtothevaluesx1
,
x2
of x respectively]
The consumption of petrol and the distance travelled by a car is a case of direct
proportion.Similarly,thetotalamountspentandthenumberofarticlespurchasedisalso
an example of direct proportion.
2021–22

DO THIS
Thinkofafewmoreexamplesfordirectproportion.CheckwhetherMohan[intheinitialexample]will
take 750 mLof water, 5 spoons of sugar,
1
2
2
spoons of tea leaves and 125 mLof milk to prepare tea for
fivepersons!Letustrytounderstandfurthertheconceptofdirectproportionthroughthefollowingactivities.
(i) • Take a clock and fix its minute hand at 12.
• Record the angle turned through by the minute hand from its original position
andthetimethathaspassed,inthefollowingtable:
TimePassed(T) (T1
) (T2
) (T3
) (T4
)
(inminutes) 15 30 45 60
Angleturned(A) (A1
) (A2
) (A3
) (A4
)
(indegree) 90 ... ... ...
T
A
... ... ... ...
What do you observe about T andA? Do they increase together?
Is
T
A
same every time?
Istheangleturnedthroughbytheminutehanddirectlyproportional
to the time that has passed?Yes!
From the above table, you can also see
T1
: T2
= A1
: A2
, because
T1
: T2
= 15 : 30 = 1:2
A1
: A2
= 90 : 180 = 1:2
Check if T2
: T3
= A2
: A3
and T3
: T4
= A3
: A4
Youcanrepeatthisactivitybychoosingyourowntimeinterval.
(ii) Ask your friend to fill the following table and find the ratio of his age to the
correspondingageofhismother.
Age Present Age
five years ago age after five years
Friend’s age (F)
Mother’s age (M)
F
M
What do you observe?
Do F and M increase (or decrease) together? Is
F
M
same every time? No!
You can repeat this activity with other friends and write down your observations.
2021–22

204 MATHEMATICS
TRY THESE
Thus, variables increasing (or decreasing) together need not always be in direct
proportion.Forexample:
(i) physicalchangesinhumanbeingsoccurwithtimebutnotnecessarilyinapredeter-
minedratio.
(ii) changesinweightandheightamongindividualsarenotinanyknownproportionand
(iii) there is no direct relationship or ratio between the height of a tree and the number
ofleavesgrowingonitsbranches. Thinkofsomemoresimilarexamples.
1. Observe the following tables and find if x and yare directly proportional.
(i) x 20 17 14 11 8 5 2
y 40 34 28 22 16 10 4
(ii) x 6 10 14 18 22 26 30
y 4 8 12 16 20 24 28
(iii) x 5 8 12 15 18 20
y 15 24 36 60 72 100
2. Principal = ` 1000, Rate = 8% per annum. Fill in the following table and find
which type of interest (simple or compound) changes in direct proportion with
timeperiod.
Timeperiod 1 year 2 years 3 years
SimpleInterest(in`)
Compound Interest (in `)
P P
1
100
+





 −
r
t
P
100
r t
× ×
If we fix time period and the rate of interest, simple interest changes proportionally
withprincipal.Wouldtherebeasimilarrelationshipforcompoundinterest?Why?
Let us consider some solved examples where we would use the concept of
direct proportion.
Example 1: The cost of 5 metres of a particular quality of cloth is ` 210.Tabulate the
cost of 2, 4, 10 and 13 metres of cloth of the same type.
Solution: Suppose the length of cloth is x metres and its cost, in `, is y.
x 2 4 5 10 13
y y2
y3
210 y4
y5
2021–22

Asthelengthofclothincreases,costoftheclothalsoincreasesinthesameratio.Itis
a case of direct proportion.
We make use of the relation of type 1 2
1 2
x x
y y
=
(i) Here x1
= 5, y1
= 210 and x2
= 2
Therefore, 1 2
1 2
x x
y y
= gives
2
5 2
210 y
= or 5y2
= 2 × 210 or 2
2 210
5
y
×
= = 84
(ii) If x3
= 4, then
3
5 4
210 y
= or 5y3
= 4 × 210 or 3
4 210
5
y
×
= = 168
[Can we use
3
2
2 3
x
x
y y
= here? Try!]
(iii) If x4
= 10, then
4
5 10
210 y
= or 4
10 210
5
y
×
= = 420
(iv) If x5
= 13, then
5
5 13
210 y
= or 5
13 210
5
y
×
= = 546
Note that here we can also use or or in the place
2
84
4
168
10
420
o
of
5
210






Example 2: An electric pole, 14 metres high, casts a shadow of 10 metres. Find the
height of a tree that casts a shadow of 15 metres under similar conditions.
Solution: Let the height of the tree be x metres. We form a table as shown below:
height of the object (in metres) 14 x
length of the shadow (in metres) 10 15
Note that more the height of an object, the more would be the length of its shadow.
Hence, this is a case of direct proportion. That is,
1
1
x
y
= 2
2
x
y
We have
14
10
=
15
x
(Why?)
or
14
15
10
× = x
or
14 3
2
×
= x
So 21 = x
Thus, height of the tree is 21 metres.
Alternately, we can write 1 2
1 2
x x
y y
= as 1 1
2 2
x y
x y
=
2021–22

206 MATHEMATICS
so x1
: x2
= y1
: y2
or 14 : x = 10 : 15
Therefore, 10 × x = 15 × 14
or x =
15 14
10
×
= 21
Example 3: If the weight of 12 sheets of thick paper is 40 grams, how many sheets of
the same paper would weigh
1
2
2
kilograms?
Solution:
Let the number of sheets which weigh
1
2
2
kg bex.We put the above information in
the form of a table as shown below:
Number of sheets 12 x
Weight of sheets (in grams) 40 2500
More the number of sheets, the more would their
weightbe.So,thenumberofsheetsandtheirweights
are directly proportional to each other.
So,
12
40
=
2500
x
or
12 2500
40
×
= x
or 750 = x
Thus, the required number of sheets of paper = 750.
Alternate method:
Two quantities x and ywhich vary in direct proportion have the relation x =ky or
x
k
y
=
Here, k =
number of sheets
weight of sheets in grams
=
12 3
40 10
=
Now x is the number of sheets of the paper which weigh
1
2
2
kg [2500 g].
Using the relation x = ky, x =
3
10
× 2500 = 750
Thus, 750 sheets of paper would weigh
1
2
2
kg.
Example 4: Atrain is moving at a uniform speed of 75 km/hour.
(i) Howfarwillittravelin20minutes?
(ii) Find the time required to cover a distance of 250 km.
Solution: Let the distance travelled (in km) in 20 minutes be x and time taken
(in minutes) to cover 250 km be y.
Distance travelled (in km) 75 x 250
Time taken (in minutes) 60 20 y
1kilogram= 1000grams
1
2
2
kilograms = 2500grams
1 hour = 60 minutes
2021–22

DO THIS
Since the speed is uniform, therefore, the distance covered would be directly
proportionaltotime.
(i) We have
75
60 20
x
=
or
75
20
60
× = x
or x = 25
So, the train will cover a distance of 25 km in 20 minutes.
(ii) Also,
75 250
60 y
=
or
250 60
75
y
×
= = 200 minutes or 3 hours 20 minutes.
Therefore,3hours20minuteswillberequiredtocoveradistanceof250kilometres.
Alternatively, when xis known, then one can determiney from the relation
250
20
x
y
= .
You know that a map is a miniature representation of a very large region.Ascale is
usually given at the bottom of the map. The scale shows a relationship between
actuallengthandthelengthrepresentedonthemap.Thescaleof themapisthusthe
ratio of the distance between two points on the map to the actual distance between
two points on the large region.
Forexample,if1cmonthemaprepresents8kmofactualdistance[i.e.,thescaleis
1 cm : 8 km or 1 : 800,000] then 2 cm on the same map will represent 16 km.
Hence, we can say that scale of a map is based on the concept of direct proportion.
Example 5: The scale of a map is given as 1:30000000. Two cities are 4 cm apart on
the map. Find the actual distance between them.
Solution: Let the map distance be x cm and actual distance be y cm, then
1:30000000 = x : y
or 7
1
3 10
×
=
x
y
Since x = 4 so, 7
1
3 10
×
=
4
y
or y = 4 × 3 × 107
= 12 × 107
cm = 1200 km.
Thus, two cities, which are 4 cm apart on the map, are actually 1200 km away from
each other.
Take a map of your State. Note the scale used there. Using a ruler, measure the “map
distance” between any two cities. Calculate the actual distance between them.
2021–22

208 MATHEMATICS
EXERCISE 13.1
1. Following are the car parking charges near a railway station upto
4 hours ` 60
8 hours ` 100
12 hours ` 140
24 hours ` 180
Check if the parking charges are in direct proportion to the parking time.
2. Amixtureofpaintispreparedbymixing1partofredpigmentswith8partsofbase.
In the following table, find the parts of base that need to be added.
Parts of red pigment 1 4 7 12 20
Parts of base 8 ... ... ... ...
3. In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much
red pigment should we mix with 1800 mL of base?
4. Amachineinasoftdrinkfactoryfills840bottlesinsixhours.Howmanybottleswill
itfillinfivehours?
5. Aphotographofabacteriaenlarged50,000times
attainsalengthof5cmasshowninthediagram.
What is the actual length of the bacteria? If the
photograph is enlarged 20,000 times only, what
wouldbeitsenlargedlength?
6. Inamodelofaship,themastis9cmhigh,while
themastoftheactualshipis12mhigh.Ifthelength
oftheshipis28m,howlongisthemodelship?
7. Suppose 2 kg of sugar contains 9 × 106
crystals.
How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a
road for 72 km. What would be her distance covered in the map?
9. A5m60cmhighverticalpolecastsashadow3m20cmlong.Findatthesametime
(i)thelengthoftheshadowcastbyanotherpole10m50cmhigh(ii)theheightofa
pole which casts a shadow 5m long.
10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far
can it travel in 5 hours?
DO THIS 1. On a squared paper, draw five squares of different sides.
Writethefollowinginformationinatabularform.
Square-1 Square-2 Square-3 Square-4 Square-5
Length of a side (L)
Perimeter (P)
L
P
2021–22

Area (A)
L
A
Find whether the length of a side is in direct proportion to:
(a) the perimeter of the square.
(b) the area of the square.
2. The following ingredients are required to make halwa for 5
persons:
Suji/Rawa = 250 g, Sugar = 300 g,
Ghee = 200 g, Water = 500 mL.
Using the concept of proportion, estimate the
changes in the quantity of ingredients, to
prepare halwa for your class.
3. Choose a scale and make a map of your
classroom, showing windows, doors,
blackboard etc. (An example is given here).
Take a few problems discussed so far under ‘direct variation’. Do you think that
they can be solved by ‘unitary method’?
13.3 Inverse Proportion
Two quantities may change in such a manner that if one quantity increases, the other
quantitydecreasesandviceversa.Forexample,asthenumberofworkersincreases,time
taken to finish the job decreases. Similarly, if we increase the speed, the time taken to
cover a given distance decreases.
Tounderstandthis,letuslookintothefollowingsituation.
Zaheedacangotoherschoolinfourdifferentways.Shecanwalk,run,cycleorgoby
car.Studythefollowingtable.
2021–22

210 MATHEMATICS
Observe that as the speed increases, time taken to cover the same distance decreases.
AsZaheedadoublesherspeedbyrunning,time
reduces to half.As she increases her speed to three
times by cycling, time decreases to one third.
Similarly, as she increases her speed to 15 times,
time decreases to one fifteenth. (Or, in other words
the ratio by which time decreases is inverse of the
ratiobywhichthecorrespondingspeedincreases).
Can we say that speed and time change inversely
inproportion?
Let us consider another example.Aschool wants to spend ` 6000 on mathematics
textbooks. How many books could be bought at ` 40 each? Clearly 150 books can be
bought. If the price of a textbook is more than ` 40, then the number of books which
couldbepurchasedwiththesameamountofmoneywouldbelessthan150.Observethe
followingtable.
Price of each book (in `) 40 50 60 75 80 100
Number of books that 150 120 100 80 75 60
can be bought
What do you observe?You will appreciate that as the price of the books increases,
the number of books that can be bought, keeping the fund constant, will decrease.
Ratiobywhichthepriceofbooksincreaseswhengoingfrom40to50is4:5,andthe
ratio by which the corresponding number of books decreases from 150 to 120 is 5 : 4.
This means that the two ratios are inverses of each other.
Notice that the product of the corresponding values of the two quantities is constant;
that is, 40 × 150 = 50 × 120 = 6000.
If we represent the price ofone book asxandthenumberof books bought asy,then
as x increases y decreases and vice-versa. It is important to note that the product xy
remainsconstant.Wesaythatx variesinverselywithy andyvariesinverselywithx.Thus
two quantities x and y are said to vary in inverse proportion, if there exists a relation
of the type xy = k between them, k being a constant. If y1
, y2
are the values of y
corresponding to the values x1
, x2
of x respectively then x1
y1
= x2
y2
(= k), or 1 2
2 1
x y
x y
= .
We say that x and y are in inverse proportion.
Hence, in this example, cost of a book and number of books purchased in a fixed
amount are inversely proportional. Similarly, speed of a vehicle and the time taken to
coverafixeddistancechangesininverseproportion.
Thinkofmoresuchexamplesofpairsofquantitiesthatvaryininverseproportion.You
maynowhavealookatthefurniture–arrangingproblem,statedintheintroductorypart
ofthischapter.
Hereisanactivityforbetterunderstandingoftheinverseproportion.
Multiplicativeinverseofanumber
is its reciprocal. Thus,
1
2
is the
inverse of 2 and vice versa. (Note
that
1 1
2 2 1
2 2
× = × = ).
2021–22

TRY THESE
DO THIS
Take a squared paper and arrange 48 counters on it in different number of rows as
shownbelow.
Number of (R1
) (R2
) (R3
) (R4
) (R5
)
Rows (R) 2 3 4 6 8
Number of (C1
) (C2
) (C3
) (C4
) (C5
)
Columns (C) ... ... 12 8 ...
What do you observe?As R increases, C decreases.
(i) Is R1
: R2
= C2
: C1
? (ii) Is R3
: R4
= C4
: C3
?
(iii) Are R and C inversely proportional to each other?
Trythisactivitywith36counters.
Observe the following tables and find which pair of variables (here x and y) are in
inverseproportion.
(i) x 50 40 30 20 (ii) x 100 200 300 400
y 5 6 7 8 y 60 30 20 15
(iii) x 90 60 45 30 20 5
y 10 15 20 25 30 35
Let us consider some examples where we use the concept of inverse proportion.
Whentwoquantitiesx andyareindirectproportion(orvarydirectly)theyare also writtenas x∝ y.
Whentwoquantitiesxandyareininverseproportion(orvaryinversely)theyarealso writtenasx∝
1
y .
2021–22

212 MATHEMATICS
Example 7: 6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it
take if only 5 pipes of the same type are used?
Solution:
Letthedesiredtimetofillthetankbexminutes.Thus,wehave
thefollowingtable.
Number of pipes 6 5
Time (in minutes) 80 x
Lesserthenumberofpipes,morewillbethetimerequiredby
it to fill the tank. So, this is a case of inverse proportion.
Hence, 80 × 6 = x × 5 [x1
y1
= x2
y2
]
or
80 6
5
x
×
=
or x = 96
Thus, time taken to fill the tank by 5 pipes is 96 minutes or 1 hour 36 minutes.
Example 8: There are 100 students in a hostel. Food provision for them is for 20
days. How long will these provisions last, if 25 more students join the group?
Solution: Suppose the provisions last for y days when the number of students is 125.
Wehavethefollowingtable.
Number of students 100 125
Number of days 20 y
Notethatmorethenumberofstudents,thesoonerwould
theprovisionsexhaust.Therefore,thisisacaseofinverse
proportion.
So, 100 × 20 = 125 × y
or
100 20
125
×
= y or 16 = y
Thus, the provisions will last for 16 days, if 25 more students join the hostel.
Alternately, we can write x1
y1
= x2
y2
as 1 2
2 1
x y
x y
= .
Thatis, x1
: x2
= y2
: y1
or 100 : 125 = y : 20
or y =
100 20
16
125
×
=
Example 9: If 15 workers can build a wall in 48 hours, how many workers will be
required to do the same work in 30 hours?
Solution:
Let the number of workers employed to build the wall in 30 hours be y.
2021–22

Wehavethefollowingtable.
Number of hours 48 30
Number of workers 15 y
Obviously more the number of workers, faster will they build the wall.
So,thenumberofhoursandnumberofworkersvaryininverseproportion.
So 48 × 15 = 30 × y
Therefore,
48 15
30
×
= y or y = 24
i.e., to finish the work in 30 hours, 24 workers are required.
EXERCISE 13.2
1. Whichofthefollowingareininverseproportion?
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.
2. In aTelevision game show, the prize money of ` 1,00,000 is to be divided equally
amongstthewinners.Completethefollowingtableandfindwhethertheprizemoney
given to an individual winner is directly or inversely proportional to the number
ofwinners?
Number of winners 1 2 4 5 8 10 20
Prize for each winner (in `) 1,00,000 50,000 ... ... ... ... ...
3. Rehmanismakingawheelusingspokes.Hewantstofixequalspokesinsuchaway
that the angles between any pair of consecutive spokes are equal. Help him by
completingthefollowingtable.
Number of spokes 4 6 8 10 12
Angle between
a pair of consecutive 90° 60° ... ... ...
spokes
2021–22

214 MATHEMATICS
(i) Are the number of spokes and the angles formed between the pairs of
consecutive spokes in inverse proportion?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15
spokes.
(iii) Howmanyspokeswouldbeneeded,iftheanglebetweenapairofconsecutive
spokes is 40°?
4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How
many would each get, if the number of the children is reduced by 4?
5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long
would the food last if there were 10 more animals in his cattle?
6. Acontractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If,
he uses 4 persons instead of three, how long should they take to complete the job?
7. A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same
batch is packed using 20 bottles in each box, how many boxes would be filled?
8. A factory requires 42 machines to produce a given number of articles in 63 days.
How many machines would be required to produce the same number of articles in
54 days?
9. Acartakes2hourstoreachadestinationbytravellingatthespeedof60km/h.How
long will it take when the car travels at the speed of 80 km/h?
10. Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job
take now?
(ii) How many persons would be needed to fit the windows in one day?
11. A school has 8 periods a day each of 45 minutes duration. How long would each
periodbe,iftheschoolhas9periodsaday,assumingthenumberofschoolhoursto
be the same?
2021–22

DO THIS
1. Takeasheetofpaper.Folditasshowninthefigure.Countthenumberofpartsand
the area of a part in each case.
Tabulate your observations and discuss with your friends. Is it a case of inverse proportion?Why?
Number of parts 1 2 4 8 16
Area of each part area of the paper
1
2
the area of the paper ... ... ...
2. Takeafewcontainersofdifferentsizeswithcircularbases.Fillthesameamountof
water in each container. Note the diameter of each container and the respective
height at which the water level stands. Tabulate your observations. Is it a case of
inverseproportion?
Diameter of container (in cm)
Height of water level (in cm)
1. Two quantities x and y are said to be in direct proportion if they increase (decrease) together in
such a manner that the ratio of their corresponding values remains constant. That is if
x
k
y
= [k is
a positive number], then x and y are said to vary directly. In such a case if y1
,y2
are the values of
y corresponding to the values x1
, x2
of x respectively then 1 2
1 2
x x
y y
= .
2021–22

216 MATHEMATICS
2. Twoquantitiesxandyaresaidtobeininverseproportionifanincreaseinxcausesaproportional
decrease in y (and vice-versa) in such a manner that the product of their corresponding values
remains constant.That is, ifxy=k, thenxand yare said to vary inversely. In this case if y1
,y2
are
the values of y corresponding to the values x1
, x2
of x respectively then x1
y1
= x2
y2
or 1 2
2 1
x y
x y
= .
2021–22

FACTORISATION 217
14.1 Introduction
14.1.1 Factors of natural numbers
YouwillrememberwhatyoulearntaboutfactorsinClassVI.Letustakeanaturalnumber,
say 30, and write it as a product of other natural numbers, say
30 = 2 × 15
= 3 × 10 = 5 × 6
Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30.
Of these, 2, 3 and 5 are the prime factors of 30 (Why?)
Anumberwrittenasaproductofprimefactorsissaidto
be in the prime factor form; for example, 30 written as
2 × 3 × 5 is in the prime factor form.
The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.
Similarly, we can express algebraic expressions as products of their factors. This is
what we shall learn to do in this chapter.
14.1.2 Factors of algebraic expressions
We have seen in ClassVII that in algebraic expressions, terms are formed as products of
factors. For example, in the algebraic expression 5xy+ 3x the term 5xy has been formed
by the factors 5, x and y, i.e.,
5xy = y
x×
×
5
Observe that the factors 5,x and y of 5xycannotfurther
be expressed as a product of factors. We may say that 5,
x andy are ‘prime’factors of 5xy. In algebraic expressions,
weusetheword‘irreducible’inplaceof‘prime’. Wesaythat
5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not
anirreducibleformof 5xy,sincethefactorxy canbefurther
expressed as a product of x and y, i.e., xy = x × y.
Factorisation
CHAPTER
14
Note 1 is a factor of 5xy, since
5xy = y
x×
×
×5
1
In fact, 1 is a factor of every term. As
in the case of natural numbers, unless
it is specially required, we do not show
1 as a separate factor of any term.
We know that 30 can also be written as
30 = 1 × 30
Thus, 1 and 30 are also factors of 30.
You will notice that 1 is a factor of any
number. For example, 101 = 1 × 101.
However, when we write a number as a
product of factors, we shall not write 1 as
a factor, unless it is specially required.
2021–22

218 MATHEMATICS
Next consider the expression 3x (x + 2). It can be written as a product of factors.
3, x and (x + 2)
3x(x + 2) = ( )
2
3 +
×
× x
x
The factors 3, x and (x +2) are irreducible factors of 3x (x + 2).
Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form
as 10x (x + 2) (y + 3) = ( ) ( )
2 5 2 3
x x y
× × × + × + .
14.2 What is Factorisation?
When we factorise an algebraic expression, we write it as a product of factors. These
factors may be numbers, algebraic variables or algebraic expressions.
Expressions like 3xy, y
x2
5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form.
Their factors can be just read off from them, as we already know.
On the other hand consider expressions like 2x + 4, 3x + 3y, x2
+ 5x, x2
+ 5x + 6.
Itisnotobviouswhattheirfactorsare.Weneedtodevelopsystematicmethodstofactorise
these expressions, i.e., to find their factors.This is what we shall do now.
14.2.1 Method of common factors
• We begin with a simple example: Factorise 2x + 4.
We shall write each term as a product of irreducible factors;
2x = 2 × x
4 = 2 × 2
Hence 2x + 4 = (2 × x) + (2 × 2)
Notice that factor 2 is common to both the terms.
Observe,bydistributivelaw
2 × (x + 2) = (2 × x) + (2 × 2)
Therefore, we can write
2x + 4 = 2 × (x + 2) = 2 (x + 2)
Thus, the expression 2x+ 4 is the same as 2 (x+ 2). Now we can read off its factors:
they are 2 and (x + 2). These factors are irreducible.
Next, factorise 5xy + 10x.
The irreducible factor forms of 5xy and 10x are respectively,
5xy = 5 × x × y
10x = 2 × 5 × x
Observe that the two terms have 5 and x as common factors. Now,
5xy + 10x = (5 × x × y) + (5 × x × 2)
= (5x × y) + (5x × 2)
Wecombinethetwotermsusingthedistributivelaw,
(5x× y) + (5x× 2) = 5x × ( y + 2)
Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.)
2021–22

FACTORISATION 219
TRY THESE
Example 1: Factorise 12a2
b + 15ab2
Solution: We have 12a2
b = 2 × 2 × 3 × a × a × b
15ab2
= 3 × 5 × a × b × b
The two terms have 3, a and b as common factors.
Therefore, 12a2
b + 15ab2
= (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)
= 3 × a × b × [(2 × 2 × a) + (5 × b)]
= 3ab × (4a + 5b)
= 3ab (4a + 5b) (requiredfactorform)
Example 2: Factorise 10x2
– 18x3
+ 14x4
Solution: 10x2
= 2 × 5 × x × x
18x3
= 2 × 3 × 3 × x × x × x
14x4
= 2 × 7 × x × x × x × x
The common factors of the three terms are 2, x and x.
Therefore, 10x2
– 18x3
+ 14x4
= (2 × x × x × 5) – (2 × x × x × 3 × 3 × x)
+ (2 × x × x × 7 × x × x)
= 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)]
= 2x2
× (5 – 9x + 7x2
) = 2 2
2 (7 9 5)
x x x
− +

Factorise: (i) 12x + 36 (ii) 22y – 33z (iii) 14pq + 35pqr
14.2.2 Factorisation by regrouping terms
Look at the expression 2xy + 2y + 3x + 3. You will notice that the first two terms have
common factors 2 and y and the last two terms have a common factor 3. But there is no
single factor common to all the terms. How shall we proceed?
Let us write (2xy + 2y) in the factor form:
2xy + 2y = (2 × x × y) + (2 × y)
= (2 × y × x) + (2 × y × 1)
= (2y × x) + (2y × 1) = 2y (x + 1)
Similarly, 3x + 3 = (3 × x) + (3 × 1)
= 3 × (x + 1) = 3 ( x + 1)
Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1)
Observe, now we have a common factor (x + 1) in both the terms on the right hand
side.Combiningthetwoterms,
2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3)
The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its
factors are (x + 1) and (2y + 3). Note, these factors are irreducible.
Note, we need to
show1 as a factor
here. Why?
Do you notice that the factor
form of an expression has only
one term?
(combiningthethreeterms)
(combiningtheterms)
2021–22

220 MATHEMATICS
What is regrouping?
Suppose, the above expression was given as 2xy + 3 + 2y + 3x; then it will not be easy to
see the factorisation. Rearranging the expression, as 2xy+ 2y+3x+ 3, allows us to form
groups (2xy + 2y) and (3x + 3) leading to factorisation. This is regrouping.
Regrouping may be possible in more than one ways. Suppose, we regroup the
expression as: 2xy + 3x + 2y + 3. This will also lead to factors. Let us try:
2xy + 3x + 2y + 3 = 2 × x × y + 3 × x + 2 × y + 3
= x × (2y + 3) + 1 × (2y + 3)
= (2y + 3) (x + 1)
The factors are the same (as they have to be), although they appear in different order.
Example 3: Factorise 6xy – 4y + 6 – 9x.
Solution:
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor 2y;
6xy – 4y = 2y (3x – 2) (a)
What about the last two terms? Observe them. If you change their order to
– 9x + 6, the factor ( 3x – 2) will come out;
–9x + 6 = –3 (3x) + 3 (2)
= – 3 (3x – 2) (b)
Step 3 Putting (a) and (b) together,
6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)
The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3).
EXERCISE 14.1
1. Findthecommonfactorsofthegiventerms.
(i) 12x, 36 (ii) 2y, 22xy (iii) 14 pq, 28p2
q2
(iv) 2x, 3x2
, 4 (v) 6 abc, 24ab2
, 12 a2
b
(vi) 16 x3
, – 4x2
, 32x (vii) 10 pq, 20qr, 30rp
(viii) 3x2
y3
, 10x3
y2
,6 x2
y2
z
2. Factorisethe followingexpressions.
(i) 7x – 42 (ii) 6p – 12q (iii) 7a2
+ 14a
(iv) – 16 z + 20 z3
(v) 20 l2
m + 30 a l m
(vi) 5 x2
y – 15 xy2
(vii) 10 a2
– 15 b2
+ 20 c2
(viii) – 4 a2
+ 4 ab – 4 ca (ix) x2
y z + x y2
z + x y z2
(x) a x2
y + b x y2
+ c x y z
3. Factorise.
(i) x2
+ x y + 8x + 8y (ii) 15 xy – 6x + 5y – 2
2021–22

FACTORISATION 221
(iii) ax + bx – ay – by (iv) 15 pq + 15 + 9q + 25p
(v) z – 7 + 7 x y – x y z
14.2.3 Factorisation using identities
We know that (a + b)2
= a2
+ 2ab + b2
(I)
(a – b)2
= a2
– 2ab + b2
(II)
(a + b) (a – b) = a2
– b2
(III)
Thefollowingsolvedexamplesillustratehowtousetheseidentitiesforfactorisation.What
wedoistoobservethegivenexpression.Ifithasaformthatfitstherighthandsideofone
of the identities, then the expression corresponding to the left hand side of the identity
givesthedesiredfactorisation.
Example 4: Factorise x2
+ 8x + 16
Solution: Observe the expression; it has three terms. Therefore, it does not fit
Identity III.Also, it’s first and third terms are perfect squares with a positive sign before
the middle term. So, it is of the form a2
+ 2ab + b2
where a = x and b = 4
suchthat a2
+ 2ab + b2
= x2
+ 2 (x) (4) + 42
= x2
+ 8x + 16
Since a2
+ 2ab + b2
= (a + b)2
,
bycomparison x2
+ 8x + 16 = ( x + 4)2
(the required factorisation)
Example 5: Factorise 4y2
– 12y + 9
Solution: Observe 4y2
= (2y)2
, 9 = 32
and 12y = 2 × 3 × (2y)
Therefore, 4y2
– 12y + 9 = (2y)2
– 2 × 3 × (2y) + (3)2
= ( 2y – 3)2
(requiredfactorisation)
Example 6: Factorise 49p2
– 36
Solution: There are two terms; both are squares and the second is negative. The
expression is of the form (a2
– b2
). Identity III is applicable here;
49p2
– 36 = (7p)2
– ( 6 )2
= (7p – 6 ) ( 7p + 6) (required factorisation)
Example 7: Factorise a2
– 2ab + b2
– c2
Solution: The first three terms of the given expression form (a – b)2
. The fourth term
is a square. So the expression can be reduced to a difference of two squares.
Thus, a2
– 2ab + b2
– c2
= (a – b)2
– c2
(ApplyingIdentityII)
= [(a – b) – c) ((a – b) + c)] (ApplyingIdentityIII)
= (a – b – c) (a – b + c) (requiredfactorisation)
Notice,howweappliedtwoidentitiesoneaftertheothertoobtaintherequiredfactorisation.
Example 8: Factorise m4
– 256
Solution: We note m4
= (m2
)2
and 256 = (16)2
Observe here the given
expression is of the form
a2
– 2ab + b2
.
Where a = 2y, and b = 3
with 2ab = 2 × 2y × 3 = 12y.
2021–22

222 MATHEMATICS
Thus,thegivenexpressionfitsIdentityIII.
Therefore, m4
– 256 = (m2
)2
– (16)2
= (m2
–16) (m2
+16) [(usingIdentity(III)]
Now, (m2
+ 16) cannot be factorised further, but (m2
–16) is factorisable again as per
IdentityIII.
m2
–16 = m2
– 42
= (m – 4) (m + 4)
Therefore, m4
– 256 = (m – 4) (m + 4) (m2
+16)
14.2.4 Factors of the form ( x + a) ( x + b)
Let us now discuss how we can factorise expressions in one variable, like x2
+ 5x + 6,
y2
– 7y + 12, z2
– 4z – 12, 3m2
+ 9m + 6, etc. Observe that these expressions are not
of the type (a + b)2
or (a – b)2
, i.e., they are not perfect squares. For example, in
x2
+ 5x + 6, the term 6 is not a perfect square. These expressions obviously also do not
fit the type (a2
– b2
) either.
They, however, seem to be of the type x2
+ (a + b) x + a b. We may therefore, try to
use Identity IV studied in the last chapter to factorise these expressions:
(x + a) (x + b) = x2
+ (a + b) x + ab (IV)
Forthatwehavetolookatthecoefficientsofxandtheconstantterm.Letusseehow
itisdoneinthefollowingexample.
Example 9: Factorise x2
+ 5x + 6
Solution: If we compare the R.H.S. of Identity (IV) with x2
+ 5x + 6, we find ab = 6,
and a + b = 5. From this, we must obtain a and b. The factors then will be
(x + a) and (x + b).
If a b = 6, it means that a and b are factors of 6. Let us try a = 6, b = 1. For these
values a + b = 7, and not 5, So this choice is not right.
Let us try a = 2, b = 3. For this a + b = 5 exactly as required.
The factorised form of this given expression is then (x +2) (x + 3).
In general, for factorising an algebraic expression of the type x2
+ px + q, we find two
factors a and b of q (i.e., the constant term) such that
ab = q and a + b = p
Then, the expression becomes x2
+ (a + b) x + ab
or x2
+ ax + bx + ab
or x(x + a) + b(x + a)
or (x + a) (x + b) which are the required factors.
Example 10: Find the factors of y2
–7y +12.
Solution: We note 12 = 3 × 4 and 3 + 4 = 7. Therefore,
y2
– 7y+ 12 = y2
– 3y – 4y + 12
= y (y –3) – 4 (y –3) = (y –3) (y – 4)
2021–22

FACTORISATION 223
Note,thistimewedidnotcomparetheexpressionwiththatinIdentity(IV)toidentify
aandb. After sufficientpracticeyoumaynotneedtocomparethegivenexpressionsfor
theirfactorisationwiththeexpressionsintheidentities;insteadyoucanproceeddirectly
as we did above.
Example 11: Obtain the factors of z2
– 4z – 12.
Solution: Here a b = –12 ; this means one of a and b is negative. Further,a + b = – 4,
this means the one with larger numerical value is negative. We try a = – 4, b = 3; but
this will not work, since a + b = –1. Next possible values are a = – 6, b = 2, so that
a + b = – 4 as required.
Hence, z2
– 4z –12 = z2
– 6z + 2z –12
= z(z – 6) + 2(z – 6 )
= (z – 6) (z + 2)
Example 12: Find the factors of 3m2
+ 9m + 6.
Solution: We notice that 3 is a common factor of all the terms.
Therefore, 3m2
+ 9m + 6 = 3(m2
+ 3m + 2)
Now, m 2
+ 3m + 2 = m2
+ m + 2m + 2 (as 2 = 1 × 2)
= m(m + 1)+ 2( m + 1)
= (m + 1) (m + 2)
Therefore, 3m2
+ 9m + 6 = 3(m + 1) (m + 2)
EXERCISE 14.2
1. Factorisethefollowingexpressions.
(i) a2
+ 8a + 16 (ii) p2
– 10 p + 25 (iii) 25m2
+ 30m + 9
(iv) 49y2
+ 84yz + 36z2
(v) 4x2
– 8x + 4
(vi) 121b2
– 88bc + 16c2
(vii) (l + m)2
– 4lm (Hint: Expand ( l + m)2
first)
(viii) a4
+ 2a2
b2
+ b4
2. Factorise.
(i) 4p2
– 9q2
(ii) 63a2
– 112b2
(iii) 49x2
– 36
(iv) 16x5
– 144x3
(v) (l + m)2
– (l – m)2
(vi) 9x2
y2
– 16 (vii) (x2
– 2xy + y2
) – z2
(viii) 25a2
– 4b2
+ 28bc – 49c2
3. Factorise the expressions.
(i) ax2
+ bx (ii) 7p2
+ 21q2
(iii) 2x3
+ 2xy2
+ 2xz2
(iv) am2
+ bm2
+ bn2
+ an2
(v) (lm + l) + m + 1
(vi) y (y + z) + 9 (y + z) (vii) 5y2
– 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2 (ix) 6xy – 4y + 6 – 9x
2021–22

224 MATHEMATICS
4. Factorise.
(i) a4
– b4
(ii) p4
– 81 (iii) x4
– (y + z)4
(iv) x4
– (x – z)4
(v) a4
– 2a2
b2
+ b4
5. Factorisethefollowingexpressions.
(i) p2
+ 6p + 8 (ii) q2
– 10q + 21 (iii) p2
+ 6p – 16
14.3 Division of Algebraic Expressions
We have learnt how to add and subtract algebraic expressions. We also know how to
multiply two expressions. We have not however, looked at division of one algebraic
expression by another.This is what we wish to do in this section.
Werecallthatdivisionistheinverseoperationofmultiplication.Thus,7×8=56gives
56 ÷ 8 = 7 or 56 ÷ 7 = 8.
Wemaysimilarlyfollowthedivisionofalgebraicexpressions.Forexample,
(i) 2x × 3x2
= 6x3
Therefore, 6x3
÷ 2x = 3x2
and also, 6x3
÷ 3x2
= 2x.
(ii) 5x (x + 4) = 5x2
+ 20x
Therefore, (5x2
+ 20x) ÷ 5x = x + 4
and also (5x2
+ 20x) ÷ (x + 4) = 5x.
We shall now look closely at how the division of one expression by another can be
carriedout.Tobeginwithweshallconsiderthedivisionofamonomialbyanothermonomial.
14.3.1 Division of a monomial by another monomial
Consider 6x3
÷ 2x
We may write 2x and 6x3
in irreducible factor forms,
2x = 2 × x
6x3
= 2 × 3 × x × x × x
Now we group factors of 6x3
to separate 2x,
6x3
= 2 × x × (3 × x × x) = (2x) × (3x2
)
Therefore, 6x3
÷ 2x = 3x2
.
Ashorterwaytodepictcancellationofcommonfactorsisaswedoindivisionofnumbers:
77 ÷ 7 =
77
7
=
7 11
7
×
= 11
Similarly, 6x3
÷ 2x =
3
6
2
x
x
=
2 3
2
x x x
x
× × × ×
×
= 3 × x × x = 3x2
Example 13: Do the following divisions.
(i) –20x4
÷ 10x2
(ii) 7x2
y2
z2
÷ 14xyz
Solution:
(i) –20x4
= –2 × 2 × 5 × x × x × x × x
10x2
= 2 × 5 × x × x
2021–22

FACTORISATION 225
TRY THESE
Therefore, (–20x4
) ÷ 10x2
=
2 2 5
2 5
x x x x
x x
− × × × × × ×
× × ×
= –2 × x × x = –2x2
(ii) 7x2
y2
z2
÷ 14xyz =
7
2 7
x x y y z z
x y z
× × × × × ×
× × × ×
=
2
x y z
× ×
=
1
2
xyz
Divide.
(i) 24xy2
z3
by 6yz2
(ii) 63a2
b4
c6
by 7a2
b2
c3
14.3.2 Division of a polynomial by a monomial
Let us consider the division of the trinomial 4y3
+ 5y2
+ 6yby the monomial 2y.
4y3
+ 5y2
+ 6y = (2 × 2 × y × y × y) + (5 × y × y) + (2 × 3 × y)
(Here, we expressed each term of the polynomial in factor form) we find that 2 ×yis
common in each term.Therefore, separating 2 × y from each term. We get
4y3
+ 5y2
+ 6y =2 × y × (2 × y × y) + 2 × y ×
5
2
×






y + 2 × y × 3
= 2y (2y2
) + 2y
5
2
y





 + 2y (3)
= 2 2
5
2
3
2
y y y
+ +





 (Thecommonfactor2yisshownseparately.
Therefore, (4y3
+ 5y2
+ 6y) ÷ 2y
=
2
3 2
5
2 (2 3)
4 5 6 2
2 2
y y y
y y y
y y
+ +
+ +
= = 2y2
+
5
2
y + 3
Alternatively, we could divide each term of the trinomial by the
monomialusingthecancellationmethod.
(4y3
+ 5y2
+ 6y) ÷ 2y =
3 2
4 5 6
2
y y y
y
+ +
=
3 2
4 5 6
2 2 2
y y y
y y y
+ + = 2y2
+
5
2
y + 3
Example 14: Divide 24(x2
yz + xy2
z + xyz2
) by 8xyz using both the methods.
Solution: 24 (x2
yz + xy2
z + xyz2
)
= 2 × 2 × 2 × 3 × [(x × x × y × z) + (x × y × y × z) + (x × y × z × z)]
= 2 × 2 × 2 × 3 × x × y × z × (x + y + z) = 8 × 3 × xyz × (x + y + z)
Therefore, 24 (x2
yz + xy2
z + xyz2
) ÷ 8xyz
=
8 3 ( )
8
xyz x y z
xyz
× × × + +
×
= 3 × (x + y + z) = 3 (x + y + z)
Here, we divide
each term of the
polynomial in the
numerator by the
monomial in the
denominator.
(Bytakingoutthe
commonfactor)
2021–22

226 MATHEMATICS
Alternately,24(x2
yz + xy2
z + xyz2
) ÷ 8xyz =
2 2 2
24 24 24
8 8 8
x yz xy z xyz
xyz xyz xyz
+ +
= 3x + 3y + 3z = 3(x + y + z)
14.4 Division of Algebraic Expressions Continued
(Polynomial ÷
÷
÷
÷
÷ Polynomial)
• Consider (7x2
+ 14x) ÷ (x + 2)
Weshallfactorise(7x2
+14x)firsttocheckandmatchfactorswiththedenominator:
7x2
+ 14x = (7 × x × x) + (2 × 7 × x)
= 7 × x × (x + 2) = 7x(x + 2)
Now (7x2
+ 14x) ÷ (x + 2) =
2
7 14
2
x x
x
+
+
=
7 ( 2)
2
x x
x
+
+
= 7x (Cancelling the factor (x + 2))
Example 15: Divide 44(x4
– 5x3
– 24x2
) by 11x (x – 8)
Solution: Factorising 44(x4
– 5x3
– 24x2
), we get
44(x4
– 5x3
– 24x2
) = 2 × 2 × 11 × x2
(x2
– 5x – 24)
(taking the common factor x2
out of the bracket)
= 2 × 2 × 11 × x2
(x2
– 8x + 3x – 24)
= 2 × 2 × 11 × x2
[x (x – 8) + 3(x – 8)]
= 2 × 2 × 11 × x2
(x + 3) (x – 8)
Therefore, 44(x4
– 5x3
– 24x2
) ÷ 11x(x – 8)
=
2 2 11 ( 3) ( – 8)
11 ( – 8)
x x x x
x x
× × × × × + ×
× ×
= 2 × 2 × x (x + 3) = 4x(x + 3)
Example 16: Divide z(5z2
– 80) by 5z(z + 4)
Solution: Dividend= z(5z2
– 80)
= z[(5 × z2
) – (5 × 16)]
= z × 5 × (z2
– 16)
= 5z × (z + 4) (z – 4) [using the identity
a2
– b2
= (a + b) (a – b)]
Thus, z(5z2
– 80) ÷ 5z(z + 4) =
5 ( 4) ( 4)
5 ( 4)
z z z
z z
− +
+
= (z – 4)
Will it help here to
divide each term of
the numerator by
the binomial in the
denominator?
We cancel the factors 11,
x and (x – 8) common to
both the numerator and
denominator.
2021–22

FACTORISATION 227
EXERCISE 14.3
1. Carryoutthefollowingdivisions.
(i) 28x4
÷ 56x (ii) –36y3
÷ 9y2
(iii) 66pq2
r3
÷ 11qr2
(iv) 34x3
y3
z3
÷ 51xy2
z3
(v) 12a8
b8
÷ (– 6a6
b4
)
2. Dividethegivenpolynomialbythegivenmonomial.
(i) (5x2
– 6x) ÷ 3x (ii) (3y8
– 4y6
+ 5y4
) ÷ y4
(iii) 8(x3
y2
z2
+ x2
y3
z2
+ x2
y2
z3
) ÷ 4x2
y2
z2
(iv) (x3
+ 2x2
+3x) ÷ 2x
(v) (p3
q6
– p6
q3
) ÷ p3
q3
3. Workoutthefollowingdivisions.
(i) (10x – 25) ÷ 5 (ii) (10x – 25) ÷ (2x – 5)
(iii) 10y(6y + 21) ÷ 5(2y + 7) (iv) 9x2
y2
(3z – 24) ÷ 27xy(z – 8)
(v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6)
4. Divideasdirected.
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1) (ii) 26xy(x + 5)(y – 4) ÷ 13x(y – 4)
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)
(iv) 20(y + 4) (y2
+ 5y + 3) ÷ 5(y + 4) (v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)
5. Factorise the expressions and divide them as directed.
(i) (y2
+ 7y + 10) ÷ (y + 5) (ii) (m2
– 14m – 32) ÷ (m + 2)
(iii) (5p2
– 25p + 20) ÷ (p – 1) (iv) 4yz(z2
+ 6z – 16) ÷ 2y(z + 8)
(v) 5pq(p2
– q2
) ÷ 2p(p + q)
(vi) 12xy(9x2
– 16y2
) ÷ 4xy(3x + 4y) (vii) 39y3
(50y2
– 98) ÷ 26y2
(5y + 7)
14.5 Can you Find the Error?
Task 1 Whilesolvinganequation,Saritadoesthefollowing.
3x + x + 5x = 72
Therefore 8x = 72
and so, x =
72
9
8
=
Where has she gone wrong? Find the correct answer.
Task 2 Appudidthefollowing:
For x = –3 , 5x = 5 – 3 = 2
Is his procedure correct? If not, correct it.
Task 3 Namrata and Salma have done the
multiplication of algebraic expressions in the
followingmanner.
Namrata Salma
(a) 3(x – 4) = 3x – 4 3(x – 4) = 3x – 12
Coefficient 1 of a
term is usually not
shown. But while
adding like terms,
we include it in
the sum.
Remember to make
use of brackets,
while substituting a
negative value.
Remember, when you multiply the
expression enclosed in a bracket by a
constant (or a variable) outside, each
term of the expression has to be
multiplied by the constant
(or the variable).
2021–22

228 MATHEMATICS
While dividing a
polynomial by a
monomial, we divide
each term of the
polynomial in the
numerator by the
monomial in the
denominator.
(b) (2x)2
= 2x2
(2x)2
= 4x2
(c) (2a – 3) (a + 2) (2a – 3) (a + 2)
= 2a2
– 6 = 2a2
+ a – 6
(d) (x + 8)2
= x2
+ 64 (x + 8)2
= x2
+ 16x + 64
(e) (x – 5)2
= x2
– 25 (x – 5)2
= x2
– 10x + 25
IsthemultiplicationdonebybothNamrataandSalmacorrect?Givereasonsforyour
answer.
Task 4 Joseph does a division as :
5
1
5
a
a
+
= +
HisfriendSirishhasdonethesamedivisionas:
5
5
a
a
+
=
And his other friend Suman does it this way:
5
1
5 5
a a
+
= +
Who has done the division correctly? Who has done incorrectly? Why?
Some fun!
Atul always thinks differently. He asks Sumathi teacher, “If what you say is true, then
why do I get the right answer for
64 4
4?’’
16 1
= = The teacher explains, “ This is so
because 64 happens to be 16 × 4;
64 16 4 4
16 16 1 1
×
= =
× . In reality, we cancel a factor of 16
and not 6, as you can see. In fact, 6 is not a factor of either 64 or of 16.” The teacher
adds further, “Also,
664 4 6664 4
,
166 1 1666 1
= = , and so on”. Isn’t that interesting? Can you
helpAtultofindsomeotherexampleslike
64
16
?
EXERCISE 14.4
Find andcorrecttheerrorsinthefollowingmathematicalstatements.
1. 4(x – 5) = 4x – 5 2. x(3x + 2) = 3x2
+ 2 3. 2x + 3y = 5xy
4. x + 2x + 3x = 5x 5. 5y + 2y + y – 7y = 0 6. 3x + 2x = 5x2
7. (2x)2
+ 4(2x) + 7 = 2x2
+ 8x + 7 8. (2x)2
+ 5x = 4x + 5x = 9x
9. (3x + 2)2
= 3x2
+ 6x + 4
Remember, when you
square a monomial, the
numerical coefficient and
each factor has to be
squared.
Make sure,
before
applying any
formula,
whether the
formula is
really
applicable.
2021–22

FACTORISATION 229
10. Substituting x = – 3 in
(a) x2
+ 5x + 4 gives (– 3)2
+ 5 (– 3) + 4 = 9 + 2 + 4 = 15
(b) x2
– 5x + 4 gives (– 3)2
– 5 ( – 3) + 4 = 9 – 15 + 4 = – 2
(c) x2
+ 5x gives (– 3)2
+ 5 (–3) = – 9 – 15 = – 24
11. (y – 3)2
= y2
– 9 12. (z + 5)2
= z2
+ 25
13. (2a + 3b) (a – b) = 2a2
– 3b2
14. (a + 4) (a + 2) = a2
+ 8
15. (a – 4) (a – 2) = a2
– 8 16.
2
2
3
0
3
x
x
=
17.
2
2
3 1
1 1 2
3
x
x
+
= + = 18.
3 1
3 2 2
x
x
=
+ 19.
3 1
4 3 4
x x
=
+
20.
4 5
5
4
x
x
+
= 21.
7 5
7
5
x
x
+
=
1. When we factorise an expression, we write it as a product of factors. These factors may be
numbers,algebraicvariablesoralgebraicexpressions.
2. An irreducible factor is a factor which cannot be expressed further as a product of factors.
3. A systematic way of factorising an expression is the common factor method. It consists of three
steps: (i) Write each term of the expression as a product of irreducible factors (ii) Look for and
separate the common factors and (iii) Combine the remaining factors in each term in accordance
withthedistributivelaw.
4. Sometimes,allthetermsinagivenexpressiondonothaveacommonfactor;butthetermscanbe
grouped in such a way that all the terms in each group have a common factor.When we do this,
there emerges a common factor across all the groups leading to the required factorisation of the
expression.Thisisthemethodofregrouping.
5. In factorisation by regrouping, we should remember that any regrouping (i.e., rearrangement) of
the terms in the given expression may not lead to factorisation. We must observe the expression
and come out with the desired regrouping by trial and error.
6. Anumberofexpressionstobefactorisedareoftheformorcanbeputintotheform:a2
+2ab+b2
,
a2
– 2ab + b2
, a2
– b2
and x2
+ (a + b) + ab. These expressions can be easily factorised using
Identities I, II, III and IV, given in Chapter 9,
a2
+ 2 ab + b2
= (a + b)2
a2
– 2ab + b2
= (a – b)2
a2
– b2
= (a + b) (a – b)
x2
+ (a + b) x + ab = (x + a) (x + b)
7. Inexpressionswhichhavefactorsofthetype(x+a)(x+b),rememberthenumericaltermgivesab.Its
factors,aandb,shouldbesochosenthattheirsum,withsignstakencareof,isthecoefficientofx.
8. Weknowthatinthecaseofnumbers,divisionistheinverseofmultiplication.Thisideaisapplicable
alsotothedivisionofalgebraicexpressions.
2021–22

230 MATHEMATICS
9. In the case of division of a polynomial by a monomial, we may carry out the division either by
dividingeachtermofthepolynomialbythemonomialorbythecommonfactormethod.
10. Inthecaseofdivisionofapolynomialbyapolynomial,wecannotproceedbydividingeachterm
inthedividendpolynomialbythedivisorpolynomial.Instead,wefactoriseboththepolynomials
andcanceltheircommonfactors.
11. In the case of divisions of algebraic expressions that we studied in this chapter, we have
Dividend=Divisor×Quotient.
Ingeneral,however,therelationis
Dividend=Divisor×Quotient+Remainder
Thus, we have considered in the present chapter only those divisions in which the remainder
is zero.
12. Therearemanyerrorsstudentscommonlymakewhensolvingalgebraexercises.Youshouldavoid
makingsucherrors.
2021–22

INTRODUCTION TO GRAPHS 231
15.1 Introduction
Haveyouseengraphsinthenewspapers,television,magazines,booksetc.?Thepurpose
of the graph is to show numerical facts in visual form so that they can be understood
quickly, easily and clearly.Thus graphs are visual representations of data collected. Data
can also be presented in the form of a table; however a graphical presentation is easier to
understand. This is true in particular when there is a trend or comparison to be shown.
We have already seen some types of graphs. Let us quickly recall them here.
15.1.1 A Bar graph
Abargraphisusedtoshowcomparisonamongcategories.Itmayconsistoftwoormore
parallelvertical(orhorizontal)bars(rectangles).
The bar graph in Fig 15.1 shows Anu’s mathematics marks in the three terminal
examinations.Ithelpsyoutocompareherperformanceeasily.Shehasshowngoodprogress.
Fig 15.1
Bar graphs can also have double bars as in Fig 15.2. This graph gives a comparative
account of sales (in `) of various fruits over a two-day period. How is Fig 15.2 different
fromFig15.1?Discusswithyourfriends.
Introduction to Graphs
CHAPTER
15
2021–22

232 MATHEMATICS
Fig 15.2
15.1.2 A Pie graph (or a circle-graph)
Apie-graphisusedtocomparepartsofawhole.Thecirclerepresentsthewhole.Fig15.3
isapie-graph.ItshowsthepercentageofviewerswatchingdifferenttypesofTVchannels.
15.1.3 A histogram
A Histogram is a bar graph that shows data in intervals. It has adjacent bars over
the intervals.
Fig 15.3
2021–22

ThehistograminFig15.4illustratesthedistributionofweights(inkg)of40personsof
alocality.
Weights (kg) 40-45 45-50 50-55 55-60 60-65
No. of persons 4 12 13 6 5
Fig 15.4
There are no gaps between bars, because there are no gaps between the intervals.
Whatistheinformationthatyougatherfromthishistogram?Trytolistthemout.
15.1.4 A line graph
A linegraphdisplays data that changes continuously over periods of time.
When Renu fell sick, her doctor maintained a record of her body temperature, taken
every four hours. It was in the form of a graph (shown in Fig 15.5 and Fig 15.6).
Wemaycallthisa“time-temperaturegraph”.
Itisapictorialrepresentationofthefollowingdata,givenintabularform.
Time 6 a.m. 10 a.m. 2 p.m. 6 p.m.
Temperature(°C) 37 40 38 35
Thehorizontalline(usuallycalledthex-axis)showsthetimingsatwhichthetemperatures
were recorded. What are labelled on the vertical line (usually called they-axis)?
In Fig 15.4 a jagged line
( ) has been used along
horizontal line to indicate
that we are not showing
numbers between 0 and 40.
2021–22

234 MATHEMATICS
Fig 15.5 Fig 15.6
Each piece of data is shown The points are then connected by line
by a point on the square grid. segments. The result is the line graph.
Whatalldoesthisgraphtellyou?Forexampleyoucanseethepatternoftemperature;
more at 10 a.m. (see Fig 15.5) and then decreasing till 6 p.m. Notice that the temperature
increased by 3° C(= 40° C – 37° C) during the period 6 a.m. to 10 a.m.
There was no recording of temperature at 8 a.m., however the graph suggests that it
was more than 37 °C (How?).
Example 1: (A graph on “performance”)
The given graph (Fig 15.7) represents the total runs scored by two batsmenAand B,
during each of the ten different matches in the year 2007. Study the graph and answer the
followingquestions.
(i) Whatinformationisgivenonthetwoaxes?
(ii) Which line shows the runs scored by batsmanA?
(iii) Were the run scored by them same in any match in 2007? If so, in which match?
(iii) Among the two batsmen, who is steadier? How do you judge it?
Solution:
(i) The horizontal axis (or the x-axis) indicates the matches played during the year
2007. The vertical axis (or the y-axis) shows the total runs scored in each match.
(ii) The dotted line shows the runs scored by BatsmanA. (This is already indicated at
the top of the graph).
2021–22

(iii) Duringthe4thmatch,bothhavescoredthesame
numberof60runs.(Thisisindicatedbythepoint
at which both graphs meet).
(iv) BatsmanAhas one great “peak” but many deep
“valleys”. He does not appear to be consistent.
B, on the other hand has never scored below a
totalof40runs,eventhoughhishighestscoreis
only 100 in comparison to 115 ofA.AlsoAhas
scored a zero in two matches and in a total of 5
matcheshehasscoredlessthan40runs.SinceA
hasalotofupsanddowns,Bisamoreconsistent
andreliablebatsman.
Example 2: The given graph (Fig 15.8) describes
the distances of a car from a city P at different times
when it is travelling from City P to City Q, which are
350kmapart.Studythegraphandanswerthefollowing:
(i) Whatinformationisgivenonthetwoaxes?
(ii) From where and when did the car begin its
journey?
(iii) How far did the car go in the first hour?
(iv) How far did the car go during (i) the 2nd hour? (ii) the 3rd hour?
(v) Was the speed same during the first three hours? How do you know it?
(vi) Did the car stop for some duration at any place? Justify your answer.
(vii) When did the car reach City Q?
Fig 15.7
Fig 15.8
2021–22

236 MATHEMATICS
Solution:
(i) Thehorizontal(x)axisshowsthetime.Thevertical(y)axisshowsthedistanceofthe
car from City P.
(ii) The car started from City P at 8 a.m.
(iii) The car travelled 50 km during the first hour. [This can be seen as follows.
At8a.m.itjuststartedfromCityP.At9a.m.itwasatthe50thkm(seenfromgraph).
Henceduringtheone-hourtimebetween8a.m.and9a.m.thecartravelled50km].
(iv) The distance covered by the car during
(a) the 2nd hour (i.e., from 9 am to 10 am) is 100 km, (150 – 50).
(b) the 3rd hour (i.e., from 10 am to 11 am) is 50 km (200 – 150).
(v) Fromtheanswerstoquestions(iii)and(iv),wefindthatthespeedofthecarwasnot
the same all the time. (In fact the graph illustrates how the speed varied).
(vi) We find that the car was 200 km away from city Pwhen the time was 11 a.m. and
also at 12 noon. This shows that the car did not travel during the interval 11 a.m. to
12 noon. The horizontal line segment representing “travel” during this period is
illustrativeofthisfact.
(vii) The car reached City Q at 2 p.m.
EXERCISE 15.1
1. The following graph shows the temperature of a patient in a hospital, recorded
every hour.
(a) What was the patient’s temperature at 1 p.m. ?
(b) When was the patient’s temperature 38.5° C?
2021–22

(c) The patient’s temperature was the same two times during the period given.
What were these two times?
(d) What was the temperature at 1.30 p.m.? How did you arrive at your answer?
(e) During which periods did the patients’ temperature showed an upward trend?
2. Thefollowinglinegraphshowstheyearlysalesfiguresforamanufacturingcompany.
(a) What were the sales in (i) 2002 (ii) 2006?
(b) What were the sales in (i) 2003 (ii) 2005?
(c) Compute the difference between the sales in 2002 and 2006.
(d) In which year was there the greatest difference between the sales as compared
to its previous year?
3. For an experiment in Botany, two different plants, plantAand plant B were grown
undersimilarlaboratoryconditions.Theirheightsweremeasuredattheendofeach
week for 3 weeks. The results are shown by the following graph.
2021–22

238 MATHEMATICS
(a) How high was PlantAafter (i) 2 weeks (ii) 3 weeks?
(b) How high was Plant B after (i) 2 weeks (ii) 3 weeks?
(c) How much did PlantAgrow during the 3rd week?
(d) How much did Plant B grow from the end of the 2nd week to the end of the
3rd week?
(e) DuringwhichweekdidPlantAgrowmost?
(f) DuringwhichweekdidPlantBgrowleast?
(g) Were the two plants of the same height during any week shown here? Specify.
4. Thefollowinggraphshowsthetemperatureforecastandtheactualtemperaturefor
each day of a week.
(a) Onwhichdayswastheforecasttemperaturethesameastheactualtemperature?
(b) Whatwasthemaximumforecasttemperatureduringtheweek?
(c) Whatwastheminimumactualtemperatureduringtheweek?
(d) On which day did the actual temperature differ the most from the forecast
temperature?
5. Use the tables below to draw linear graphs.
(a) Thenumberofdaysahillsidecityreceivedsnowindifferentyears.
Year 2003 2004 2005 2006
Days 8 10 5 12
(b) Population(inthousands)ofmenandwomeninavillageindifferentyears.
Year 2003 2004 2005 2006 2007
Number of Men 12 12.5 13 13.2 13.5
Number of Women 11.3 11.9 13 13.6 12.8
2021–22

6. A courier-person cycles from a town to a neighbouring suburban area to deliver a
parcel to a merchant. His distance from the town at different times is shown by the
followinggraph.
(a) What is the scale taken for the time axis?
(b) How much time did the person take for the travel?
(c) How far is the place of the merchant from the town?
(d) Did the person stop on his way? Explain.
(e) Duringwhichperioddidheridefastest?
7. Can there be a time-temperature graph as follows? Justify your answer.
(i) (ii)
(iii) (iv)
2021–22

240 MATHEMATICS
15.2 Linear Graphs
A line graph consists of bits of line segments
joinedconsecutively.Sometimesthegraphmay
beawholeunbrokenline.Suchagraphiscalled
a linear graph.To draw such a line we need to
locate some points on the graph sheet. We will
now learn how to locate points conveniently on
a graph sheet.
15.2.1 Location of a point
The teacher put a dot on the black-board. She asked the
students how they would describe its location. There were
several responses (Fig 15. 9).
The dot is
very close to
the left
upper corner
of board
The dot is
near the left
edge of the
board
Fig 15.9
The dot is
in the upper
half of the
board
Can any one of these statements help fix the position of the dot? No! Why not?
Thinkaboutit.
Johnthengaveasuggestion.Hemeasuredthedistanceofthedotfromtheleftedgeof
the board and said, “The dot is 90 cm from the left edge of the board”. Do you think
John’ssuggestionisreallyhelpful?(Fig15.10)
Fig 15.10 Fig 15.11
A, A1
, A2
, A3
are all 90 cm away A is 90 cm from left edge and
from the left edge. 160 cm from the bottom edge.
2021–22

Rekha then came up with a modified statement : “The dot is 90 cm from the left edge
and160cmfromthebottomedge”.Thatsolvedtheproblemcompletely!(Fig15.11)The
teachersaid,“Wedescribethepositionofthisdotbywritingitas(90,160)”.Willthepoint
(160, 90) be different from (90, 160)? Think about it.
The 17th century mathematician Rene Descartes, it is said, noticed the movement
of an insect near a corner of the ceiling and began to think of determining the
position of a given point in a plane. His system of fixing a point with the help of
two measurements, vertical and horizontal, came to be known as Cartesian system,
in his honour.
15.2.2 Coordinates
Suppose you go to an auditorium and
search for your reserved seat.You need to
know two numbers, the row number and
the seat number. This is the basic method
for fixing a point in a plane.
Observe in Fig 15.12 how the point
(3, 4) which is 3 units from left edge and 4
unitsfrombottomedgeisplottedonagraph
sheet.Thegraphsheetitselfisasquaregrid.
We draw the xand yaxes conveniently and
then fix the required point. 3 is called the
x-coordinate of the point; 4 is the
y-coordinate of the point.We say that the
coordinates of the point are (3, 4).
Example 3:Plotthepoint(4,3)onagraph
sheet. Is it the same as the point (3, 4)?
Solution: Locate the x, y axes, (they are
actually number lines!). Start at O (0, 0).
Move4unitstotheright;thenmove3units
up, you reach the point (4, 3). From
Fig 15.13, you can see that the points (3, 4)
and (4, 3) are two different points.
Example 4: From Fig 15.14, choose the
letter(s)thatindicatethelocationofthepoints
givenbelow:
(i) (2, 1)
(ii) (0, 5)
(iii) (2, 0)
Alsowrite
(iv) The coordinates ofA.
(v) The coordinates of F. Fig 15.14
Fig 15.13
Fig 15.12
Rene Descartes
(1596-1650)
2021–22

242 MATHEMATICS
Solution:
(i) (2, 1) is the point E (It is not D!).
(ii) (0, 5) is the point B (why? Discuss with your friends!). (iii) (2, 0) is the point G.
(iv) PointAis (4, 5) (v) F is (5.5, 0)
Example 5: Plotthefollowingpointsandverifyiftheylieonaline.Iftheylieonaline,
nameit.
(i) (0, 2), (0, 5), (0, 6), (0, 3.5) (ii) A (1, 1), B (1, 2), C (1, 3), D (1, 4)
(iii) K (1, 3), L (2, 3), M (3, 3), N (4, 3) (iv) W (2, 6), X (3, 5),Y(5, 3), Z (6, 2)
Solution:
(i) (ii)
These lie on a line. These lie on a line. The line is AD.
The line is y-axis. (You may also use other ways
of naming it). It is parallel to the y-axis
(iii) (iv)
These lie on a line. We can name it as KL These lie on a line. We can name
or KM or MN etc. It is parallel to x-axis it as XY or WY or YZ etc.
Note that in each of the above cases, graph obtained by joining the plotted points is a
line. Such graphs are called linear graphs.
Fig 15.15
2021–22

EXERCISE 15.2
1. Plotthefollowingpointsonagraphsheet.Verifyiftheylieonaline
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5)
2. Drawthelinepassingthrough(2,3)and(3,2).Findthecoordinatesofthepointsat
which this line meets thex-axis andy-axis.
3. Write the coordinates of the vertices
ofeachoftheseadjoiningfigures.
4. State whether True or False. Correct
that are false.
(i) Apointwhosexcoordinateiszero
and y-coordinateisnon-zerowill
lie on the y-axis.
(ii) Apointwhoseycoordinateiszero
and x-coordinate is 5 will lie on
y-axis.
(iii) The coordinates of the origin
are (0, 0).
15.3 Some Applications
In everyday life, you might have observed that the more you use a facility, the more you
pay for it. If more electricity is consumed, the bill is bound to be high. If less electricity is
used,thenthebillwillbeeasilymanageable.Thisisaninstancewhereonequantityaffects
another.Amountofelectricbilldependsonthequantityofelectricityused.Wesaythatthe
quantity of electricity is an independent variable (or sometimes control variable) and
theamountofelectricbillisthedependentvariable.Therelationbetweensuchvariables
can be shown through a graph.
Thenumberoflitresofpetrolyoubuytofillacar’spetroltankwilldecidetheamount
you have to pay.Which is the independent variable here?Think about it.
Example 6: (Quantity and Cost)
Thefollowingtablegivesthequantityofpetrolanditscost.
No. of Litres of petrol 10 15 20 25
Cost of petrol in ` 500 750 1000 1250
Plot a graph to show the data.
2021–22

244 MATHEMATICS
TRY THESE
Solution: (i) Let us take a suitable scale on both the axes (Fig 15.16).
Fig 15.16
(ii) Marknumberoflitresalongthehorizontalaxis.
(iii) Mark cost of petrol along the vertical axis.
(iv) Plot the points: (10,500), (15,750), (20,1000), (25,1250).
(v) Jointhepoints.
Wefindthatthegraphisaline.(Itisalineargraph).Whydoesthisgraphpassthrough
theorigin?Thinkaboutit.
This graph can help us to estimate a few things. Suppose we want to find the amount
needed to buy 12 litres of petrol. Locate 12 on the horizontal axis.
Followtheverticallinethrough12tillyoumeetthegraphatP(say).
FromPyoutakeahorizontallinetomeettheverticalaxis.Thismeetingpointprovides
the answer.
Thisisthegraphofasituationinwhichtwoquantities,areindirectvariation.(How?).
Insuchsituations,thegraphswillalwaysbelinear.
In the above example, use the graph to find how much petrol can be purchased
for ` 800.
Cost
(in
`
`
`
`
`
)
2021–22

Example 7: (Principal and Simple Interest)
A bank gives 10% Simple Interest (S.I.) on deposits by senior citizens. Draw a graph to
illustrate the relation between the sum deposited and simple interest earned. Find from
yourgraph
(a) the annual interest obtainable for an investment of ` 250.
(b) the investment one has to make to get an annual simple interest of ` 70.
Solution:
Sum deposited Simple interest for a year
` 100 `
100 1 10
100
× ×
= ` 10
` 200 `
200 1 10
100
× ×
= ` 20
` 300 `
300 1 10
100
× ×
= ` 30
` 500 `
500 1 10
100
× ×
= ` 50
` 1000 ` 100
We get a table of values.
Deposit (in `) 100 200 300 500 1000
Annual S.I. (in `) 10 20 30 50 100
(i) Scale : 1 unit = ` 100 on horizontal axis; 1 unit = ` 10 on vertical axis.
(ii) MarkDepositsalonghorizontalaxis.
(iii) MarkSimpleInterestalongverticalaxis.
(iv) Plot the points : (100,10), (200, 20), (300, 30), (500,50) etc.
(v) Join the points. We get a graph that is a line (Fig 15.17).
(a) Corresponding to ` 250 on horizontal axis, we
get the interest to be ` 25 on vertical axis.
(b) Corresponding to ` 70 on the vertical axis,
we get the sum to be ` 700 on the horizontal
axis.
Steps to follow:
1. Find the quantities to be
plotted as Deposit and SI.
2. Decide the quantities to be
taken on x-axis and on
y-axis.
3. Choose a scale.
4. Plot points.
5. Join the points.
TRY THESE
Is Example 7, a case of direct variation?
2021–22

246 MATHEMATICS
Fig 15.17
Example 8: (Time and Distance)
Ajit can ride a scooter constantly at a speed of 30 kms/hour. Draw a time-distance graph
forthissituation.Useittofind
(i) the time taken byAjit to ride 75 km. (ii) thedistancecoveredbyAjitin3
1
2
hours.
Solution:
Hours of ride Distance covered
1 hour 30 km
2 hours 2 × 30 km = 60 km
3 hours 3 × 30 km = 90 km
4 hours 4 × 30 km = 120 km and so on.
We get a table of values.
Time (in hours) 1 2 3 4
Distance covered (in km) 30 60 90 120
(i) Scale: (Fig 15.18)
Horizontal:2units=1hour
Vertical: 1 unit=10km
(ii) Marktimeonhorizontalaxis.
(iii) Mark distance on vertical axis.
(iv) Plot the points: (1, 30), (2, 60), (3, 90), (4, 120).
2021–22

(v) Join the points. We get a linear graph.
(a) Correspondingto75kmontheverticalaxis,wegetthetimetobe2.5hourson
the horizontal axis. Thus 2.5 hours are needed to cover 75 km.
(b) Corresponding to 3
1
2
hours on the horizontal axis, the distance covered is
105 km on the vertical axis.
EXERCISE 15.3
1. Drawthegraphsforthefollowingtablesofvalues,withsuitablescalesontheaxes.
(a) Cost of apples
Number of apples 1 2 3 4 5
Cost (in `) 5 10 15 20 25
(b) Distance travelled by a car
Time (in hours) 6 a.m. 7 a.m. 8 a.m. 9 a.m.
Distances (in km) 40 80 120 160
Fig 15.18
2021–22

248 MATHEMATICS
(i) Howmuchdistancedidthecarcoverduringtheperiod7.30a.m.to8a.m?
(ii) What was the time when the car had covered a distance of 100 km since
it’sstart?
(c) Interest on deposits for a year.
Deposit (in `) 1000 2000 3000 4000 5000
Simple Interest (in `) 80 160 240 320 400
(i) Does the graph pass through the origin?
(ii) Use the graph to find the interest on ` 2500 for a year.
(iii) Togetaninterestof`280peryear,howmuchmoneyshouldbedeposited?
2. Drawagraphforthefollowing.
(i) Side of square (in cm) 2 3 3.5 5 6
Perimeter (in cm) 8 12 14 20 24
Is it a linear graph?
(ii) Side of square (in cm) 2 3 4 5 6
Area (in cm2
) 4 9 16 25 36
Is it a linear graph?
1. Graphical presentation of data is easier to understand.
2. (i) A bar graph is used to show comparison among categories.
(ii) A pie graph is used to compare parts of a whole.
(iii) A Histogram is a bar graph that shows data in intervals.
3. A linegraph displays data that changes continuously over periods of time.
4. Alinegraphwhichisawholeunbrokenlineiscalledalineargraph.
5. For fixing a point on the graph sheet we need, x-coordinate and y-coordinate.
6. Therelationbetweendependentvariable andindependentvariableisshownthroughagraph.
2021–22

PLAYING WITH NUMBERS 249
16.1 Introduction
You have studied various types of numbers such as natural numbers, whole numbers,
integers and rational numbers.You have also studied a number of interesting properties
aboutthem.InClassVI,weexploredfindingfactorsandmultiplesandtherelationships
amongthem.
Inthischapter,wewillexplorenumbersinmoredetail.Theseideashelpinjustifying
testsofdivisibility.
16.2 Numbers in General Form
Let us take the number 52 and write it as
52 = 50 + 2 = 10 × 5 + 2
Similarly,thenumber37canbewrittenas
37 = 10 × 3 + 7
In general, any two digit number ab made of digits a and b can be written as
ab = 10 × a + b = 10a + b
What about ba? ba = 10 × b + a = 10b + a
Let us now take number 351.This is a three digit number. It can also be written as
351 = 300 + 50 + 1 = 100 × 3 + 10 × 5 + 1 × 1
Similarly 497 = 100 × 4 + 10 × 9 + 1 × 7
In general, a 3-digit number abc made up of digits a, b and c is written as
abc = 100 × a + 10 × b + 1 × c
= 100a + 10b + c
In the same way,
cab = 100c + 10a + b
bca = 100b + 10c + a and so on.
Playing with Numbers
CHAPTER
16
Here ab does not
mean a × b!
2021–22

250 MATHEMATICS
TRY THESE
1. Writethefollowingnumbersingeneralisedform.
(i) 25 (ii) 73 (iii) 129 (iv) 302
2. Writethefollowingintheusualform.
(i) 10 × 5 + 6 (ii) 100 × 7 + 10 × 1 + 8 (iii) 100 × a + 10 × c + b
16.3 Games with Numbers
(i) Reversing the digits – two digit number
Minakshi asks Sundaram to think of a 2-digit number, and then to do whatever she asks
himtodo,tothatnumber.Theirconversationisshowninthefollowingfigure.Studythe
figure carefully before reading on.
It so happens that Sundaram chose the number 49. So, he got the reversed number
94; then he added these two numbers and got 49 + 94 = 143. Finally he divided this
number by 11 and got 143 ÷ 11 = 13, with no remainder. This is just what Minakshi
had predicted.
2021–22

TRY THESE
TRY THESE
Check what the result would have been if Sundaram had chosen the numbers shown
below.
1. 27 2. 39 3. 64 4. 17
Now, let us see if we can explain Minakshi’s “trick”.
Suppose Sundaram chooses the number ab, which is a short form for the 2-digit
number10a+ b.On reversing the digits, he gets the numberba= 10b + a.When he adds
the two numbers he gets:
(10a + b) + (10b + a) = 11a + 11b
= 11 (a + b).
So, the sum is always a multiple of 11, just as Minakshi had claimed.
Observeherethatifwedividethesumby11,thequotientisa+b,whichisexactlythe
sum of the digits of chosen numberab.
You may check the same by taking any other two digit number.
ThegamebetweenMinakshiandSundaramcontinues!
Minakshi: Thinkofanother2-digitnumber,butdon’ttellmewhatitis.
Sundaram: Alright.
Minakshi: Nowreversethedigitsofthenumber,andsubtractthesmallernumberfrom
the larger one.
Sundaram: I have done the subtraction. What next?
Minakshi: Now divide your answer by 9. I claim that there will be no remainder!
Sundaram: Yes,youareright.Thereisindeednoremainder!ButthistimeIthinkIknow
how you are so sure of this!
In fact, Sundaram had thought of 29. So his calculations were: first he got
the number 92; then he got 92 – 29 = 63; and finally he did (63 ÷ 9) and got 7 as
quotient, with no remainder.
Check what the result would have been if Sundaram had chosen the numbers shown
below.
1. 17 2. 21 3. 96 4. 37
LetusseehowSundaramexplainsMinakshi’ssecond“trick”.(Nowhefeelsconfident
of doing so!)
Suppose he chooses the 2-digit number ab=10a +b. After reversing the digits, he
gets the number ba = 10b + a. Now Minakshi tells him to do a subtraction, the
smallernumberfromthelargerone.
• If the tens digit is larger than the ones digit (that is, a b), he does:
(10a + b) – (10b + a) = 10a + b – 10b – a
= 9a – 9b = 9(a – b).
2021–22

252 MATHEMATICS
TRY THESE
• If the ones digit is larger than the tens digit (that is, b a), he does:
(10b + a) – (10a + b) = 9(b – a).
• And, of course, if a = b, he gets 0.
Ineachcase,theresultingnumberisdivisibleby9.So,theremainderis0.Observe
herethatifwedividetheresultingnumber(obtainedbysubtraction),thequotientis
a – b or b – a according as a b or a b.You may check the same by taking any
othertwodigitnumbers.
(ii) Reversing the digits – three digit number.
Now it is Sundaram’s turn to play some tricks!
Sundaram: Think of a 3-digit number, but don’t tell me what it is.
Minakshi: Alright.
Sundaram: Nowmakeanewnumberbyputtingthedigitsinreverseorder,andsubtract
thesmallernumberfromthelargerone.
Minakshi: Alright, I have done the subtraction. What next?
Sundaram: Divide your answer by 99. I am sure that there will be no remainder!
In fact, Minakshi chose the 3-digit number 349. So she got:
• Reversednumber:943; • Difference: 943 – 349 = 594;
• Division: 594 ÷ 99 = 6, with no remainder.
Check what the result would have been if Minakshi had chosen the numbers shown
below. In each case keep a record of the quotient obtained at the end.
1. 132 2. 469 3. 737 4. 901
Let us see how this trick works.
Let the 3-digit number chosen by Minakshi be abc = 100a + 10b + c.
After reversing the order of the digits, she gets the numbercba=100c + 10b + a.On
subtraction:
• If a c, then the difference between the numbers is
(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a
= 99a – 99c = 99(a – c).
• If c a,then the difference between the numbers is
(100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).
• And, of course, if a = c, the difference is 0.
In each case, the resulting number is divisible by 99. So the remainder is 0. Observe
that quotient is a – c or c– a.You may check the same by taking other 3-digit numbers.
(iii) Forming three-digit numbers with given three-digits.
Now it is Minakshi’s turn once more.
2021–22

TRY THESE
Minakshi: Thinkofany3-digitnumber.
Sundaram: Alright, I have done so.
Minakshi: Now use this number to form two more 3-digit numbers, like this: if the
number you chose is abc, then
• ‘thefirstnumberiscab(i.e.,withtheonesdigitshiftedtothe“leftend”of
thenumber);
• theothernumberisbca (i.e.,withthehundredsdigitshiftedtothe“right
end”ofthenumber).
Nowaddthemup.Dividetheresultingnumberby37.Iclaimthattherewill
be no remainder.
Sundaram: Yes.Youareright!
Infact,Sundaramhadthoughtofthe3-digitnumber237. AfterdoingwhatMinakshihad
asked, he got the numbers 723 and 372. So he did:
2 3 7
+ 7 2 3
+ 3 7 2
1 3 3 2
Thenhedividedtheresultingnumber1332by37:
1332 ÷ 37 = 36, with no remainder.
Check what the result would have been if Sundaram had chosen the numbers
shownbelow.
1. 417 2. 632 3. 117 4. 937
Willthistrickalwayswork?
Let us see. abc = 100a + 10b + c
cab = 100c + 10a + b
bca = 100b + 10c + a
abc + cab + bca = 111(a + b + c)
= 37 × 3(a + b + c), which is divisible by 37
16.4 Letters for Digits
Here we have puzzles in which letters take the place of digits in an arithmetic ‘sum’, and
theproblemistofindoutwhichletterrepresentswhichdigit;soitislikecrackingacode.
Herewesticktoproblemsofadditionandmultiplication.
Form all possible 3-digit numbers using all the digits 2, 3 and
7 and find their sum. Check whether the sum is divisible by
37! Is it true for the sum of all the numbers formed by the
digits a, b and c of the number abc?
2021–22

254 MATHEMATICS
Herearetworuleswefollowwhiledoingsuchpuzzles.
1. Each letter in the puzzle must stand for just one digit. Each digit must be
represented by just one letter.
2. The first digit of a number cannot be zero. Thus, we write the number “sixty
three” as 63, and not as 063, or 0063.
Arule that we would like to follow is that the puzzle must have just one answer.
Example 1: Find Q in the addition.
3 1 Q
+ 1 Q 3
5 0 1
Solution:
There is just one letter Q whose value we have to find.
Studytheadditionintheonescolumn:fromQ+3,weget‘1’,thatis,anumberwhose
ones digit is 1.
For this to happen, the digit Q should be 8. So the puzzle can be solved as shown below.
3 1 8
+ 1 8 3
5 0 1
That is, Q = 8
Example 2: FindAand B in the addition.
A
+ A
+ A
B A
Solution: This has two letters A and B whose values are to be found.
Study the addition in the ones column: the sum ofthree A’s is a number whose ones digit
isA. Therefore, the sum of two A’s must be a number whose ones digit is 0.
This happens only forA = 0 andA = 5.
IfA = 0, then the sum is 0 + 0 + 0 = 0, which makes B = 0 too. We do not want this
(as it makes A = B, and then the tens digit of BA too becomes 0), so we reject this
possibility. So,A = 5.
Therefore, the puzzle is solved as shown below.
5
+ 5
+ 5
That is, A= 5 and B = 1. 1 5
2021–22

DO THIS
Example 3: Find the digitsAand B.
B A
× B 3
5 7 A
Solution:
This also has two letters A and B whose values are to be found.
Since the ones digit of 3 ×A isA, it must be thatA = 0 orA = 5.
Now look at B. If B = 1, then BA × B3 would at most be equal to 19 × 19; that is,
itwouldatmostbeequalto361.Buttheproducthereis57A, whichismorethan500.So
we cannot have B = 1.
If B = 3, then BA × B3 would be more than 30 × 30; that is, more than 900. But 57A
is less than 600. So, B can not be equal to 3.
Putting these two facts together, we see that B = 2 only. So the multiplication is either
20 × 23, or 25 × 23.
The first possibility fails, since 20 × 23 = 460. But, the
second one works out correctly, since 25 × 23 = 575.
So the answer isA = 5, B = 2.
Writea2-digitnumberabandthenumberobtainedbyreversingitsdigitsi.e.,ba.Find
their sum. Let the sum be a 3-digit number dad
i.e., ab + ba = dad
(10a + b) + (10b + a) = dad
11(a + b) = dad
The sum a + b can not exceed 18 (Why?).
Is dad a multiple of 11?
Is dad less than 198?
Writeallthe3-digitnumberswhicharemultiplesof11 upto 198.
Find the values of a and d.
EXERCISE 16.1
Findthevaluesofthelettersineachofthefollowingandgivereasonsforthestepsinvolved.
1. 2. 3.
3 A
+ 2 5
B 2
4 A
+ 9 8
C B 3
2 5
× 2 3
5 7 5
1 A
× A
9 A
2021–22

256 MATHEMATICS
4. 5. 6.
7. 8. 9.
10.
16.5 Tests of Divisibility
InClassVI,youlearnthowtocheckdivisibilitybythefollowingdivisors.
10, 5, 2, 3, 6, 4, 8, 9, 11.
You would have found the tests easy to do, but you may have wondered at the same
timewhytheywork.Now,inthischapter,weshallgointothe“why”aspectoftheabove.
16.5.1 Divisibility by 10
This is certainly the easiest test of all!We first look at some multiples of 10.
10, 20, 30, 40, 50, 60, ... ,
and then at some non-multiples of 10.
13, 27, 32, 48, 55, 69,
From these lists we see that if the ones digit of a number is 0, then the number is a
multipleof10;andiftheonesdigitisnot0,thenthenumberisnotamultipleof10.So,we
getatestofdivisibilityby10.
Of course, we must not stop with just stating the test; we must also explain why it
“works”. That is not hard to do; we only need to remember the rules of place value.
Take the number. ... cba; this is a short form for
... + 100c + 10b + a
Here a is the one’s digit, b is the ten’s digit, c is the hundred’s digit, and so on. The
dots are there to say that there may be more digits to the left of c.
Since 10, 100, ... are divisible by 10, so are 10b, 100c, ... .And as for the number a
is concerned, it must be a divisible by 10 if the given number is divisible by 10. This is
possible only when a = 0.
Hence, a number is divisible by 10 when its one’s digit is 0.
Look at the multiples of 5.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50,
A B
+ 3 7
6 A
A B
× 3
C A B
A B
× 6
B B B
A 1
+ 1 B
B 0
A B
× 5
C A B
2 A B
+ A B 1
B 1 8
1 2 A
+ 6 A B
A 0 9
2021–22

TRY THESE
TRY THESE
We see that the one’s digits are alternately 5 and 0, and no other digit ever
appears in this list.
So, we get our test of divisibility by 5.
If the ones digit of a number is 0 or 5, then it is divisible by 5.
Let us explain this rule.Any number ... cbacan be written as:
... + 100c + 10b + a
Since 10, 100 are divisible by 10 so are 10b, 100c, ... which in turn, are divisible
by 5 because 10 = 2 × 5.As far as numbera is concerned it must be divisible by 5 if the
number is divisible by 5. So a has to be either 0 or 5.
(The first one has been done for you.)
1. If the division N ÷ 5 leaves a remainder of 3, what might be the ones digit of N?
(Theone’sdigit,whendividedby5,mustleavearemainderof3.Sotheone’sdigit
must be either 3 or 8.)
2. If the division N ÷ 5 leaves a remainder of 1, what might be the one’s digit of N?
3. If the division N ÷ 5 leaves a remainder of 4, what might be the one’s digit of N?
Here are the even numbers.
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, ... ,
and here are the odd numbers.
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ... ,
We see that a natural number is even if its one’s digit is
2, 4, 6, 8 or 0
Anumberisoddifitsone’s digit is
1, 3, 5, 7 or 9
Recallthetestofdivisibilityby2learntinClassVI,whichisasfollows.
If the one’s digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.
Theexplanationforthisisasfollows.
Any number cba can be written as 100c + 10b + a
Firsttwotermsnamely100c,10bare divisible by 2 because 100 and 10 are divisible
by2.Sofarasaisconcerned,itmustbedivisibleby2ifthegivennumberisdivisibleby
2. This is possible only when a = 0, 2, 4, 6 or 8.
(The first one has been done for you.)
1. IfthedivisionN÷ 2leavesaremainderof1,whatmightbetheone’sdigitof N?
(Nisodd;soitsone’sdigitisodd.Therefore,theone’sdigitmustbe1,3,5,7or9.)
2021–22

258 MATHEMATICS
2. IfthedivisionN÷2leavesnoremainder(i.e.,zeroremainder),whatmightbethe
one’s digit of N?
3. Suppose that the division N ÷ 5 leaves a remainder of 4, and the division N ÷ 2
leaves a remainder of 1. What must be the one’s digit of N?
16.5.4 Divisibility by 9 and 3
Look carefully at the three tests of divisibility found till now, for checking division by
10, 5 and 2. We see something common to them: they use only the one’s digit of the
given number; they do not bother about the ‘rest’of the digits. Thus, divisibility
is decided just by the one’s digit. 10, 5, 2 are divisors of 10, which is the key
number in our place value.
But for checking divisibility by 9, this will not work. Let us take some number say 3573.
Its expanded form is: 3 × 1000 + 5 × 100 + 7 × 10 + 3
This is equal to 3 × (999 + 1) + 5 × (99 + 1) + 7 × (9 + 1) + 3
= 3 × 999 + 5 × 99 + 7 × 9 + (3 + 5 + 7 + 3) ... (1)
We see that the number 3573 will be divisible by 9 or 3 if (3 + 5 + 7 + 3) is divisible
by 9 or 3.
We see that 3 + 5 + 7 + 3 = 18 is divisible by 9 and also by 3. Therefore, the number
3573isdivisiblebyboth9and3.
Now, let us consider the number 3576.As above, we get
3576 = 3 × 999 + 5 × 99 + 7 × 9 + (3 + 5 + 7 + 6) ... (2)
Since (3 + 5 + 7 + 6) i.e., 21 is not divisible by 9 but is divisible by 3,
therefore 3576 is not divisible by 9. However 3576 is divisible by 3. Hence,
(i) AnumberNisdivisibleby9ifthesumofitsdigitsisdivisibleby9.Otherwiseitis
notdivisibleby9.
(ii) AnumberNisdivisibleby3ifthesumofitsdigitsisdivisibleby3.Otherwiseitis
notdivisibleby3.
If the number is ‘cba’, then, 100c + 10b + a = 99c + 9b + (a + b + c)
=
divisible by 3 and 9
9(11 ) ( )
c b a b c
+ + + +

Hence, divisibility by 9 (or 3) is possible if a + b + c is divisible by 9 (or 3).
Example 4: Check the divisibility of 21436587 by 9.
Solution: The sum of the digits of 21436587 is 2 + 1 + 4 + 3 + 6 + 5 + 8 + 7 = 36.
Thisnumberisdivisibleby9(for36÷9=4).Weconcludethat21436587isdivisibleby9.
We can double-check:
21436587
9
= 2381843 (thedivisionisexact).
2021–22

TRY THESE
Solution: The sum of the digits of 152875 is 1 + 5 + 2 + 8 + 7 + 5 = 28. This number
is not divisible by 9. We conclude that 152875 is not divisible by 9.
Checkthedivisibilityofthefollowingnumbersby9.
1. 108 2. 616 3. 294 4. 432 5. 927
Example 6: If the three digit number 24x is divisible by 9, what is the value of x?
Solution: Since 24x is divisible by 9, sum of it’s digits, i.e., 2 + 4 + x should be
divisible by 9, i.e., 6 + x should be divisible by 9.
This is possible when 6 + x = 9 or 18, ....
But, since x is a digit, therefore, 6 + x = 9, i.e., x = 3.
1. Youhaveseenthatanumber450isdivisibleby10.Itisalsodivisibleby2and5
whicharefactorsof10.Similarly,anumber135isdivisible9.Itisalsodivisible
by 3 which is a factor of 9.
Can you say that if a number is divisible by any number m, then it will also be
divisible by each of the factors of m?
2. (i) Write a 3-digit number abc as 100a + 10b + c
= 99a + 11b + (a – b + c)
= 11(9a + b) + (a – b + c)
If the number abc is divisible by 11, then what can you say about
(a – b + c)?
Is it necessary that (a + c – b) should be divisible by 11?
(ii) Write a 4-digit number abcd as 1000a + 100b + 10c + d
= (1001a + 99b + 11c) – (a – b + c – d)
= 11(91a + 9b + c) + [(b + d) – (a + c)]
If the number abcd is divisible by 11, then what can you say about
[(b + d) – (a + c)]?
(iii) From (i) and (ii) above, can you say that a number will be divisible by 11if
thedifferencebetweenthesumofdigitsatitsoddplacesandthatofdigitsat
the even places is divisible by 11?
Solution: The sum of the digits of 2146587 is 2 + 1 + 4 + 6 + 5 + 8 + 7 = 33. This
number is divisible by 3 (for 33 ÷ 3 = 11).We conclude that 2146587 is divisible by 3.
2021–22

260 MATHEMATICS
TRY THESE
Solution: The sum of the digits of 15287 is 1 + 5 + 2 + 8 + 7 = 23. This number is not
divisible by 3.We conclude that 15287 too is not divisible by 3.
Checkthedivisibilityofthefollowingnumbersby3.
1. 108 2. 616 3. 294 4. 432 5. 927
EXERCISE 16.2
1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z?
You will find that there are two answers for the last problem. Why is this so?
3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one
of these numbers: 0, 3, 6, 9, 12, 15, 18, ... . But since x is a digit, it can only be that
6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of
fourdifferentvalues.)
4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
1. Numbers can be written in general form. Thus, a two digit number ab will be written as
ab = 10a + b.
2. Thegeneralformofnumbersarehelpfulinsolvingpuzzlesornumbergames.
3. The reasons for the divisibility of numbers by 10, 5, 2, 9 or 3 can be given when numbers are
writteningeneralform.
2021–22

ANSWERS 261
EXERCISE 1.1
1. (i) 2 (ii)
11
28
−
2. (i)
2
8
−
(ii)
5
9
(iii)
6
5
−
(iv)
2
9
(v)
19
6
4. (i)
1
13
−
(ii)
19
13
−
(iii) 5 (iv)
56
15
(v)
5
2
(vi) –1
5. (i) 1isthemultiplicativeidentity (ii) Commutativity
(iii) Multiplicativeinverse
6.
96
91
−
7. Associativity 8. No, because the product is not 1.
9. Yes, because 0.3 ×
1
3
3
=
3 10
1
10 3
× =
10. (i) 0 (ii) 1 and (–1) (iii) 0
11. (i) No (ii) 1, –1 (iii)
1
5
−
(iv) x (v) Rationalnumber
(vi) positive
EXERCISE 1.2
1. (i) (ii)
2.
3. Some of these are 1,
1
2
, 0, –1,
1
2
−
4.
7 6 5 4 3 2 1 1 2
, , , , , , ,0, ... , ,
20 20 20 20 20 20 20 20 20
− − − − − − −
(There can be many more such rational numbers)
5. (i)
41 42 43 44 45
, , , ,
60 60 60 60 60
(ii)
8 7 1 2
, ,0, ,
6 6 6 6
− −
(iii)
9 10 11 12 13
, , , ,
32 32 32 32 32
(Therecanbemanymoresuchrationalnumbers)
ANSWERS
2021–22

262 MATHEMATICS
6.
3 1 1
, 1, , 0,
2 2 2
−
− − (There can be many more such rational numbers)
7.
97 98 99 100 101 102 103 104 105 106
, , , , , , , , ,
160 160 160 160 160 160 160 160 160 160
(Therecanbemanymoresuchrationalnumbers)
EXERCISE 2.1
1. x = 9 2. y = 7 3. z = 4 4. x = 2 5. x = 2 6. t = 50
7. x = 27 8. y = 2.4 9. x =
25
7
10. y =
3
2
11. p = –
4
3
12. x = –
8
5
EXERCISE 2.2
1.
3
4
2. length = 52 m, breadth = 25 m 3.
2
1
5
cm 4. 40 and 55
5. 45, 27 6. 16, 17, 18 7. 288, 296 and 304 8. 7, 8, 9
9. Rahul’s age: 20 years; Haroon’s age: 28 years 10. 48 students
11. Baichung’sage:17years;Baichung’s father’sage:46years;
Baichung’s grandfather’s age = 72 years 12. 5 years 13
1
2
−
14. ` 100 → 2000 notes; ` 50 → 3000 notes; ` 10 → 5000 notes
15. Number of ` 1 coins = 80; Number of ` 2 coins = 60; Number of ` 5 coins = 20
16. 19
EXERCISE 2.3
1. x = 18 2. t = –1 3. x = –2 4. z =
3
2
5. x = 5 6. x = 0
7. x = 40 8. x = 10 9. y =
7
3
10. m =
4
5
EXERCISE 2.4
1. 4 2. 7, 35 3. 36 4. 26 (or 62)
5. Shobo’s age: 5 years; Shobo’s mother’s age: 30 years
6. Length = 275 m; breadth = 100 m 7. 200 m 8. 72
9. Grand daughter’s age: 6 years; Grandfather’s age: 60 years
10. Aman’s age: 60 years; Aman’s son’s age: 20 years
2021–22

ANSWERS 263
EXERCISE 2.5
1. x =
27
10
2. n = 36 3. x = –5 4. x = 8 5. t = 2
6. m =
7
5
7. t = – 2 8. y =
2
3
9. z = 2 10. f = 0.6
EXERCISE 2.6
1. x =
3
2
2. x =
35
33
3. z = 12 4. y = – 8 5. y = –
4
5
6. Hari’s age = 20 years; Harry’s age = 28 years 7.
13
21
EXERCISE 3.1
1. (a) 1, 2, 5, 6, 7 (b) 1, 2, 5, 6, 7 (c) 1, 2
(d) 2 (e) 1
2. (a) 2 (b) 9 (c) 0 3. 360°; yes.
4. (a) 900° (b) 1080° (c) 1440° (d) (n – 2)180°
5. Apolygonwithequalsidesandequalangles.
(i) Equilateraltriangle (ii) Square (iii) Regularhexagon
6. (a) 60° (b) 140° (c) 140° (d) 108°
7. (a) x + y + z = 360° (b) x + y + z + w = 360°
EXERCISE 3.2
1. (a) 360° – 250° = 110° (b) 360° – 310° = 50°
2. (i)
360
9
°
= 40° (ii)
360
15
°
= 24°
3.
360
24
= 15 (sides) 4. Number of sides = 24
5. (i) No; (Since 22 is not a divisor of 360)
(ii) No; (because each exterior angle is 180° – 22° = 158°, which is not a divisor of 360°).
6. (a) The equilateral triangle being a regular polygon of 3 sides has the least measure of an interior
angle = 60°.
(b) By (a), we can see that the greatest exterior angle is 120°.
EXERCISE 3.3
1. (i) BC(Opposite sides are equal) (ii) ∠ DAB (Opposite angles are equal)
2021–22

264 MATHEMATICS
(iii) OA (Diagonals bisect each other)
(iv) 180° (Interior opposite angles, since AB || DC )
2. (i) x = 80°; y = 100°; z = 80° (ii) x = 130°; y = 130°; z = 130°
(iii) x = 90°; y = 60°; z = 60° (iv) x = 100°; y = 80°; z = 80°
(v) y = 112°; x = 28°; z = 28°
3. (i) Can be, but need not be.
(ii) No; (in a parallelogram, opposite sides are equal; but here,AD ≠ BC).
(iii) No; (in a parallelogram, opposite angles are equal; but here, ∠A ≠ ∠C).
4. Akite,forexample 5. 108°; 72°; 6. Eachisarightangle.
7. x = 110°; y = 40°; z = 30°
8. (i) x = 6; y = 9 (ii) x = 3; y = 13; 9. x = 50°
10. NM || KL (sum of interior opposite angles is 180°). So, KLMN is a trapezium.
11. 60° 12. ∠P = 50°; ∠S = 90°
EXERCISE 3.4
1. (b), (c), (f), (g), (h) are true; others are false.
2. (a) Rhombus;square. (b) Square;rectangle
3. (i) A square is 4 – sided; so it is a quadrilateral.
(ii) A square has its opposite sides parallel; so it is a parallelogram.
(iii) A square is a parallelogram with all the 4 sides equal; so it is a rhombus.
(iv) A square is a parallelogram with each angle a right angle; so it is a rectangle.
4. (i) Parallelogram;rhombus;square;rectangle.
(ii) Rhombus;square (iii) Square;rectangle
5. Bothofitsdiagonalslieinitsinterior.
6. AD || BC; AB ||DC. So, in parallelogramABCD, the mid-point of diagonal AC is O.
EXERCISE 5.1
1. (b), (d). In all these cases data can be divided into class intervals.
2. Shopper Tally marks Number
W | | | | | | | | | | | | | | | | | | | | | | | 28
M | | | | | | | | | | | | 15
B | | | | 5
G | | | | | | | | | | 12
2021–22

ANSWERS 265
3. Interval Tally marks Frequency
800 - 810 | | | 3
810 - 820 | | 2
820 - 830 | 1
830 - 840 | | | | | | | | 9
840 - 850 | | | | 5
850 - 860 | 1
860 - 870 | | | 3
870 - 880 | 1
880 - 890 | 1
890 - 900 | | | | 4
Total 30
4. (i) 830 - 840 (ii) 10
(iii) 20
5. (i) 4 - 5 hours (ii) 34
(iii) 14
EXERCISE 5.2
1. (i) 200 (ii) Lightmusic (iii) Classical - 100, Semi classical - 200, Light - 400, Folk - 300
2. (i) Winter (ii) Winter - 150°, Rainy - 120°, Summer - 90° (iii)
3.
2021–22

266 MATHEMATICS
4. (i) Hindi (ii) 30 marks (iii) Yes 5.
EXERCISE 5.3
1. (a) Outcomes → A, B, C, D
(b) HT, HH, TH, TT (Here HTmeans Head on first coin and Tail on the second coin and so on).
2. Outcomesofaneventofgetting
(i) (a) 2, 3, 5 (b) 1, 4, 6
(ii) (a) 6 (b) 1, 2, 3, 4, 5
3. (a)
1
5
(b)
1
13
(c)
4
7
4. (i)
1
10
(ii)
1
2
(iii)
2
5
(iv)
9
10
5. Probability of getting a green sector =
3
5
; probability of getting a non-blue sector =
4
5
6. Probabilityofgettingaprimenumber=
1
2
; probabilityofgettinganumberwhichisnotprime=
1
2
Probability of getting a number greater than 5 =
1
6
Probability of getting a number not greater than 5 =
5
6
EXERCISE 6.1
1. (i) 1 (ii) 4 (iii) 1 (iv) 9 (v) 6 (vi) 9
(vii) 4 (viii) 0 (ix) 6 (x) 5
2. Thesenumbersendwith
(i) 7 (ii) 3 (iii) 8 (iv) 2 (v) 0 (vi) 2
(vii) 0 (viii) 0
3. (i), (iii) 4. 10000200001, 100000020000001
5. 1020304030201, 1010101012
6. 20, 6, 42, 43
7. (i) 25 (ii) 100 (iii) 144
8. (i) 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9. (i) 24 (ii) 50 (iii) 198
2021–22

ANSWERS 267
EXERCISE 6.2
1. (i) 1024 (ii) 1225 (iii) 7396 (iv) 8649 (v) 5041 (vi) 2116
2. (i) 6,8,10 (ii) 14,48,50 (iii) 16,63,65 (iv) 18,80,82
EXERCISE 6.3
1. (i) 1, 9 (ii) 4, 6 (iii) 1, 9 (iv) 5
2. (i),(ii),(iii) 3. 10, 13
4. (i) 27 (ii) 20 (iii) 42 (iv) 64 (v) 88 (vi) 98
(vii) 77 (viii) 96 (ix) 23 (x) 90
5. (i) 7; 42 (ii) 5; 30 (iii) 7, 84 (iv) 3; 78 (v) 2; 54 (vi) 3; 48
6. (i) 7; 6 (ii) 13; 15 (iii) 11; 6 (vi) 5; 23 (v) 7; 20 (vi) 5; 18
7. 49 8. 45 rows; 45 plants in each row 9. 900 10. 3600
EXERCISE 6.4
1. (i) 48 (ii) 67 (iii) 59 (iv) 23 (v) 57 (vi) 37
(vii) 76 (viii) 89 (ix) 24 (x) 32 (xi) 56 (xii) 30
2. (i) 1 (ii) 2 (iii) 2 (iv) 3 (v) 3
3. (i) 1.6 (ii) 2.7 (iii) 7.2 (iv) 6.5 (v) 5.6
4. (i) 2; 20 (ii) 53; 44 (iii) 1; 57 (iv) 41; 28 (v) 31; 63
5. (i) 4; 23 (ii) 14; 42 (iii) 4; 16 (iv) 24; 43 (v) 149; 81
6. 21 m 7. (a) 10 cm (b) 12 cm
8. 24 plants 9. 16children
EXERCISE 7.1
1. (ii) and (iv)
2. (i) 3 (ii) 2 (iii) 3 (iv) 5 (v) 10
3. (i) 3 (ii) 2 (iii) 5 (iv) 3 (v) 11
4. 20 cuboids
EXERCISE 7.2
1. (i) 4 (ii) 8 (iii) 22 (iv) 30 (v) 25 (vi) 24
(vii) 48 (viii) 36 (ix) 56
2. (i) False (ii) True (iii) False (iv) False (v) False (vi) False
(vii) True
3. 11, 17, 23, 32
2021–22

268 MATHEMATICS
EXERCISE 8.1
1. (a) 1 : 2 (b) 1 : 2000 (c) 1 : 10
2. (a) 75% (b) 66
2
3
% 3. 28% students 4. 25 matches 5. ` 2400
6. 10%, cricket → 30 lakh; football → 15 lakh; other games → 5 lakh
EXERCISE 8.2
1. ` 1,40,000 2. 80% 3. ` 34.80 4. ` 18,342.50
5. Gain of 2% 6. ` 2,835 7. Loss of ` 1,269.84
8. ` 14,560 9. ` 2,000 10. ` 5,000 11. ` 1,050
EXERCISE 8.3
1. (a) Amount = ` 15,377.34; Compound interest = ` 4,577.34
(b) Amount = ` 22,869; Interest = ` 4869 (c) Amount = ` 70,304, Interest = ` 7,804
(d) Amount = ` 8,736.20, Interest = ` 736.20
(e) Amount = ` 10,816, Interest = ` 816
2. ` 36,659.70 3. Fabina pays ` 362.50 more 4. ` 43.20
5. (ii) ` 63,600 (ii) ` 67,416 6. (ii) ` 92,400 (ii) ` 92,610
7. (i) ` 8,820 (ii) ` 441
8. Amount = ` 11,576.25, Interest = ` 1,576.25Yes.
9. ` 4,913 10. (i) About 48,980 (ii) 59,535 11. 5,31,616 (approx)
12. ` 38,640
EXERCISE 9.1
1. Term Coefficient
(i) 5xyz2
5
–3zy –3
(ii) 1 1
x 1
x2
1
(iii) 4x2
y2
4
– 4x2
y2
z2
– 4
z2
1
(iv) 3 3
– pq –1
qr 1
– rp –1
(v)
2
x 1
2
2
y 1
2
–xy –1
(vi) 0.3a 0.3
– 0.6ab – 0.6
0.5b 0.5
2021–22

ANSWERS 269
Firstmonomial →
Secondmonomial↓
2. Monomials: 1000,pqr
Binomials: x + y, 2y – 3y2
, 4z – 15z2
, p2
q + pq2
, 2p + 2q
Trinomials :7 + y + 5x, 2y – 3y2
+ 4y3
, 5x – 4y + 3xy
Polynomials that do not fit in these categories: x + x2
+ x3
+ x4
, ab + bc + cd + da
3. (i) 0 (ii) ab + bc + ac (iii) – p2
q2
+ 4pq + 9
(iv) 2(l2
+ m2
+ n2
+ lm + mn + nl)
4. (a) 8a – 2ab + 2b – 15 (b) 2xy – 7yz + 5zx + 10xyz
(c) p2
q – 7pq2
+ 8pq – 18q + 5p + 28
EXERCISE 9.2
1. (i) 28p (ii) – 28p2
(iii) – 28p2
q (iv) –12p4
(v) 0
2. pq; 50 mn; 100 x2
y2
; 12x3
; 12mn2
p
3.
2x –5y 3x2
– 4xy 7x2
y –9x2
y2
2x 4x2
–10xy 6x3
–8x2
y 14x3
y –18x3
y2
–5y –10xy 25y2
–15x2
y 20xy2
–35x2
y2
45x2
y3
3x2
6x3
–15x2
y 9x4
–12x3
y 21x4
y –27x4
y2
– 4xy –8x2
y 20xy2
–12x3
y 16x2
y2
–28x3
y2
36x3
y3
7x2
y 14x3
y –35x2
y2
21x4
y –28x3
y2
49x4
y2
– 63x4
y3
–9x2
y2
–18x3
y2
45x2
y3
–27x4
y2
36x3
y3
– 63x4
y3
81x4
y4
4. (i) 105a7
(ii) 64pqr (iii) 4x4
y4
(iv) 6abc
5. (i) x2
y2
z2
(ii) – a6
(iii) 1024y6
(iv) 36a2
b2
c2
(v) – m3
n2
p
EXERCISE 9.3
1. (i) 4pq + 4pr (ii) a2
b – ab2
(iii) 7a3
b2
+ 7a2
b3
(iv) 4a3
– 36a (v) 0
2. (i) ab + ac + ad (ii) 5x2
y + 5xy2
– 25xy
(iii) 6p3
– 7p2
+ 5p (iv) 4p4
q2
– 4p2
q4
(v) a2
bc + ab2
c + abc2
3. (i) 8a50
(ii)
3
5
− x3
y3
(iii) – 4p4
q4
(iv) x10
4. (a) 12x2
– 15x + 3; (i) 66 (ii)
3
2
−
(b) a3
+ a2
+ a + 5; (i) 5 (ii) 8 (iii) 4
5. (a) p2
+ q2
+ r2
– pq – qr – pr (b) – 2x2
– 2y2
– 4xy + 2yz + 2zx
(c) 5l2
+ 25ln (d) – 3a2
– 2b2
+ 4c2
– ab + 6bc – 7ac
2021–22

270 MATHEMATICS
EXERCISE 9.4
1. (i) 8x2
+ 14x – 15 (ii) 3y2
– 28y + 32 (iii) 6.25l2
– 0.25m2
(iv) ax + 5a + 3bx + 15b (v) 6p2
q2
+ 5pq3
– 6q4
(vi) 3a4
+ 10a2
b2
– 8b4
2. (i) 15 – x – 2x2
(ii) 7x2
+ 48xy – 7y2
(iii) a3
+ a2
b2
+ ab + b3
(iv) 2p3
+ p2
q – 2pq2
– q3
3. (i) x3
+ 5x2
– 5x (ii) a2
b3
+ 3a2
+ 5b3
+ 20 (iii) t3
– st + s2
t2
– s3
(iv) 4ac (v) 3x2
+ 4xy – y2
(vi) x3
+ y3
(vii) 2.25x2
– 16y2
(viii) a2
+ b2
– c2
+ 2ab
EXERCISE 9.5
1. (i) x2
+ 6x + 9 (ii) 4y2
+ 20y + 25 (iii) 4a2
– 28a + 49
(iv) 9a2
– 3a +
1
4
(v) 1.21m2
– 0.16 (vi) b4
– a4
(vii) 36x2
– 49 (viii) a2
– 2ac + c2
(ix)
2 2
3 9
4 4 16
x xy y
+ +
(x) 49a2
– 126ab + 81b2
2. (i) x2
+ 10x + 21 (ii) 16x2
+ 24x + 5 (iii) 16x2
– 24x + 5
(iv) 16x2
+ 16x – 5 (v) 4x2
+ 16xy + 15y2
(vi) 4a4
+ 28a2
+ 45
(vii) x2
y2
z2
– 6xyz + 8
3. (i) b2
– 14b + 49 (ii) x2
y2
+ 6xyz + 9z2
(iii) 36x4
– 60x2
y + 25y2
(iv)
4
9
m2
+ 2mn +
9
4
n2
(v) 0.16p2
– 0.4pq + 0.25q2
(vi) 4x2
y2
+ 20xy2
+ 25y2
4. (i) a4
– 2a2
b2
+ b4
(ii) 40x (iii) 98m2
+ 128n2
(iv) 41m2
+ 80mn + 41n2
(v) 4p2
– 4q2
(vi) a2
b2
+ b2
c2
(vii) m4
+ n4
m2
6. (i) 5041 (ii) 9801 (iii) 10404 (iv) 996004
(v) 27.04 (vi) 89991 (vii) 6396 (viii) 79.21
(ix) 99.75
7. (i) 200 (ii) 0.08 (iii) 1800 (iv) 84
8. (i) 10712 (ii) 26.52 (iii) 10094 (iv) 95.06
EXERCISE 10.1
1. (a) →(iii) →(iv) (b) →(i) →(v) (c) →(iv) →(ii)
(d) →(v) →(iii) (e) →(ii) →(i)
2. (a) (i) →Front, (ii) →Side, (iii) →Top (b) (i) →Side, (ii) →Front, (iii) →Top
(c) (i) →Front, (ii) →Side, (iii) →Top (d) (i) →Front, (ii) →Side, (iii) →Top
3. (a) (i) →Top, (ii) →Front, (iii) →Side (b) (i) →Side, (ii) →Front, (iii) →Top
(c) (i) →Top, (ii) →Side, (iii) →Front (d) (i) →Side, (ii) →Front, (iii) →Top
(e) (i) →Front, (ii) →Top, (iii) →Side
2021–22

ANSWERS 271
EXERCISE 10.3
1. (i) No (ii)Yes (iii) Yes 2. Possible, only if the number of faces are greater than or equal to 4
3. only(ii)and(iv)
4. (i) Aprism becomes a cylinder as the number of sides of its base becomes larger and larger.
(ii) Apyramid becomes a cone as the number of sides of its base becomes larger and larger.
5. No. It can be a cuboid also 7. Faces → 8, Vertices → 6, Edges → 30
8. No
EXERCISE 11.1
1. (a) 2. ` 17,875 3. Area = 129.5 m2
; Perimeter = 48 m
4. 45000 tiles 5. (b)
EXERCISE 11.2
1. 0.88 m2
2. 7 cm 3. 660 m2
4. 252 m2
5. 45 cm2
6. 24 cm2
, 6 cm 7. ` 810 8. 140 m
9. 119 m2
10. Area using Jyoti’s way =
2 2
1 15
2 (30 15) m 337.5 m
2 2
× × × + = ,
Area using Kavita’s way =
2
1
15 15 15 15 337.5 m
2
× × + × =
11. 80 cm2
, 96 cm2
, 80 cm2
, 96 cm2
EXERCISE 11.3
1. (a) 2. 144 m 3. 10 cm 4. 11 m2
5. 5 cans
6. Similarity→Bothhavesameheights.Difference→oneisacylinder,theotherisacube.Thecubehas
largerlateralsurfacearea
7. 440 m2
8. 322 cm 9. 1980 m2
10. 704 cm2
EXERCISE 11.4
1. (a) Volume (b) Surface area (c) Volume
2. Volume of cylinder B is greater; Surface area of cylinder B is greater.
3. 5 cm 4. 450 5. 1 m 6. 49500 L
7. (i) 4times (ii) 8times 8. 30 hours
EXERCISE 12.1
1. (i)
1
9
(ii)
1
16
(iii) 32
2021–22

272 MATHEMATICS
2. (i) 3
1
(– 4)
(ii) 6
1
2
(iii) (5)4
(iv) 2
1
(3)
(v) 3
1
( 14)
−
3. (i) 5 (ii)
1
2
(iii) 29 (iv) 1 (v)
81
16
4. (i) 250 (ii)
1
60
5. m = 2 6. (i) –1 (ii)
512
125
7. (i)
4
625
2
t
(ii) 55
EXERCISE 12.2
1. (i) 8.5 × 10– 12
(ii) 9.42 × 10– 12
(iii) 6.02 × 1015
(iv) 8.37 × 10– 9
(v) 3.186 × 1010
2. (i) 0.00000302 (ii) 45000 (iii) 0.00000003
(iv) 1000100000 (v) 5800000000000 (vi) 3614920
3. (i) 1 × 10– 6
(ii) 1.6 × 10–19
(iii) 5 × 10– 7
(iv) 1.275 × 10–5
(v) 7 × 10–2
4. 1.0008 × 102
EXERCISE 13.1
1. No 2. Parts of red pigment 1 4 7 12 20
Parts of base 8 32 56 96 160
3. 24 parts 4. 700 bottles 5. 10– 4
cm; 2 cm 6. 21 m
7. (i) 2.25 × 107
crystals (ii) 5.4 × 106
crystals 8. 4 cm
9. (i) 6 m (ii) 8 m 75 cm 10. 168 km
EXERCISE 13.2
1. (i),(iv),(v) 2. 4 → 25,000; 5 → 20,000; 8 → 12,500; 10 → 10,000; 20 → 5,000
Amountgiventoawinnerisinverselyproportionaltothenumberofwinners.
3. 8 → 45°, 10 → 36°, 12 → 30° (i) Yes (ii) 24° (iii) 9
4. 6 5. 4 6. 3 days 7. 15 boxes
8. 49machines 9.
1
1 hours
2
10. (i) 6 days (ii) 6 persons 11. 40 minutes
EXERCISE 14.1
1. (i) 12 (ii) 2y (iii) 14pq (iv) 1 (v) 6ab (vi) 4x
(vii) 10 (viii) x2
y2
2021–22

ANSWERS 273
2. (i) 7(x – 6) (ii) 6(p – 2q) (iii) 7a(a + 2) (iv) 4z(– 4 + 5z2
)
(v) 10 lm(2l + 3a) (vi) 5xy(x – 3y) (vii) 5(2a2
– 3b2
+ 4c2
)
(viii) 4a(– a + b – c) (ix) xyz(x + y + z) (x) xy(ax + by + cz)
3. (i) (x + 8) (x + y) (ii) (3x + 1) (5y – 2) (iii) (a + b) (x – y)
(iv) (5p + 3) (3q + 5) (v) (z – 7) (1 – xy)
EXERCISE 14.2
1. (i) (a + 4)2
(ii) (p – 5)2
(iii) (5m + 3)2
(iv) (7y + 6z)2
(v) 4(x – 1)2
(vi) (11b – 4c)2
(vii) (l – m)2
(viii) (a2
+ b2
)2
2. (i) (2p – 3q) (2p + 3q) (ii) 7(3a – 4b) (3a + 4b) (iii) (7x – 6) (7x + 6)
(iv) 16x3
(x – 3) (x + 3) (v) 4lm (vi) (3xy – 4) (3xy + 4)
(vii) (x – y – z) (x – y + z) (viii) (5a – 2b + 7c) (5a + 2b – 7c)
3. (i) x(ax + b) (ii) 7(p2
+ 3q2
) (iii) 2x(x2
+ y2
+ z2
)
(iv) (m2
+ n2
) (a + b) (v) (l + 1) (m + 1) (vi) (y + 9) (y + z)
(vii) (5y + 2z) (y – 4) (viii) (2a + 1) (5b + 2) (ix) (3x – 2) (2y – 3)
4. (i) (a – b) (a + b) (a2
+ b2
) (ii) (p – 3) (p + 3) (p2
+ 9)
(iii) (x – y – z) (x + y + z) [x2
+ (y + z)2
] (iv) z(2x – z) (2x2
– 2xz + z2
)
(v) (a – b)2
(a + b)2
5. (i) (p + 2) (p + 4) (ii) (q – 3) (q – 7) (iii) (p + 8) (p – 2)
EXERCISE 14.3
1. (i)
3
2
x
(ii) – 4y (iii) 6pqr (iv)
2
2
3
x y (v) –2a2
b4
2. (i)
1
(5 6)
3
x − (ii) 3y4
– 4y2
+ 5 (iii) 2(x + y + z)
(iv)
2
1
( 2 3)
2
x x
+ + (v) q3
– p3
3. (i) 2x – 5 (ii) 5 (iii) 6y (iv) xy (v) 10abc
4. (i) 5(3x + 5) (ii) 2y(x + 5) (iii)
1
( )
2
r p q
+ (iv) 4(y2
+ 5y + 3)
(v) (x + 2) (x + 3)
5. (i) y + 2 (ii) m – 16 (iii) 5(p – 4) (iv) 2z(z – 2) (v)
5
( )
2
q p q
−
(vi) 3(3x – 4y) (vii) 3y(5y – 7)
EXERCISE 14.4
1. 4(x – 5) = 4x – 20 2. x(3x + 2) = 3x2
+ 2x 3. 2x + 3y = 2x + 3y
4. x + 2x + 3x = 6x 5. 5y + 2y + y – 7y = y 6. 3x + 2x = 5x
2021–22

274 MATHEMATICS
7. (2x)2
+ 4(2x) + 7 = 4x2
+ 8x + 7 8. (2x)2
+ 5x = 4x2
+ 5x
9. (3x + 2)2
= 9x2
+ 12x + 4
10. (a) (–3)2
+ 5(–3) + 4 = 9 – 15 + 4 = –2 (b) (–3)2
– 5(–3) + 4 = 9 + 15 + 4 = 28
(c) (–3)2
+ 5(–3) = 9 – 15 = – 6
11. (y – 3)2
= y2
– 6y + 9 12. (z + 5)2
= z2
+ 10z + 25
13. (2a + 3b) (a – b) = 2a2
+ ab – 3b2
14. (a + 4) (a + 2) = a2
+ 6a + 8
15. (a – 4) (a – 2) = a2
– 6a + 8 16.
2
2
3
1
3
x
x
=
17.
2 2
2 2 2 2
3 1 3 1 1
1
3 3 3 3
x x
x x x x
+
= + = + 18.
3
3 2
x
x +
=
3
3 2
x
x +
19.
3
4 3
x +
=
3
4 3
x +
20.
4 5 4 5 5
1
4 4 4 4
x x
x x x x
+
= + = +
21.
7 5 7 5 7
1
5 5 5 5
x x x
+
= + = +
EXERCISE 15.1
1. (a) 36.5° C (b) 12 noon (c) 1 p.m, 2 p.m.
(d) 36.5° C; The point between 1 p.m. and 2 p.m. on the x-axis is equidistant from the two points
showing 1 p.m. and 2 p.m., so it will represent 1.30 p.m. Similarly, the point on the y-axis,
between 36° C and 37° C will represent 36.5° C.
(e) 9 a.m. to 10 a.m., 10 a.m. to 11 a.m., 2 p.m. to 3 p.m.
2. (a) (i) ` 4 crore (ii) ` 8 crore
(b) (i) ` 7 crore (ii) ` 8.5 crore (approx.)
(c) ` 4 crore (d) 2005
3. (a) (i) 7 cm (ii) 9 cm
(b) (i) 7 cm (ii) 10 cm
(c) 2 cm (d) 3 cm (e) Second week (f) First week
(g) At the end of the 2nd week
4. (a) Tue,Fri,Sun (b) 35° C (c) 15° C (d) Thurs
6. (a) 4 units = 1 hour (b)
1
3
2
hours (c) 22 km
(d) Yes; This is indicated by the horizontal part of the graph (10 a.m. - 10.30 a.m.)
(e) Between 8 a.m. and 9 a.m.
7. (iii) isnotpossible
EXERCISE 15.2
1. Points in (a) and (b) lie on a line; Points in (c) do not lie on a line
2. The line will cut x-axis at (5, 0) and y-axis at (0, 5)
2021–22

ANSWERS 275
3. O(0, 0), A(2, 0), B(2, 3), C(0, 3), P(4, 3), Q(6, 1), R(6, 5), S(4, 7), K(10, 5), L(7, 7), M(10, 8)
4. (i) True (ii) False (iii) True
EXERCISE 15.3
1. (b) (i) 20 km (ii) 7.30 a.m. (c) (i) Yes (ii) ` 200 (iii) ` 3500
2. (a) Yes (b) No
EXERCISE 16.1
1. A = 7, B = 6 2. A = 5, B = 4, C = 1 3. A = 6
4. A = 2, B = 5 5. A = 5, B = 0, C = 1 6. A = 5, B = 0, C = 2
7. A = 7, B = 4 8. A = 7, B = 9 9. A = 4, B = 7
10. A = 8, B = 1
EXERCISE 16.2
1. y = 1 2. z = 0 or 9 3. z = 0, 3, 6 or 9
4. 0, 3, 6 or 9
JUST FOR FUN
1. More about Pythagorean triplets
We have seen one way of writing pythagorean triplets as 2m, m2
– 1, m2
+ 1.
A pythagorean triplet a, b, c means a2
+ b2
= c2
. If we use two natural numbers m and n(m n), and
take a = m2
– n2
, b = 2mn, c = m2
+ n2
, then we can see that c2
= a2
+ b2
.
Thusfordifferentvaluesofmandnwith mnwecangeneratenaturalnumbersa,b,c suchthatthey
formPythagoreantriplets.
For example: Take, m = 2, n = 1.
Then, a = m2
– n2
= 3, b = 2mn = 4, c = m2
+ n2
= 5, is a Pythagorean triplet. (Check it!)
For, m = 3, n = 2, we get,
a = 5, b = 12, c = 13 which is again a Pythagorean triplet.
Take some more values for m and n and generate more such triplets.
2. When water freezes its volume increases by 4%. What volume of water is required to make 221 cm3
ofice?
3. If price of tea increased by 20%, by what per cent must the consumption be reduced to keep the
expense the same?
2021–22

276 MATHEMATICS
4. CeremonyAwards began in 1958.There were 28 categories to win an award. In 1993, there were 81
categories.
(i) The awards given in 1958 is what per cent of the awards given in 1993?
(ii) The awards given in 1993 is what per cent of the awards given in 1958?
5. Outofaswarmofbees,onefifthsettledonablossomofKadamba,onethirdonaflowerofSilindhiri,
and three times the difference between these two numbers flew to the bloom of Kutaja. Only ten
bees were then left from the swarm. What was the number of bees in the swarm? (Note, Kadamba,
Silindhiri and Kutaja are flowering trees. The problem is from the ancient Indian text on algebra.)
6. In computing the area of a square, Shekhar used the formula for area of a square, while his friend
Maroof used the formula for the perimeter of a square. Interestingly their answers were numerically
same.Tell me the number of units of the side of the square they worked on.
7. Theareaofasquareisnumericallylessthansixtimesitsside.Listsomesquaresinwhichthishappens.
8. Is it possible to have a right circular cylinder to have volume numerically equal to its curved surface
area? If yes state when.
9. Leela invited some friends for tea on her birthday.Hermotherplacedsomeplatesandsomepuris on
a table to be served. If Leela places 4purisineachplate1platewouldbeleftempty.Butifsheplaces
3puris in each plate 1puriwouldbeleft.Findthenumberofplatesandnumberofpurison the table.
10. Is there a number which is equal to its cube but not equal to its square? If yes find it.
11. Arrangethenumbersfrom1to20inarowsuchthatthesumofanytwoadjacentnumbersisaperfect
square.
Answers
2. 212
1
2
cm3
3.
2
16 %
3
4. (i) 34.5% (ii)289%
5. 150
6. 4units
7. Sides = 1, 2, 3, 4, 5 units
8. Yes, when radius = 2 units
9. Number of puris = 16, number of plates = 5
10. – 1
11. One of the ways is, 1, 3, 6, 19, 17, 8 (1 + 3 = 4, 3 + 6 = 9 etc.). Try some other ways.
2021–22

Math Class 8.pdf

More Related Content

Similar to Math Class 8.pdf (20)

Recently uploaded (20)

Math Class 8.pdf