Mathematics mathematics elementary

Prescribed by National Curriculum and Textbook Board as a Textbook (Bangla version) for
Class Five from 2006, English version of the book from academic year 2007.
Elementary Mathematics
Book Five
Written by
Dr. Munibur Rahman Chowdhury
A. M. M. Ahsan Ullah
Hamida Banu Begum
Md. Rafiqul Islam
Edited by
A.F.M. Khodadad Khan
Translated by
A.M.M. Ahsan Ullah
Arjumand Ara
NATIONAL CURRICULUM & TEXTBOOK BOARD, DHAKA

Published by
National Curriculum and Textbook Board
69-70, Motijheel Commercial Area, Dhaka-1000.
[All rights reserved by the publisher]
Is permitted to print and sell 500 (Five hundred)
Copies of the book for the year 2009 Only.
Circular No.317 Date: 15/12/08 Allotment Letter No. 717/08/09 Date : 23/02/09
Rahmania Offest Press, Banglabazer, Dhaka- 1100
First Print (Bangla version) : September 2005
First Print (English Version) : February 2007
2nd Print (English version) : March 2008
Reprint (English Version) : April 2009
Computer Graphic
Laser Scan Limited
15/2, Topkhana Road, BMA Bhaban.
Cover & Illustrations:
Domion Neoton Pinaru
Picture Coloured by
Ahmed Ullah
Printing Supervision
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Design
Website version developed by Mars Solutions Limited

Preface
For improving the existing quality of Primary Education in Bangladesh, National
Curriculum and Textbook Board (NCTB) in collaboration with PEDP-II initiated an
extensive program for development of curriculum and teaching learning materials in
2002. In the light of this program the curriculum, textbooks and other teaching
learning materials of Primary levels have been prepared, revised and evaluated.
The textbook entitled, 'Elementary Mathematics' has been prepared on the basis of
attainable competencies for the students of Class Five. The subject matter of the
textbook is derived from the basic elements of mathematics following the attainable
learning outcomes as depicted in the curriculum. This will facilitate our young
learners to know how they can make best use of this knowledge.
The contents of the book are analyzed and explained in such a manner with practical
examples, illustrations and system of planned activities, that students are inspired to
study the subject with a keen interest.
This book is originally published in Bangla. From this year NCTB is publishing the
English version of the textbook. English is the language of choice in today's
globalized world. To facilitate the verbal and written communication skills of our
future citizens and suitably prepare them for international competition, we decided to
translate the original Bangla textbooks into English. It's pleasant to note that the
number of English medium schools in Bangladesh is increasing very fast. In this
context NCTB decided to publish all the textbooks of Primary level in English. This
was a big endeavour for us. Despite our all efforts the first edition may not be totally
error free. However, in the future editions we shall try to remove all errors and
discrepancies.
Finally, I would like to express my heartfelt thanks and gratitude to those who have
made their valuable contributions in writing, editing, evaluating and translating this
book. I sincerely hope that the book will be useful to those for whom it has been
prepared.
Prof. Md. Mostafa Kamaluddin
Chairman
Dhaka

Contents
Chapter Subject Page
One Addition and Subtraction 1 - 9
Two Multiplication and Division 10 - 19
Three Problems Involving the First Four Rules 20 - 23
Four The Unitary Method 24 - 31
Five Simplification 32 - 35
Six Average 36 - 40
Seven H.C.F. and L.C.M. 41 - 50
Eight Mathematical Symbols and Sentences 51 - 57
Nine Common Fractions 58 - 100
Ten The Decimal Fraction 101 - 115
Eleven Percentage 116 - 127
Twelve Measurement 128 - 148
Thirteen Time 149 - 164
Forteen Income, Expenditure and Cash Memo 165 - 171
Fifteen The Arrangement of Data 172 - 175
Sixteen Graph 176 - 180
Seventeen Geometry 181 - 192
Eighteen The Usage of Calculator 193 - 197
Nineteen Population 198 - 203
Answer 204 - 215

Addition and Subtraction
Example 1. Add : 23041, 10532, 31210, 22000, 2104 and 102.
Solution : 2 3 0 4 1 Ajuts Thousands Hundreds Tens Ones
1 0 5 3 2 2 3 0 4 1
3 1 2 1 0 1 0 5 3 2
2 2 0 0 0 3 1 2 1 0
2 1 0 4 2 2 0 0 0
1 0 2 2 1 0 4
8 8 9 8 9 1 0 2
8 8 9 8 9
Answer : 88989.
Example 2. Add : 312024, 103001, 150310, 201200, 131141 and
102203.
Solution : 3 1 2 0 2 4 L Aj Th H T O
1 0 3 0 0 1 3 1 2 0 2 4
1 5 0 3 1 0 1 0 3 0 0 1
2 0 1 2 0 0 1 5 0 3 1 0
1 3 1 1 4 1 2 0 1 2 0 0
1 0 2 2 0 3 1 3 1 1 4 1
9 9 9 8 7 9 1 0 2 2 0 3
9 9 9 8 7 9
Answer : 999879.
The numbers are placed as ones under ones, tens under tens,
hundreds under hundreds.
Addition is started from the place of ones.
The sum of each place is written below in absence of
carrying number.
Chapter - 1

2 Addition and Subtraction
Example 3. Add : 572104, 650489, 948531, 886350, 95448 and
749065.
Solution : Ni L Aj Th H T O
3 4
5
6
9
8
7
3
7
5
4
8
9
4
1
2
0
8
6
5
9
2
1
4
5
3
4
0
2
0
8
3
5
4
6
4
9
1
0
8
5
Answer :
5 7 2 1 0 4
6 5 0 4 8 9
9 4 8 5 3 1
8 8 6 3 5 0
9 5 4 4 8
7 4 9 0 6 5
3 9 0 1 9 8 7
3901987. 3 9 0 1 9 8 7
The numbers are arranged in the table.
The sum of the digits of ones is 27. 27 is 2 tens and 7 ones, so
7 of 27 is placed under the line in the ones place and 2 of 27 is
written within a circle on the top of tens place.
In the same way, adding the digits of tens place, the carrying
number 2 is written on the top of the hundreds place.
Similarly, the digits of the hundreds place, thousands place,
ajuts place and lacs place are added.
Adding the digits of lacs place, the carrying number is written
on the top of nijuts place within a circle.
As there is no other digit in the nijuts place, the digit is written
as sum in nijuts place, within the circle.
Example 4. Add the numbers writing them side by side.
(a) 31420, 50261, 47523, 9431, 23084 and 62100.
(b) 425731, 653802, 749243, 86010, 72301 and 930211.

3Elementary Mathematics
Solution :
(a) 31420 + 50261 + 47523 + 9431 + 23084 + 62100
= 223819
Answer : 223819.
(b) 425731 + 653802 + 749243 + 86010 +72301 + 930211
= 2917298
Answer : 2917298.
The numbers are written side by side putting plus sign (+) in
between them.
The result of the addition is written on the line below putting
equal sign (=).
Before adding the digits of ones, tens, hundreds, thousands,
...........places successively the digits are marked with ( ).
The carrying digits are added to the digit on the left side.
Remarks : It is convenient to mark on the digits of the same place
with ( ) at the time of adding.
Add (One is worked out) :
(a) 6 3 2 0 5 4 (b) 4 6 8 5 3 (c) 5 3 2 8 6 1 (d) 7 3 2 8 4 6
2 8 4 7 1 8 7 2 0 4 1 7 0 4 1 0 5 8 5 3 9 2
4 0 5 6 5 3 5 3 9 0 5 2 9 5 0 3 1 9 0 8 7 0 5
5 2 0 8 9 0 2 0 7 1 4 3 2 0 5 2 4 6 2 4 1 5 7
7 4 1 5 0 6 1 6 3 0 0 8 1 3 0 1 6 5 7 0 2 3 4
8 1 3 2 3 2 8 1 4 6 7 9 0 6 3
3 3 9 8 0 5 3
Fill in the boxes :
(a) 645317 + 8297 + 53086 + 264309 + 708321 =
(b) 526394 + 230087 + 630801 + 319500 + 460009
+ 809625 =

Example 5. Mr. Mizan sold rice from three pieces of land for
Tk. 25087, Tk. 16920 and Tk. 30725 repectively and gram from
another piece of land for Tk. 9872. What is the total price of the
crops that Mr. Mizan sold ?
Solution : Mr. Mizan sold rice from three pieces of land :
Tk. 2 5 0 8 7
Tk. 1 6 9 2 0
Tk. 3 0 7 2 5
He sold gram : Tk. 9 8 7 2
Total Tk. 8 2 6 0 4 (Adding)
Answer : Mr. Mizan sold crops worth Tk. 82604.
Example 6. Subtract 713241 from 839642.
Solution : 8 3 9 6 4 2 L Aj Th H T O
− 7 1 3 2 4 1 8 3 9 6 4 2 minuend
1 2 6 4 0 1 − 7 1 3 2 4 1 subtrahend
1 2 6 4 0 1 difference
Answer : 126401.
Solution : L Aj Th H T O
7
− +1
5+10
9
8
4
3
2+1
4+10
7
6
3
Answer :
7 5 8 3 4 6
− 9 4 2 7 3
6 6 4 0 7 3
664073. 6 6 4 0 7 3
Solution : L Aj Th H T O
9
− 7+1
6+10
8+1
0+10
4+1
7+10
9
5
0+1
2+10
6
Answer :
9 6 0 7 5 2
− 7 8 4 9 0 6
1 7 5 8 4 6
175846. 1 7 5 8 4 6
Example 9. Subtract by writing the digits side by side
(a) 95742 from 650263
(b) 579406 from 836041

Solution :
(a) 650263 − 95742 = 554521
Here, 650263 is minuend, 95742 is subtrahend and the difference is
554521.
Answer : 554521.
(b) 836041 − 579406 = 256635
Here, 836041 is minuend, 579406 is subtrahend and the difference is
256635.
Answer : 256635.
Minus sign (−) is placed in between subtrahend and minuend
and equal sign (=) is infront of difference.
To subtract conveniently, the digits of ones, tens, hundreds,
thousands places are marked with ( ) .
Subtract (one is worked out) :
(a) 8 5 3 2 0 4 (b) 6 5 0 4 8 7 (c) 7 2 8 5 1 3 (d) 8 6 0 5 2 1
4 7 8 5 2 6 7 3 2 9 5 5 4 3 2 7 9 3 9 4 0 7 6
3 7 4 6 7 8
Fill in the boxes (one is worked out) :
(a) 628713 − 372946 = 255767
(b) 462051 − 94608 =
(c) 750432 − 364853 =
(d) 804706 − 525376 =

Example 10 12804 litchies were plucked from MinaÊs litchi trees.
From these her father sold 7925 litchies. How many litchies are left?
Solution : Number of litchies plucked : 1 2 8 0 4
Number of litchies sold : (−) 7 9 2 5
The number of litchies left : 4 8 7 9
Answer : 4879 litchies are left.
Example 11. The sum of three numbers is 845076. Of them two
numbers are 321674 and 286539. What is the third number?
Solution : The two numbers 3 2 1 6 7 4
2 8 6 5 3 9
The sum of two numbers 6 0 8 2 1 3
The sum of three numbers 8 4 5 0 7 6
and the sum of two numbers 6 0 8 2 1 3
∴ The other number is 2 3 6 8 6 3(by subtraction)
Answer : The third number is 236863.
Example 12. Abdul Latif had Tk. 621345. From that money he
gave his elder daughter Tk. 85924, Tk. 84790 to younger daughter
and to his son he gave Tk. 9830 more than his elder daughter. He
gave the rest of the money to his wife. How much will his wife get?
Solution: Elder daughter got Tk.8 5 9 2 4
Younger daughter got Tk.8 4 7 9 0
Two daughters got Tk.1 7 0 7 1 4 (by addition)
The son got Tk. 9830 more than elder daughter
8 5 9 2 4
(+) Tk. 9 8 3 0
∴ The son got Tk. 9 5 7 5 4
Three persons got Tk. 1 7 0 7 1 4
(+) Tk. 9 5 7 5 4
Tk. 2 6 6 4 6 8
Total amount was Tk. 6 2 1 3 4 5
Three persons together got Tk. 2 6 6 4 6 8
The rest amount Tk. 3 5 4 8 7 7 (by subtraction)
Answer : Abdul LatifÊs wife got Tk. 354877.
Tk.

Exercise - 1
1. Add :
(a) 8 3 5 7 2 (b) 2 3 0 1 4 0 (c) 5 3 6 7 0 9
4 0 2 1 3 3 1 2 1 0 3 4 7 0 9 8 5
2 1 0 0 4 1 2 4 3 1 5 3 2 5 0 7 1
6 5 3 2 2 0 0 2 1 0 6 0 8 2 1 3
8 4 5 0 1 1 2 3 0 1 2 8 9 0 5 0
(d) 1 3 2 0 2 3 (e) 6 4 8 7 5 3 (f) 8 7 2 0 9 4
2 1 0 2 4 0 2 6 9 1 0 7 4 3 5 8 0 9
3 0 1 1 0 2 8 5 3 2 6 4 5 6 9 2 7 5
1 2 3 0 1 2 5 9 8 6 0 3 3 2 5 6 8 0
1 2 0 3 0 0 8 7 5 6 2 6 5 0 9 4 2
1 0 2 3 1 2 6 9 8 7 5 7 9 8 0 5 6
2. Fill in the boxes:
(a) 43570 + 26095 + 40591 + 9207 + 8754 =
(b) 560829 + 427108 + 82075 + 6429
+ 325790 + 53008 =
(c) 325708 + 246805 + 508362 + 730096
+ 92346 + 64008 =
(d) 410527 + 356807 + 512934 + 239006
+ 614500 + 800357 =
3. Subtract :
(a) 53205 from 865497
(b) 462136 from 798546
(c) 96837 from 875320
(d) 479568 from 962053
(e) 658379 from 927405.

4. Fill in the boxes :
(a) 763569 − 51407 =
(b) 972403 − 48705 =
(c) 860523 − 285306 =
(d) 732645 − 428759 =
5. Fill in the boxes :
(a) 5 3 6 1 4 (b) 8 2 7 5 3
2 7 9 8 5 6 9 8 6 7
6 9 5 4 6 5 4 7 9 8
7 6 8 9 3 7 5 6 9
2 2 8 2 6 4 2 9 7 1 5 0
(c) 8 7 5 0 6 (d) 9 6 4 2 5
− 9 6 7 8 − 5 7 4 6 8
7 7 8 2 4 3 9 4 5 7
6. 240851 mangoes are plucked from a tree of RajuÊs mango
garden. From another tree 92105 mangoes are plucked and from
another tree 55008 mangoes are plucked. How many mangoes
are plucked from that garden ?
7. Minued is 985214 and subtrahend is 97465. What is the
difference?
8. If 68975 is subtracted from a number, the difference is 794768.
What is the number?
9. Subtract the greatest number of five digits from the least
number of six digits.
10. To get the sum 9 lacs, what is the number that should be added
to the greatest number of 4 digits?
11. Find the difference of the greatest and least number of 6 digits
that are formed using the digits 6, 8, 9, 5, 0, 4 once.

12. RajuÊs grandfather sold paddy for Tk. 25830, wheat for
Tk. 30645 and gram for Tk. 9786. He bought jute for Tk. 45927.
How much money does he have left ?
13. There were 6785 students in a school. At the beginning of the
year 542 students left the school and 950 new students got
admitted. How many students are there at present in that
school?
14.
15. Mina has Tk. 45987. Raju has Tk. 8250 less than that of Mina.
Rani has Tk. 985 more than that of Raju. How much money do
the three persons have altogether?
16. The sum of four numbers is 468520. Two of them are 73584
and 64209. The third number is 9485 less than the first one.
What is the fourth number?
17. Rahela Begum has Tk. 985320 in a bank. She drew Tk. 172500,
Tk. 150430 and Tk. 286250 respectively in three instalments.
From the rest of the money she bought saving certificates for
Tk. 2 lacs. How much money does she have left in the bank.
18. SimaÊs uncle sold fish from his pond for Tk. 29850 and eggs for
Tk. 20678. From that money he gave Tk. 15600 to his son and
Tk. 10890 to his daughter. From the rest of money he bought a
cow for Tk. 15000. How much money does he have left?
Mr. Farid sold wheat for Tk. 8604 and mustard for
Tk. 5798. From that money he gave Tk. 4250 to his son and paid
Tk. 5000 against loan. He deposited the rest of the money in the
bank. How much money did he deposit in the bank?

Chapter-2
Multiplication and Division
Example 1. Multiply : 5 3 8
× 1 2 5
Solution : 5 3 8 multiplicand
× 1 2 5 multiplier
2 6 9 0 5 3 8 × 5
1 0 7 6 0 5 3 8 × 2 ten
5 3 8 0 0 5 3 8 × 1 hundred
6 7 2 5 0 product
Answer : 67250.
Example 2. Multiply : 2465 by 234. .
Solution : 2 4 6 5 multiplicand
× 2 3 4 multiplier
9 8 6 0 2 4 6 5 × 4
7 3 9 5 0 2 4 6 5 × 3 ten
4 9 3 0 0 0 2 4 6 5 × 2 hundred
5 7 6 8 1 0 product
Answer : 576810.
Example 3. Multiply: 3271 by 420.
Solution : 3 2 7 1 multiplicand
× 4 2 0 multiplier
6 5 4 2 0 3 2 7 1 × 2 ten
1 3 0 8 4 0 0 3 2 7 1 × 4 hundred
1 3 7 3 8 2 0 product
Answer : 1373820.
Since zero (0) is in the ones place of multiplier, multiplication of ones
place is not shown. So in the first step, putting zero in the ones place
multiplication of tens place has been written moving one step to the left
and in the second step putting zero in the ones and tens places
multiplication of hundreds has been written moving two steps to the left.

Example 4. Multiply : 6582 by 308.
Solution : 6 5 8 2
× 3 0 8
5 2 6 5 6 6 5 8 2 × 8
1 9 7 4 6 0 0 6 5 8 2 × 3 hundred
2 0 2 7 2 5 6
Answer : 2027256.
Since zero (0) is in the tens place of multiplier, multiplication of
tens place is not shown. So in the second step, putting zero in the
ones and tens places, multiplication of hundreds has been written
moving two steps to the left.
Example 5. Multiply: 7596 by 600.
Solution : 7 5 9 6
× 6 0 0
4 5 5 7 6 0 0 7 5 9 6 × 6 hundred
Answer : 4557600.
Since zero (0) is in the ones and tens places of multiplier,
multiplications of ones and tens places are not shown. So putting
zero in the ones and tens places, multiplication of hundreds has
been written moving two steps to the left.
Example 6. Multiply: 7 6 9 Example 7. Multiply: 4 5 8 6
× 2 4 8 × 4 7 9
Solution : 7 6 9 Solution : 4 5 8 6
× 2 4 8 × 4 7 9
6 1 5 2 4 1 2 7 4
3 0 7 6 0 3 2 1 0 2 0
1 5 3 8 0 0 1 8 3 4 4 0 0
1 9 0 7 1 2 2 1 9 6 6 9 4
Answer : 190712. Answer : 2196694.
Remarks : Students will do the sum of multiplication as usual.
No explanation will be asked from them.

12 Multiplication and Division
Example 8. Multiply : 3 7 9 2 Example 9. Multiply : 9 2 0 7
× 8 5 0 × 6 4 5
Solution : 3 7 9 2 Solution : 9 2 0 7
× 8 5 0 × 6 4 5
1 8 9 6 0 0 4 6 0 3 5
3 0 3 3 6 0 0 3 6 8 2 8 0
3 2 2 3 2 0 0 5 5 2 4 2 0 0
5 9 3 8 5 1 5
Answer : 3223200. Answer : 5938515.
Example 10. Multiply :4 9 0 8 Example 11. Multiply : 8 0 7 5
× 4 0 0 × 7 0 0
Solution : 4 9 0 8 Solution : 8 0 7 5
× 4 0 0 × 7 0 0
1 9 6 3 2 0 0 5 6 5 2 5 0 0
Answer : 1963200. Answer : 5652500.
Example 12. Multiply by easy method :
(a) 9999 × 425 (b) 8254 × 999
Solution : (a) 9999 × 425 (b) 8254 × 999
= (10000−1) × 425 = 8254 × (1000−1)
= (10000 × 425) − (1×425) = (8254×1000) − (8254×1)
= 4250000 − 425 = 8254000 − 8254
= 4249575 = 8245746
Answer : 4249575. Answer : 8245746.

Example 13. A person earns Tk. 216 a day. How much money does
he earn in a year? [1 year = 365 days]
Solution : 1 year = 365 days
In 1 day he earns Tk. 216
∴ In 365 days he earns Tk. (216 × 365)
Here, 3 6 5
× 2 1 6
2 1 9 0
3 6 5 0
7 3 0 0 0
7 8 8 4 0
∴ In 1 year he earns Tk. 78840
Answer : Tk. 78840.
Remark : For the convenience of multiplication, the small number
is taken as multiplier.
Example 14. There are 2605 betel-nut trees in a garden. How many
such betel-nut trees are there in 316 gardens?
Solution : In 1 garden there are 2605 betel-nut trees
∴ " 316 " " " (2605× 316) "
Here, 2 6 0 5
× 3 1 6
1 5 6 3 0
2 6 0 5 0
7 8 1 5 0 0
8 2 3 1 8 0
Answer : There are 823180 betel-nut trees.

14
Example 15. A goat costs Tk. 3925. What will be the cost of 250
such goats?
Solution : 1 goat costs Tk. 3925
∴ 250 goat costs Tk. (3925 × 250)
Here, 3 9 2 5
× 2 5 0
1 9 6 2 5 0
7 8 5 0 0 0
9 8 1 2 5 0
Answer : Tk. 981250.
Exercise - 2 (a)
1. Find the product :
(a) 2 9 8 (b) 4 2 7 (c) 8 0 5 (d) 2 6 9 0
× 1 6 9 × 2 0 8 × 3 2 9 × 2 1 9
(e) 3 6 9 5 (f) 5 0 4 6 (g) 8 3 6 4 (h) 7 6 9 2
× 3 8 7 × 6 2 0 × 7 0 9 × 8 0 0
2. Multiply:
(a) 835 by 570 (b) 946 by 632 (c) 2694 by 623
(d) 7091 by 780 (e) 9246 by 905 (f) 6953 by 900
3. Multiply by easy method :
(a) 9999 × 321 (b) 9999 × 207
(c) 6258 × 999 (d) 4593 × 999
4. The multiplicand is 6381 and the multiplier is 215. What is the
product?
5. A bicycle costs Tk. 9825. What will be the cost of 135 such bicycles?
6. There are 500 sheets of paper in a rim. How many sheets of paper
are there in 198 rims?
7. A godown has 8326 kg of rice. How much rice is there in 602 such
godowns?
8. A person earns Tk. 275 every day. How much money does he earn
in a year? [1 year = 365 days]
9. A chair costs Tk. 800. What is the cost of 2614 such chairs?
10. There are 539 pages in a book. How many pages are there in 4038
such books?
11. There are 5834 plants in a nursery. How many plants are there in
486 such nurseries?
Multiplication and Division

Division
Example 1. Divide 8304 by 16. Example 2. Divide 64896 by 48.
Solution : 1 6) 8 3 0 4 (5 1 9 Solution : 48) 6 4 8 9 6 (1352
8 0 4 8
3 0 1 6 8
1 6 1 4 4
1 4 4 2 4 9
1 4 4 2 4 0
0 9 6
Answer : Quotient is 519. 9 6
0
Answer : Quotient is 1352..
Example 3. Divide 79213 by 25.
Solution :25) 7 9 2 1 3 (3168
7 5
Example 4. Divide 89250 by 125.4 2
Solution : 125) 8 9 2 5 0 (7142 5
8 7 51 7 1
17 51 5 0
1 2 52 1 3
5 0 02 0 0
5 0 01 3
0Answer : Quotient is 3168,
Answer : Quotient is 714.remainder is 13.
Here, dividend is 8304, divisor
is 16 and quotient is 519.
Here, divisor is 48, dividend is
64896, quotient is 1352 and
remainder is 0.
Here, divisor is 25 ,dividend is
79213, quotient is 3168 and
remainder is 13

Example 5. Divide 98027 by 132. Example 6. Divide 49735 by 245.
Solution: Solution:132) 9 8 0 2 7 (742 245) 4 9 7 3 5 (203
9 2 4 4 9 0
5 6 2 7 3 5
5 2 8 7 3 5
3 4 7 0
2 6 4 Answer : Quotient is 203.
8 3
Answer : Quotient is 742,
remainder is 83.
In example 6, in the second step of
division, divisor 245 is greater than the
dividend 73. So in the quotient, zero (0)
is placed in the tens place. Then 5 of
dividend is written on the right side of
73 and 735 is divisible by 245.
Division by 10 or 100
Solution : 10) 2 4 6 5 7 (2465
2 0
4 6
4 0
6 5
6 0
5 7
5 0
7
Answer : Quotient is 2465, remainder is 7.
There is one zero (0) in the ones place of divisor. In the dividend,
putting a comma after one place from the right, we get 2465, 7.
Here, the number on the left of comma is the quotient and the
number on the right is the remainder. So, whenever we need to
divide any number by 10, it is easier to find the quotient and the
remainder in this way.

Solution : 3981, 2
Answer : Quotient is 3981 and remainder is 2.
Solution : 100 ) 3 0 8 4 5 ( 308
3 0 0
8 4 5
8 0 0
4 5
There are two zeros in the divisor. In the dividend, putting a
comma after 2 digits from the right, we get 308, 45. Here, the
number on the left of the comma is the quotient and the number on
the right is the remainder. So, whenever we need to divide any
number by 100, it is easier to find the quotient and the remainder in
this way.
Solution : Here, 425, 78.
So, quotient is 425 and remainder is 78.
Solution : Here, the divisor is 100. In the divident putting a comma
after two digits from the right we get 635,00.
So, 635 is quotient and 00 is remainder.
To divide by 10 or 100 easily, put comma in the dividend counting
the digits from the right that equal the zeros in the divisor from the
right. Then the number on the left of comma will be the quotient
and the number on the right will be the remainder.

Example 12. In a sum, the dividend is37024, the quotient is 89 and
the remainder is 0. What is the divisor?
Solution : 89) 3 7 0 2 4 (416
3 5 6
1 4 2
8 9
5 3 4
5 3 4
0
Answer : divisor is 416.
In case of exactly
divisible
Dividend ÷ divisor = quotient
Dividend ÷ quotient = divisor
Divisor × quotient = dividend
In case of not exactly
divisible
Dividend = divisor × quotient + remainder
Divisor = (dividend − remainder) ÷ quotient
Quotient = (dividend − remainder) ÷ divisor
Example 13. In a sum, the dividend is 28087, the divisor is 264 and
the remainder 103. What is the quotient?
Solution : We know,
Quotient = (dividend − remainder) ÷ divisor
Here, dividend − remainder = 28087 − 103 = 27984
∴ Quotient = 27984 ÷ 264
= 106
Answer : Quotient is 106.
Example 14. If 325 kg. of rice cost Tk. 5200, what is the cost of
1 kg. of rice?
Solution : 325) 5 2 0 0 (16
3 2 5
1 9 5 0
1 9 5 0
0 ∴ The cost of 1 kg. of rice is Tk.16.
Answer : Tk. 16.

Exercise -2 (b)
1. Divide :
(a) 76695 ÷ 15 (b) 54871 ÷ 37 (c) 25748 ÷ 98
(d) 52895 ÷ 149 (e) 42835 ÷ 197 (f) 52185 ÷ 213
(g) 75089 ÷ 325 (h) 52889 ÷ 289 (i) 43702 ÷ 342
(j) 14970 ÷ 365
2. Divide :
(a) 24135 ÷ 10 (b) 87520 ÷ 10 (c) 43560 ÷ 100
(d) 65900 ÷ 100 (e) 67500 ÷ 100 (f) 39768 ÷ 100
(g) 83090 ÷ 100 (h) 93500 ÷ 100
3. (a) Dividend 37920, quotient 12 and remainder 0. What is the divisor?
(b) Dividend 73635, quotient 152 and remainder 67. What is the divisor?
(c) Dividend 35792, divisor 47 and remainder 25. What is the quotient?
(d) Divisor 325, quotient 72 and remainder 9. What is the dividend?
4. How many days are there in 10000 hours?
5. Divide the greatest number of five digits by 87.
6. There are 355 mangoes in a basket. How many such baskets are
needed for 25560 mangoes?
7. A shopkeeper buys 246 kg. of lentils for Tk. 11808. What is the
cost of 1 kg. of lentils?
8. From the greatest number of 5 digits using the digits 8, 3, 5, 0, 7
once and divide it by the greatest number of 3 digits?
9. Tk. 10500 is required when each person is paid Tk. 140. How
many persons can be paid by this amount?
10. The product of two numbers is 43290. One number is 555. What
is the other number?
11. Divide by 165 the least number formed by taking 9, 4, 0, 2, 6
once.
12 The product of two numbers is 89262. One number is 342.
What is the other number ?

Chapter - 3
Problems Involving the First Four Rules
Example 1. 8 dozens of khatas cost Tk. 2400. What is the cost of
one khata?
Solution : 1 dozen = 12 pieces
∴ 8 dozens = (8 ×12 ) pieces = 96 pieces
Here, 8 dozens i.e., 96 khatas cost Tk. 2400
∴ 1 khata costs Tk. (2400 ÷ 96)
Now, 96) 2 4 0 0 (25
1 9 2
4 8 0
4 8 0
0
∴ 1 khata costs Tk. 25
Answer : Tk. 25.
Example 2. The combined age of father and daughter is 80 years.
The father is 4 times older than the daughter. How
old are they?
Solution : The age of daughter = 1 times of daughterÊs age
FatherÊs age = 4 times of daughterÊs age
∴ Combined age of the father and the daughter = 5 times of
daughterÊs age
5 times of daughterÊs age = 80 years
∴ DaughterÊs age = 80 years ÷ 5 = 16 years
∴ FatherÊs age = 16 years × 4 = 64 years.
Answer : FatherÊs age is 64 years and daughterÊs age is 16 years.

Example 3. In a sum, the divisor is 78, the quotient is 25 and the
remainder is 63. What is the dividend ?
Solution : We know,
Dividend = Divisor × quotient + remainder
= 78 × 25 + 63
= 1950 + 63
= 2013
Answer : Dividend is 2013.
Example 4. The total cost of 6 chairs and 4 tables is Tk. 5570. The
price of a chair is Tk. 675. What is the price of a table?
Solution : 1 chair costs Tk. 675
∴ 6 chairs cost Tk. (675 × 6) = Tk. 4050
6 chairs and 4 tables cost = Tk. 5570
Q 6 chairs cost = Tk. 4050
∴ 4 tables cost = Tk. 1520 (by subtracting)
∴ 1 table costs = Tk. (1520 ÷ 4)
= Tk. 380
Answer : Tk. 380.
Example 5. Mina and Rina together have Tk. 7532. Mina has
Tk. 560 more than Rina. How much money does each
of them have?
Solution : Mina has Tk. 560 more. If Tk. 560 is taken away
from the total amount, then both of them have the
equal amount.
Tk. 7532 ă Tk. 560 = Tk. 6972
∴ Rina has Tk. (6972 ÷ 2)
or, Tk. 3486
Therefore, Mina has Tk. (3486 + 560) = Tk. 4046
Answer : Mina has Tk. 4046 and Rina has Tk. 3486.

22 Problems Involving the First Four Rules
Example 6. Mr. AltafÊs monthly salary is Tk. 9870. He spends
Tk. 3800 for house rent and Tk. 5650 for family
expenditure. He keeps the rest of the money in the
bank. How much money does he save in a year?
Solution : Expenditure for house rent and family affairs
Tk. (3800 + 5650) = Tk. 9450
Deposit in the bank in every month
Tk. (9870 − 9450) = Tk. 420
∴ Savings in 1 year Tk. (420 × 12) = Tk. 5040
Answer : Tk. 5040.
Exercise - 3
1. 9 dozens of pencils cost Tk. 1620. What is the price of a pencil?
2. 38 halis of mangoes cost Tk. 1216. What is the price of a mango?
3. 12 dozens of khatas cost Tk. 2304. What is the price of 8 khatas?
4. The age of father and son together is 96 years. If the fatherÊs age is
3 times than that of the sonÊs, what is the age of the father and
the son?
5. There are 168 mangoes in a basket. From the mangoes that are
there in 15 such baskets, 780 were given to Mina and 750 to Rani.
The rest were given to Raju. How many mangoes did Raju get?
6. In a sum, the dividend is 8903, divisor is 87 and remainder is 29,
what is the quotient?
7. In a sum, the divisor is 12 times greater than the remainder and
the dividend is 9896. The remainder is 8. What is the quotient?
8. The product of two numbers is 6272. 4 times of one of these
numbers is 256. What is the other number?
9. The price of 2 cows and 3 goats together is Tk. 9924. One goat
costs Tk. 1280. What is the price of a cow?

10. The cost of 14 chairs and 6 tables together is Tk. 8890. If one
table costs Tk. 350, what is the cost of a chair?
11. Raju and Rani together have 8690 litchies. Rani has 786 litchies
less than Raju. How many litchies do Raju and Rani have?
12. FaridaÊs and FatemaÊs salaries together is Tk. 9905. FatemaÊs
salary is Tk. 489 more than that of Farida. Find the salary of
each.
13. Mr. Jalal's monthly salary is Tk. 8765. He spends Tk. 3225
as house rent and Tk. 4850 for other expenses. He deposits the
rest of the money in the bank. How much money does he
deposit in the bank in 8 months?
14. Rokeya Begum gets Tk. 3500 as house rents and Tk. 5085 as
pension per month. She spends Tk. 7948 for the education of
her son and other expenses. She saves the rest of the money in
the bank. How much does she save in a year?
15. Zahidul Hasan bought 40 kg. of rice from a departmental store,
soyabene oil for Tk. 265 and fish for Tk. 588. Each kg. of
rice costs Tk. 18. He gave the shopkeeper Tk. 2000. How
much money did the shopkeeper return him?

Chapter - 4
The Unitary Method
Example 1. The cost of 1 khata is Tk. 7. What is the cost of 5 khatas?
Solution : 1 Khata costs Tk. 7
∴ 5 khatas cost Tk. (7 × 5)
= Tk. 35
Answer : Tk. 35.
Let us notice
The price of 5 khatas is obtained by multiplying the price of 1
khata by 5. Multiplying the price of 1 object by a fixed number,
the price of that fixed number of objects can be obtained.
Tick ( ) the correct answer :
(1) 1 pencil costs Tk. 3. What is the price of 6 pencils?
a. Tk. 2 b. Tk. 9 c. Tk. 18 d. Tk.12
(2) 1 khata costs Tk. 5. What is the price of 10 khatas?
a. Tk. 15 b. Tk. 50 c. Tk. 2 d. Tk. 20
Example 2. 3 khatas cost Tk. 21. What is the price of 1 khata?
Solution : 3 khatas cost Tk. 21
∴ 1 " " Tk.
3
21
= Tk. 7
Answer : Tk. 7
Let us notice : The price of 1 khata is obtained by dividing Tk. 21
(which is the price of 3 khatas) by 3.
If the price, weight and length of some objects of the same kind
are given, then by dividing them by the number of objects, the
price, weight and length of an object can be obtained.
√

Tick (√) the correct answer :
(1) The price of 6 pencils is Tk. 12. What is the price of 1 pencil?
a. Tk. 72 b. Tk. 3 c. Tk. 2 d. Tk. 24
(2) The price of 5 mangoes is Tk. 35. What is the price of 1 mango?
a. Tk. 7 b. Tk. 175 c. Tk. 15 d. Tk. 3
(3) 3 kg. of rice cost Tk. 60. What is the cost of 1 kg.?
a. Tk.15 b. Tk. 20 c. Tk. 30 d. Tk. 12
Example3. 10 pieces of pencils cost Tk. 30. What is the cost of 8
pencils ?
Solution : Cost of 10 pieces of pencils is Tk. 30
∴ " 1 " " Tk. (30 ÷ 10)
= Tk. 3
∴ " 8 " " Tk. (3 × 8)
= Tk. 24
Answer : Tk. 24.
Let us notice:
(1) First, dividing the price of 10 pencils by 10, the price of 1 pencil
is obtained. Then multiplying the price of 1 pencil by 8, the cost
of 8 pencils is obtained.
(2) Here, we need to determine the cost of pencils. That is why, while
arranging the numbers the cost of pencil is written at the end.
Since the problem is solved by finding the cost of an object, the
method is known as the Unitary Method.
In the Unitary Method at the time of arranging the sum, the
expression which is to be determined must be written at the end
(right side) of the first line.
Example 4. 9 books cost Tk. 108. What is the cost of 12 books?
Solution : 9 books cost Tk. 108.
∴ 1 " " Tk. (108 ÷ 9)
= Tk. 12
∴ 12 " " Tk. (12 × 12)
= Tk. 144
Answer : Tk. 144.

26 The Unitary Method
Example 5. 100 litchies cost Tk. 80. What is the cost of 25 litchies?
Solution : 100 litchies cost Tk. 80, or 8000 Paisa.
∴ 1 " " (8000 ÷ 100) Paisa.
= 80 Paisa
∴ 25 " " " (80 × 25) Paisa
= 2000 Paisa
= Tk. 20
Answer : Tk. 20.
Example 6. The weight of 1 dozen of apple each of the same
weights is 72 grams. What is the weight of 9 such apples ?
Solution : 12 apples weight 72 grams
∴ 1 (72÷12) grams
= 6 grams
∴ 9 " "
" "
(6 × 9) grams
= 54 grams.
Answer : 54 grams.
Example 7. A labourer earns Tk. 875 per week. How much taka
does he earn in 12 days?
Solution : In 7 days he earns Tk. 875
∴ 1 " " Tk. (875 ÷ 7)
= Tk. 125
∴ 12 "" Tk. (125 × 12)
= Tk. 1500
Answer : Tk. 1500.

Example 8. A hostel of 40 students has food for 20 days. If 10 new
students come to the hostel, how many days will the food last?
Solution : After arrival, the total number of students become = (40 +10)
= 50
40 students have food for 20 days
∴ 1 " " (20 × 40) days
= 800 days
∴ 50 " "
"
" (800 ÷ 50) days
= 16 days
Answer : 16 days.
Let us notice : In the case of consuming the same quantity of food,
if the number of students decreases then the number of days
increases. So in the second step, multiplication is done. Again, if the
number of students increases, the number of days decreases. So in
the third step division is done.
Example 9. 6 persons can cut the crops of a land in 21 days. In how
many days will 18 persons cut the crops of the same land?
Solution : 6 persons can cut crops in 21 days
∴ 1 person " " (21 × 6) days
= 126 days
∴ 18 persons " " (126 ÷ 18) days
= 7 days
Answer : 7 days.

Example 10. 200 men can dig a pond in 15 days. How many more
men are to be appointed to dig the pond in 10 days?
Solution : In 15 days a pond is dug by 200 men
∴ " 1 day " " (200 × 15) men
= 3000 men
∴ " 10 days " " (3000 ÷ 10) men
= 300 men
(300 − 200) = 100 men are required in addition.
Answer : Additional 100 men are to be appointed.
Exercise - 4
1. Put tick (√ ) on the correct answer :
(1) If the cost of a khata is Tk. 8, then to find the cost of 4
khatas, 8 and 4 are to be added / multiplied / divided / subtracted.
(2) If the cost of a pair of pigeon is Tk. 100, then to find the
cost of 3 pairs of pigeon, 100 and 3 are to be multiplied / divided /
added / subtracted.
(3) If the cost of 8 pencils is Tk. 24, then to find out the cost of a
pencil, 24 and 8 are to be multiplied / added / subtracted / divided.
(4) If the cost of 6 khatas is Tk. 24, then to find the cost of 3
khatas, it is necessary first to find the cost of 2 khatas /
5 khatas / 1 khata / 3 khatas from that of 6 khatas.
(5) If the cost of 6 khatas is Tk. 24, then the cost of 1 khata will
be less / equal / more / double.

(6) 3 baskets can contain 48 mangoes. To find out how many mangoes
can contain in 9 baskets, we need to find out first the number of
mangoes contain in 2 / 1 / 3 / 9 basketÊs.
(7) 3 baskets contain 48 mangoes. Then to find the number of
mangoes contained in 1 basket, 48 is to be divided / added / multiplied
/ subtracted by 3 .
2. Fill in the gaps :
(1)  the cost of 1 object by a fixed number, the cost of that fixed
number is obtained.
(2) If the cost of 1 dozen bananas is Tk. 30, then the cost of 3 dozen
bananas is Tk.  .
(3) If the cost of 10 eggs is Tk. 40, then the cost of an egg is
Tk.  .
(4) If the cost of 6 pencils is Tk. 24, then the price of 1 pencil is
Tk.  .
(5) If the cost of 1 pencil is Tk. 4, then the cost of 3 pencils is
Tk.  .
(6) If 3 baskets contain 48 mangoes, then 1 basket contains 
mangoes.
(7) 1 basket contains 8 mangoes. 9 baskets contain  mangoes.
(8) If the number of basket decreases, the number of mangoes .
(9) It the number of basket increases, the number of mangoes also  .
3. Tick (√) the correct answer :
(1) If the cost of 5 pencils is Tk. 20, then what is the cost of 1 pencil ?
(a) Tk. 15 (b) Tk. 4 (c) Tk. 25 (d) Tk. 12
(2) If the cost of 1 khata is Tk. 8, then what is the cost of 8 khatas?
(a) Tk. 1 (b) Tk. 16 (c) Tk. 64 (d) Tk. 20
(3) If the cost of 8 pencils is Tk. 24, then what is the cost of 4 pencils?
(a) Tk. 12 (b) Tk. 32 (c) Tk. 16 (d) Tk. 28
(4) If the cost of 1 dozen eggs is Tk. 48, then what is the cost of 6
eggs?
(a) Tk. 24 (b) Tk. 42 (c) Tk. 72 (d) Tk. 64

4. If the cost of 15 kg. of rice is Tk. 300, then what is the cost of
8 kg. rice?
5. If the cost of 12 metre cloth is Tk. 120, then what is the cost of
15 metre cloth.
6. If the cost of 4 khatas is Tk. 28, then what is the cost of
11 khatas?
7. If the cost of 1 dozen bananas is Tk. 60, then what is the cost of
22 bananas?
8. If the cost of 100 litchies is Tk. 90, then what is the cost of
20 litchies?
9. If the cost of 24 kg. pulse is Tk. 1200, then what is the cost of
13 kg.?
10. If the cost of 5 kg. mutton is Tk. 600, then what is the cost of
8 kg. mutton?
11. If the cost of 6 chairs is Tk. 1200, then what is the cost of
9 chairs?
12. 20 persons can do a work in 15 days. In how many days 15
persons can do the work ?
13. 16 persons can do a work in 5 days. In how many days 20
persons can do the work?
14. 8 persons can reap the crops of a land in 21 days. In how many
days 14 persons can reap the crops of the land?
15. 200 persons have food for 20 days. How many persons will eat
the food in 40 days?
16. A labourer earns Tk. 490 in a week. In how many days will he
earn Tk. 1050?
17. 45 men require a quantity of rice for 20 days. How many men
will consume that quantity of rice in 25 days?

20. A hostel of 500 students has enough food for 50 days. After 10
days 300 students came into the hostel. How long would the
food last?
21. In a family of 8 persons have food for 26 days. After 5 days
one person left the house. How long would the remaining food
last?
22. 200 persons can dig a pond in 25 days. How many more
persons are to be appointed to dig the pond in 20 days?
23. 30 workers can finish a piece of work in 18 days. How many
more workers will be required to complete the work in 12
days?
24. 16 men require 56 kg. of rice for a week. How many kg. of rice
are required for 24 men for six weeks?
25. A hostel had 15 days food for 400 students. After 5 days a few
students left the hostel. The remaining food lasted for 25 more
days. How many students left the hostel?
18. A hostel of 16 students has enough food for 25 days. Some
new students came into the hostel and the food was consumed
in 20 days. What is the number of new students?
19. 1200 soldiers in a fort have enough food to last for 28 days.
400 soldiers left the fort. How long would the food last?

Chapter - 5
Simplification
Example 1. Simplify : 48 − 65 + 378 + 39 − 116 − 7 − 147 + 248
Solution : 48 − 65 + 378 + 39 − 116 − 7 − 147 + 248
= (48 + 378 + 39 + 248) − (65 + 116 + 7 + 147)
= 713 − 335
= 378
Answer : 378.
Let us notice :
(a) The positive numbers are added together ;
48 + 378 + 39 + 248 = 713
(b) The negative numbers are added together ;
65 + 116 + 7 + 147 = 335
(c) Then the second added number is subracted from the first
added number ; 713 − 335 = 378
(d) To add the numbers of the same signs brackets ( ) are used.
Example 2. Simplify : 12 + 29 × 64 ÷ 8 × 3 − 52 ÷ 13 − 8 × 30 ÷ 6
Solution : 12 + 29 × 64 ÷ 8 × 3 − 52 ÷ 13 − 8 × 30 ÷ 6
= 12 + 29 × 8 × 3 − 4 − 8 × 5
= 12 + 232 × 3 − 4 − 40
= (12 + 696) − (4 + 40)
= 708 − 44
= 664
Answer : 664.
Let us notice :
In the sum there are functions of addition, subtraction,
multiplication and division.
First the division, later the multiplication and at the end the
addition and subtraction have been carried out.

Example 3. Simplify : 5 × 24 ÷ 8 × 3 + 5 − 17 + 18 ÷ 3
Solution : 5 × 24 ÷ 8 × 3 + 5 − 17 + 18 ÷ 3
= 5 × 3 × 3 + 5 − 17 + 6
= 5 × 9 + 5 − 17 + 6
= 45 + 5 − 17 + 6
= (45 +5 + 6) − 17
= 56 − 17
= 39
Answer : 39.
Example 4. Simplify : 25 ÷ 5 × [5 × {24 ÷ (18 − 15)}]
Solution : 25 ÷ 5 × [5 × {24 ÷ (18 − 15)}]
= 25 ÷ 5 × [5 × {24 ÷ 3}]
= 25 ÷ 5 × [5 × 8]
= 25 ÷ 5 × 40
= 5 × 40
= 200
Answer : 200.
Let us notice :
• In the sum there are functions of first brackets ( ), second
brackets { }, third brackets [ ] , division, multiplication and
subtraction.
• The functions inside the first brackets, second brackets and third
brackets have been carried out successively.
Example 5. Simplify : 38− [13× (73− 46 ÷ 2×3) − {45 − 3 (7 +21 ÷7)}]
Solution : 38 − [13 × (73 − 46 ÷ 2 × 3) − {45 − 3 (7 + 21 ÷ 7)}]
= 38 − [13 × (73 − 23 × 3) − {45 − 3 (7 + 3)}]
= 38 − [13 × (73 − 69) − {45 − 3 × 10}]
= 38 − [13 × 4 − {45 − 30}]
= 38 − [13 × 4 − 15]
= 38 − [52 − 15]
= 38 − 37
= 1
Answer : 1.

34 Simplification
Example 6. Simplify : 78 − [56 + {165 − (48 ÷ 6 × 9) × 2}]
Solution : 78 − [56 + {165 − (48 ÷ 6 × 9) × 2}]
= 78 − [56 + {165 − (8 × 9) × 2}]
= 78 − [56 + {165 − 72 × 2}]
= 78 − [56 + {165 − 144}]
= 78 − [56 + 21]
= 78 − 77
= 1
Answer : 1.
Example 7. Simplify : 48 ÷ [4 + 28 ÷ {4 + 12 ÷ (7 − 3)}]
Solution : 48 ÷ [4 + 28 ÷ {4 + 12 ÷ (7 − 3)}]
= 48 ÷ [4 + 28 ÷ {4 + 12 ÷ 4}]
= 48 ÷ [4 + 28 ÷ {4 + 3}]
= 48 ÷ [4 + 28 ÷ 7]
= 48 ÷ [4 + 4]
= 48 ÷ 8
= 6
Answer : 6.
Exercise - 5
Simplify :
1. 80 + 47 − 43 + 24 − 49
2. 688 − 589 + 321 − 301 + 81
3. 56 ÷ 7 − 9 + 121 ÷ 11
4. 2 × 3 × 18 ÷ 6 × 45 ÷ 9
5. 16 × 4 + 168 ÷ 8 − 65 + 48
6. 13 × 4 + 132 ÷ 12 − 7 × 52 ÷ 13 − 12 ÷ 3 × 5
7. 49 ÷ 7 − 64 ÷ 8 + 81 ÷ 9 − 42 ÷ 6
8. 12 × 14 + 85 × 5 − 41 × 4 − 6 × 3

9. 32 − 2 × 4 + 12 × 3 − 6 × 5 × 7 + 15 × 8 × 3
10. 12 ÷ 6 − 6 ÷ 2 + 15 ÷ 5
11. 16 ÷ 4 + 15 ÷ 3 × 2
12. 9 × 37 − 6 × 42 + 4 × 82 − 3 × 32 + 8 − 2 × 3
13. (18 − 9) × 5 + (20 + 3 − 15)
14. (36 ÷ 3) × {4 × (5 + 4 − 8 + 1)}
15. 98 − [25 + 3 (11 + 3) − {36 − 3 (12 − 7)}]
16. 405 − {146 + (62 × 4 − 93 + 15 + 3)}
17. 7 − {502 ÷ (25 ÷ 5 × 5 − 23)} + 251
18. 36 ÷ 6 × {7 × (15 − 9 − 11 + 7)}
19. 65 − [25 + 2 (13 + 8) − {25 − 2 (13 − 8)}]
20. 81 ÷ [6 + 39 ÷ {4 + 90 ÷ (33 − 23)}]
21. 270 − 5 [4 + 81 ÷ {93 − (60 ÷ 5) (35 − 28)}]
22. 107 + 91 ÷ 13 − {77 − (85 ÷ 5 × 3 + 21)} − 105
23. 82 + [3 {(99 ÷ 11) (18 ÷ 6)}] − 140
24. 49 ÷ [5 + 24 ÷ {5 + 70 ÷ (33 − 23)}]
25. 67 − [13 × (73 − 46 ÷ 2 × 3) − {45 − 3 (7 + 21 ÷ 7)}]
26. [105 + 91 ÷ 13 − {77 − (85 ÷ 5 × 3 + 21)}] −106
27. 65 5 [4 + 81 ÷ {93 − (60 ÷ 5) (35 − 28)}]
28. [36 ÷ {2 + 28 ÷ (4 + 12 ÷ 4)}] ÷ 6

It is seen in the above picture that there are 11 mangoes in the first
basket, 6 mangoes in the second basket and 13 mangoes in the third
basket. If the mangoes of the three baskets are kept in a big basket,
then the number of mangoes in the big basket becomes (11 + 6 +13) = 30.
If there are 10 mangoes in each of the three baskets then the total
number of mangoes will be 10 + 10 + 10=30. In this case we say, there
are 10 mangoes in the three baskets in average.
Here, 10 = 30 ÷ 3
= Total number of mangoes in three baskets ÷ number of baskets.
Generally,
Average of more than one expression of the same kind = Sum of the
expressions ÷ number of the expressions.
In the above example the total number of mangoes in the three baskets
was 30 and in average there were 10 mangoes in a basket.
Here, 10 × 3 = 30.
Generally,
Average × number of expressions = sum of the expressions
Chapter - 6
Average
11 6
13
30
10
10 10
1st basket 2nd basket 3rd basket

Determination of Average
Example 1. The ages of Mina, Mithu, Runu, Rani and Nila are 10,
11, 12, 13 and 14 years respectively. What is the average age of
them?
Solution : Here the sum of expressions = (10+11+12+13+14) years
= 60 years
Number of the expressions = 5
∴ The average of age = 60 years ÷ 5
= (60 ÷ 5) years = 12 years.
Answer : The average age is 12 years.
Example 2. A cricket player scored 23, 12, 37, 3, 14, 13 runs in six
Solution : Total runs = 23+12+37+3+14+13 = 102
Number of matches = 6
Therefore, the average of run = 102 ÷ 6 = 17
Answer : He scored 17 runs in average.
Example 3. The rainfall in 12 months of a place in Bangladesh in
2004 is as follows :
Month Quantity of rainfall
23 m.m.
06 m.m.
18 m.m.
57 m.m.
31 m.m.
108 m.m.
296 m.m.
271 m.m.
104 m.m
33 m.m.
21 m.m.
04 m.m.
What is the average rainfall in a month of 2004 of that place?
one day matches series. What is his average run in that series?
January
February
March
April
May
June
July
August
September
October
November
December

38 Average
Solution : Total rainfall
=(23+06+18+57+31+108+296+271+104+33+21+04) m.m.
= 972 m.m.
Number of months = 12
Monthly average of railfall = (972 m.m.) ÷ 12
= (972 ÷ 12) m.m.
= 81 m.m.
Answer : Monthly average of rainfall of that place in 2004 is 81 m.m.
Example 4. In a joint business Helen invested Tk. 15,000, Kamal
Tk. 20,500, Dalim Tk. 18,000, Titas Tk. 13,500 and Shishir Tk. 17,000.
How much did they invest in average?
Solution : 15,000
20,500
18,000
13,500
17,000
Total : 84,000
Therefore, total investment = Tk. 84,000
Number of investment = 5
Therefore, average investment = Tk. 84,000 ÷ 5
= Tk. (84,000 ÷ 5)
= Tk. 16,800
Answer : Tk. 16,800
Example 5. In a test series of five matches, the average run of six
batsmen of the visiting team was 76 and the average run of four bowlers
was 21. In that series how many runs in average did the players
score?
Solution : Total runs of the batsmen in 5 games.
= 6×76×5 = 76×30 = 2280
Total runs of the bowlers in 5 games
= 4×21×5 = 21×20 = 420
∴ Grand total of runs = 2280 + 420 = 2700
∴

The number of expressions to be considered to determine the average
= Number of players × Number of matches.
= 10 × 5 = 50
Therefore, average = 2700 ÷ 50 = 54
Answer : 54 runs.
Example 6. The sum of 13 numbers is 1924 and the average of 7
numbers is 172. What is the average of remaining 6
numbers? What is the average of all the numbers?
Solution : The sum of 7 numbers = 172×7 = 1204
So, the sum of remaining 6 numbers = 1924−1204 = 720
Hence the average of 6 numbers = 720 ÷ 6 = 120
The average of all numbers i.e. of 13 numbers
= 1924 ÷ 13 = 148
Answer : The average of other 6 numbers is 120. The average of
all numbers is 148.
Example 7. A bus is hired for Tk. 2400 by 60 employees of an
industry to go to a picnic on the condition that everybody
will bear the expenses equally. But some of the employee
did not go to the picnic and thus each employee spent an
extra amount of Tk. 8
.
00. How many employees went to
the picnic?
Solution : If 60 employees had gone to the picnic then the fare
given by each employee would have been Tk. 2400 ÷ 60
= Tk. 40.
Some employees did not go to the picnic, then fare per
employee was Tk. 40 + Tk. 8 = Tk. 48.
Fare per head × Number of the participants in the picnic
= Tk. 2400
Therefore, the number of participants in the picnic = 2400÷ 48 = 50
Answer : 50 employees went to the picnic.

40 Average
Exercise - 6
1. Find the average :
(a) 23, 37, 47, 61 (b) 22, 46, 60, 72
(c) Tk. 364, Tk. 541, Tk. 775.
(d) 47 c.m, 49 c.m, 54 c.m, 52 c.m, 53 c.m.
(e) 42 kg, 32 kg, 37 kg, 29 kg, 41 kg, 35 kg.
(f) 90 litre, 92 litre, 89 litre, 86 litre, 93 litre, 85 litre, 95 litre.
2. Lily bought 8 tapes of various prices. The average price of the tapes
is Tk. 5
.
75. What is the total cost of the tapes Lily bought?
3. In annual examination Molly obtained 68 in Bangla, 96 in
Mathematics, 81 in English, 77 in Environmental Studies-Science, 73
in Environmental Studies-Social. What was her average number in
each subject ?
4. The heights of Koli, Doli, Poli, Moli and Lily are respectively 123
c.m, 131 c.m, 135 c.m, 126 c.m and 130 c.m. What is their average
height ?
5. The average age of Apu and Dipu is 22 years and that of Dipu and
Tipu is 23 years. The age of Apu is 21 years. What are the ages of
Dipu and Tipu?
6. The sum of 7 numbers is 401. The average of their first 3 numbers is
56 and that of their last 3 numbers is 58. What is the fourth number?
7. Among 11 numbers, the average of first 6 numbers is 87 and the
average at last 5 numbers is 131. What is the average of all
numbers?
8. In a union, the population of 5 villages are respectively 1327, 1872,
2187, 2516 and 2943. What is the average population of the
villages?
9. In annual examination the marks ten students secured in
mathematics are as follows: 76, 61, 87, 56, 42, 64, 73, 68, 50, 73.
What is their average number?
10. The average age of three children and their father is 17 years. The
average age of those three children and their mother is 16. The
motherÊs age is 34 years. What is the age of the father?
11. In Dhaka in the month of Ashar the daily average rainfall in first 10
days is 23 m.m, in second 10 days is 27 m.m, in third 10 days is 29
m.m and in the last day is 16 m.m. What is the average rainfall in
that month in Dhaka?

Chapter - 7
H.C.F. and L.C.M.
Let us consider the factors (divisors) of three numbers 18, 24, 30.
All factors of 18 : 1 , 2 , 3 , 6 , 9, 18
All factors of 24 : 1 , 2 , 3 , 4, 6 , 8, 12, 24
All factors of 30 : 1 , 2 , 3 , 5 , 6 , 10, 15, 30
So, the common factor / divisor of the three numbers are 1, 2, 3, 6.
Among them 6 is the greatest. So, 6 is the highest common factor or
H.C.F. of 18, 24, 30.
The common prime factors of the three numbers are 2 and 3, whose
product is equal to H.C.F.
The greatest common divisor (factor) of more than one
number is their Highest Common Factor (H.C.F.)
H.C.F. of more than one number = the product of common
prime factors. If the numbers do not have any common prime
factor then their H.C.F. is 1.
The other name of divisor is factor.
Determination of H.C.F. by prime factorization :
Example 1. Find the H.C.F. of 40, 60 and 75.
Solution : 40 = 2×20 = 2×2×10 = 2×2×2×5
60 = 3×20 = 3×2×2×5
75 = 3×25 = 3×5×5
It is seen that, 5 is the only prime factor (divisor) of the three
numbers 40, 60, 75.
∴ Their H.C.F. is 5.
Answer : 5.
Highest Common Factor (H.C.F.)

42 H.C.F. and L.C.M.
Example 2. Find the H.C.F. of 48, 72, 168.
Solution : 48 = 6×8 = 2×3×2×2×2 = 2×2×2×2×3
72 = 8×9 = 2×2×2×3×3
168 = 8×21 = 2×2×2×3×7
It is seen that, 2, 2, 2, 3 are the prime factors of the three
numbers 48, 72, 168.
So, the H.C.F. of the three numbers = 2×2×2×3 = 24
Answer : 24.
Example 3. Find the H.C.F. of 24, 30, 77.
Solution : 24 = 3×8 = 3×2×2×2 = 2×2×2×3
30 = 5×6 = 5×2×3 = 2×3×5
77 = 7×11
Answer : 1.
If more than 1 numbers do not have common prime
factor then their H.C.F. is 1
Determination of H.C.F. by the method of division
To determine the H.C.F. of 18 and 84, first, we divide the large
number by the smaller number. Here, dividing 84 by 18 we get the
remainder 12. In the second step, we divide the smaller number by
the remainder 12. Now the remainder is 6. In the third step, we
divide the remainder of the previous step by the remainder 6 of that
step, 12 = 6 2, thus the remainder becomes zero and the process of
division is finished. The last divisor is the H.C.F. of the given
numbers.
It is seen that, the three numbers 24, 30, 77 have no common
prime factor. So, 1 is the only common prime factor of the given
numbers. So 1 is their H.C.F.

This is shown below in a normal process :
18) 84 (4
72
12) 18 (1
12
6) 12 (2
12
0
Explanation : Always, dividend = quotient × divisor + remainder.
The above example has three steps :
(a) 84 = 4×18 +12 (b) 18 = 1×12 + 6 (c) 12 = 2× 6
It is seen from (c) that 6 is the factor of 12. So from (b) it is seen that 6
is one of the factors of 18. Then from (a) it is seen that 6 is a factor of
84. Hence, 6 is the common factor of 18 and 84.
Conversely, it can be understood from (a) that the common factor of
84 and 18 will also be a factor of 12. So it can be observed from (b)
that the factor will also be a factor of 6 i.e any common factor of 18
and 84 is also a common factor of 6. Hence, 6 is the highest common
factor (or H.C.F.) of 18 and 84.
The method of finding H.C.F. by division is known as EuclidÊs
Method.
Example 4. Find the H.C.F. of 171 and 561.
Solution : 171) 561 (3
513
48) 171(3
144
27) 48 (1
27
21 ) 27 (1
21
6 ) 21 (3
18
3 ) 6 (2
6
0
Required H.C.F. = Last divisor = 3
Answer : 3

In finding the H.C.F. by EuclidÊs Method, the quotients obtained in
different steps have no role; the last remainder is the required H.C.F.
In one application of EuclidÊs Method the H.C.F. of only two,
numbers can be determined.
Solution : At first find the H.C.F. of 231 and 385.
231 ) 385 (1
231
154 ) 231 (1
154
77 ) 154 (2
154
0
Therefore, the H.C.F. of 231 and 385 is 77.
Now we find the H.C.F. of 77 and 143.
77 ) 143 (1
77
66 ) 77 (1
66
11 ) 66 (6
66
0
Answer : 11.
Explanation : Any common factor of 143, 231 and 385 is also factor
of 77 and 385. Again, any factor of 77 and 385 is also factor of 143,
231 and 385. So, H.C.F. of 143, 231 and 385 = H.C.F. of 77 and 385.
Remarks : Example 5. could be solved by two other methods :
First by finding the H.C.F. of 143 and 231;
or, by finding the H.C.F. of 143 and 385.
As we have at first determined the H.C.F. of two greater numbers, the
work has become easier.

Solution : First, we determine the H.C.F. of 304 and 418.
304 ) 418 (1
304
114) 304 (2
228
76) 114 (1
76
38) 76 (2
76
∴ The H.C.F. of 304 and 418 is 38. 0
Now, we find the H.C.F. of 38 and 231.
38 ) 231 (6
228
3 ) 38 (12
3
8
6
2) 3 (1
2
1) 2 (2
2
0
So, the H.C.F. of 38 and 231 is 1. Hence, the H.C.F. of the three
given numbers is 1.
Answer : 1.
Lowest Common Multiple (L.C.M)
Some of multiples of 18, 24 and 27 are written :
Multiples of18 : 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198,
216, 234, etc.
Multiples of 24: 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, etc.
Multiples of 27: 27, 54, 81, 108, 135, 162, 189, 216, 243, etc.
216 is found in every row of the multiples. So, the number 216 is
common multiple of 18, 24 and 27. There is no number in any of the
three rows which is less than 216. So, 216 is the Lowest or
Least Common Multiple (L.C.M) of 18, 24 and 27.

Find the L.C.M. by Prime Factors.
Example 7.
Solution :
Solution :
Find L.C.M. of 18, 24 and 30.
18 = 2 × 3 × 3
24 = 2 × 2 × 2 × 3
30 = 2 × 3 × 5
In the prime factors of three numbers 2 occurs in maximum 3 times
(as factors of 24) and 3 occurs twice (as factors of 18) and 5 occurs
once (as factors of 30). So, the L.C.M. of 18, 24, 30
= 2 × 2 × 2 × 3 × 3 × 5
= 4 × 9 × 10 = 360
Answer : 360.
Determination of L.C.M. by Short Method.
In this method all prime factors of the given numbers are obtained
one by one. The product of these prime factors is the L.C.M. of the
given numbers.
Example 8. Find the L.C.M. of 18, 24 and 40.
2 18, 24, 40
2 9, 12, 20
2 9, 6, 10
3 9, 3, 5
3, 1, 5
So the required L.C.M. = 2×2×2×3×3× 5 = 360
Answer : 360.
Explanation : 24 is divided by 2 thrice. The last quotient is 3 which
is the prime number. So, 24 = 2 × 2 × 2 × 3.
18 is divided by 2 once and the quotient 9 is divided by 3 twice. The
last quotient is 1. So, 18 = 2 × 3 × 3.

40 is divided by 2 thrice, the last quotient is 5, which is the prime
number. So, 40 = 2 × 2 × 2 × 5.
The method of finding L.C.M. by short method is the unified form
of finding the prime factor.
In each step at least two of the given numbers are divided by a
suitable prime number. The quotients of the numbers exactly
divisible by the factors are written below the lines and indivisible
numbers are written again. Similar work has been done with the
numbers of the next one.
Example 9. Find the L.C.M. of 16, 24, 30, 42, 45.
Solution : 2 16, 24, 30, 42, 45
2 8, 12, 15, 21, 45
2 4, 6, 15, 21, 45
3 2, 3, 15, 21, 45
5 2, 1, 5, 7, 15
2, 1, 1, 7, 3
So required L.C.M. = 2 × 2 × 2 × 3 × 5 × 2 × 7 × 3 = 5040
Answer : 5040.
Easy Problems Concerning H.C.F. and L.C.M.
Example 10. What is the greatest number by which dividing 138,
215, 457 leaves remainders of 3, 5, 7 respectively?
Solution : 138 − 3 = 135, 215 − 5 = 210, 457 − 7 = 450.
135, 210 and 450 will be divisible by the required number.
This number is the H.C.F. of 135, 210, 450.
210) 450 (2
420
30) 210 (7
210
So, H.C.F. of 210 and 450 is 30. 0
∴ The greatest number is 15. Again, H.C.F. of 30 and 135 is 15.
Answer : 15.
30) 135(4
120
15) 30 (2
30
0
Here, L.C.M. of 18, 24, 40 = 2×2×2×3×3×5 = 360.

Example 11. Four bells rang together at first and then they started
to ring at an interval of 6, 9, 12 and 15 seconds respectively. When
will they ring together again?
Solution : Minimum time to be determind is equal to the L.C.M. of 6, 9,
12, 15.
3 6, 9, 12, 15
2 2, 3, 4, 5
1, 3, 2, 5
So, the L.C.M. of the four numbers = 3×2×3×2×5 = 18×10 = 180.
Hence, four bells will ring after minimum 180 seconds = 3 minutes.
Answer : 3 minutes.
Example 12. Find the least number which when divided by 6, 10,
15 and 21 will leave remainders 4 in each case.
Solution : The least number to be determind will be 4 more than
the L.C.M. of 6, 10, 15 and 21.
3 6, 10, 15, 21
2 2, 10, 5, 7
5 1, 5, 5, 7
1, 1, 1, 7
So, the L.C.M. of the numbers =3 × 2 × 5 × 7 = 21 × 10 = 210.
Hence, the required least number = 210 + 4 = 214.
Answer : 214.
Example 13 : Find the least number which when divided by 4, 6,
10, 16 will leave remainders 2, 4, 8, 14 respectively.
Solution : 4 − 2 = 2, 6 − 4 = 2, 10 − 8 = 2, 16 − 14 = 2
i.e. in each case, divisor − remainder = 2; so the required
least number of 4, 6, 10 and 16 will be 2 less than the L.C.M.
2 4, 6, 10, 16
2 2, 3, 5, 8
1, 3, 5, 4
So, the L.C.M. of the numbers = 2×2×3×5×4 = 12×20 = 240.
Hence, the required least number = 240 − 2 = 238.
Answer : 238

Example 14. A rectangular room is of 7.20 metres in length and
4.40 metres in breadth. What is the biggest size of
marble slab that can be used to cover the floor without
breaking any part of it?
Solution : The length of the room = 7.20 m. = 72 d.m., breadth =
4.40 m. = 44 d.m. As the stone is square and
nonbreakable, the length of its side should be common
factor of length and breadth . So the length of the
greatest square size stone in d.m. will be equal to the
H.C.F. of 72 and 44.
72 = 8×9 = 2×2×2×3×3
44 = 4 ×11 = 2×2×11
∴ H.C.F. of 72 and 44 = 2×2 = 4
4 d.m. = 40 c.m.
Answer : The length of the side of the greatest square size stone is 40 c.m.
Exercise - 7
1. Find the H.C.F. by prime factor :
(a) 144, 192 (b) 160, 275 (c) 112, 343
(d) 24, 60, 72 (e) 45, 75, 120 (f) 22, 77, 110
(g) 48, 72, 160, 240 (h) 35, 56, 84, 119 (i) 30, 50, 90, 140
(j) 36, 48, 84, 156
2. Find the H.C.F. by EuclidÊs Method :
(a) 90, 325 (b) 76, 361 (c) 115, 810
(d) 154, 484 (e) 336, 730 (f) 1769, 2378
(g) 77, 231, 352 (h) 117, 183, 303 (i ) 56, 147, 357
(j ) 75, 225, 390
3. Find the L.C.M. by prime factor :
(a) 51, 85 (b) 76, 95 (c) 42, 112
(d) 32, 48, 80 (e) 35, 55, 75 (f) 28, 42, 70

4. Find the L.C.M. by short method :
(a) 24, 36, 40 (b) 15, 33, 45 (c) 25, 45, 85
(d) 21, 35, 49, 56 (e) 24, 36, 48, 72 (f) 18, 27, 45, 54
(g) 26, 44, 77, 99 (h) 12, 18, 24, 30, 42 (i) 15, 25, 35, 45, 55
5. What is the greatest number which if divided by 76, 114 and 228
will leave no remainder?
6. What is the highest number of boys and girls among whom 60
mangoes and 150 litchies may be distributed equally. How many
mangoes and litchies will each get?
7. Two drums have the capacity to contain 228 litres and 348 litres
of liquid. What should be the highest capacity of a pitcher that can
be used in full number of times to fill both the drums. How many
pitchers of water each drums can contain?
will leave 2 in each case?
will leave remainder of 3, 5 and 7 respectively.
10. A rectangular hall room is 12.25 metres in length an d 7.75 metres
in breadth What is the biggest square size tile that can be used to.
cover the floor of the room without breaking any of them?
11. What is the least number which if divided by 18, 24, 30, 36 will
leave no remainder?
12. What is the least number which when divided by 16, 24, 32, 40
gives the same remainder 6?
13. What is the least number which when divided by 12, 18, 30
gives the remainders 6, 12, 24 respectively ?
14. Four bells rang at the same time and then in an interval of 5, 7,
12 and 15 seconds respectively. After what interval of time will
they again ring together?
15.
16. What is least number which is exactly divisible by 15, 18, 20, 24
and 32 after adding 7 to it ?
Some plants are planted in rows each contains 3, 5, 6, 8, 10
and 15 numbers, in each case two remains left. What is the least
number of the plants?

Chapter - 8
Mathematical Symbols and Sentences
Various types of symbols are used in mathematics. e.g.
Numeral symbols
0 1 2 3 4 5 6 7 8 9
Zero One Two Three Four Five Six Seven Eight Nine
all numbers can be written by these ten symbols. The numeral
symbols used in writing number are known as digits. As for examples,
hundreds digit is 7.
It is noticed : 1, 2, 3, 4, 5, 6, 7, 8, 9 each of these numbers is a
number of one digit.
Procedural Symbols :
+ − × ÷
Plus sign Minus sign Multiplication
sign
Division sign
Relation Symbols :
= > < =/ >/ </
Equal Greater Less Not equal Not greater Not less
Bracket Symbols :
( )) { }{ } [ ]]
First bracket Second bracket Third bracket
Special Symbols : e.g.
Letter Symbols : e.g., x, y, z.
Example 1. Express the problem by symbol and find out the
unknown number.
(a) If we add a number with 19 the sum is 72. What is
that number?
(b) If we divide a number by 7, the quotient is 15. What
is that number?
786 is the number of 3 digits whose ones digit is 6, tens digit is 8 and
To indicate unknown quantity, the special symbol or letter symbol is used.

52 Mathematical Symbols and Sentences
Solution : (a) Choosing the symbol for the unknown number the
problem becomes :
+ 19 = 72 then, = what ?
Subtraction is the reverse process of addition.
So, the number will be the difference of 19 from 72.
∴ = 72 − 19 = 53
Answer : 53.
(b) Choosing the letter symbol X for the unknown number the
problem becomes :
X ÷ 7 = 15 then, X = what ?
Multiplication is the reverse process of division, so the
number X will be the product of 15 and 7.
∴ X = 15 × 7 = 105
Answer : 105.
Numeral Expression :
Combining various numbers by procedural symbol and
necessary brackets a numeral expression is formed e.g.,
(36 ÷ 4) × 5 − 7
Mathematical sentence containing special symbols or letter
symbols is known as Open Sentence.
Example 2. Form mathematical open sentence from the following
sentences using letter symbols . „33 litchies are distributed among a few
boys and girls. Everyone has got 6 litchies and 3 remained
undistributed‰.
Solution: Here, the number of boys and girls is unknown.
Let us suppose that X is the number of boys and girls.
Every one has got 6 litchies;
So X persons have got 6 × X litchies.
According to the question 6 × X + 3 = 33
If the 3 undistributed litchies are left then the number of litchies
is = 33 − 3 = 30. ∴ 6 × X = 30
30 is the 6 times of X. So X is one sixth of 30. i.e.
X = 30 ÷ 6 = 5
Answer : Required open sentence : 6 × X + 3 = 33
The value of X (number of boys and girls) = 5.

Example 3. Form Mathematical open sentence from the following
sentences using symbol ≤ (small and equal) the same symbol as >/
and letter symbols. Determine whether the sentence is true in terms
of the different values of the letter symbol.
„21 lozenges are equally distributed among 5 children.‰
Solution : Let us suppose that x number of lozenges is ditributed to
each child. Here, the value of x may be 1, 2, 3... .... So 5
children got 5 × x lozenges. According to the question this
number is either less than 21 or equal to 21. i.e. 5× x ≤ 21
Now 5 × 1 = 5, 5 × 2 = 10, 5 × 3 = 15 and 5 × 4 = 20.
Every one of these multiples is less than 21. But 5 × 5 = 25,
5 × 6 = 30,...; So, other multiples of 5 are greater than 21.
So the open sentence is true for x. when its values are only
1, 2, 3, 4.
Answer : Required open sentence 5 × x ≤ 21.
This is true if and only if x = 1, 2, 3 or 4
Example 4. Form the indicated numeral expression using bracket
and find the expression .
(a) Dividing 72 by 8, the quotient is multiplied by 3.
(b) 72 is divided by the product of 8 and 3.
Solution : (a) Required numeral expression (72 ÷ 8) × 3.
Determination of value : 72 ÷ 8 = 9 and 9 × 3 = 27 (Answer)
(b) Required numeral expression 72 ÷ (8 × 3).
Determination of value : 8 × 3 = 24 and 72 ÷ 24 = 3 (Answer)
Mathematical Sentence and Open Sentence
Mathematical sentence is such a statement containing, numbers,
symbols, expressions or mathematical conception which can be
easily ascertained true or false. e.g., (72 ÷ 8) × 3 = 72 ÷ (8 × 3)
From example 4 we see that this mathematical sentence is false.
The example of true mathematical sentence is „Every triangle has
three sides and three angles‰

The statement concerning mathematical affairs which can be ascertained
easily as false or true is known as mathematical statement, e.g.,
Statement True / False
87 is a prime number. False
The triangle in the picture False
is a right angle triangle.
81 is a square number. True
The mathematical statement containing letter symbol or unknown
number or expression is known as open sentence. e.g.,
72 ÷ (8 × 3) = (x ÷ 8) × 3
Example 5. What is the value of the letter symbol for which the
open sentence (x ÷ 8) × 3 = 72 ÷ (8 × 3) is true ?
Solution : 72 ÷ (8 × 3) = 72 ÷ 24 = 3.
∴ (x ÷ 8) × 3 = 3 (This should be)
As 1 × 3 = 3, so x ÷ 8 = 1
∴ 8 ÷ 8 = 1. Hence, x = 8
Answer : 8
Successive addition
Example 6. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = How much ?
Solution : Let us take the required summation as = x
On one side, x = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
On the other side, x = 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1
Adding two expressions of right side termwise we get
x + x = 16 + 16 + 16 + 16 + 16 + 16 +16 + 16
or, 2 × x = 16 × 8 = 128
∴ x = 128 ÷ 2 = 64
Answer : 64.

Explanation : The expression of the example is a series. The terms
of the original series are written in reverse order. Then adding them
term wise we have got the same number 16 in each term. In that case
the number of terms of the series = Number of terms of the original
series = 8.
So, 2 × x = 16 × 8.
It can be noticed that every term of the given series has 2 in excess
from the previous term i.e. the difference of a term from previous
term is a fixed number. Such a series is known as arithmetic
progression. The sum of terms of any arithmetic progression may
be obtained by the method shown below :
Example 7. 1 + 3 + 9 + 27 + 81 = How much?
Solution : This series, each term is three times of the previous term.
Let the required summation = x, then
x = 1 + 3 + 9 + 27 + 81
Therefore, 3× x = 3 + 9 + 27 + 81 + 243
Again, 1 × x = 1 + 3 + 9 + 27 + 81
Subtracting two expressions to the right side
3 × x − 1 × x = − 1 + 243 [ Equal term is deleted ]
or, 2 × x = 242
∴ x = 242 ÷ 2 = 121
Answer : 121.
Explanation : In this example, each term is three times of the
previous term. For this reason, 3×x series is formed and from this
series x = 1×x series is subtracted. Eventually from the last term of
3×x series first term of x series is subtracted as the other terms of the
two series are arranged as indicated. This type of series is called
geometric progression. The summation of the terms of any
geometric progression may be obtained by the shown method.

Exercise - 8
1. Express by symbol :
(a) Three lac thirty.
(b) Five crore fifty two lac seventy thousand seven hundred
seven.
(c) Dividing three hundred forty three by seven and
subtraction of twenty nine from the quotient.
(d) Division of eight times of forty by the difference of fifty
one and twenty six.
(e) Subtracting seven thousand nine hundred seventy two from
eight thousand five hundred eighty seven, multiplication of
that difference by sum of one hundred twenty seven and
eighty two.
2. Express by using procedural symbol and relation symbol:
(a) The difference of sum of three hundred forty seven and
four hundred twenty three from eight hundred eighteen is
equal to difference of three hundred forty seven and then
that of four hundred twenty three from eight hundred
eighteen.
(b) The product obtained by multiplying one hundred twenty
seven by sum of forty three and sixteen is equal to sum of
the products of one hundred twenty seven with forty three
and one hundred twenty seven with sixteen.
(c) The number obtained by multiplying the quotient of
seventy five and five with three is not equal to the number
obtained by dividing seventy five by the product of five
and three.
3. Find out whether the following statements are true and write
True / False at the end of each statement :
(a) 347 − (189 − 65) = 347 − 189 + 65
(b) (285 ÷ 5) × 3 = 285 ÷ (5 × 3)
(c) (134 + 78 + 49) × 7 = 134 × 7 + 78 × 7 + 49 × 7

4. What are the values of letter symbol for the open sentences
below are true ?
(a) (x + 15) × 3 = 90
(b) 7 × x < 50
(c) (x ÷ 6) + 5 = 12
5. Form open sentence putting letter symbol in the place of
unknown number and find the unknown number :
(a) Montu had a few marbles. He gave four marbles to Pintu.
Thus Montu had eight marbles left.
(b) Piya has some money. Her mother gave her taka fifty in
addition. The money she has now is ten taka more than
the double of the first amount of money.
(c) TusharÊs father gave him some money for shopping. His
mother gave him Tk. 50 in addition. Tushar bought fish
for Taka one hundred sixty, chicken for Taka seventy
five and vegetables for Taka thirty eight. After shopping
he had Taka twenty seven left.
6. Find the successive sum :
(a) 1 + 4 + 7 + 10 + 13 + 16 + 19
(b) 2 + 6 + 10 + 14 + 18 + 22 + 26 + 30
(c) 3 + 8 + 13 + 18 + 23 + 28
(d) 1 + 2 + 4 + 8 + 16 + 32 + 64
(e) 2 + 6 + 18 + 54 + 162

Chapter - 9
Common Fractions
Fractions with the Same Denominator
5
3
is a fraction having 3 as numerator and 5 as denominator.
5
2
The two fractions have the same denominator 5. These are
fractions with the same denominator.
The denominators of the two fractions
5
3
and
6
1
are different.
These fractions may be expressed with the same denominator.
5
3
=
65
63
×
×
=
30
18
6
1
=
56
51
×
×
=
30
5
The two fractions have been reduced to fractions with the same
denominator 30.
Example 1. Reduce
4
3
,
8
7
and
12
11
to fractions with denominator 48.
Solution : 48 ÷ 4 = 12 ∴
4
3
=
124
123
×
×
=
48
36
48 ÷ 8 = 6 ∴
8
7
=
68
67
×
×
=
48
42
48 ÷ 12 = 4 ∴
12
11
=
412
411
×
×
=
48
44
.
Answer :
4
3
=
48
36
,
8
7
=
48
42
,
12
11
=
48
44
.
Here 48 is a common multiple of the
denominators of the given fractions.

Fractions with the Same Numerator
7
6
13
6
The numerators of the two fractions are 6. These are fractions with
the same numerator.
7
5
and
8
3
have different numerators. They can be expressed as
fractions with the same numerator.
7
5
=
37
35
×
×
=
21
15
8
3
=
58
53
×
×
=
40
15
The two fractions are reduced to fractions with the same numerator 15.
Example 2.
9
2
,
10
3
,
17
4
as fractions with numerator 12.
Solution : 12 ÷ 2 = 6 ∴
9
2
=
69
62
×
×
=
54
12
12 ÷ 3 = 4 ∴
10
3
=
410
43
×
×
=
40
12
12 ÷ 4 = 3 ∴
17
4
=
317
34
×
×
=
51
12
Answer :
9
2
=
54
12
,
10
3
=
40
12
,
17
4
=
51
12
Here 12 is a common multiple of the numerators of the
given fractions.

Lowest Form of Fractions
It is seen that,
8
6
=
4
3
i.e.,
8
6
and
4
3
are equivalent fractions.
Omitting the common factors of numerator and denominator of
8
6
we get,
8
6
=
2221
321
××/×
×/×
=
22
3
×
=
4
3
The lowest form of
8
6
is
4
3
The lowest form of a fraction is obtained by omitting the
common factors of the numerator and the denominator of
the fraction.
Example 3. Express
72
48
into lowest form.
Solution :
72
48
=
33222
32222
/××/×/×/
/××/×/×/
=
3
2
Answer :
3
2
.
8
6
4
3

Example 4. Express
3
2
,
6
5
,
9
4
and
12
7
into lowest form of fraction
with the same denominator.
Solution : Here, the denominators of the fractions are 3, 6, 9 and 12.
Now, 2 3, 6, 9, 12
3 3, 3, 9, 6
1, 1, 3, 2
∴ The L.C.M. of 3, 6, 9 and 12 = 2 × 3 × 3 × 2 = 36
36 ÷ 3 = 12 ∴
3
2
=
123
122
×
×
=
36
24
36 ÷ 6 = 6 ∴
6
5
=
66
65
×
×
=
36
30
36 ÷ 9 = 4 ∴
9
4
=
49
44
×
×
=
36
16
36 ÷ 12 = 3 ∴
12
7
=
312
37
×
×
=
36
21
Answer :
3
2
=
36
24
,
6
5
=
36
30
,
9
4
=
36
16
,
12
7
=
36
21
.
The least denominator of more than one fractions is
the common multiple of their denominators. So at
first L.C.M. of the denominators is obtained then
considering this L.C.M. as common denominator the
fractions are expressed as fractions with the same
denominator.

62 Common Fractions
Comparison of Fractions
6
1
6
3
6
1
and
6
3
both have the same denominator 6; but numerator 3 > numerator 1.
Among fractions with the same denominator the fraction
with greater numerator is the greater fraction.
Example 5. Using symbol, show which one of the two fractions
24
5
and
24
11 is greater.
Solution : Here the denominators of the two fractions are the same.
Their numerators are 5 and 11.
Since 11 > 5 So,
24
11
>
24
5
Example 6. Using symbol, show which one of the fractions
12
7 and
4
3 is greater.
Solution : Here the denominators of the two fractions are 12 and 4.
Now, 2 12, 4
2 6, 2
3, 1
∴ L.C.M. of 12 and 4 = 2 × 2 × 3 = 12
12 ÷ 12 = 1 ∴
12
7
=
112
17
×
×
=
12
7
; 12 ÷ 4 = 3 ∴
4
3
=
34
33
×
×
=
12
9
Since 9 >7. So,
12
9
>
12
7
i.e.,
4
3
>
12
7
Answer :
4
3
>
12
7
.
Expressing the two fractions as fractions with same denominator, their
numerators are compared. The fraction with greater numerator is greater.
6
3
>
6
1

Example 7. Arrange the fractions
9
7
,
12
5
,
18
11
in ascending order of
the values.
Solution : The denominators of the given fractions are 9, 12 and 18.
Now, 2 9, 12, 18
3 9, 6, 9
3 3, 2, 3
1, 2, 1
∴ L.C.M. of 9, 12, 18 = 2 × 3 × 3 × 2 = 36
36 ÷ 9 = 4 ∴
9
7
=
49
47
×
×
=
36
28
36 ÷ 12 = 3 ∴
12
5
=
312
35
×
×
=
36
15
36 ÷ 18 = 2 ∴
18
11
=
218
211
×
×
=
36
22
Since 15<22< 28 So,
36
15
<
36
22
<
36
28
or,
12
5
<
18
11
<
9
7
∴ Arranging the fractions in ascending order of values,
12
5
,
18
11
,
9
7
.
Answer :
12
5
,
18
11
,
9
7
.
Remarks : Arrangement of the fractions after writing them from
smaller to greater successively is called „ascending order‰
7
3
,
14
9
,
21
5
in descending order
of the values.
Solution : The denominators of the given fractions are 7, 14, 21.
Now, 7 7, 14, 21
1, 2, 3
∴ L.C.M. of 7, 14, 21 = 7 × 2 × 3 = 42
42 ÷ 7 = 6 ∴
7
3
=
67
63
×
×
=
42
18
we get,

64 Common Fractions
42 ÷ 14 = 3 ∴
14
9
=
314
39
×
×
=
42
27
42 ÷ 21 = 2 ∴
21
5
=
221
25
×
×
=
42
10
Since, 27 > 18 > 10 So,
42
27
>
42
18
>
42
10
or ,
14
9
>
7
3
>
21
5
Arranging the fractions in descending order of values,
7
3
,
21
5
.
Answer :
14
9
,
7
3
,
21
5
.
Let us notice the picture :
4
3
>
16
3
.
4
3
and
16
3
have the same numerator i.e. numerator 3.
4 is the denominator of
4
3 , 16 is the denominator of
16
3 and 4 < 16.
Among fractions with the same numerator the fraction with
the smaller denominator is the greater fraction.
Example 9. Using symbol, show which one of the fractions
13
4 ,
29
4
is greater.
Solution : The numerators of the fractions are the same
i.e. numerator 4 and denominators are 13 and 29.
Since 13 < 29. So,
13
4 >
29
4 .
Answer :
13
4
>
29
4
.
4
3
16
3
we get,
14
9
,

9
8
,
33
8
,
15
8
,
19
8
in ascending
order of the values.
Solution : The numerators of the fractions is 8 and denominator
are 9, 33, 15 and 19.
Since, 9<15<19<33. So,
33
8
<
19
8
<
15
8
<
9
8
.
So, arranging the fractions in ascending order, we get,
33
8 ,
19
8
,
15
8
,
9
8
Answer :
33
8
,
19
8
,
15
8
,
9
8
.
Exercise - 9 (a)
1. Expressing the fraction in its lowest term insert the correct
number in the box :
a)
24
12
= b)
65
26
= c)
88
22
=
d)
99
66
= e)
75
55
= f)
72
42
=
2. Express the fractions as the lowest form of some denominator :
a)
8
5
,
12
1
b)
9
7
4
3
, c)
15
11
,
25
9
d)
6
1
,
7
3
,
21
19
e)
5
3
,
20
7
,
15
13
f)
16
11
,
2
1
,
24
7
,
3
2
,
8
3
3. Which one is the greatest and lowest fraction :
a)
3
1
,
12
7
,
18
13
b)
7
2
,
14
5
,
70
51
,
35
22
c) ,
8
3
32
21
,
3
1
,
24
17
,
16
9
d)
75
16
,
25
11
,
50
23
,
15
8
,
10
7
4. Arrange the fractions in ascending order of values :
a)
20
19
,
15
7
,
2
1 b)
50
49
,
25
21
,
5
2 c)
43
6
,
37
2
,
29
3
,
13
1
d)
60
59
,
6
5
,
5
4
,
4
3 e)
31
22
,
79
66
,
71
33
,
24
11
,
5
3
.

66 Common Fractions
5. Arrange the fractions in descending order of values :
a)
75
49
,
25
21
,
15
13
,
5
4
b)
78
61
,
26
15
,
39
25
,
13
9
c)
17
2
,
93
14
,
41
2
,
23
7
,
10
1
d)
14
3
,
39
20
,
23
12
,
11
2
,
9
4
6. Drawing a circular area Apu coloured
5
1
of it with blue,
4
1
of it
with yellow and
20
11
of it with red. Which one is the greater
coloured area?
7. Mr. Belal left his residence in the morning with some money.
He takes lunch with
8
3 of his money, gave a begger
24
1 and gave
to rickshaw puller
3
1
. He had some money left. On which
work did he spend more?
8. Mr. Jamir Ali cultivated tomatoes in
4
1
portion of his land,
potatoes in
12
5
portion and radish in
3
1
portion. Which is the
vegetable cultivated in the least portion of land?
9. The numbers of white ducks in farms No. 1, No. 2 and No. 3
are respectively
16
1
,
32
3
and
71
6
portions of the total ducks.
What is the farm that contains the least number of ducks?
10. In a village among the total number of people capable of
earning,
5
3
are cultivators,
10
1
are service holders,
50
3
are
engaged in other professions and
25
6
are unemployed. Which
is the profession taken by the largest number of people?

Proper, Improper and Mixed Fractions
one third or
3
1
two third or
3
2
three third or
3
3
or whole
3
1
,
3
2
are proper fractions. The numerator is less than the
denominator in these fractions.
Together 1
3
1
Together four third or,
3
4
It is seen that, 1
3
1
=
3
4
This is a mixed fraction This is an improper fraction
whose numerator is greater
The integer of the fraction is 1. than the denominator.
9
4
2,
7
2
5,
4
3
3 are mixed fractions. In these fractions, proper fractions
are attached with integers. 3
7
4 is a mixed fraction. This is read as „3
whole four seventh.‰
8
19
,
3
7
,
5
11 etc. are improper fractions. In these fractions the numerator
is greater than the denominator.
• The fraction in which the numerator is smaller than the denominator is a
proper fraction.
• The fraction in which the numerator is greater than the denominator is an
improper fraction.
• The fraction with integer and proper fraction is mixed fraction.
• Generally the integer of the mixed fraction is read as „whole‰.
1 whole
3
1
three-third
one -third
which has an integer portion.

68 Common Fractions
∴
9
83
= 9
9
2
∴
15
79
= 5
15
4
∴
3
7 = 2
3
1
Mutual Conversion of Mixed and Improper Fractions
1 whole three - third
1 whole three - third
3
1
one - third
Total 2 whole
3
1
Total seven third
or, 2
3
1
or,
3
7
It is seen that, 2
3
1
=
3
7
We observe that :
3
7
3) 7 ( 2 → 2 whole
6
1 → 1 third
The denominator of the proper fraction which is related with the
improper and mixed fractions is the same. The quotient which we get
after dividing the numerator of the improper fraction with this
denominator is the whole part of the conversed mixed fraction. And the
remainder is the numerator of the proper fraction which is related to the
conversed mixed fraction.
Example 1. Express the following improper fractions into mixed
fractions:
a)
9
83
b)
15
79
c)
19
91
d)
27
92
e)
32
97
Solution : a) 9 ) 8 3 ( 9
8 1
2
b) 15 ) 7 9 ( 5
7 5
4

∴
19
91
= 4
19
15
∴
27
92
= 3
27
11
∴
32
97
= 3
32
1
c) 19 ) 9 1 ( 4
7 6
1 5
d) 27 ) 9 2 ( 3
8 1
1 1
e) 32 ) 9 7 ( 3
9 6
1
7 =
1
7
, 4 =
1
4
, 11 =
1
11
etc.
Any integer (whole number) can be written as denominator with 1.
We observe that :
1
2
3
2
121
2
1
2
21
2
1
21
21
2
1
1
1
2
1
1
2
1
=
+×
=+
×
=+
×
×
=== ++
2
5
11
5
152
5
1
5
52
5
1
51
52
5
1
1
2
5
1
2
5
1
=
+×
=+
×
=+
×
×
=+=+=
7
4
31
4
347
4
3
4
47
4
3
41
47
4
3
1
7
4
3
7
4
3
=
+×
=+
×
=+
×
×
==+= +
Shortly, 1 ;
5
11
5
152
5
1
2;
22
121
2
1 3
=
+×
==
+×
= 7
4
3
=
4
347 +×
=
4
31
Mixed fraction =
denominator
whole portion × denominator + numerator
= Improper fraction
Example 2. Express the following mixed fractions into improper
fractions :
a)
6
1
4 b)
11
7
8 c)
13
4
5 d)
10
3
6 e)
17
11
5
Answer : (a) 9
9
2
5
15
4(b) 4
19
15(c) 3
27
11(d) 3
32
1(e) .

70 Common Fractions
.
Solution : a)
6
25
6
164
6
1
4 =
+
=
×
b)
11
95
11
7118
11
7
8 =
+
=
×
c)
13
69
13
4135
13
4
5 =
+
=
×
d)
10
63
10
3106
10
3
6 =
+
=
×
e)
17
96
17
11175
17
11
5 =
+
=
×
Answer : a)
6
25
b)
11
95
c)
13
69
d)
10
63
e)
17
96
Exercise - 9 (b)
1. Choose proper, improper and mixed fractions from the
following fractions :
25
7
9,
71
1
,
12
3
8,
41
42
,
7
60
,
3
2
17,
23
13
,
15
14
,
4
27
.
2. Express the following improper fractions into mixed fractions :
a)
5
29 b)
9
85 c)
11
78 d)
37
89 e)
15
23
f)
51
73
g)
59
147
h)
25
103
i)
43
521
j)
203
1416
3. Express the following mixed fractions into improper fractions:
a)
16
3
3 b)
44
52 c)
48
71 d)
16
310
e)
7
3
88 f)
11
7
112 g)
23
11
27 h)
33
29
98
i)
49
1
100 j)
19
18103

4. Insert the correct number in the box given :
a)
9
85 =
49 b)
20
9
11 =
20
c)
13
115 =
13
11 d)
25
267 = 10
25
e) 12
71
38 =
71
f) 291
11
10 =
11
g)
351
4670
=
351
107
5. Robin had 25 marbles. From these he lost 7 marbles. What is
the portion of the total marbles he lost ?
6. Helal Mia has 5 bighas of land. He planted rice in 4 bighas
and in half of another bigha. In the rest portion he planted
sugercane. In how many bighas did he plant sugercane?
7. Kamal had Tk. 300. From this money he bought a shirt for
Tk. 75. What portion of total money did he spend to buy
the shirt ?

72 Common Fractions
Addition and Subtraction of Fractions
Addition of Fractions
Example 1. =+++
75
23
75
13
75
4
75
1 What ?
Solution :
75
23
75
13
75
4
75
1
+++
=
75
231341+ ++ =
75
41
Answer :
75
41
The denominator of the summation of some fractions with
the same denominator is the common denominator of the
fractions. The numerator of this summation is the summation
of the numerators of the given fractions.
Example 2.
25
11
15
7
5
2 ++ = What?
Solution : The denominators of the fractions are 5, 15, 25.
Now, 5 5, 15, 25
1, 3, 5
∴ L.C.M. of 5, 15, 25 = 5 × 3 × 5 = 75
75 ÷ 5 = 15 ∴
75
30
155
152
5
2 =
×
×=
75 ÷ 15 = 5 ∴
75
35
515
57
15
7
=
×
×
=
75 ÷ 25 = 3 ∴
75
33
325
311
25
11
=
×
×
=
∴
75
33
75
35
75
30
25
11
15
7
5
2 ++=++ =
75
333530 ++
=
75
98
= 1
75
23
Answer : 1
75
23
Before addition the fractions are to be expressed as fractions
with the same denominator.

Example 3.
9
10
9
13
9
11
++ = What ?
Solution :
9
10
9
13
9
11
++
=
9
101311 ++
=
9
34
= 3
9
7
Answer : 3
9
7
If the summation is an improper fraction, then generally
this is to be expressed as mixed fraction.
Example 4.
11
5
11
2
11
1
243 ++ = What ?
Solution :
11
5
11
2
11
1
243 ++
=
11
27
11
46
11
34
++
=
11
274634 ++
=
11
107
= 9
11
8
Answer : 9
11
8
Addition is done after expressing the mixed fractions
into improper fractions.
.

74 Common Fractions
Explanation :
Here denominators are 4, 3, 9.
Now, 3 4, 3, 9
4, 1, 3
L.C.M. of 4, 3, 9 = 3×4×3 = 36
36 ÷ 4 = 9
36 ÷ 3 = 12
36 ÷ 9 = 4
Explanation :
Here donominators are 4, 2, 8.
Now, 2 4, 2, 8
2 2, 1, 4
1, 1, 2
L.C.M. of 4, 2, 8 = 2 × 2 × 2 = 8
∴ 8 ÷ 4 = 2
8 ÷ 2 = 4
8 ÷ 8 = 1
Example 5.
9
5
1
3
2
3
4
1
5 ++ = What ?
Solution :
9
51
3
23
4
15 ++
=
9
14
3
11
4
21
++
=
36
4141211921 ×+×+×
=
36
56132189 ++
=
36
377
= 10
36
17
Answer : 10
36
17
.
Example 6.
8
3
2
2
1
5
4
1
2 ++ = What?
Solution :
8
3
2
2
1
5
4
1
2 ++
=
8
19
2
11
4
9
++
=
8
11941129 ×+×+×
=
8
194418 ++
=
8
81
= 10
8
1
Answer : 10
8
1
.

Example 7. Bijoy Babu went to market and bought rice with
2
1
portion of his money, vegetables with
10
3 portion and
salt with
20
1 portion. What portion of his total money
did he spend?
Solution :
20
1
10
3
2
1
++
=
20
1610 ++
=
20
17
∴ Bijoy Babu spent
20
17 portion of his money.
Answer :
20
17 portion.
Subtraction of Fractions
Example 8.
28
11
28
19
− = What ?
Solution :
28
11
28
19
−
=
28
1119 −
=
728
28 =
7
2
Answer :
7
2
Let us notice:
• The denominator of the difference of two fractions with the
same denominator is the common denominator of the given
fractions. The numerator of their difference is the difference
of the numerators of the given fractions.
• Summation or difference of fractions is to be always reduced
into least form.

76 Common Fractions
Explanation :
= 9 4 = 36
∴ 36 ÷ 9 = 4
36 ÷ 4 = 9
Explanation :
Here, denominators of the two
fractions are 8 and 5.
The L.C.M. of 8 and 5 = 8 5 = 40
∴ 40 ÷ 8 = 5
40 ÷ 5 = 8
Example 9.
9
7 −
4
3 = What ?
Solution :
9
7 −
4
3
=
49
47
×
×
−
94
93
×
×
=
36
28 −
36
27
=
36
2728 −
=
36
1
Answer :
36
1 .
Example 10.
7
61 −
7
23 = What ?
Solution :
7
61
−
7
23
=
7
2361−
=
7
38 = 5
7
3
Answer 5
7
.3
Example 11. 4
8
1 − 2
5
3 =
4
8
1 − 2
5
3
What ?
=
8
33 −
5
13
=
40
813533 ×−×
=
40
104165 −
=
40
61 = 1
40
21
Answer : 1
40
21
The two fractions do not have the
same denominator. They are
expressed with the same denominater.
Their denominators are 9 and 4.
The L.C.M. of 9 and 4
Solution :
.

Example 12. Sharif Mia cultivated jute in
4
1 portion of his land and
paddy in
3
1 portion and wheat in the rest of the land. In
what portion of his total land did he cultivate wheat ?
Solution :
12
43
3
1
4
1 +
=+
=
12
7
Sharif Mia cultivated jute and p
That is, total portion of the land ( )
addy in
12
7 portion of his land.
The rest of the land (1 land −
12
7 portion of the land).
=
12
712 −
portion
=
12
5 portion
∴ He cultivated wheat in
12
5 portion of his land.
Answer :
12
5
portion.
Exercise - 9 (c)
1. Find the summation ;
a)
25
12
25
7
25
1
++ b)
18
11
18
5
18
1
++
c)
4
1
3
1
2
1 ++ d)
10
7
4
3
5
1 ++
e)
14
9
7
4
4
1
++ f)
9
23
9
19
9
10
++
12
7
1−

78 Common Fractions
g)
22
81
22
45
22
23 ++ h)
8
5
6
8
32
8
14 ++
i)
3
2
16
13
16
5
3
12
710
8
77 ++++
j) 3
10
12
5
4
4
32
4
11 ++++
k) 1 +
36
13
18
5
1
2
17
9
23 +++
2. Find the difference :
a)
17
6
17
9
− b)
45
14
45
19
− c)
32
15
4
3
− d)
40
11
20
13
−
e)
42
19
14
9
− f)
6
13
6
41
− g)
10
79
10
103
− h)
12
5
4
12
7
5 −
i)
9
7
4
9
1
6 − j)
5
1
3
10
9
4 − k)
5
2
2
4
3
5 − l) 1 −
15
7
m) 6
13
9
2−
3. Simplify :
a)
12
7
6
5
2
1
−+ b)
2
1
5
2
1
4
3
6 +−
c)
4
1
1
8
1
5
1
5 −− d)
24
11
1
6
1
4
4
3
7 −−
e)
15
4
2
10
7
4
5
3
8 +−
4. Mina started to read a book. On the first day she read
5
2 portion of
the book and the next day she read
15
8 portion. What portion of
the book did she read in two days?
5.
5
1 portion of a bamboo is coloured black,
6
1 portion is red and
10
3
portion green. What portion of the bamboo is coloured?
6. After buying a basket of 240 mangoes it is seen that
16
1
portion of the total mangoes was rotten. How many fresh
mangoes are there in the basket?

7. From a land a farmer used
8
3 portion to construct a house,
5
2
portion for garden and dug a pond in the remaining land. What
portion of the land did he use to dig the pond?
8. A gardener planted rose in
5
2 portion of his garden, in
15
1
portion he planted tube rose and in
25
3 portion china rose.
What portion of land did he use in total to plant the flowers?
9. Jalil Mia had Tk. 200. He gave Kamal
5
2
portion of it and
4
1
portion to Mukul. How much money did he have left?
Multiplication of Fractions
The multiplication of fractions by integer.
9
8
9
8
9
2
9
2
9
2
9
2
=+++
9
2 × 4 =
9
8 =
9
2 4×
Let us observe that : =+++
9
2
9
2
9
2
9
2
9
2222 +++
=
9
42×
=
9
8
Fraction × Integer =
Numerator of fraction × Integer
Denominator of fraction

80 Common Fractions
Example 1.
8
7 × 5 = What ?
Solution :
8
7 × 5 =
8
57×
=
8
35
= 4
8
3
Answer : 4
8
3.
Example 2.
6
13
× 7 = What ?
Solution :
6
13 × 7 =
6
713 ×
=
6
91 = 15
6
1
Answer : 15
6
1
.
Example 3. 6
10
1 × 8 = What ?
Solution : 6
10
1 × 8 =
10
61× 8
=
10
861×
=
5
10
244
488
=
5
244
= 48
5
4
Answer : 48
5
4 .

Multiplication of fraction by fraction :
* Area of the square = 1 sq. metre
* The square is divided into 15
small rectangles.
Area of each rectangle =
15
1 sq. metre
* The area of the deep coloured
portion
5
4
×
3
2
sq. metre
* There are 8 small rectangles in
deep coloured portion
* The area of deep coloured
portion =
15
1
× 8 sq. metre
=
15
8
sq. metre
∴
5
4
×
3
2
=
15
8
Let us observe that :
5
4
×
3
2
=
15
8
or,
5
4
×
3
2
=
35
24
×
×
The product of two fractions =
fractionstheofrsdenominatotheofproductThe
fractionstheofnumeratorstheofproductThe
Example 4.
6
5 ×
8
3 = What ?
Solution :
6
5
×
8
3
=
8
2
6
1
35
×
× =
16
5
Answer :
16
5 .
Example 5.
7
11
×
9
4
= What ?
Solution :
7
11×
9
4 =
97
411
×
× =
63
44
Answer :
63
44
.
Breadth
=
3
2
Metre 1
m
e
t
r
e
1 Metre
Length =
5
4
Metre

82 Common Fractions
Example 6.
10
13 ×
17
25 = What ?
Solution :
10
13 ×
17
25 =
17210
52513
×
×
=
34
65
=
12
49
7
3
= 1
34
31
Answer : 1
34
31
Example 7. 4
12
1
×
7
3
= What ?
Solution : 4
12
1 ×
×
7
3
=
1
1
7124
3497
×
×
(Expressing into improper fractions)
=
4
7
= 1
4
3
Answer : 1
4
3 .
Example 8.
17
2
2
8
3
6 × = What ?
Solution :
17
2
2
8
3
6 × =
1
17
9
36
2
8
3
51
× =
2
27 = 13
2
1
Answer : 13
2
1
.

The meaning of “of ”
Let us observe that : * The area is divided into 15 small equal areas.
* Coloured portion =
5
2 portion of the whole area.
* Deep coloured portion =
3
2 of coloured portion
=
3
2 portion of
5
2 of the whole area.
But deep coloured portion =
15
4 portion of the whole area
∴
∴
3
2 of
5
2 portion =
15
4
Briefly,
3
2 of
5
2 =
15
4
But
15
4 =
35
22
×
×
∴
3
2 of
5
2 =
35
22
×
×
Example 9.
9
8 portion of the mathematics book of class V is
Arithmetic.
16
5 portion of Arithmetic is a fraction. What
portion of mathematics book is a fraction?
Solution :
16
5 of
9
8 =
2
169
5
1
8
×
×
=
29
51
×
×
=
18
5
18
5
portion of mathematics book is a fraction.
Answer :
18
5
portion.

84 Common Fractions
Example 10.
Example 11. Write down the reciprocal fractions of
4
3
portion of one metre cloth is red. A handkerchief is
made with
2
1 of the red portion of the cloth. What
portion of one metre cloth was used to make the
handkerchief ?
Solution : To make handkerchief (
2
1 of
4
3 ) portion of one metre
cloth is used.
Now,
2
1 of
4
3 =
24
13
×
×
=
8
3
To make handkerchief
8
3 portion of one metre cloth is used.
Answer :
8
3 portion.
Reciprocal fraction
Every fraction has numerator and denominator. A reciprocal fraction
is obtained by changing the denominator of a fraction to numerator
and numerator to denominator.
4
3 is a fraction whose denominator is 4 and numerator is 3.
Interchanging the places of 3 and 4 we get
3
4 .
3
4 is the reciprocal fraction of
4
3 .
or,
4
3 is the reciprocal fraction of
3
4 .
Let us observe that :
4
3
3
4
× = 1
Any fraction × its reciprocal fraction = 1
9
5
6
1
5
9
10
11
7, , , , .

Solution : The reciprocal fraction of
9
5 is
5
9
‰ ‰
6
1 is
1
6 or 6
‰ ‰
10
11 is
11
10
‰ ‰ 7 is
7
1
‰
‰
‰
‰
‰ ‰
5
9 is
9
5
Answer :
5
9 , 6,
11
10 ,
7
1 ,
9
5 .
Division of fraction :
• The area is divided into 20 equal small areas.
• Among 20 small areas 12 areas are coloured.
i.e., coloured portion =
5
20
3
12 or,
5
3 portion of the area.
Deep coloured portion is 3 portions of 20 portion of the area.
i.e., Deep coloured portion =
20
Again, Deep coloured portion is 1 portion of 4 coloured portions.
i.e., Deep coloured portion = coloured portion ÷ 4
=
5
3 portions of the area ÷ 4
= (
5
3 ÷ 4) portions of the area.

86 Common Fractions
Again, deep coloured portion
=
4
1 portion of coloured portion
=
4
1 portion of
5
= (
5
3
×
4
1
) portion of the area.
5
3 ÷ 4 =
5
3 ×
4
1
It is seen that, dividing
5
3 by 4 is the same as multiplying by .
4
1
So, to divide
5
3
by 4, it is multiplied by
4
1
To divide a fraction by a fraction, we need to multiply the
first fraction by the reciprocal fraction of the second fraction.
Example 12.
25
3 ÷ 4 = What ?
Solution :
25
3 ÷ 4 =
25
3 ÷
1
4
=
25
3 × (reciprocal fraction of
1
4 )
=
25
3 ×
4
1
=
25 × 4
3 × 1
=
100
3
Answer :
100
3 .
.

Example 13.
7
8 ÷ 9 = What ?
Solution :
7
8 ÷ 9 =
7
8 ÷
1
9
=
7
1
9 )
=
7
8
×
9
1
=
97
18
×
×
=
63
8
Answer :
63
8 .
Example 14. 3
5
3 ÷ 15 = What ?
Solution : 3
5
3 ÷ 15
=
5
1
15 )
= =
5
6
18 ×
5
15
1
5 × 5
6 × 1
=
25
6
Answer :
25
6 .
Example 15. 13 ÷ 2
4
1 = What?
Solution : 13 ÷ 2
4
1 =
1
13 ÷
4
9
=
1
13 ×
9
4
=
9
52 = 5
9
7
Answer : 5
9
7 .

88 Common Fractions
Example 16.
11
4
7 = What ?
Solution :
11
4
7 = 7 ÷
11
4
=
4
11
1
7 ×
×
=
4
77
= 19
4
1
Answer : 19
4
1 .
Example 17.
20
7
18
13
= What ?
Solution :
20
7
18
13
=
18
13 ÷
20
7
=
9
18
13
7
10
20
=
63
130
= 2
63
4
Answer : 2
63
4 .

Mathematics mathematics elementary

_
property
that property?
The cost of dozens of egg is Tk. 21. What is the const of
dozens of eggs?

Exercise - 9 (e)
Simplify :
1.
24
1
6
1
12
1
4
1
3
1
−+−+
2.
27
16
2
3
5
5
3
4
3
2
1
×÷+÷
3.
63
25
7
1
2
5
1
4
7
1
8
3
×÷×+
4. 75 ÷
21
8
7
3
6 × of
2
1
5
3
×
5.
5
4
of
10
9
8
7
÷ of
9
5
2
1
4
3
×−
6.
3
2
1
4
3
3 ÷ of 1
10
3
5
4
1
9
1
2
1
2 −÷×+
7.
33
31
1
2
1
5 × of 2
3
1
5
8
1
1 +÷
8.
4
1
6 of
12
1
2
3
2
3
2
4
3
1
2 −+÷
9.
44
9
2
1
2
4
1
11
2
1
13 +×÷ of
8
3
3
1
7 ×
10.
9
1
1)
6
1
3
1
(
5
2
2
25
13
÷−×+
11. (
2
11
2
1(3)
2
11
2
12
2
13 ÷÷×÷ of )
2
12
12.
15
1 of 20 {( 1}
5
41
5
3)
3
2
3
14 −÷×+
13.
2
1
{5
16
5[4
23
201 ÷× of )}]
14
3
7
5
(
8
3
1 −+
14.
2
11
2
1(5
3
2{3
4
3[1
3
17 −−+− of )}]
4
3
3
12 +

98 Common Fractions
15. [10 − {
17
6
(2
3
2
20
9
1
11
10
1
3 +÷× of )}]
2
1
50
11 ×
16.
4
38
5
1{4
15
7[14
7
5 ÷−× of )}]
3
1
5
35
15
1(7
5
17 +++
17. ]
3
24
5
24
7
13)}
8
73
12
1(6)
9
48
3
1{(3
32
5[
5
2 ×÷+−÷+××
Problems relating divisions
Example 1.
4
3
8 dozens of bananas are distributed among 21
persons. How many bananas will each person get?
Solution :
4
3
8 dozens =
4
35 dozens.
1 dozen = 12
∴
4
35
dozens =
1
4
31235× = 105
The total number of bananas are 105.
21 persons will get 105 bananas
∴ 1 person " "
21
5105
= 5
1
bananas
Answer
Each person will get 5 bananas
: 5 bananas
Example 2. Mr. Sajjad had Tk. 24000 from which he gave
12
5
portions to an orphanage and
8
3 portions to an educational
institution. How much money did he have left ?

Solution : He donated to orphanage and educational institution
(
12
5 +
8
3 ) of total money.
or, )
24
910( + portion or,
24
19 portion
The rest is ( )
24
191− portion of total money.
or,
24
1924 − portion or,
24
5 portion
Mr. Sajjad had total Tk. 24000.
∴
24
5 portion of total money= ( )
24
5
1
1000
24000 ×
= Tk. 5000
Now, he has Tk. 5000.
Answer : Tk. 5000
Exercise - 9 (f)
1. Babul Mia, Badal Mia and Lipi Begum got respectively
6
1
portion,
3
1 portion and
4
1 portion of the yearly income of a
fishery. What portion of the total income did they get?
2.
5
2 portion of a bamboo is coloured red,
10
1 portion green and
5
1
portion yellow and 3 metres of the bamboo is not coloured. What
is the length of the whole bamboo?
3. Molly bought paper with
5
1 portion of her money, pen with
3
1
portion and khata with
4
1 portion. She had Tk. 26 left . How
much money did she have first?
4. Nogen Babu kept
8
1 portion of his property for himself and gave
8
1
portion to his wife. He distributed the rest of the property equally
among his 4 sons and each son got Tk. 15000. What is the price of
the total property?

100 Common Fractions
5. Kamal Babu spent
5
1 portion of his monthly income for house
rent,
8
3 portion for education expenses of his sons and daughters
and
10
3 portions for other expenses. He keeps the rest of the
money in a bank. What is his monthly income if he saves
Tk. 14400 at the end of the year?
6. Rony and Panna got respectively
4
3 portion and
3
2 portion of
the total marks secured in the annual examination. Rony got 50
marks more than Panna. What was the total marks and what
mark did each of them get?
7. The divisor is 12
2
1 times the quotient. The quetient is 4
5
1
What is the dividend ?
8. The product of two fractions is 74
6
1 . One fraction is
9. In Hatirdiya Primary School there are 800 students.
20
13 portion
of them are boys and the rest are girls.
14
13 portion of the girls
passed in the annual examination. How many girls failed in the
examination?
10. Mr. Motin had some money. He bought a cycle with
2
1 portion
of his money. He bought a radio with
5
4 portion of the cost of
the cycle and distributed the rest of the money among his two
daughters. Each daughter got Tk. 300. How much money did
Mr. Motin have?
11. There was 260 kg. of rice in a rice shop. The shopkeeper sold
5
3 portion of his rice and kept the rest equally in 4 sacks. How
many kilograms of rice did he keep in each sack?
What is the other ?
28
8
1

The Decimal Fraction
The Summation of Decimal Fraction
Example 1. Add : 5&03 and 8&64.
Solution : Explanation
Tens Ones Tenths Hundredths
5
8
0
6
3
4
5&03
8&64
13&67
Answer : 13&67. 1 3 6 7
One number is placed below the other in such a way that the
decimal points remain in the same line vertically. After the
summation has been carried out normally, the decimal point has
been placed in the gap of decimal point.
Example 2. Add : 7&89 and 9&48.
7+1
9
8+1
4
9
8
7&89
9&48
17&37
Answer : 17&37. 1 7 3 7
Example 3. Add : 12&04, 0&5, 0&03, 0&009.
Solution : 12&040
0&500
0&030
0&009
12&579
Answer : 12&579.
Chapter-10

102 The Decimal Fraction
The numbers have been arranged in such a way that the decimal
points remain vertically in the same line. For the convenience of
the summation, to the extreme right of the digits, extra zeros have
been added where necessary to equalize the number of digits after
the decimal points. In other words, the digits have been made
three. Arranging the numbers in this way, the summation has been
carried out simply and right below the line of the decimal points, a
point has been given.
* The value of fraction does not change if zeros are added to the
right of the last digit of the decimal point.
* If there is no digit on the left of the decimal point, sometimes a
zero is added to write the number.
Subtraction of Decimal Fraction
Example 4. Subtract : 2&69 from 8&15.
8
2+1
1+10
6+1
5+10
9
8&15
2&69
5&46
Answer : 5&46. 5 4 6
The digits to be subtracted have been arranged below the
subtraction in such a way that the one decimal points remain just
below the others. It is not possible to deduct 9 hundredth from 5
hundredth. That is why10 hundredth has been added to 5 hundredth to
make it 15 hundredth and then it has been added to 6 as 1 tenth. Similarly,
10 tenth has been added to 1 and 1 ones has been added to 2.
Example 5. Subtract : 0&
0&
001 from 5.
Solution : 5&000
&001
4&999
Answer : 4&999.

The subtraction 5 does not have any decimal fraction. In the
number to be subtracted, there are three digits on the right of the
decimal point. For this reason, three digits have been added after
placing a decimal point on the right of the digit 5. Finally, they
have been subtracted normally.
Example 6. Simplify : 1&1 + 0&1 − 0&001 + 0&101 − 0&01
Solution : 1&1 + 0&1 − 0&001 + 0&101 − 0&01
= (1&1 + 0&1 + 0&101) − (0&001 + 0&01)
= (1&100 + 0&100 + 0&101) − (0&001 + 0&010)
= 1&301 − 0&011
= 1&290
= 1&29
Answer : 1&29.
Example 7. Simplify : 2002 − 0&101 + 77&077 − 33&003
Solution : 2002 − 0&101 + 77&077 − 33&003
= (2002 + 77&077) − (0&101 + 33 003)
= 2079&077 − 33&104
= 2045&973
Answer : 2045&973.
Observe that :
a. The digits with plus sign have been put within brackets
before carrying out the summation.
b. The digits with minus sign have been put within brackets
before carrying out the summation.
c. The summation of the numbers with minus sign has been
subtracted from the summation of the numbers with plus
sign.

Example 8. Mina had Tk. 50. She bought a pencil for Tk. 5.50,
biscuits for Tk. 12.75 and chocolates for Tk. 2.60. How
much money did she have left?
Solution : The price of pencil = Tk. 5.50
The price of biscuit = Tk. 12.75
The price of chocolates = Tk. 2.60
Total amount of money spent = Tk. 20.85
The amount Mina had = Tk. 50.00
The amount spent = Tk. 20.85
The amount left = Tk. 29.15
Answer : Tk. 29.15.
Exercise - 10 (a)
1. 1&02, 2&002, 3&0003
2. 0&09, 9&009, 99&0009
3. 3&081, 0&1042, 0&5, 7&002
4. 8&005, 200&09, 1&234, 0&003
5. 4&379, 25&257, 9&123, 1&999
6. Tk. 5&05, Tk. 8&07, Tk. 29&90.
7. 60&75 , 30&25 , 25&20 , 63&80
8. Tk. 9&20, Tk.17&50, Tk. 23&75, Tk.0&05.
9. 5&125 metre, 2&005 metre, 3 metre, 22&02 metre.
Subtract : (Questions 10 -15)
10. 0&0003 from 0&03
11. 1&11 from 100&01
Add : (Questions 1-9)

12. 0.0003from 0.0007
13. Tk. 2.92 from Tk. 4.01.
14. 29.0008 seconds from 30.256 seconds.
15. 14.895 km from 17.575 km.
Simplify (Questions 16 -20) :
16. 80.0018 ă 17.999 + 2.005 ă 4.001
17. 200 ă 0.291 ă 9.6 + 5.0005
18. 28.30 minutesă3.135 minutes+ 9.0105 minutesă 17.92 minutes.
19. Tk. 303.05 ă Tk. 125.28 ă Tk. 0.96ă Tk. 29.75.
20. 500.003 metres + 125.625 metres ă 130.202 metres ă 25 metres.
21.
22.
23.
24.
25.
BadalÊs school is 2 km. away from his home. After starting from
his home he reached his school walking 0.25 km. on foot and the
rest of the way riding on a rickshaw. How much distance did he
go by rickshaw?
Mr. Hakim went to market and bought fish for Tk. 50.50,
vegetables for Tk.10.25 and oil for Tk. 7.20. How much money
did he spend buying those things?
Toma had Tk. 50. She gave Tk. 15.50 to her younger brother and
Tk. 12.75 to her friend. How much money did she have left?
Monu Mia had Tk. 200. He spent Tk. 110.50 to buy a shirt and
Tk. 40.75 for an undergarment. How much money did he have
left?
Parul Begum has a land of 10 kathas. She cultivates paddy in 4.5
kathas, chilli in 2.2 kathas, potatoes in 1.75 kathas and eggplants
in the rest of the land. In how much land did she cultivate
eggplants?

Multiplication of an integer by a decimal fraction and a
decimal fraction by an integer.
Example 9. Multiply 5&304 by 25.
Solution : 5 3 0 4
× 2 5
2 6 5 2 0
1 0 6 0 8 0
1 3 2 6 0 0
In the multiplicand there are 3 digits
on the right of the decimal point.
∴ 5&304 × 25 = 132&600
= 132&6
Answer : 132&6.
Explanation :
5&304 = 5 3 0 4 thousandth
× 2 5
1 3 2 6 0 0 thousandth
∴ 5&304 × 25 = 132&600
= 132&6
Example 10. Multiply 35 by 2&215.
Solution : 35 × 2&215 = 2&215 × 35
2 2 1 5
× 3 5
1 1 0 7 5
6 6 4 5 0
7 7 5 2 5
The multiplicand has 3 digits on the
right of the decimal point.
∴ 35 × 2&215 = 77&525
Answer : 77&525.
Explanation :
2&215 = 2215 thousandth
3 5
× 2 2 1 5
7 7 5 2 5 thousandth
∴ 35 × 2&215 = 77&525
It is convenient to multiply a decimal fraction by an integer.
The decimal point of the given number has been omitted and
multiplied normally as an integer. The number to be multiplicand
has three digits on the right of the decimal point. The product of the
multiplication has been obtained after placing a decimal point before
the last three digits. As the product has 2 zeros on the extreme right
of the decimal point, they have been omitted. This however, makes
no change in the product.

The multiplication of decimal fractions by 10, 100
Solution :
6&123 × 10
= 61&23
Answer : 61&23
Explanation :
6&123 = 6123 thousandth
× 10
61230 thousandth
= 6123 hundredth
= 61&23
Solution :
8&947 × 100
= 894&7
Answer : 894&7.
Explanation :
8&947 × 100
= 100
1000
8947 ×
=
10
8947
= 894&7
There is one zero on the right of the multiplier 1. The product can
be obtained if the decimal point of the multiplicand is moved one
place to the right.
The multiplier 1 has two zeros on its right. The product can be
obtained if the decimal point of the multiplicand is moved two
places to the right.

The multiplication of a decimal fraction by a decimal fraction.
Example 13. Multiply 0&67 by 0&29.
Solution : 6 7
× 2 9
6 0 3
1 3 4 0
1 9 4 3
The multiplicand has 2 and the
multiplier has 2 total 4 digits on the
∴ 0&67 × 0&29 = 0&1943
Answer : 0&1943.
Explanation :
0&67 × 0&29
=
100
29
100
67
×
=
10000
2967×
=
10000
1943
= 0&1943
Example 14. Multiply 0&37 by 0&172.
Solution : 3 7
× 1 7 2
7 4
2 5 9 0
3 7 0 0
6 3 6 4
To the right of the decimal point, there are 2 digits in the multiplicand
and 3 in the multiplier (5 digits in total)
∴ 0&37× 0&172 = 0&06364
Answer : 0&06364.
First, the decimal point has been omitted and the numbers have been
multiplied normally as integers. The number of digits after the decimal
points of both the multiplier and the multiplicand has been added. Finally,
the decimal point has been placed be the same total of digits from the
right of the product, i.e. four digits from the right of the result.
There are 2 digits after the decimal point of the multiplicand and 3 after
that of the multiplier. Therefore, in the product there must be 5 digits
on the right of the decimal point. As the product has four digits, one
zero has been added on the left before placing the decimal point so that
there remains 5 digits on the right of the decimal point.

Example 15. An orange costs Tk. 4 75. How much does a dozen&
cost ?
Solution :
The cost of 1 dozen oranges = Tk.(4&75 × 12).
The multiplicand has 2 digits on the
∴ 4&75 × 12 = 57&00
= 57
∴ The price of one dozen oranges is Tk. 57.
Answer : Tk. 57.
1 dozen = 12
4 7 5
x1 2
9 5 0
4 7 5 0
5 7 0 0
Exercise - 10 (b)
1. Find out the product :
a) 4&34 × 14 b) 9&789 × 15 c) 29 × 2&0303
d) 35 × 6&1001 e) 32&48 × 10 f) 2&007 × 100
g) 0&0008 × 100 h) 18&251 × 4&6 i) 26&51 × 3&2
j) 0&2 × 2&2× 2&22 k) 0&1 × 0&01 × 0&001 l) 0&04× 0&004 × 25
2. One pen costs Tk. 20&25. What is the cost of 8 pens ?
3. The cost of one litre of soyabean oil is Tk. 49.75. What is the
cost of 5 litres?
4. A man walks 2&25 km. everyday during his morning walk. He
walked the same distance everyday all through the month of
April. How much distance did he walk in the month of April?
5. A car goes 45&5 km. in an hour. How far will it go in 12&2
hours ?
6. If one inch is equal to 2&45 cm. (centimetres), how many
centimetres are equal to 8&5 inches ?

The division of a decimal fraction by an integer
Example 16. Divide 50&4 by 8.
Solution :
8) 5 0&4 ( 6&3
4 8
2 4
2 4
0
Answer : 6&3
The placemet of decimal
point in quotient.
8) 5 0&4 ( 6&3
4 8
2 4
2 4
0
While dividing whenever the digit after the decimal point is taken,
a decimal point is placed in the quotient.
Explanation : 50&4 = 504 tenth
8) 5 0 4 tenth ( 63 tenth
4 8
2 4
2 4
0
∴ Quotient = 63 tenth
= 6&3

Solution : 16) 2 9&5 ( 1&84375
1 6
1 3 5
1 2 8
7 0
6 4
6 0
4 8
1 2 0
1 1 2
8 0
8 0
0
Answer : 1&84375.
* During division a zero has been added on the right of the
remainder till the part on the right of the decimal point
becomes completely divisible.
* 29&5 = 29&50000
Therefore, 29&5 ÷ 16 = 29&50000 ÷ 16
The division of an integer by a decimal fraction
Example 18. Divide 28 by 2&5.
Solution : In 28 ÷ 2&5, the divisor is 2&5 and by removing its
decimal point once to the right we get 25.
By adding a zero on the right of the divisible 28, we get 280.
25) 2 8 0 (11&2
2 5
3 0
2 5
5 0
5 0
0
∴ 28 ÷ 2&5 = 11&2
Answer : 11&2.

Explanation : 28 ÷ 2&5 = 25280
25
280
1052
1028
52
28 ÷==
×
×=
&&
The division of a decimal fraction by 10, 100
Solution :
29&32 ÷ 10 = 2&932
Answer : 2&932
Explanation :
29&32 ÷ 10 = 2932 hundredth ×
10
1
= 2932 thousandth
= 2&932
The divisor has one zero (0) on its right. The quotient has been obtained
by shifting the decimal point to the left side once on the right side.
The divisible and the quotient have the same digits. Only the place of
the decimal points has been shifted.
Solution : 154&9 ÷ 100 = 1&549.
Answer : 1&549.
Explanation : 154&9 ÷ 100 = 1549 tenth ×
100
1
10
1
1549
100
1
××=
1000
1
1549×=
= 1549 thousandth = 1&549.
• In the divisor on the right of 1 there are 2 zeros. The quotient has been
obtainted by shifting the decimal point of the divisible twice to
the left.
• In the divisor 1 if there are 3 zeros on the right, the quotient can be
obtained by shifting the decimal point of the divisible thrice to the left.
• We need to add zero (0) if needed.
The decimal point of the divisor has been removed in such a way
that the divisor becomes an integer. On the right of the divisible, a
zero is added as many times as the decimal point is shifted in the
divisor. The quotient has been obtained by dividing this changed
divisible with the changed divisor.

The division of a decimal fraction by a decimal fraction.
Example 21. Divide 85&1 by 4&6.
Solution : In 85&1 ÷ 4&6 the divisor is 4&6 and by shifting the decimal
point once to the right, we get 46 and by shifting the decimal point of the
divisible 85&1 once on the right, we get 851 i.e. 851&00.
46) 8 5 1&00 (18&5
4 6
3 9 1
3 6 8
2 3 0
2 3 0
0
∴ 85&1 ÷ 4&6 = 18&5
Answer : 18&5.
Explanation :
To remove the decimal point of the divisor it has been multiplied by
10. In other words, the decimal point has been shifted once to the
right. To keep the value of the number, the decimal point of the
divisible is also shifted once to the right.
While dividing the number on the right of the decimal point, a
zero needs to be added to the right of the remainder until it
becomes exactly divisible.
Example 22. Divide 8&51 by 4&6.
Solution : 8&51 ÷ 4&6 by shifting the decimal point of the divisor 4&6
once to the right, we get 46 and by shifting that of the divisible 8&51
once to the right, we get 85&1.
46) 8 5&1 (1&85
4 6
3 9 1
3 6 8
2 3 0
2 3 0
0∴ 8&51 ÷ 4&6 = 1&85
Answer : 1&85.

Example 23. The cost of 5 kg. potatoes isTk. 63&75. What is the
cost of 1 kg. potatoes?
Solution : The cost of 5 kg. potatoes is Tk.
Tk.
63&75.
∴ 1" " " "
5
7563&
Here, 5) 6 3&7 5 (12&75
5
1 3
1 0
3 7
3 5
2 5
2 5
0
∴ The cost of one kg. potatoes is Tk. 12&75.
Answer : Tk. 12&75.
Example 24. Amena bought 5 notebooks for Tk. 12&50 and 4
ballpen for Tk. 6&50. She gave a Tk. 100 note to the shopkeeper.
How much money will he give her back?
Solution : The cost of 5 notebooks is (12&50 × 5) = Tk. 62&50
The cost of 4 ballpen is (6&50 × 4) = Tk. 26&00
Total cost = Tk. 88&50
Amena gave the shopkeeper Tk. 100&00
The total cost of 5 notebooks and
Tk. 88&504 ballpen
The shopkeeper will return Tk. 11&50 (by subtracting)
Answer : Tk. 11&50.

Exercise - 10 (c)
1. Find out the quotient :
a) 6&45 ÷ 5 b) 18&56 ÷ 4 c) 56&25 ÷ 12 d) &85 ÷ 17
e) 24&6 ÷ 8 f) &005 ÷ 25 g) 15 ÷ &04 h) 48 ÷ 7&5
i) 2 ÷ 1&25 j) 7&89 ÷ 10 k) 14&3 ÷ 10 l) 86&1 ÷ 100
m) 72&35 ÷ 100 n) 56&25 ÷ 1&25 o) 6&4 ÷ &25 p) &027 ÷ 18
q) &0075 ÷ 12&5 r) 3&25 ÷ &04
2. 11 ballpens cost Tk. 27&50. How much does one ballpen cost?
3. The cost of 4 oranges is Tk. 12&80. How much does one cost?
4. A worker earns Tk. 492&50 in 5 days. How much does he earn per
day?
5. The product of two numbers is 8&8. One number is 2&75. What is
the other?
6. The product of two numbers is 19 &845. One number is 0&9. What
is the other?
7. The divisor is 0&008; divisible is 0&204; what is the quotients?
8. By what number 0&006 should be divide to get the quotient 0
.
12?
9. If 6&5 metres equals to 255&905 inches, how many inches equals to one
metre?
10. The bill of 28 telephone calls is Tk. 48&16. What is the bill of one call?
11. 25 litres of fuel cost Tk. 393&75. How much does 1 litre cost?
12. A car goes 45.6 km. per hour. How many hours will it take to go
319&2 km.?
13. The cost of 24&5 kg wheat is Tk. 367&50. How much does 1 kg. cost?
14. The amount of money which was collected after each of 10 students
donated Tk. 15&50 was distributed equally among 4 poor people.
How much did one poor person get?
15. A teacher bought oranges for Tk. 722.15 at the rate of 60
.
60 and
distributed them equally among 13 students. How many oranges will
one student get?
16. The sum of two numbers is 70&60. The greater number is 4&50
more than the smaller one. What are these numbers?

Chapter - 11
Percentage
The big area at the right has been
divided into 100 small areas.
Among these 100 small areas 5 are
deep coloured. It can be seen that the
deep coloured areas cover
100
5 of
the total area. The fraction
100
5 is called 5 per hundred
or, 5 percent. It is written as 5%.
The marks obtained in mathematics in the last annual examination
Subject Total marks Mithu Rina Raju Rani Rita
Math 100 85 80 82 65 60
Mithu obtained 85 in 100. In other words, he obtained the marks
„85 per hundred‰ or '85 percent' or 85%.
85% =
100
1
85
100
85
×=
Rina obtained the marks 80% and Raju 82%.
Similarly, Rani obtained 65% and Rita 60%.
Mizan bought some raw materials for Tk. 100 and sold them for
Tk. 115. He made a profit of (Tk. 115 ă Tk. 100) = Tk. 15 in
Tk. 100. His profit was Tk. 15 percent. It means, his profit
is 15%.
The word percentage is briefly expressed with the symbol %.
The percentage is a fraction which has the denominator 100
in each case.

2
1
% =
100
2
1
=
2
1
×
100
1
1% =
100
1
= 1 ×
100
1
10% =
100
10
= 10 ×
100
1
20% =
100
20
= 20 ×
100
1
30% =
100
30
= 30 ×
100
1
40% =
100
40
= 40 ×
100
1
50% =
100
50
= 50 ×
100
1
60% =
100
60
= 60 ×
100
1
70% =
100
70
= 70 ×
100
1
80% =
100
80
= 80 ×
100
1
90% =
100
90
= 90 ×
100
1
100% =
100
100
= 100 ×
100
1
Example 1. Express 35% as a common fraction.
Solution : 35% =
20
100
17
35×
=
20
7
Answer :
20
7
Example 2. Express
4
3
in percentage.
Solution :
4
3
=
100
1
4
1003
1004
1003
×
×
=
×
×
= 75 ×
100
1
= 75%
Answer : 75%.
.

118 Percentage
Example 3. Express
10
15
in percentage.
Solution :
10
15
=
100
1
10
10015
10010
10015
×
×
=
×
×
= 150 ×
100
1
= 150%
Answer : 150%
10
15
is an improper fraction. Since it has been expressed in
percentage, it exceeded 100%.
Example 4. Express in percentage how much is Tk. 10 of Tk. 25
in portion.
Solution : Tk. 10 is
25
10
portion of Tk. 25.
Now,
25
10
=
10025
10010
×
×
=
100
1
25
10010
×
×
= 40×
100
1
= 40%
Answer : 40%.
Example 5. 75% of 4 = how much ?
Solution : 75% of 4 = 75×
100
1
of 4
=
100
75
of 4
=
100
75
× 4 = 3
Answer : 3.
Example 6. A government employee gets 45% of his salary as house
rent. What portion of his salary does he get as house rent?

Solution : 45% =
100
45
=
20
9
Answer :
20
9
portion of salary.
Example 7. Mr. Anisul Islam deposits 5% of his salary in a bank.
In one month he deposited Tk. 400. What is his monthly salary?
Solution : If Tk. 5 is deposited the monthly salary is Tk. 100
∴ 1 "" " " " Tk.
5
100
∴ 400 "" " " " Tk.
1
5
80
400100×
= Tk. 8000
∴ The monthly salary of Mr. Anisul Islam is Tk. 8000.
Answer : Tk. 8000.
Example 8. The population of Ibrahimpur village increased by 4%
and became 1872. What was the number of people before?
Solution : The population has increased by 4%. In other words, if
the population was 100 earlier, now it has become (100 + 4) or 104.
If the present population is 104 previously it was 100
∴ 1" " "
104
100
∴ 1872"
"
" " "
26
104
72
936
1872
25
100×
= 1800 people
∴ Earlier the population of the village was 1800.
Answer : 1800 people .
13 1

120 Percentage
Example 9. A pen is bought for Tk. 50 and sold for Tk. 56. What is
the percentage of profit?
Solution : Selling price = Tk. 56
Cost price = Tk. 50
∴ Profit = Tk. 56 ă Tk. 50 = Tk. 6
Now, in Tk. 50 the profit is Tk. 6
∴ 1 " " Tk.
50
6
∴ 100 "
"
"
"
" " Tk.
50
2
1006×
= Tk. 12
∴ The profit is 12 percent.
Answer : 12%.
Profit is made if the selling price is greater than the
cost price.
Profit = selling price ă cost price.
Profit is calculated on the cost price.
Alternate method :
Selling price = Tk. 56
Cost price = Tk. 50
∴ Profit =
50
6
portion of the cost price
Here,
50
6
=
10050
1006
×
×
=
50
600
×
100
1
= 12%
Answer : 12%.
1

forma-16, Math - V
Example 10. A pen is bought for Tk. 50. What is the percentage of
loss if it is sold for Tk. 45?
Solution : Cost price = Tk. 50
Selling price = Tk. 45
∴ Loss = Tk. 50 ă Tk. 45 = Tk. 5
Now, for Tk. 50 the loss is Tk. 5
∴ 1 """ " "
50
5
∴ 100 """ " "
1
10
50
10
100
1
5×
= Tk. 10
∴ The loss is 10 percent.
Answer : 10%.
If the selling price is lower than the cost price a loss is made.
Loss = cost price ă selling price.
Loss is calculated on the cost price.
Example 11. The cost price of a pants is Tk. 280. What will
be the selling price if a profit of 25% is to be made?
Solution : If the cost price is Tk. 100 and the target profit is 25%,
the selling price of the pants will be Tk. (100 +25) = Tk. 125
Now, if Tk. 100 is the cost price, the selling price will be Tk. 125
∴ "1 " " Tk.
100
125
∴ ""
"
280 " "
"
"
"
"
"
" Tk.
4
100
70
280
5
125×
= Tk. 350
∴ The selling price of the pants is Tk. 350
.
Answer : Tk. 350.
1

122 Percentage
Example 12. A goat is sold at a loss of 6%. If the selling price of
the goat is Tk. 2256, what is the cost price?
Solution : If the cost price is Tk. 100, then with 6% loss the selling
price will be Tk. (100 - 6) or, Tk. 94.
Now, if the selling price is Tk. 94, the cost price is Tk. 100
∴ " 1 " " Tk.
94
100
∴ " 2256 "
"
"
"
" " Tk.
94
2256100×
= Tk. 2400
∴ The cost price of the goat is Tk. 2400
Answer : Tk. 2400.
Example 13. Mr. Habib received Tk. 96 as interest of in 4 years by
depositing Tk. 800 in a bank. What is the rate of interest in
percentage per annum?
Solution : The interest on Tk. 800 in 4 years is Tk. 96
∴ 1" 1 " Tk.
8004
96
×
∴ 100"
"
"
"
" 1 " Tk.
8004
100
3
12
96
×
×
= Tk. 3
∴The rate of interest in percentage per annum Tk. 3
Answer : Rate of interest 3%.
The rate of interest in percentage =
PrincipalTime
100Interest
×
×
The money deposited in a bank is called the principal.
The money in addition to the principal given by the
bank is called the interest.
The interest, which is considered per Tk. 100 in 1
year, is called the rate of interest.
1 8 1
1
47 1
1128
24

Example 14. The rate of annual interest in percentage is Tk. 6.
What is the interest on Tk. 600 in 3 years ?
Solution : The interest on Tk. 100 in 1 year is Tk. 6
∴ 1 1 " Tk.
100
6
∴ 600 1 " Tk.
100
6006×
∴ 600
"
"
"
"
"
" 3 " Tk.
100
36006 ××
= Tk. 108
Answer : Tk. 108.
Interest =
100
TimeRatePrincipal ××
Example 15. What is the principal which will yield an interest of
Tk. 180 in 4 years at 5% interest?
Solution : The interest is Tk. 5 in 1 year on the principal Tk. 100
∴ 1 1 " " Tk.
5
100
∴ 180 1 " " Tk.
5
180100×
∴ 180
"
"
"
"
"
" 4 "
"
"
" " Tk.
45
180
5
20
100
×
×
= Tk. 900
Answer : The principal is Tk. 900.
Principal =
TimeRate of interest
100Interest
×
×
Let us notice : If the number of year increases, the amount of
principal will decrease for the same amount of interest. So, it is
divided by 4.
1
6
1 1

124 Percentage
Example 16. In how many years the interest on Tk. 850 at 6%,
interest rate will be Tk. 255?
Solution : The interest on Tk. 100 for 1 year is Tk. 6
∴ 1 1 " Tk.
100
6
∴ 850 1 " Tk.
"
"
"
"
"
"
20
100
170
850
3
6×
= Tk. 51
Now, the interest amounts to Tk. 51 in 1 year.
∴ 1 """"
51
1
∴ 255 """"
51
5
2551×
= 5 years.
Answer : 5 years.
Time =
RatePrincipal
100Interest
×
×
Exercise - 11
1. Express in common fractions :
(a) 5% (b) 15% (c) 25% (d) 35% (e) 60%
(f) 50% (g) 75% (h) 85% (i) 8
3
1
% (j) 16
3
2
%
2. Express as percentages :
(a)
4
1
(b)
5
2
(c)
8
5
(d)
10
1
(e)
12
5
(f)
20
1
(g)
15
2
(h)
7
3
(i)
15
18
(j)
20
25
(k)
50
65
(l)
65
88
34 17
4 2 1
1

3. Find the values of :
(1) 4% of Tk. 75 (2) 15% of Tk. 90 (3) 4% of Tk. 3.20&
(4) 20% of 20 kg. (5) 30% of 50 gm. (6) 8
3
1
% of 80 km.
(7) 20% of Tk. 550 (8) 15% of Tk. 300 (9) 40% of Tk. 60
4. Express the first numbers as percentage of the second :
(a) Tk. 5, Tk. 20 (b) Tk. 16, Tk. 25
(c) 21 kg, 35 kg (d) 15 m., 45 m.
(e) 45 kg, 72 kg (f) Tk. 70 , Tk. 80
5. In the annual examination Jasim secured 420 marks out of 700
marks. What percentage of marks did he get?
6. In your class among 60 students 30% are female students.
How many are female students?
7. At present the monthly income of Mr. Hanif is Tk. 1550.
Earlier his income was Tk. 1500. What is the percentage of his
increased income?
8. In a school among 560 students 224 are females. What is the
percentage of female students in that school?
9. In a class among 150 students 6 are absent. What is the
percentage of absent students?
10. 8% of the total population of Narayanpur die every year. If the
number of population of that village is 6250, how many
people die each year?
11. RahimaÊs monthly income is Tk. 420 and expenditure is
Tk. 336. What is the percentage of expenditure relating to her income?
12. In your area
25
18 portion people are educated. What is the
percentage of educated people in the area?
13. Last year from Ekhlashpur High School 64 students appeared
in the S.S.C. examination. 56 among them passed. What
percentage of students passed?

126 Percentage
14. 45% of people in Alenia village are educated. The total
population of that village is 900. What is the number of educated
people in that village?
15. Among 450 soldiers 63 were martyr. What is the percentage of
martyred soldiers?
16. Mr. Proshanto spends 40% of his salary on house rent. His
monthly income is Tk 6550. How much does he spend on house
rent?
17. A fisherman caught 150 fish from a pond. Among these 30 were
Ruhi. What was the percentage of Ruhi fish?
18. 250 Fazli mangoes were sent from Chapai Nawabgonj. Among
these 25 were rotten. What is the percentage of fresh mangoes?
19. Anwara Begum gets Tk. 1000 as her monthly salary. She saves
10% of her salary in providend fund. How much does she save
every month?
20. Mr. Monower gets Tk. 2000 as his monthly salary. 40% of his
salary covers his house rent. What is the amount of money he gets
as house rent?
21. In Bangladesh 50 babies are born per thousand each year and 20
of them die. What is the percentage of population increase in
Bangladesh?
22. In Santoshpur village with 3% increase the number of population
turned 1236. What was the number of population earlier?
23. Of Bangladesh in the year 1981 the population was 9 crores and
in year 1982 it became 9 crores and 27 lacs. What is the
percentage of population increase in 1982?
24. A trader named Halim purchased rice for Tk. 250.00 and sold it
for Tk. 275.00. What is the percentage of his profit?
25. The cost price of a chair is Tk. 175 and the selling price is Tk. 140.
What is the percentage of profit or loss?

26. The cost price of a commodity is Tk. 150. What will be its
selling price if a profit of 25% is to be made?
27. A pen has been bought for Tk. 60 and sold at a profit of 25%.
What is the selling price?
28. A pants was sold for Tk. 588 at a loss of 2%. What is the cost of
the pants?
29. A shopkeeper bought a shirt for Tk. 240 and sold it for Tk. 210.
What is the percentage of profit or loss?
30. Apples worth Tk. 1000 were sold at a loss of 2%. What is the
total amount of loss?
31. Monowara deposited an amount of Tk. 900 in a bank and got
Tk. 216 as interest in 3 years. What is the percentage of interest
in that bank?
32. Mithu deposited an amount of Tk. 1200 in a bank. He got Tk. 432
as interest after 4 years. What is the percentage of interest in that
bank?
33. The interest of Tk. 600 in 3 years is Tk. 144. Find out the rate of
interest.
34. How much will Tk. 350 yield in 4 years at a rate of 5% simple
interest per annum?
35. How much will Tk. 1200 yield in 3 years at the rate of 7 percent
interest ?
36. In how many years will Tk. 425 become Tk. 510 as principal
including interest, if the rate of interest is 5%?
37. In how many years Tk. 350 will yield an interest of Tk. 168 if
the rate of interest is 8%?
38. An amount yields Tk. 360 in 5 years at the rate of interest 6%.
Find out the principal.

Measurement
Measurement of Length
The basic unit of measurement of length : Metre
The following table shows the relation between metre and the
other units of measuring length in Metric System.
1 Kilometre (k.m.) = 1000 metre
1 Hectometre (h.m.) = 100 metre
1 Decametre (decam.) = 10 metre
1 Metre (m.) = 1 metre
1 Decimetre (decim.) =
10
1 m. or,0.1 metre
1 Centimetre (c.m.) =
100
1
m. or, 0.01 metre
1 Millimetre (m.m.) =
1000
1
m. or,0.001 metre
Kilo is thousand times (10 ×10 ×10)
Hecto is hundred times (10 ×10)
Deca is ten times (10)
Deci is tenth ( )
10
1
or,
Centi is hundredth
Milli is thousandth
or, 0.01
0.1
Chapter - 12
In case of transforming to greater unit from basic unit of metre,
each preceding unit is to be multiplied by 10. Again in case of
transforming to smaller unit from basic unit of metre, each
preceding unit is to be multiplied by 10
1
( )
100
1
or, 0.001( )1000
1

Measurement of Weight
The basic unit of measurement of weight : Gram
Metric Units of Measurement of Weight :
The table of metric units for measuring weight is found by writing
gram instead of in the table of metric units for measuring
length.
Two more units which are used for measuring weight in metric
system:
100 kilogram (k.g.) = 1 quintal
10 quintal or 1000 kilogram = 1metric ton
Example 4. Express 6 kilogram 25 gram 59 centigram 5 milligram
in milligrams.
Solution : 6 kilogram = 6 × 1000 gram
= 6000 × 1000 m.g. = 6000000 m.g.
25 gram = 25 × 1000 m.g. = 25000 m.g.
59 centigram = 59 × 10 m.g. = 590 m.g.
5 milligram = 5 × 1 m.g. = 5 m.g.
Summation = 6025595 m.g.
Answer : 6025595 milligram
Example 5. Express 29 kilogram 206 gram 129 milligram in
kilograms.
Solution : 29 kilogram = 29 × 1 k.g. = 29 k.g.
206 gram = 206 × 0&001 k.g. = 0&206 k.g.
129 milligram = 129 × 0&001 gram
= 0&129 × 0&001 k.g.= 0&000129 k.g.
Summation = 29&206129 k.g.
Answer : 29&206129 k.g.
.

132 Measurement
National Units for measuring weight :
Relationship between Metric and National Systems for
measuring weight :
(approx.)
(approx.)
(approx.)
(approx.)
(approx.)
(approx.)
(approx.)
maund
maund
maund
maund

maund
maund
maund
maund
maund
maund
maund

Divide
Divide
by
by

136 Measurement
Solution : The potatoes of Mr. Aziz had : 4 5 k.g.
he sold : 2 8 k.g. 425 g.
The potatoes left : 1 6 k.g. 575 gram
∴ Mr, Aziz has 16 k.g. 575 gram of potatoes left.
Answer : 16 k.g. 575 gram.
Measurement of the volume of liquid
The basic unit for measurement of the volume of liquid: Litre
Metric units for measuring the volume of liquids :
The table of metric units for measuring the volume of liquid is found
by writing ' litre ' instead of 'metre' in the table of metric units for
measuring length.
International unit of measrement for liquids.
1000 millilitre
or = 1 litre
1000 cubic centimetre
1000 litre = 1 cubic metre (Cubic metre is the
international unit or S.I. Unit)
Exercise - 12 (a)
1. Express in millimetres:
(a) 36 centimetre 8 millimetre;
(b) 99 metre 30 centimetre 5 millimetre;
(c) 8 kilometre 95 centimetre 2 millimetre;
(d) 8 kilometre 245 metre 96 millimetre;
(e) 7 kilometre 250 metre 85 centimetre.
Example16. Mr. Aziz sold 28 k.g. 425 gram of potatoes from 45
k.g. of potatoes he had. How much potatoes does he have left?

2. Express in metre :
(a) 58 centimetre 8 millimetre;
(b) 43 metre 25 centimetre 7 millimetre;
(c) 90 metre 36 centimetre 6 millimetre;
(d) 8 kilometre 9 metre 125 centimetre;
(e) 6 kilometre 237 centimetre.
3. Express in kilometre :
(a) 22 kilometre 35 metre 27 centimetre ;
(b) 19 kilometre 87 metre 90 millimetre ;
(c) 125 kilometre 75 metre 250 millimetre ;
(d) 78 kilometre 40 metre 15 centimetre 4 millimetre ;
(e) 308 metres 96 millimetre.
4. Express in milligram :
(a) 18 gram 15 centigram 6 milligram;
(b) 4 kilogram 27 gram 45 centigram 2 milligram;
(c) 9 kilogram 38 gram 10 centigram;
(d) 5 kilogram 65 gram 37 centigram;
(e) 4 kilgram 236 centigram.
5. Express in gram :
(a) 29 kilogram 515 gram; (b) 60 kilogram 182 gram;
(c) 82 kilogram 98 gram 45 centigram;
(d) 9 kilogram 65 gram 54 centigram 9 milligram.
6. Express in kilogram :
(a) 45 kilogram 25 gram 59 centigram ;
(b) 12 kilogram 98 gram 145 milligram;
(c) 9 kilogram 427 gram 505 milligram;
(d) 27 kilogram 306 gram 97 centigram.
7. 385000 gram equal how many quintal ? How many metric ton?
8. 85321 kilogram equal how many metric ton?
9. 4 maund 22 seer equal how many kilogram?
10. Express 37 kilogram in seer.

138 Measurement
11. 80 kilogram equal how many seer ?
12. Add :
(a) k.m. h.m. deca m. metre (b) k.g. h.g. deca g. gram
21 7 2 8 28 6 0 9
9 0 6 5 34 8 9 2
28 9 0 7 20 0 5 0
10 2 8 1 8 2 8 7
13. Subtract :
(a) k.m. h.m. deca m. metre (b) k.g. h.g. deca g. gram
98 4 3 5 54 6 2 3
28 8 5 7 39 5 7 9
14. Multiply :
(a) 6 k.m. 8 h.m. 5 metre by 8 ;
(b) 8 k.m. 11 deca m. 6 metre by 7;
(c) 9 k.g. 5 hectogram 17 gram by 8;
(d) 5 k.g. 4 h.gram 8 deca g. 9 gram by 8.
15. Divide :
(a) 28 kilometre 9 hectometre 6 decametre 2 metre by 9;
(b) 49 kilometre 7 decametre 7 metre by 9;
(c) 69 k.g. 9 hectogram 6 decagram 8 gram by 8.
16. Sumi's weighs 32 k.g. 8 h.g. 9 deca g. 5 gram, Mitu's weighs
29 k.g. 7 h.g. 9 gram. Raju's weighs 35 k.g. 7 deca g. 8 gram
and Saju's weighs 30 k.g. 7 gram. What is their total weight?
17. Hashem Mian has sold 37 k.g. 500 grams of rice from 98 k.g.
of rice. How much rice does he have left?
18. A drum can contain 35 k.g. 3 hectogram, 7 decagram 9 gram
of flour. How much can 9 similar drums contain?
19. Babul can go 78 kilometre riding on a bicycle for 5 hours.
How much distance will he go per hour?

Measurement of Areas
The amount of surface with a particular
measure enclosed by a borderline is
called its area.
The area of some particular shapes is shown below :
Triangular
area
Quadrilateral
area
Rectangular
area
Square
area
In each figure the deep coloured space is an area.
A particular unit is used to measure an area. An area with a
particular unit of length and breadth is considered as a square unit.
Here, the deep space the borderline is an
area. This area has a particular
measurement. This measurement is called
its area.
Each of these is an enclosed area with a particular borderline
having a particular measure. These measurement are the areas of
the figures.
In the figure of the triangle, the deep space inside is the triangular area.
In the figure of the quadrilateral, the deep space inside is the quadrilateral
area.
In the figure of the rectangle, the deep space inside is the rectangular
area.
In the figure of the square, the deep space inside is the square area.

140 Measurement
3c.m.
(breadth)
A B
D C
1 c.m.
1 c.m.
1 c.m. 1 c.m. 1 c.m. 1 c.m.
1 c.m.
1 c.m.
1 c.m.
A
B
D C
The Metric Units of Land Measurement :
1 square metre = 100 square decimetres = 10,000 square c.m.
1 square decametre = 100 square metres = 1 acre
1 square hectometre = 100 square decametre = 10,000 square metres
1 hectre = 10,000 square metres = 1 square hectometre
The Area of the Square
This is a square. The lenght of its each side is
1 centimetre.
The area of this square = 1 c.m. × 1 c.m.
= 1 square centimetre.
That means, Area = Length × Breadth
The Area of the Rectangle :
Therefore, the area of the rectangle = The area of 12 small squares
= (12 × 1) square c.m.
= (4 × 3×1) square c.m.
= (4 × 3) square c.m.
= 12 square c.m.
Therefore, the area of the rectangle = 12 square c.m.
A B C D is a rectangle. Its length is 4 c.m. and breadth is 3 c.m.
Along with its length and breadth in every 1 c.m. lines have been
drawn to divide up the rectangle. By this, the rectangle has been
divided into (4 3) = 12 smaller areas. Each small area has the
lenght 1 c.m. and breadth 1 c.m. All these small areas are
squares and each of them has the area = 1 square c.m.
4 c.m. (length)

Let us notice : To find the area, we multiply the length of the
rectangle by its breadth.
Example 1. The length of a rectangle is 35 c.m. and breadth 25 c.m.
What is its area?
Solution : We know,
The area of a rectangle = length × breadth
∴ The area = 35 c.m. × 25 c.m.
= (35× 25) square c.m.
= 875 square c.m.
Answer : 875 square c.m.
Example 2. What is the area of a rectangle of 5 c.m. length and
2 c.m. 5.m.m. breadth?
Solution : Length = 5 c.m. = 5 × 10 m.m. = 50 m.m.
Breadth = 2 c.m. 5 m.m. = 25 m.m.
∴ The area of the rectangle = length × breadth
= 50 m.m. × 25 m.m.
= 1250 square m.m.
= 12&50 square c.m. [Q 100 sq. m.m. = 1 sq. c.m.]
Answer : 12&50 square c.m.
When measuring an area, if the length and breadth are in different
units, then to transform them in the same unit.
The Area of Triangle :
A B C D is a rectangle. The diagonal line
AC has divided it into two triangles of equal
measure. The area of each of these triangle
is the half of that of the rectangle.
A
B C
D
Fig. 1
The measurement of the area of rectangle = The measurement of
length × measurement of breadth
In short, the area of rectangle = length × breadth

142 Measurement
Since A B C D is a rectangle, the side AB is perpendicular to side BC.
That means, the side A B is the height of the triangle A B C.
Therefore, in Fig. 1, the area of the triangle A B C
=
2
1
(the area of rectangle A B C D)
=
2
1
(BC×AB) [QThe area of rectangle ABCD = length×breadth= BC×AB]
=
2
1
(the base of the triangle × the height of the triangle)
=
2
1
(base × height)
∴ The area of the triangle ABC =
2
1
(base × height)
In Fig. 2, the area of the triangle ABC
= The area of triangle ABD
+ the are of triangle ACD.
=
2
1
(the area of rectangle AFBD) +
2
1
(the area of rectangle AECD)
=
=
2
1
2
1
(BD
(BD + DC) × AD
× AD)
2
1
(DC AD)
=
2
1
(BC × AD [Q BD + DC = BC ]
=
2
1
=
2
1
(base × height)
2
1
(base × height)
Fig. 2
D
A
B C
F E
+

In Fig. 3, The area of the triangle ABC
= the area of triangle ABD
− the area of triangle ACD
=
2
1
(the area of rectangle ADBF) −
2
1
(the area of rectangle ADCE)
=
2
1
(BD ×AD)−
2
1
(CD × AD)
=
2
1
(BD − CD) × AD
=
2
1
(BC × AD) [QBD − CD = BC]
=
2
1
=
2
1
(base × height)
2
1
(base × height)
Generally, the area of triangle =
2
1
(base × height)
Example 3. The length of the land of a triangle is 8 c.m. and height
5 c.m. What is the area of the triangle?
Solution : We know, the area of triangle =
2
1
(base × height)
=
2
1
(8 c.m. × 5 c.m.)
=
1
(8×5) square c.m.
=
2
1
× 40 square c.m.
= 20 square c.m.
∴ The area of the triangle = 20 square c.m .
A
B DC
EF
Fig. 3

144 Measurement
A
B
D
CE
A
B C
D
E
F
Example 4. A rectangle formed by two triangles has the length 12
metre and breadth 8 metre. What is the area of the
rectangle? What is the area of one triangle?
Solution : We know, the area of rectangle = length × breadth
∴ the area of rectangle = 12 metre × 8 metre
= (12 ×
×
8) square metre
= 96 square metre
We know, the area of triangle =
2
1
(the area of rectangle)
=
2
1
96 square metre
= 48 square metre
Answer : The area of rectangle = 96 square metre.
and the area of triangle = 48 square metre.
Area of quadrilateral :
In Fig. 1, ABCD is a quadrilateral.
AC is the diagonal line which
divides the quadrilateral into two
trangles ABC and ACD.
Therefore, the area of the quadrilateral ABCD
= the area of triangle ABC + the area of triangle ACD.
∴ The area of the quadrilateral ABCD
= the area of the rectangle ABED + the area of the triangle DEC.
Fig.-1
Fig.-2
In Fig. 2, ABCD is a quadrilateral. Let us
draw the line DE from the point D which
is equal and parallel to AB. The line DE
divides the quadrilateral into two areas.
One is the rectangle ABED and the other
is the triangle DEC.

Here, in the triangle ABC the base is AC and height BF.
∴The area of the triangle ABC =
2
1
(base × height)
=
2
1
(AC × BF)
=
2
1
(5 × 4) sq. c.m.
=
2
1
× 20 sq. c.m. = 10 sq. c.m.
Again, in the triangle ACD the base is AC and height DE .
∴ The area of the triangle ACD =
2
1
(base × height)
=
2
1
(AC × DE)
=
2
1
(5 × 2) sq. c.m.
=
2
1
× 10 sq. c.m.
= 5 sq. c.m.
∴ The area of the quadrilateral ABCD
= the area of the triangle ABC+ the area of the triangle ACD
= 10 square c.m. + 5 square c.m.
= 15 square c.m.
Answer: 15 square c.m.
Example 5. In a quadrilateral, a diagonal line is 5 c.m. long.
The perpendicular distances of the vertex opposite to the
diagonal line are 4 c.m. and 2 c.m. Find out the area of the
quadrilateral.
Solution : The area of the quadrilateral
ABCD = the area of the triangle ABC +
the area of the triangle ACD.
A
F
B C
D
4 c.m.
2 c.m.
5c.m.
E

146 Measurement
Solution: ABCD is a quadrilateral.
Here, the side BC = 6 c.m.
side AD = 4 c.m. ABC = 90
Then height AB = 3 c.m.
Let us draw a perpendicular line DF on BC from point D so
that, AD = BF.
Now, ABFD is a rectangle and DFC is a triangle.
∴ AB = DF = 3 c.m.
and AD = BF = 4 c.m.
it is given, BC = 6 c.m.
∴ FC = BCă BF
= (6 − 4) c.m. = 2 c.m.
∴ the area of quadrilateral ABCD
= The area of rectangle ABFD + the area of triangle DCF.
= (BF × AB) +
2
1
(FC × DF)
= (4 c.m. × 3 c.m.) +
2
1
(2 c.m. × 3 c.m.)
= 12 square c.m. +
2
1
× 6 square c.m.
= 12 square c.m. + 3 square c.m.
= 15 square c.m.
∴ The area of the quadrilateral = 15 square c.m..
Example 6. In a quadrilateral the lengths of both parallel sides are 6
c.m. and 4 c.m. and one angle is rightangle respectively. If the distance
between the sides is 3 c.m., what is the area of the quadrilateral?
A D
B F C
4 c.m
3 c.m
6 c.m

Exercise -12 (B)
1. Find out the area of the following rectangles :
(a) Length 36 c.m. and breadth 30 c.m.
(b) Length 2 m. and breadth 65 c.m.
(c) Length 45 m. and breadth 37 m.
(d) Length 82 m. and breadth 63.25 m.
2. The lengths of one side of several squares are given below.
Find out their area :
(a) 27 metre (b) 76 metre (c) 6 metre 44 c.m.
3. The base and height of several triangles are given below. Find
out their area :
(a) base 10 metre and height 15 metre.
(b) base 25 metre and height 18 metre.
(c) base 7 metre 35 centimetre and height 2 metre.
4. The area of a rectangle is 5400 square metre and breadth 20
metre. What is its length?
5. The area of a rectangular garden is 3392 square c.m. Its length
is 64 c.m. What is its breadth?
6. The area of a rectangular land is 1092 square metre. If its
length is 39 metre, what is its breadth?
7. The length of one side of a square is 45 metre. What is its
area?
8. In a triangle the base is 28 metre and the height is 12 metre.
What is its area?

148 Measurement
9. The area of a triangular land is 8544 square metre. Its base
measures 100 metre. What is its height?
10. The area of a triangle is 189 square metre and height 14
metre. What is the measure of its base ?
11. The base of a triangular land is 280 metre and height 150
metre. Find out its area in hector.
12. The length of a diagonal line of a quadrilateral is 9 c.m. The
perpendicular distances of the vertex opposite the diagonal
line are 6 c.m. and 8 c.m. Find out the area of the quadrilateral?
13. The lengths of two parallel sides of a quadrilateral are 8 c.m.
and 6 c.m. respectively and the distance between the sides is
5 c.m. What is the area of the quadrilateral?
14. The lengths of two parallel sides of a quadrilateral land are 40
metre and 50 metre respectively and the distance between the
sides is 25 metre. What is the area of the land?

Time
The names of twelve Bengali months and the number of days :
Months Number of days Months Number of days
Baishak 31 Kartik 30
Jaistha 31 Agrahayan 30
Ashar 31 Poush 30
Shravan 31 Magh 30
Bhadra 31 Falgun 30
Ashwin 30 Chaitra 30
According to Bengali practice the day and date starts when the sun rises.
The names of twelve English months and the number of days :
Months Number of days Months Number of days
January 31 July 31
February 28 August 31
March 31 September 30
April 30 October 31
May 31 November 30
June 30 December 31
According to international or English practice the day and date
starts after 12 pm.
The names of twelve Hijri months :
1. Muharram 5. Jamadiul Awal 9. Ramjan
2. Safar 6. Jamadius Sani 10. Shawal
3. Rabiul Awal 7. Rajab 11. Jilkadh
4. Rabius Sani 8. Shaban 12. Jilhajj
On the 20 September, 622 A.D. Hazrat Muhammad (Sm)
migrated from Macca to Madina. Since then the Hijri year
has been counted.
Chapter- 13

150 Time
Depending on the duration of the moon, the days in Hijri month
can be 29 or 30.
The Hijri year is counted on a lunar month. A Hijri year has 11
days less than a solar year.
The first month of the Hijri year is Muharram and the last one is
Jilhajj.
A Hijri day and date is counted after the sunset.
Leap year
February 2004
Saturday Sunday Monday Tuesday Wednesday Thursday Friday
1 2 3 4 5 6
7 8 9 10 11 12 13
14 15 16 17 18 19 20
21 22 23 24 25 26 27
28 29
In the year 2004 the month of February had 29 days. The year
was a leap year.
Normally a year has 365 days. However, the earth takes
approximately 365&2422 days to move around the sun once. As
a result, in a normal year 0.2422 days or
4
1
of a day is not
counted. To solve this problem after each 4 years a day is added
and it is considered as an extra day in the month of February.
The year in which 29 days instead of 28 days are counted in the
month of February is called a leap year.
If an extra day is added in every four years then in four years
(1-0.2422 × 4) = 0.0312 or in four hundred years 3.12 or 3
days (approx.) are added in addition. For this reason, in every
four hundred years 3 leap years are dropped.
From the year 8 A.D. the leap years are regularly counted based
on the reformed calender. According to it,

Example 1. What is the number of days in the month of February of
1996?
Solution : 4) 9 6 (24
8
1 6
1 6
0
Answer : 29 days.
Example 2. Is 1998 a leap year?
The year in which the last two digits are not 00, can be a leap year
if its last two digits make a number that is divisible by 4.
The year in which the last two digits are 00, can be a leap year if
its first two digits make a number that is divisible by 4. Otherwise, it
is not considered a leap year.
Observe that : If the number formed by the last two digits of a
number is divisible by 4, the number itself is divisible by 4. Therefore,
each leap year is divisible by 4.
However, the years 1500, 1700, 1800, 1900, 2100 are not leap years,
although they are divisible by 4. The first two digits of these numbers
are not divisible by 4. In other words, they are not divisible by 400.
The years 1600, 2000, 2400 are leap year, since they are divisible by
400.
In the number 1996, the number formed with the digits in ones and
tens places i.e. the number 96 is divisible by 4. So, the given number is
also divisible by 4. Therefore, the year is a leap year.
Therefore, in 1996 there are 29 days in the month of February.
Solution : In the number 1998, the number formed with the digits in
ones and tens places, i.e. the number 98 is not divisible by 4. So, the
given number is not divisible by 4.
Therefore, 1998 is not a leap year.
Answer : It is not a leap year.

152 Time
Exam ple 3. Is 2000 a leap year?
Solution : In the number 2000 the digits in ones and tens places are
0 and the first two digits form 20 which is divisible by 4. So, the
number 2000 is divisible by 400.
Therefore, 2000 is a leap year.
Answer : It is a leap year.
Decade, Era, Century
Units of measuring time :
The period of successive 10 years is considered as a decade. For
example, the period from the year 1991 to 2000 is considered a
decade.
The period of successive 12 years is considered as an era. For
example, the period from the year 1991 to 2002 is considered an
era.
The period of successive 100 years is considered as a century. For
example, the period from the year 1901 to 2000 is considered a
century and it is called the twentieth century.
According to English calender, the period from year 1 to year 100
is considered as the first century. The period from the year 1801 to
1900 is called the nineteenth century.
According to Bengali calender, the period from the year 1101 to
year 1200 is considered as the twelfth century and the period from
the year1301 to year 1400 is considered as the fourteenth century.
60 seconds = 1 minute
60 minutes = 1 hour
24 hours = 1 day
7 days = 1 week
30 days (generally) = 1 month
12 months = 1 year
365 days = 1 year
12 years = 1 era
100 years = 1 century

The Addition and Subtraction of Numbers Related
with Time.
Example 4. Add : 4 years 5 months 15 days 12 hours and 7 years
8 months 20 days 16 hours.
Solution : Years Months Days Hours
4 5 15 12
7 8 20 16
12 2 6 4
Answer : 12 years 2 months 6 days 4 hours.
Let us notice that :
12 hours + 16 hours = 28 hours = 1 day 4 hours
[Q 24 hours = 1 day]
15 days + 20 days + (in hand) 1 day = 36 days = 1 month 6 days
[Q 30 days = 1 month]
5 months + 8 months + (in hand) 1 month = 14 months
= 1 year 2 months.
[Q
Q
12 months = 1 year]
Example 5. Add : Months Days Hours Minutes
5 12 20 25
8 6 8 10
6 5 14 30
Solution : Years Months Days Hours Minutes
5 12 20 25
8 6 8 10
6 5 14 30
1 7 24 19 5
Answer : 1 year 7 months 24 days 19 hours 5 minutes.
[ 60 minutes
= 1 hour
24 hours = 1 day
12 months = 1 year]

154 Time
Example 6. Add : Years Months Days Hours Minutes Seconds
7 6 12 15 20 10
4 2 10 20 15 30
10 8 6 12 25 40
Solution :
Years Months Days Hours Minutes Seconds
7 6 12 15 20 10
4 2 10 20 15 30
10 8 6 12 25 40
22 5 0 0 1 20
Answer : 22 years 5 months 0 day 0 hour 1 minute 20 seconds.
Example 7. Subtract : Years Months Days Hours
18 6 15 12
11 9 10 5
Solution : Years Months Days Hours
18 6 15 12
11 9 10 5
6 9 5 7
Answer : 6 years 9 months 5 days 7 hours.
Example 8. Subtract : Hours Minutes Seconds
16 20 15
12 30 40
[Q60 seconds = 1 minute
60 minutes = 1 hour
24 hours = 1 day
30 days = 1 month
12 months = 1 year ]
Explanation : It is not possible to subtract 9 months from 6 months.
For this reason, from the place of years, 1 year = 12 months is
taken and added with 6 months. The result is 18 months. Now by
subtracting 9 months from 18 months we get 9 months. Again, we
added 1 year in hand with 11 years which came to 12 years. Again, by
subtracting 12 years from 18 years we get 6 years.

Solution : Hours Minutes Seconds
16 20 15
12 30 40
3 49 35
Answer : 3 hours 49 minutes 35 seconds.
[Explanation :By adding 15 seconds with 1 minute= 60 seconds we get 75
seconds and by subtracting 45 seconds from 75 seconds we get 35 seconds.
Again, by adding 1 hour = 60 minutues with 20 minutes we get 80 minutes
and by adding 1 minute in hand with 30 minutes we get 31 minutes. Now,
we get 49 minutes by subtracting 31 minutes from 80 minutes.
Again, 1 hour in hand has been added with 12 hours and then it has been
subtracted from 16 hours.]
Example 9 : Multiply 4 years 6 months 11 days by 4.
Solution : 4 years 6 months 11 days
× 4
18 1 14
Answer : 18 years 1 month 14 days.
Explanation : 11 days × 4 = 44 days = 1 month14 days.
In the place of days 14 has been written while 1 month in hand.
6 months × 4 = 24 months, with it if 1 month is added the result becomes
25 months or 2 years 1 month. In the place of month 1 has been written
and 2 years in hand.
4 years × 4 = 16 years, with it if 2 years is added, the result becomes 18
years.
Example 10. Multiply 8 days 7 hours 30 minutes 18 seconds by 5.
Solution : Days Hours Minutes Second
8 7 30 18
× 5
1 month 11 days 13 hours 31 minutes 30 seconds
Answer : 1 month 11 days 13 hours 31 minutes 30 seconds.
The Multiplication and Division of Numbers Related with Time

156 Time
Example 11. Transform 2 years 3 months 15 days into hours.
Solution: 2 years = 2 × 365 days = 7 3 0 days [Q 1 year = 365 days]
3 months = 3 × 30 days = 9 0 days [Q 1 month= 30 days]
15 days = 15 days = 1 5 days
Summation =
8 3 5 daysNow,
8 3 5 days
× 2 4 [Q 24 hours = 1day]
3 3 4 0
1 6 7 0 0
2 0 0 4 0 hours.
Answer : 20040 hours.
Example 12 Transform 2 years 6 months 12 days into minutes.
Solution: 2 years = 2 × 365 days = 7 3 0 days
6 months = 6 × 30 days = 1 8 0 days
12 days = 1 2 days
Summation = 9 2 2 days
Now, 9 2 2 days
× 2 4 [Q 1 day = 24 hours]
3 6 8 8
1 8 4 4 0
2 2 1 2 8 hours
× 6 0 [Q 1 hour = 60 minutes]
1 3 2 7 6 8 0 minutes
Answer : 1327680 minutes.

Example 13. Transform 69835 hours into years, months and days.
Solution : 2 4) 6 9 8 3 5 hours ( 2 9 0 9 days.
4 8
2 1 8
2 1 6
2 3 5
2 1 6
1 9 hours
Answer : 7 years 11months 24 days 19 hours.
Example 14. Transform 97860 seconds into days, hours and
minutes.
Solution : 6 0 ) 9 7 8 6 0 seconds ( 1 6 3 1 minutes
6 0
3 7 8
3 6 0
1 8 6
1 8 0
6 0
6 0
0
6 0 ) 1 6 3 1 minutes ( 27 hours.
1 2 0
4 3 1
4 2 0
1 1 minutes
Answer : 1 day 3 hours 11 minutes.
365) 2 9 0 9 days
days
(7 years
2 5 5 5
3 5 4
30) 3 5 4 days
days
days
(11months
3 0
5 4
3 0
2 4
2 4 ) 2 7 hours ( 1 day
2 4
3 hours
Let us notice that:One year is determined on the basis of 365 days
to findout years from days.
[N.B. If the days exceeds 365, then we have to divide it by 365. The
remaining days exceeding 30 have to be divided by 30 to turn into
months.]

158 Time
Example 15. Divide 24 years 8 months 6 days 12 hours by 16.
Solution : 1 6 ) 2 4 years 8 months 6 days 1 2 hours ( 1 year
1 6
8 years
× 1 2 [Q 1 year = 12 months]
9 6 months
+ 8 months
1 6 ) 1 0 4 months ( 6 months
9 6
8 months
× 3 0 [Q 1 month = 30 days.]
2 4 0 days
+ 6 days
1 6 ) 7 2 0 minutes ( 4 5 minutes
6 4
8 0
8 0
0
Answer : 1 year 6 months 15 days 9 hours 45 minutes.
1 6 ) 2 4 6 days ( 1 5 days
1 6
8 6
8 0
6 days
× 2 4 [Q 1 day = 24 hours]
1 4 4 hours
+ 1 2 hours
1 6) 1 5 6 hours ( 9 hours
1 4 4
1 2 hours
× 6 0 [Q 1 hour = 60 minutes]

International Time Table
According to International time table a day is counted as 24 hours
from midnight i.e. after 12:00 p.m. to 12:00 p.m. of the next midnight.
However, when it is 2:15 at night in our country, in International
time it is also 2:15. Again, when it is 2:15 in the afternoon in our
country, it is 14:15 in the International time.
International time table is expressed below :
The time table of our country
12:25 at night 7:15 in the morning 4:45 in the afternoon 9:10 at night
International time table
0 :25 7 : 15 16 : 45 21 : 10
Express the times shown in the clocks in International time (one is
done) :
night noon morning night
22 : 05 hours.
In International time we do not call the time as morning, noon,
afternoon, evening, night, etc.

160 Time
Time table (Railway)
Dhaka - Chittagong
Station
Distance
from
Dhaka
k.m.
704
Moha-
nagar
Provati
4
Karna-
phuli
Express
722
Moha-
nagar
Godhuli
702
Suborno
Express
2
Chitta-
gong
Mail
742
Turna
Express
Dhaka Dep. 00 0740 0530 1515 1630 2230 2300
Dhaka Airport Dep. 19 0821 0625 1558 1705 2315 2338
Tongi Dep. 23 0747 2330
Ghorashal Flag. Dep. 47 0828 0003
Norshindi Dep. 58 0850 0028
Bhairab Bazar Dep. 87 1017 1045 1735 0115
Ashugonj Dep. 91 1100 0142
Brahmonbaria Dep. 104 1050 1129 1805 0210 0150
Akhaura Dep. 120 1215 0300
Comilla Dep. 168 1211 1350 1928 0400 0335
Laksam Dep. 192 1445 0435 0410
Hasanpur Dep. 210 1523
Feni Dep. 232 1322 1606 2039 0527 0458
Chittagong Arrival 321 1515 1840 2255 2235 0755 0705
Let us notice the time table of railway above :
The Karnaphuli Express Train leaves the Dhaka Station at 5:30 a.m.
and reaches the Chittagong Station at 6:40 p.m.
Fill in the blanks following the time table above:
(a) The Suborno Express Train leaves the Dhaka Station at
and reaches the Chittagong Station at
(b) The Mohanagar Provati Train leaves the Bhairab Bazar
Station at
(c) The Chittagong Mail Train leaves the Comilla Station at
and reaches the Chittagong Station at
(d) The Turna Express Train leaves the Dhaka at and
reaches the Chittagong Station at
(e) The Mohanagar Godhuli Train reaches the Chittagong Station
at

Example 16. Raju was born on 26 March in 1991. What will be his
age on 5 June 1997?
Solution : Years Months Days
1997 06 05
(−) 1991 03 26
06 02 10
+ 1
06 02 11
∴ Raju will be 6 years 2 months 11 days old.
Answer : 6 years 2 months 11 days.
26 - 31 March in 1991→ 6 days
From 1 April in 1991 to
31 March in 1997→ 6 years
From 1 April to 31 May 1997 → 2 months
From 1 June to 5 June in 1997 → 5 day
From 26 March in 1991 to 5 June in 1997→ 6 years 2 months 11 days
Example 17. How many days are there from 1 February 1998 to 26
September?
Solution : The number formed with the digits in ones and tens
places of 1998 year is 98. This is not divisible by 4.
Therefore, 1998 is not divisible by 4.
So, 1998 is not a leap year. That means, in 1998 the month of
February has 28 days.
The number of days to be found: 28 days
31 days
30 days
31 days
30 days
31 days
31 days
26 days
Total 238 daysAnswer : 238 days.
As the month he was born in
March (31 days)
∴(31+5) days = 36 days
− 26 days
10 days[To find out the age
the day of birth is added]
February
March
April
May
June
July
August
September
Expplanation :

162 Time
Example 18. BabyÊs date of birth is 26 February, 1980. On which
date her age will be 17 year 2 months 24 days?
Solution: Years Months Days
1980 02 26
(+) 17 02 24
1997 05 21
(−) 1 [To count age the date of birth
is added. So, here 1 is subtracted]1997 05 20
Answer : 20 May, 1997.
Explanation: As the month of birth is February and a leap year, the
sum has been solved by counting 29 days in the month.
Exercise - 13
1. Find out which of the years below are leap years :
(a) 1922 (b) 1930 (c) 1952 (d) 1964 (e) 1978
(f) 1984 (g) 1996 (h) 2000 (i) 2006 (j) 2008
2. Add :
(a) Years Months Days Hours Minutes
8 6 12 7 30
9 2 20 8 25
11 15 10 45
(b) Years Months Days Hours Minutes Seconds
5 7 16 8 40 20
12 3 4 15 50 10
6 9 25 20 9 45
10 5 1 6 20 55
(It is assumed that the month
February has 29 days)
26 +24 = 50 days
1 month = 29 days
21 days

3. Subtract :
(a) Days Hours Minutes Seconds
20 8 45 20
7 9 25 40
(b) Years Months Days Hours Minutes seconds
25 4 26 12 30 35
18 9 23 15 55 10
4. Multiply: (a) 4 months 12 days 8 hours 20 minutes by 7
(b) 8 months 22 days 25 minutes 40 seconds by 5
(c) 5 years 2 months 10 days 6 hours by 9.
5. Divide: (a) 22 years 8 months 10 days by 15
(b) 16 years 7 months 15 days 20 hours 12 minutes by12.
6. Transform 4 days 9 hours 10 minutes into seconds.
7. Transform 3 years 4 months 8 days into hours.
8. Thansform 3 years 2 days into minutes.
9. Transform into years, months, days etc.
(a) 84735 minutes.
(b) 16180 hours
(c) 20960 hours
(d) 93540 seconds
10. Mithu was born on 25 January, 1985. How old will he be on 20
October in 2005?
11. Rima was born on 10 April, 1976. How old will she be on 6
September in 1998?

164 Time
13. How many days are there from 1 January to 25 August in
2004?
14. Tuhin was 8 years 1 month 12 days old on 31 March, 1997.
What was his date of birth?
15. Ritu was 15 years 2 months 5 days old on 2 April, 1996. What
was her date of birth?
16. Kajol was born on 26 March, 1992. On which date will she be
8 years 2 months 14 days old?
17. Moushumi was born on 30 May, 1986. On which date will she
be 20 years old ?
18. Anjana Sarkar was born on 20 February in 1971. On which
date will she be 57 years old?
19. Mina studies at 7: 00 p.m. everyday. When in international time
does she study?
20. RaniÊs father returns home from his office at 5:20 p.m. At
which hour in international time does he return home?
21. Pintu reached from Dhaka to Chittagong station by Mohanagar
Godhuli Train at 2255 international time. At which hour did he
reach Chittagong?
12. ShamimaÊs date of birth is 26 February, 1980. What will be
her age on 15 May, 2000?

Everybody has to take a profession to earn his/her living. What a man
earns by his profession has to be spent to meet the needs of his daily
life. We need to keep a record of our income and expenditure. This is
the register of income and expenditure. We can understand the proverb
„cut your coat according to your cloth‰ only when we keep a record of
our income and expenditure.
On Monday, the 30 May, 2005 Mr. Sukkur Ali had Tk. 835.70 in his
funds. On that day he sold chillis for Tk. 445.50, mustard for
Tk 315.80 and turmeric for Tk. 225.00 and bought the following things:
Rice 5 k.g. Tk. 110
.
00
Fish Tk. 80
.
50
Meat Tk. 110
.
00
Soyabean Oil 1 litre Tk. 54
.
00
Sugar
2
1
k.g. Tk. 18
.
00
Salt 1 k.g. Tk. 9
.
00
Stationary Tk. 15
.
75
Below is the record of income and expenditure of Mr. Sukkur
Ali for that day.
Date 30-05-2005
Income Tk. Expenditure Tk.
Openning balance 835
.
70 Rice 5 k.g. 110
.
00
Chillis sold 445
.
50 Fish 80
.
50
Mustard sold 315
.
80 Meat 110
.
00
Turmeric sold 225
.
00 Soyabean oil 1 litre 54
.
00
Total income 1822
.
00 Sugar
2
1
k.g. 18
.
00
Total expenditure 397
.
25 Salt 1 k.g. 9
.
00
Closing balance 1424
.
75 Stationary 15
.
75
397
.
25
On 31 May, 2005 the amount in funds Tk. 1424.75
(One thousand 4 hundred 24 taka and 75 Paisa).
Chapter -14
Income, Expenditure and Cash Memo

166 Income, Expenditure and Cash Memo
Record of income and expenditure
Example 1. Rita went to a shop with Tk. 50.00. She bought 2 note
books for Tk. 18.00, 1 pencil for Tk. 6.00, 1 book for Tk. 8.00 and 1
eraser for Tk. 2.50. Write down her income and expenditure.
Solution : Income & Expenditure
Expenditure
Income Tk. 50
.
00 Notebooks 2 nos. Tk. 18
.
00
Expenditure Tk. 34
.
50 Pencil 1 nos. Tk. 6
.
00
Closing balance Tk. 15
.
50 Book 1 nos. Tk. 8
.
00
Eraser 1 nos. Tk. 2
.
50
Tk. 34
.
50
Answer : Tk. 15.50.
3 Days’ Sample of Monthly Income and Expenditure
An account for the income and expenditure of Mr. Jasim in the
month of May, 2005.
The name of the month and year : May, 2005.
Date and Day Income Tk. Expenditure Tk.
01-5-2005 Openning balance 224
.
00 Shopping 305
.
00
Sunday Salary 3225
.
00 Tution fee of son 50
.
00
Selling of Chillis 300
.
00 Rickshaw fare 8
.
00
Given by son 1200
.
00 363
.
00
Total 4949
.
00
Expenditure 363
.
00
.
00
Let us notice that : The income on 30 May 2005 was written on
the left and the expenditure on the right. The funds of the next
day closing balance can be found by subtracting the expenditure
from income.

02-5-2005 Openning balance 4586
.
00 Tax for land 325
.
00
Monday Dues Doctors fee 50
.
00
with Swapon 200
.
00 Medicine 60
.
00
Total 4786
.
00 Clothe for daughter 215
.
00
Expenditure 650
.
00 650
.
00
.
00
03-5-2005
Tuesday
Openning balance 4136
.
00 Rice 325
.
00
Selling of pulses 250
.
00 Oil 50
.
00
Total 4386
.
00 Fish 80
.
00
Expenditure 567
.
00 Other groceries 12
.
00
Clossing Balance 3819
.
00 Repayment of loan
to Shamsu
100
.
00
567
.
00
Observe that :The income and expenditure of the month of May can be
found if we write down the account of 31 days following
the accounts of the dates above.
Cash Memo A cash memo is a receipt which contains quantity,
description, rate and price of goods sold. The
shopkeeper or the seller gives it to a buyer when
something is sold.
Example 2. From Manjur Store Al Amin bought 3 notebooks at the
rate of Tk. 6, one eraser at Tk. 2
.
50, one pencil at Tk. 5, a ballpen at
Tk. 3
.
50. The receipt of the sold goods is like the following:

168 Income, Expenditure and Cash Memo
Cash Memo
Manjur Store
No. 8 Station Road, Tongi Date : 10/5/05
Name of the Client : Al Amin
6 Cherag Ali Road, Tongi.
Description of
Items
Quantity Rate in
Taka
Price in Taka
Notebook
Eraser
Pencil
Ballpen
3 nos.
1 nos.
1 nos.
1 nos.
6
.
00
2
.
50
5
.
00
3
.
50
18
.
00
2
.
50
5
.
00
3
.
50
Total = 29
.
00
In words : Twenty nine taka only.
Signature of the sales person.
Example 3. On 12 May 2005 from Hat Bazar store Mithu bought
5 k.g. of rice at the rate of Tk. 22
.
00 per k.g., 2 k.g. of lentils at the
rate of Tk. 44
.
00 per k.g., 2 k.g. of beans at the rate of Tk. 40
.
00 per
k.g., 1 k.g. of salt at the rate of Tk. 8
.
00 per k.g., 2 k.g. of sugar at
the rate of Tk. 36
.
00 per k.g. and litre of milk at the rate of
Tk. 24
.
00 per litre. Prepare a cash memo showing how much money
he spent.
2
1

Solution : Cash Memo
No. 135
Hat Bazar Store
Date : 12/5/05Sat Masjit Road, Dhaka
Name of the Client : Mithu
Address : 13/2 Jigatola
Dhaka- 1209.
Description of
Items
Quantity Rate (Taka) Price (Taka)
Rice
Lentils
Beans
Salt
Sugar
Milk
5 k.g.
2 k.g.
2 k.g.
1 k.g.
2 k.g.
2
1
litre
22
.
00
44
.
00
40
.
00
8
.
00
36
.
00
24
.
00
110
.
00
88
.
00
80
.
00
8
.
00
72
.
00
12
.
00
Total = 370
.
00
In words : Three hundred and seventy taka only.
Signature of the sales person.

170 Income, Expenditure And Cash Memo
Exercise - 14
1. Answer in brief :
(a) In which place do we write the openning balance in the
statement of income and expenditure?
(b) How do we get the openning balance in the statement of
income and expenditure?
(c) In which place should we write the amount for buying pulses?
(d) In which place should we write the selling price of pulses?
(e) Prepare a sample of monthly income and expenditure of your
family?
(f) What is cash memo?
(g) Who prepares the cash memo?
(h) On whose name do we write the cash memo?
2. You went to market with Tk. 200.00. You bought 2 books for
Tk. 25.50, a quire of paper for Tk. 12.75 and an eraser for Tk. 2
.
25
Prepare a statement of income and expenditure.
3.
4.
Rice Tk. 132.00
Pulses Tk. 40.00
Soyabean oil Tk. 25.00
Flour Tk. 36.00
Matches Tk. 2.00
Prepare his record of income and expenditure for that day.
Raju went to market with Tk. 250.00. He bought fish for
Tk. 25.50, vegetables for Tk. 8.50, green chillies for Tk. 2.00 and
betel nut for Tk. 6.25. He also spent Tk. 4.00 as rickshaw fare.
Write an account of his income and expenditure for the day.
On 18 Baishak, 1412 Bengali year, Monday, Kalu Byapari sold
jute for Tk. 570.00, mustard for Tk. 350.00, green chillies for
Tk. 150.00 and bought the following things:

5. On Sunday the 8 May, 2005 Mr. Jasim had Tk. 324
.
60 as his closing
balance for the day before. On that day he sold potatoes for Tk.
200
.
00, onions for Tk 140
.
00 and vegetables for Tk. 80
.
75. He also
lend Tk. 50
.
00 to a worker named Moti. On that day he bought fish
for Tk. 50
.
00, pulses for Tk.16
.
00, salt for Tk. 4
.
00, a dress for his
daughter for Tk. 200
.
00, a lungi for himself for Tk. 85
.
00. He also
received Tk. 500
.
00 sent by his eldest son. Prepare a record of
income and expenditure for that day.
6. On 28 April, 2005 Mr. Asgar bought a shirt for himself for Tk. 250
.
00,
a pants for his son for Tk. 325
.
00, a saree for his wife for Tk. 225
.
00
and a kameeez for his daughter for Tk. 400
.
00 from Amrit Garments
Shop, Narshingdi. Prepare a cash memo of the items he bought.
7. On 25 April, 2005 from 'Boi Bitan' New market, Chittagong, Mina
bought a Bengali book for Tk. 17
.
75, an English book for Tk. 8
.
75, a
book on Mathematics for Tk. 12
.
50 . a book on Science for Tk. 7
.
25, a
book on Social science for Tk. 8
.
15, a book on Religion for Tk.8
.
30
and an Atlas book for Tk. 14
.
60. Prepare a cash memo showing the
total price of the books.
8. Tick ( ) the correct answer :
(a) In how many parts do we divide the statement of income and
expenditure?
1 / 2 / 3 / 4
(b) On which side of the khata do we write 'Income'?
above / below / left / right
(c) Which is the daily income and expenditure?
1 day's / 15 day's / 1 week's / 1 month's
(d) Which of these will go to income?
School fee of son / father's salary / shopping expenditure /
payment of loan.
(e) What is cash memo?
rate / price / the total price / the list of price.
(f) Who writes the cash memo?
the buyer / seller / teacher / student.

Chapter - 15
The Arangement of Data
Arranged and disarranged data
Often we describe different information by number. For example,
following are the marks which 20 students obtained in mathematics
in their annual examination:
69 62 41 52 45
44 55 32 58 38
65 63 39 48 57
68 49 60 40 47
The classification of number The number of students
30  39 3
40  49 7
50  59 4
60  69 6
Total = 20
In this way, with the classification of numbers, the table prepared
with the number of students in each class are arranged data.
These numbers represent the data of their achievements in
mathematics. These data are not well arranged according to any
particular order.
If the data are arranged in order, we can get the necessary
information very easily. For example, we can arrange these data of
achievements in mathematics according to the order of merit or class.
The data arranged in this way are arranged data.
Among the marks secured 32 is the lowest and 69 is the highest.
The table below is prepared by arranging the marks from 30 to 69
with 10 as intervening number.

The class interval
of numbers
Tally sign The number of students /
occuring numbers.
30 - 39  3
40 - 49   7
50 - 59  4
60 - 69   6
Total = 20
The second number is 62; it also belongs to the 60 - 69 class. Now,
another tally has been given to that class. In this way when 4 tallies
are given the fifth one is not given separately. Rather it is given
diagonally  on the other 4 tallies.
Example 1. The daily sales (in taka) of 20 shops are as follows:
2230 2215 2225 2230 2225 2220
2230 2225 2220 2225 2220 2225
2215 2230 2225 2215 2220 2225
2230 2220.
Arrange the data.
Solution :
The daily
sale (in taka)
Tally The number of
shops
2215  3
2220  5
2225   7
2230  5
Total = 20
Starting from 30 a class is formed with 10 as the intervening
number. First, the classes are written on the left. The first number is
69 and it belongs to the 60 - 69. Therefore, a tally sign is given.

174 The Arangement of Data.
Example 2. Below are the weights (k.g.) of 22 people :
60, 30, 50, 20, 40, 30, 40, 30, 60, 20,
60, 50, 60, 30, 60, 20, 30, 20, 40, 30,
60, 20.
Arrange the data.
Solution :
Weights(k.g.) Tally The number of
people
20  5
30   6
40  3
50  2
60   6
Total = 22
Exercise - 15
1. Following are the marks secured in mathematics by 30
students. Arrange the data.
71, 75, 63, 75, 51, 75, 57, 75, 52.,
70, 75, 72, 58, 63, 69, 71, 60, 61,
69, 53, 52, 72, 72, 72, 52, 63, 54,
55, 55, 69.
2. Following are the weekly income (in taka) of 15 families.
Arrange the data.
420, 260, 290, 480, 275, 475,
325, 280, 480, 330, 375, 430,
270, 340, 360.

3. Following are the numbers of benches (seat and high) of five
classrooms. Express the number of benches with tally signs.
Classes Number of
Benches
Class One 30 Pairs
Class Two 25 Pairs
Class Three 22 Pairs
Class Four 15 Pairs
Class Five 12 Pairs
4. Following are the weights (k.g.) of 20 people. Arrange the
data.
60 30 40 50 20 40 30 50 20 30
40 60 20 40 30 50 20 30 20 30.
5. Following are the wages of 28 labourers. Arrange the data.
154 132 149 138 149 137 144
138 162 157 140 150 143 148
145 146 136 150 157 164 156
132 152 138 144 137 134 150.

Graph is a useful means of easily understanding and visualizing any
given information or collected data.
Example 1. The numbers of students studying in different classes of
Joypara Government Primary School are as follows :
Classes The number of
students
75
67
58
53
49
Following the information of the table draw a bar graph.
Solution :
Chapter - 16
Graph
Let us draw a horizontal line AB and a vertical line AC. Keeping
equal distance, let us draw the straight line AB several times. Consider
each alternative portion as class one, two, three, four and five.
On the line AC let us draw sufficient lines that have 1 c.m. distance
from one another. On these lines draw that are parallel to AB.
Let us think, along the uppersides of the line AC each centimetre
represents 10 students. Therefore, along the line AC each millimetre
represents one student.
Class One
Class Two
Class Three
Class Four
Class Five

The class wise number of students in Joypara Government Primary
School (Along the vertical line AC, 1 c.m. represents 10 students)
Example 2. In a one day cricket match a bowler bowled 10 overs.
The bar graph below shows the number of runs given by him in
different overs (i.e. the number of runs scored by the batsmen of the
other team). Answer the following questions observing the graph :
(a) In which over did he give the highest number of runs?
(b) In which over did he give the lowest number of runs?
(c) How many total runs did he give in 10 overs and what is the
average runs in each over?
(d) Arrange the number of runs in descending order and write the
corresponding overs under the number of runs.
12
11
10
9
8
7
6
5
4
3
2
1
0
C
A BClass One Class Two Class Three Class Four Class Five
On the two marked places for class one along the line AB, let us draw
two lines of 75 m.m. = 7.5 c.m. height that remain parallel to line AC
and connect their two vertices. The narrow rectangle which has thus
been obtained, is the bar graph of 75 students of class one. To make the
graph beautiful and clear, let us coloured the column. Similarly, draw
6.7 c.m., 5.8 c.m., 5.3 c.m. and 4.9 c.m. with hight of bar graph of
students for classes two, three, four and five respectively.
The whole figure obtained in this way is the graph of the given data.

178 Graph
(a) In the 4th over he gave 12 runs which is the highest.
(b) In the 7th over he gave 2 runs which is the lowest.
(c) In 10 overs he gave 5+7+3+12+4+7+2+6+4+5 = 55 runs in
total. The average runs = 55 ÷10 = 5&5
Therefore, in an average he gave 5.5 runs in each over.
(d) Runs : 12 7 7 6 5 5 4 4 3 2
Overs : 4th 2nd 6th 8th 1st 10th 5th 9th 3rd 7th
13
12
11
10
9
8
7
6
5
4
3
2
1
0
A
C
1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th B
Solution

Exercise - 16
1. In Jhawtola Government Primary School, the numbers of
students in five classes are as follows :
Classes Number of students
98
87
79
66
55
Draw a bar graph following the data.
2. Poran Mondol, a cultivator under the terms of lease, harvested
5 kinds of crops last year. The table below shows the quantity
of crops. Following the information of the table draw a bar graph.
Crops Quantity (Quintals)
105
92
86
73
58
3. The distances between Dhaka and several other cities
(In the nearest multiple of 10 kilometre) are given in the table
below. Draw a bar graph following the data.
Places Distances
330 k.m.
290 k.m.
260 k.m.
310 k.m.
440 k.m.
410 k.m.
Class One
Class Two
Class Three
Class Four
Class Five
Paddy
Jute
Wheat
Gram
Pulses
Khulna
Chittagong
Kushtia
Rajshahi
Rangpur
Sylhet

180 Graph
4. The following bar graph shows the marks Pamela secured in
different subjects in the annual examination. Observing the
graph answer the following questions :
(a) In which subject did she get the highest marks and how
much is that?
(b) In which subject did she get the lowest marks and how
much is that?
(c) How much did she get in Bengali?
(d) How much total marks did she get and what is the average
marks in each subject?
The bar graph of the marks secured by Pamela.
(Along the line AC each space represents 10 marks)
0
1
2
3
4
5
6
7
8
9
10
C
A Bengali English Math Science Social
Science
B

Chapter - 17
Geometry
Angle and Measurement of Angle
Right Angle Acute Angle Obtuse Angle
A
C
B
A B
C
A A X Y
Z
B BD
C C
90Ĉ
In the adjancent figure, the rays AB and AC
formed the angle BAC or CAB in their
common end point A. The vertex of the
angle is A and the angle is expressed by
writing ∠BAC or, ∠CAB.
We use the protractor
to measure angle in degree as
unit. In the figure,
∠ CAB = 60 degree or, 60Ĉ.
The angle which is smaller than a right angle is called an acute
angle. The angle which is greater than a right angle is called an
obtuse angle.
By measuring an angle, it can be determined whether it is right or
acute or obtuse angle.
The angle which is less than 90Ĉ is an acute angle.
The angle which is 90Ĉ is a right angle.
The angle which is greater than 90Ĉ is an obtuse angle.

182 Geometry
Complementary Angle
Supplementary Angle
D
C
A B
B
H
GA
In the figure, ∠BAC is a right angle. The ray
AD stands between the two arms of the
angle. As a result, ∠BAD and ∠DAC two
angles were formed. Measuring the two
angles, it was observed that the total
measurement of the two angles is equal to the
measurement of ∠BAC i.e. 90°. We say,
∠BAD and ∠DAC are complementary
angles to each other; or, they are mutually
complementary angles.
If the sum of the two angles is 90°, then the angles are
complementary to each other; or, they are mutually
complementary angles.
If the two complementary angles are drawn as adjacent angles,
they will produce a right angle.
AB is a straight line. On this line G is a
point between the points A and B. H is
a point which is not on the line
AGB. GH is a ray which is different
from GA and GB. Thus, we get two
angles ∠AGH and ∠BGH. Measuring the angles, we find that the
sum of these two angles is equal to ∠AGB i.e. 180° as ∠AGB is
a straight angle. We say, ∠AGH and ∠BGH are supplementary
angles to each other; or, they are mutually supplementary angles.
If the sum of the two angles is 180°, then the angles are
supplementary to each other; or, they are mutually supplementary
angles.
If the two supplementary angles are drawn as adjacent angles,
they produce a straight angle.

B
A
D
M
E NF O
Vertically Opposite Angles
Observe that : The measurments of an angle and its vertically opposite
angle are equal.
:
Triangle : Acute, Right and Obtuse-angled Triangle
Acute-angled triangle Right-angled triangle Obtuse-angled triangle
C
A
B
C
D
O
Two straight lines AB and CD intersected each other
at the point O in the figure. Four angles, such as
∠AOD, ∠DOB, ∠BOC and ∠COA are formed at O.
Among these, ∠AOD is vertically opposite to ∠BOC
or, they are mutually vertically opposite angles.
Again, ∠AOC and ∠BOD are vertically opposite angles or they are
mutually vertically opposite angles. Again rays OA, OB are opposite to
each other as the points A,O,B rest on the same straight line. Again, OC
and OD are opposite rays. The sides of any of the four vertically
opposite angles at the point O are opposite to the rays of the sides of
the main angle.
An angle produced by the two rays opposite to the two sides of an
angle is called the vertically opposite angle of the given angle.
If two straigth lines intersect each other at a point, two pairs of
vertically opposite angles are obtained at the point.
The sides of a pair of angles vertically opposite to each other produce
two straight lines that intersect each other. The meeting point is the
common vertex of the pair of angles.
The space closed by the three segments is a triangle. These three
segments are the three sides of the triangle. Each triangle has three
angles.
Here, we have drawn three triangles. If we measure the angles
∠BAC, ∠CBA and ∠ACB of the triangle with a protractor, we will see
that each of these is an acute angle. We call the triangle ABC an acute
angled triangle.

184 Geometry
A B
C
D
If each angle of a triangle is acute, then the triangle is an
acute - angled triangle.
If one angle of a triangle is a right angle, the triangle
is called a right-angled triangle.
If one angle of a trianlge is obtuse, the triangle is
called an obtuse-angled triangle.
In an acute-angled triangle, all three angles are acute angles.
In a right-angled triangle, only one angle is a right angle, the other
two angles are acute angles.
In an obtuse-angled triangle, only one angle is an obtuse angle,
the other two angles are acute angles.
Do Yourself
1. Draw an acute, an obtuse and a right angled
triangle.
(a) Measure, in each case, the length of the three sides and
write them.
(b) Measure, in each case, the angles and find out whether the
summation of the three angles remain equal every time.
Quadrilateral
In the triangle DEF, ∠DEF is a right angle and ∠EDF and ∠DFE
are acute angles. We call DEF a right-angled triangle.
In the triangle MON, ∠MON is an obtuse angle and ∠OMN and
∠MNO are both acute angles. We call MON an obtuse-angled triangle.
The figure is a quadrilateral
formed by four segments of line.
AB, BC, CD and DA are the four
sides of this figure. AC and BD
are two diagonals.

Α Β
CD
O
Parallelogram
In this figure, the quadrilateral
ABCD is a parallelogram. If we
measure the lenght of its sides we
will find that the length of any of its
two opposite sides are equal;
AB = CD and BC = DA.
By measuring,∠DAB, ∠ABC, ∠DCB, ∠CDA with the help of a protractor,
we find, ∠DAB = ∠BCD and ∠ABC = ∠CDA.
∠DAB and ∠BCD and ∠ABC and ∠CDA are two pairs of opposite
angles of the parallelogram.
Now, let us draw the two diagonals of the parallelogram which
intersect each other at the point O. By measuring the lengths of AO and
OC, we find that these two segments are equal; again, the lengths of BO
and OD are also equal. Therefore, the diagonal lines intersect each other
equally at their meeting point.
The quadrilateral in which the opposite sides are equal and parallel, is a
parallelogram.
The opposite angles of a parallelogram are equal.
The diagonals of a parallelogram intersect each other equally.
Rhombus D
A
O
C
B
Rhombus is such a quadrilateral whose sides are
equal but none of the angles is a right angle.
In figure, ABCD is a rhombus. Its diagonals
intersected at O. After measuring, AO = CO and
BO = DO. So, AC and BD diagonals intersect each
other equally. Now, measuring the four angles,
∠AOB, ∠ΒΟC, ∠COD, ∠DOA with a protractor,
it is found that each angle measures 90°. That means, diagonals intersect each
other equally at a right angle, Again, measuring, it is found that ∠ADC =
∠ABC and ∠DAB = ∠BCD. That means, the opposite angles are equal.
Rhombus is a quadrilateral with equal sides none of whose angle is not a
right angle.
In a rhombus the opposite sides are equal each other and opposite angles also
are equal.
The diagonals of a rhombus intersect each other equally at a right angle.

186 Geometry
Rectangle
Rectangle is a kind of parallelogram in
which each angle is a right anlge. In the
figure, ABCD is a rectangle. If we
measure its two diagonals AC and BD ,
we will find that their lengths are equal.
It can be mentioned that, if an angle of a parallelogram is a right angle,
the other three angles are also right angles.
The quadrilation in which the opposite sides are equal and paralled and
angles are right angle, it is called a rectangle. In a rectangle each angle
is a right-angle and the opposite sides are equal.
In a rectangle the diagonals are equal and they intersect each other
equally.
Square
D
A B
C
A B
CD
Square is a rectangle in which the sides are
all equal i.e. a square is a rectangle in which
each angle is a right angle and the sides are
equal. In the figure, ABCD is a square.
Since each square is a rectangle, the
lengths of both the diagonals are equal. If
we measure the four angles obtained at the
meeting point, we will find that each of
these is a right angle.
Since the opposite sides of a rectangle are equal, so if any two adjacent
sides of it are equal, it will be a square.
The quadrilateral in which each sides are equal and paralled and angles
are is called a square. right-angle, is called a square.
All the sides of a square are equal and each angle is a right angle.
The two diagonals of a square are equal and they intersect each
other at a right angle.

Do yourself
1. Draw two quadrilaterals with two sides of unequal length.
(a) In each case, measure the length of the four sides and two
diagonals and note them.
(b) Measure the four angles and say whether the summation of
the four angles that you have noted are equal in both cases.
2. Draw a parallelogram, a rhombus and a rectangle.
(a) In each case find out whether the length of the opposite sides are
equal or not.
(b) In each case find out whether the opposite angles are equal or
not.
(c) Find out if the diagonals have intersected themselves
equally at the meeting point.
(d) In the case of rhombus, measure the angles obtained at the
meeting point of the diagonals and find out whether they are
perpendicular to each other.
3. Draw a square with each side of 4 c.m. length.
(a) Measure the length of each diagonals and note.
(b) Mark the middle points of the sides and connect them
successively. What kind of quadrilateral has been produced?
Measure the lengths of its sides and the angles.
Circle
Place a one or two taka Bangladeshi coin on a
white piece of paper. Use your left index
finger to press the coin tightly on the paper.
Now using a sharp pencil with your right hand
draw a line round the edge of the coin. If you
remove the coin, you will see an enclosed
round curved line on the paper. This is a
circle.
A
B
C
D
E
F
G
Inside the circle, the point A is placed exactly in the middle of the circle.
Place the points B, C, D, E, F, G on the circle and draw the lines AB, AC,
AD, AE, AF, AG. If we measure these lines, we will find that all of them are
equal in length. The point A is called the centre and the lines AB, AC, AD,
etc. are each a radius of the circle. The length of the radius is also called the
radius of the circle.

188 Geometry
cm
A circle is an enclosed curved line having a point inside, from
which every point on the line remains at the same distance. this point
is the center of the circle
The line from the centre of a circle which connects any point on
the circle is a radius of the circle. Every radius of the circle is
equal in length. This length is also called the radius.
Drawing a Circle
A pair of pencil compasses is
used to draw a perfect circle.
The nib/end of one of the
metal legs of the compasses
is pointed; in the other leg a
pencil can be tightly fixed
using a screw.
As shown in the above picture, a circle is drawn by placing the
pointed leg on the paper and moving round the pencil fixed in the
other leg. To draw a circle of a particular radius, the pointed end of
the compasses should be placed on the 0 (zero) marked point of the
scale and the nib of the pencil fixed in the other leg should be placed
on the point marking a particular length. Then, by lifting the
compasses carefully (so that the legs remain at the same distance)
from the scale, it should be placed on the paper to draw the circle.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Chord and Arc of a Circle
The circle in the figure has the
centre A. Take any two points M, N
on the circle and then draw the line
MN connecting the two points. MN
is a chord of the circle which divides
it into two parts.
If we take the points O, P on the two opposite parts of the circle, then
they can be called the MON and MPN parts. The part of the circle
divided by the chord is called the arc of the circle or the arc. In the
figure, the chord MN produces the arcs MON and MPN.
The line that connects any two points on the circle is called
a chord.
Every chord divides the circle into two arcs.
Diameter
In the figure, the chord BC is drawn
over the centre A. In this case, the
chord is called a diameter of the circle.
The two arcs produced by the diameter BC are equal. Each of them
is a semi-circle.
Any chord that moves towards the centre is a diameter of the
circle.
Every diameter divides the circle into two semi-circles.
The length of the diameter is also called diameter .
Diameter is two times the radius.
A
C
B
M
P
N
A
O

190 Income, Expenditure And Cash Memo
N
O
BA
M
Do yourself
1. Taking any centre and radius draw a circle with the help of a
compasses. Draw a number of radiuses on the circle. Measure
them to find out if all the radiuses are equal in length.
2. Draw a circle with the diameter BC. Take a point D on the
semi-circle; extend the lines BD and CD and measure ∠BDC using
a protractor. Measure ∠BDC for different places of the point B.
Find, if all of them are equal.
3. Draw a line AB. Take A as the centre and draw an arc whose
radius is appoximately three- fourth of the length of AB. Take B as
the centre and on the same side of the line AB, draw another arc of
the same radius. The arcs cross each other at the point M.
Following the process, draw two more arcs on the other side of the
line AB. These two arcs cross each other at the point N. Draw the
line M N; the lines AB and MN cross each other at the point O.
(a) Measure, if AO = BO and
MO = NO or not.
(b) Measure,∠AOM, ∠AON,
∠BOM, ∠BON to find
out if all of them are
right angles.
4. Using a ruler draw three lines of any length you like. Place the two
legs of the compasses on the two ends of the first line and lift them
carefully so that the distance between them remains the same. Place the
compass on another piece of white paper and mark the two ends of the
compasses. Connect the points to draw the line AB which equals the first
line. Now, taking A as the centre, on any side of the line AB, draw an arc
with the same radius as the second line. Again, taking B as the centre, on
the same side of the line AB, draw another arc with the same radius as
the third line. If needed, stretch the third line so that the arcs cross each
other at the point C. Draw the line AC and BC. ABC becomes a triangle
in which the length of three sides equal the length of the three lines
drawn earlier.

Exercise - 17
1. Fill in the blanks :
(a) The measure of the right angle is .
(b) The measure of an acute angle is  than the measure of a
right angle.
(c) The measure of an obtuse angle is  than the measure of a
right angle.
(d) One angle of a right angle is  and the other two angles are
 .
(e) A angle has  obtuse angle and  acute angles.
(f) A triangle in which each angle is less than  is an acute
angled triangle.
2. Following are the measures of some angles; draw them.
(a) 30° (b) 45° (c) 60° (d) 85°
(e) 95° (f) 120° (g) 135° (g) 160°.
3. Draw an acute- angled, right- angled and an obtuse
angled triangle. In each case, measure the three angles to find out
their sum.
4. Following are the measures of some angles. Mention in each case, the
measure of the complementary angle and draw the complementary angle.
(a) 60° (b) 45° (c)72° (d) 25° (e) 50°
5. Following are the measures of some angles. In each case, draw the
angle and its supplementary and vertically opposite angles in the
same figure mentioning their degrees.
Mark also the vertically opposite angle of the supplementary angle.
(a) 45° (b) 120° (c) 72° (d)110° (e) 85°
6. Draw some right- angled triangles. In each case, measure the two
angles other than the right angle and find the sum. What is the total
measure of the three angles in each case?

192 Geometry
7. Draw a quadrilatreal. Measure the length of its 4 sides and two
diagonals. Measure also its four angles and find out their sum.
8. Draw a paralletlogram with a side of 4. c.m. length and
another beside it of 3 c.m. length. Measure their two opposite sides
and also the pair of opposite angles. Draw the two diagonals of the
parallelogram. Measure the four divided parts of the diagonals at their
meeting point.
9. Draw a rhombus and a square with a side 3&5 c. m. length. In each
case measure the other three sides and the four angles. Draw two
diagonals in each case and find out whether the lines intersect
themselves in acute angles.
10. Draw a circle of 2 c.m. radius. Draw a chord that is a diameter and
not a diameter of the circle. Measure their lengths.

7 8 9
4 5 6
1 2 3
0 . = +
-
x
÷
%
CE
ON/AC
+/-
MRC M M OFF
Chapter - 18
The Usage of Calculator
Calculator : A calculator is a small electronic device to help us
calculate numbers. It helps us to do the arithmetical calculations
easily. Different models of calculators are available.
A calculator normally has 25 buttons and a display box. The
machine has to be activated first by pressing ON/AC and after its
use it has to be closed by pressing OFF .
Example 1. Add : 8 + 7 = What ?
Solution : The following buttons of the calculator have been
pressed one after another.
ON/AC 8 + 7 = 15
8 + 7 = 15Answer : 15.
-+

194 The Usage of Calculator
First, the machine has been activated by pressing .
Afterwards, the necessary buttons have been pressed. After the
result had been obtained, the calculator was shut down by
pressing OFF .
Example 2. Add : 12 + 5 = What ?
ON/AC
ON/AC
1 2 + 5 = 17
12 + 5 = 17
Answer : 17.
Example 3. Add : 10 + 7 + 13 = What ?
Solution : The following buttons of the calculator have been pressed
one after another.
ON/AC 1 0 + 7 + 1 3 = 30
10 + 7 + 13 = 30
Answer : 30.
Example 4. Add : 9 + 27 + 48 = What?
AC 9 + 2 7 + 4 8 = 84
9 + 27 + 48 = 84
Answer : 84.
Example 5. Subtract : 252 − 50 = What ?
AC 2 5 2 − 5 0 = 202
252 − 50 = 202
Answer : 202.

Example 6. Subtract : 490 −
−
281 = What?
AC 4 9 0 2 8 1 = 209
490 − 281 = 209
Answer : 209.
Example 7. Multiply : 253 × 78 = What ?
AC 2 5 3 × 7 8 = 19,734
253 × 78 = 19,734
Answer : 19734.
Example 8. Simplify (with the help of a calculator) :
92 + 13 − 8 × 9 = What ?
Solution : AC 9 2 + 1 3 = 105
92 + 13 = 105
AC 8 × 9 = 72
8
105 - 72 = 33
× 9 = 72
AC 105 − 72 = 33
Answer : 33.
Let us notice that :
• 92 + 13 ă 8 × 9 = What? To solve the problem the calculator
has been used in three steps.
• It can also be solved in one step.
AC 9 2 + 1 3 = M+
C
+/− 8 × 9 − MRC =

Example 9. Simplify : 45 ÷ 5 × 8 − 71 (with the help of a calculator)
pressed in one after another.
AC 4 5 ÷ 5 × 8 − 7 1 = 1
45 ÷ 5 × 8 − 71 = 1
Answer : 1.
Solution : It is given,
Fish Tk. 220
Vegetables Tk. 45
Oil Tk. 52
The following buttons of the calculator have been pressed one after
another.
ON/AC 2 2 0 + 4 5 + 5 2 = 317
220 + 45 + 52 = 317
The following buttons of the calculator have been pressed one after
another.
AC 5 0 0 − 3 1 7 = 183
500 − 317 = 183
Answer : Tk. 183.
Exercise - 18
1. Add using a calculator :
(a) 8 + 7 = What ? (b) 12 + 15 = What ?
(c) 20 + 0 = What ? (d) 128 + 38 + 44 = What ?
(e) 215 + 15 + 25 + 40 = What ?
Example 9. Mr. Anwar went to a market and bought fish for
Tk. 220, vegetables for Tk. 45 and oil for Tk. 52. He gave
Tk. 500 to the shopkeeper. How much will the shopkeeper return
him? Find out the result using a calculator.

2. Subtract using a calculator :
(a) 18 −10 = What ? (b) 52−0 = What? (c) 324−124 = What?
(d) 525−425 = What? (e) 200 −100 = What ?
3. Multiply using a calculator :
(a) 8 × 3 = What ? (b) 125 × 25 = What ? (c) 220 × 0 = What ?
(d) 4 × 444 = What? (e) 580 × 8 = What ?
4. Simplify using a calculator :
(a) 324−55− 62 + 65 + 68 −10
(b) 2 × 3 × 4 + 5 × 6
(c) 6 × 4 ÷ 3
(d) 8 × 3 − 2 + 2 × 3
5. Mr. Shafiq went to a market and bought fish for Tk. 225,
vegetables for Tk. 30, onions for Tk. 25 and oil for Tk. 180. He
gave Tk. 500 to the shopkeeper. How much will the shopkeeper
return him? Find out the result using a calculator.
6. Bakul went to a shop and bought books for Tk. 227, notebooks
for Tk. 75 and an eraser for Tk. 8. He gave Tk. 400 to the
shopkeeper. Use a calculator to find out how much the
shopkeeper will return him.

Chapter -19
Population
The table of the population in the year 2001
Total population 123851120
Total number of males 63894740
Total number of females 59956380
The number of Sayeed’s family members :
SayeedÊs father SayeedÊs mother Sayeed SayeedÊs sister Total
1 1 1 1 4
The number of SayeedÊs family members is 4. This is a small family.
The number of Salma’s family members :
SalmaÊs
father
SalmaÊs
mother
SalmaÊs
brother
Salma and her
sisters
Total
number
The number
of family
members
1 1 7 3 12
The number of SalmaÊs family members is 12. This is a large family.
Bangladesh is our motherland. The total area of this country is
1,47,570 square kilometre. In the year 2001** the population of
this country was as follows :
According to the calculation in 2001, out of our total population 6
crore 38 lac 94 thousand 7 hundred are males and 5 crore 99 lac 56
thousand 3 hundred 80 are females. The population of males is a bit
higher than that of the females. The number of our population was
found out by counting. This counting is called census. Census is a
process of officially counting a countryÊs population and recording
various facts. Generally, census is undertaken after every 10 years.
The number
of family
members
**Population census-2001. National Report, Bangladesh Bureau of Statistics. Planning
Division, Page-15.

Let us observe the table below :
Now, find out which family in the table above is smaller?
Find out the number of members of 2 families of your neighbourhood
Father Mo
the
r
Brot
hers
Sisters Uncl
es
Aunts Others Total number of
family members
1.
2.
The families of 2
neighbours
Grand
father
Grand
mother
The population in the last several years in Bangladesh
Year Population
1974 7 crores 63 lacs
(approximately)
In terms of area Bangladesh holds the 90th position in the world.
However, in terms of population it holds the 9th position.
Exercise - 19 (a)
(1) According to the census of 1991 what was the population of
Bangadesh?
(a) 11&14 crores (b) 12&60 crores
(c) 12&81 crores (d) 13&00 crores.
(2) What is the total area of Bangladesh?
(a) 1,47,181 sq. kilometre (b) 1,47,570 sq, kilometre
(c) 6,78,000 sq. kilometre (d) 32,88,000 sq. kilometre.
(3) How many years does it take to hold a census?
(a) 10 years (b) 15 years
(c) 20 years (d) 25 years
2. In the table below write down the number of family members of three
of your classmates.
Families Number
of
members
Which is the
largest family
Which is the
smallest family

Population Related Problems
Density :
The area of Rasulpur village is 4 square kilometre. 3200
people live in that village. How many people live in each square
kilometre?
In 4 square kilometre live 3200 people
∴ 1 " " " (3200 ÷ 4) "
= 800 people
Let us notice that : In Rasulpur village in each square kilometre live
800 people. It means, in that village density of population is 800.
The number of people living in each square kilometre
of an area is called the density of population.
Example 1. The population in Bhutan is about 9 lacs and its total
area is about 47 thousand square kilometre. Find
out the density of population of Bhutan.
Solution: In 47000 square kilometre live 9,00000 people.
1 " " "
47000
900000
= 19&2 people
∴
∴
∴
The density of population is 19&2.
Answer : 19&2.
Example 2. The population of Narayangonj District is about 21 lacs
70 thousands and its area is 701 square kilometre. Find
out the density of population of Narayangonj..
Solution : In 701 square kilometre live 21,70,000 people
1
701
2170000
= 3095&58 people
Therefore, the density of population is 3095&58.
Answer : 3095&58.
" "

F - 1. Bu - 5
Urban Population
The populations of several cities are given below :
populationcity
year 1961 1974 1981 1991 1998
comment
Chittagong 3,54,206 8,89,760 13,90,684 20,40,663 64,42,000
Khulna 1,27,970 4,37,344 6,52,000 8,77,388 24,58,000
Rajshahi 56,885 90,909 2,53,726 5,17,136 23,18,000
Question : Form the above table, find out how many times the
population of Chittagong, Khulna and Rajshahi have increased in the last
37 years.
The following table shows the area, population and density of
different divisions in 2001 :
Bangladesh /
Division
Area
square
k.m.
The number
of males and
females
Males Females Density
square
k.m.
Bangladesh 1,47,570 12,38,51,120 6,38,94,740 5,99,56,380 839
Barisal
Division
13,297 81,53,960 41,56,280 39,97,680
Chittagong
Division
33,771 2,41,19,660 1,23,04,540 1,18,15,120
Dhaka
Division
31,120 3,89,87,140 2,03,80,260 1,86,06,880 1,253
Khulna
Division
22,273 1,46,04,900 75,32,140 70,72,760 656
Rajshahi
Division
34,514 3,00,88,740 1,54,80,300 1,46,08,440
Sylhet
Division
12,596 78,96,720 40,41,220 38,55,500 627
Question : Find out the density of population of Barisal, Chittagong and
Rajshahi division.

202 Population
The density of populations of different divisions in the year 1991
and 2001 are given below :
DensityBangladesh /
Division 1991 2001
Bangladesh 720 839
Barisal Division 561 613
Chittagong Division 608 714
Dhaka Division 1050 1253
Khulna Division 570 656
Rajshahi Division 759 872
Sylhet Division 537 627
Question : Find out the increase rate of the density of population of
Dhaka and Chittagong Division from 1991 to 2001.
Following are the districts with the highest and lowest density of
population :
(a) Highest : In Dhaka District with 5,887 people per square kilometre.
(b) Lowest : In Bandarban District with 67 people per square kilometre.
The density of population per square kilometre of different countries
in Asia:
India 300 people
Pakistan 184 people
Srilanka 189 people
Nepal 165 people
Bhutan 19 people
Question : Find out the difference between the density of population of
India and Bhutan
Exercise – 19 (B)
(1) Which district of Bangladesh comes first in terms of density of
population ?
(a) Khulna (b) Rangpur
(c) Dhaka (d) Bogra

(2) Which district has the lowest density of population ?
(a) Bandarban (b) Rajshahi
(c) Chittagong (d) Noakhali
(3) The average number of population per square kilometre of
a country is called-
(a) population (b) total population
(c) density of populaton (d) average population
(4) How do we determine the density of population ?
(a) By dividing the total number of adult population by the total
area.
(b) By dividing the total population by the area of total
cultivable land.
(c) By dividing the total population by total area.
(d) By dividing the total number of population capable of working
by total area.
2. The population of Narsingdi District is 19 lacs and its area is 1141
square kilometres. Find out the density of population of this district.
3. The area of Rangamati District is 6116 square kilometres and its
population is 5 lac 25 thousand. Find out the density of
population of that district.
4. Write down in the following table how many students are there
in different classes of your school :
Class Numbers of
male students
Number of female
students
Total number of
students
One
Two
Three
Four
Five
Total
Following the table above answer these questions :
Which class has the highest number of students ?
Which class has the lowest number of students ?
5. What is the density of population ? How do we determine it ?
6. Which district has the highest number of population and how many
of them are per kilometre?

204 Population
Answer
Exercise – 1
1. (a) 159771 (b) 979069 (c) 2230028 (d) 988989
(e) 2527164 (f) 3651856
2. (a) 128217 (b) 1455239 (c) 1967325 (d) 2934131
3. (a) 812292 (b) 336410 (c) 778483 (d) 482485
(e) 269026
4. (a) 712162 (b) 923698 (c) 575217 (d) 303886
5. (a) (b)
(c) (d)
6. 387964, 7. Difference 887749
8. 863743 9. 1
10. 890001 11. 580851 12. Tk. 20334 13. 7193
14. Tk. 5152 15. Tk. 122446 16. 266628
17. Tk. 176140 18. Tk. 9038
Exercise – 2 (a)
1. (a) 50362 (b) 88816 (c) 264845 (d) 589110
(e) 1429965 (f) 3128520 (g) 5930076 (h) 6153600
2. (a) 475950 (b) 597872 (c) 1678362 (d) 5530980
(e) 8367630 (f) 6257700
3. (a) 3209679 (b) 2069793 (c) 6251742 (d) 4588407
4. 1371915 5. Tk. 1326375 6. 99000 sheets
3
8 5
2
8
4
7
1
42
7
5
2
7
9
6
7

7. 5012252 k.g. 8. Tk. 100375 9. Tk. 2091200
10. 2176482 11. 2835324
Exercise – 2 (b)
1. (a) quotient 5113 (b) quotient 1483 (c) quotient 262, remainder 72
(d) quotient 355 (e) quotient 217, remainder 86 (f) quotient 245
(g) quotient 231, remainder 14 (h) quotient 183, remainder 2
(i) quotient 127, remainder 268 (j) quotient 41, remainder 5.
2. (a) quotient 2413, remainder 5 (b) quotient 8752, remainder 0,
(c) quotient 435, remainder 60 (d) quotient 659, remainder 0,
(e) quotient 675, remainder 0 (f) quotient 397, remainder 68
(g) quotient 830, remainder 90 (h) quotient 935, remainder 0.
3. (a) divisor 3160 (b) divisor 484 (c) quotient 761 (d) dividend 23409
4. 416 days 16 hours 5. quotient 1149, remainder 36,
6. 72 7. Tk. 48 8. quotient 87, remainder 617
9. 75 persons 10. Other number is 78
11. quotient 124, remainder 9. 12. Other number is 261.
Exercise – 3
1. Tk. 15 2. Tk. 8 3. Tk. 128
4. Age of father 72 years, age of son 24 years, 5. 990
6. quotient 102, 7. Quotient 103
8. Other number is 98, 9. Tk. 3042, 10. Tk. 485,
11. Raju 4738, Roni 3952, 12. Farida Tk. 4708, Fatema Tk. 5197
13. Tk. 5520, 14. Tk. 7644, 15. Tk. 427.

206 Anser
Exercise – 4
4. Tk. 160 5. Tk. 150 6. Tk. 77 7. Tk. 110
8. Tk. 18 9. Tk. 650 10. Tk. 960 11. Tk. 1800
12. 20 days 13. 4 days 14. 12 days 15. 100 persons
16. 15 days 17. 36 person 18. 4 persons 19. 42 days
20. 25 days 21. 24 22. 50 persons 23. 15
24. 504 k.g. 25. 240 persons.
Exercise – 5
1. 59 2. 200 3. 10 4. 90
5. 68 6. 15 7. 1 8. 411
9. 210 10. 2 11. 14 12. 315
13. 53 14. 96 15. 52 16. 86
17. 7 18. 84 19. 13 20. 9
21. 205 22. 4 23. 23 24. 7
25. 30 26. 1 27. 0 28. 1
Exercise – 6
1. (a) 42 (b) 50 (c) Tk. 560 (d) 51 c.m. (e) 36 k.g.
(f) 90 litres
2. Tk. 46 3. 79 4. 129 c.m.
5. Dipu and Tipu each 23 years old.
6. 59 7. 107 8. 2169 persons
9. 65 10. 38 years 11. 26 m.m.

Exercise – 7
1. (a) 48 (b) 5 (c) 7 (d) 12 (e) 15
(f) 11 (g) 8 (h) 7 (i) 10 (j) 12
2. (a) 5 (b) 19 (c) 5 (d) 22 (e) 2
(f) 29 (g) 11 (h) 3 (i) 7 (j) 15
3. (a) 255 (b) 380 (c) 336
(d) 480 (e) 5775 (f) 420
4. (a) 360 (b) 495 (c) 3825
(d) 5880 (e) 144 (f) 270
(g) 36036 (h) 2520 (i) 17325
5. 38 6. 30 persons, 2 mangoes, 5 litchies
7. 12 litres; in first drum 19 pitchers, in second drum 29 pitchers.
8. 15 9. 21 10. 25 c.m. square size
11. 360 12. 486 13. 174 14. 7 minutes
15. 122 16. 1433
Exercise – 8
1. (a) 300030 (b) 55270707 (c) (343 Á 7) ă 29
(d) (40 8) Á (51 ă 26) (e) (8587 ă 7972) (127 + 82)
2. (a) 818 ă (347 + 423) = 818 ă 347 ă 423
(b) 127 (43 + 16) = 127 43 + 127 16
(c) (75 Á 5) 3 =/ 75 Á (5 3)
3. (a) true (b) false (c) true
4. (a) 15 (b) 1, 2, 3, 4, 5, 6, 7 (c) 42
5. (a) 12 (b) Tk. 40 (c) Tk. 250
6. (a) 70 (b) 128 (c) 93 (d) 127 (e) 242

208 Answer
Exercise – 9 (a)
1. (a)
2
1
(b)
5
2
(c)
4
1
(d)
3
2
(e)
15
11
(f)
12
7
2. (a)
24
2
12
1
,
24
15
8
5
== (b)
36
27
4
3
,
36
28
9
7
== .
(c)
75
27
25
9
,
75
55
15
11
== (d)
42
38
21
19
,
42
18
7
3
,
42
7
6
1
===
(e)
60
52
15
13
,
60
21
20
7
,
60
36
5
3
===
(f)
48
18
8
3
,
48
32
3
2
,
48
14
24
7
,
48
24
2
1
,
48
33
16
11
===== .
3. (a) greatest
18
13
, least
3
1
(b) greatest
70
51
, least
7
2
(c) greatest
24
17
, least
3
1
(d) greatest
10
7
, least
75
16
4. (a)
20
19
,
2
1
,
7
15
(b)
50
49
,
25
21
,
5
2
(c)
43
6
,
29
3
,
13
1
,
37
2
(e)
60
59
,
6
5
,
5
4
,
4
3
(f)
79
66
,
31
22
,
5
3
,
71
33
,
24
11
.
5. (a)
75
49
,
5
4
,
25
21
,
15
13
(b)
26
15
,
39
25
,
13
9
,
78
61
(c)
41
2
,
10
1
,
17
2
,
93
14
,
23
7
(d))
11
2
,
14
3
,
9
4
,
39
20
,
23
12
6. Red colour 7. To eat 8. Tomato
9. In farm no. 1 10. Cultivation
Exercise - 9 (b)
1. Proper fractions :
71
1
,
23
13
,
15
14
; Improper fractions :
41
42
,
7
60
,
4
27
Mixed fractions:
25
7
9,
12
5
8,
3
2
17

2. (a) 5
5
4
(b) 9
9
4
(c) 7
11
1
(d) 2
37
15
(e) 1
15
8
(f) 1
51
22
(g) 2
59
29
(h) 4
25
3
(i) 12
43
5
(j) 6
203
198
.
3. (a)
16
51
(b)
44
93
(c)
48
55
(d)
16
163
(e)
7
619
(f)
11
1239
(g)
23
632
(h)
33
3263
(i)
49
4901
(j)
19
1975
.
4. (a) 9
9
4 (b)
20
229
(c) 8
13
11
(d) 10
25
17
(e)
71
890
(f)
11
3211
(g) 13
351
107
.
5.
25
7
portion 6.
2
1
bigha 7.
4
1
portion.
Exercise – 9 (c)
1. (a)
5
4
(b)
18
17
(c) 1
12
1
(d) 1
20
13
(e) 1
28
13
(f) 5
9
7
(g) 6
22
17
(h) 13
8
1
(i) 23
4
1
(j) 9
10
9
(k) 13
36
13
2. (a)
17
3
(b)
9
1
(c)
32
9
(d)
8
3
(e)
21
4
(f) 4
3
2
(g) 2
5
2
(h) 1
6
1
(i) 1
3
1
(j) 1
10
7
(k) 3
20
7
(l)
15
8
(m) 3
40
33
.
3. (a)
4
3
(b) 5
20
17
(c) 4
40
33
(d) 2
8
1
(e) 6
6
1
4.
15
14
5.
3
2
portion
portion
6. 225 7.
40
9
portion
8.
75
44 9. Tk. 70.

210 Answer
Exercise – 9 (d)
1. (a) 7
2
1
(b) 3
13
3
(c)
16
5
(d)
34
11
(e) 2
35
2
(f)
4
1
(g) 13
2
1
(h) 24
49
43
(i) 54
23
9
.
2. (a)
8
1
(b) 21
3
2
(c)
20
3
(d)
4
3
(e) 5
7
1
(f)
15
2
(g)
30
7
(h) 4
10
9
(i) 3
9
5
.
3. (a)
45
44
(b)
32
15
(c)
32
15
(d) 6
3
2
(e)
8
1
(f)
3
2
4. Tk. 77 5. 195 kilometre 6. Tk. 30000.
7. 7 metres 8. 3 metre 9. 112
4
3
10. 9 metre 11. 8
3
2
.
Exercise – 9 (e)
1.
8
5
2. 1
5
3
3.
72
35
4. 1
3
1
5.
54
41
6.
30
17
7. 4
4
1
8. 1
24
17
9. 3
16
9
10.
25
22
11. 2
4
1
12. 1
9
2
13. 1 14. 4
3
2
15. 7
15
2
16. 1 17. 1
3
2
.
Exercise – 9 (f)
1.
4
3
portion 2. 10 metre 3. Tk. 120 4. Tk. 80,000
5. Tk. 9600
6. Total 600 marks, Roni gets 450 marks, Panna gets 400 marks.
7. Dividend 220
2
1
8. 2
135
86
9. 20 persons 10. Tk. 6000 11. 26 k.g

Exercise – 10 (a)
1. 6&0223 2. 108&0999 3. 10&6872 4. 209&332
5. 40&758 6. Tk. 43&02 7. 1800
8. Tk. 50&50
9. 32&15 metre 10. 0&0297 11. 98&9 12. 0&0004
13. Tk. 1&09 14. 1&2552 seconds 15. 2&68 k.m.
16. 60&0068 17. 195&1095 18. 16&2555 minutes
19. Tk. 147&06 20. 470&426 metre 21. 1&75 k.m.
22. Tk. 67&95 23. Tk. 21&75 24. Tk. 48&75 25. Tk. 1&55
Exercise – 10 (b)
1. (a) 60&76 (b) 146&835 (c) 58&8787 (d) 213&5035
(e) 324&8 (f) 200&7 (g) 0&08 (h) 83&9546
(i) 84&832 (j) 0&9768 (k) 0&000001 (l) 0&004
2. Tk. 162 3. Tk. 248&75 4. 67&5 kilometre
5. 555&1 kilometre 6. 20&825 centimetre
Exercise – 10 (c)
1. (a) 1&29 (b) 4&64 (c) 4&6875 (d) 0&05 (e) 3&075
(f) 0&0002 (g) 375 (h) 6&4 (i) 1&6 (j) 0&789
(k) 1&43 (l) 0&861 (m) 0&7235 (n) 45 (o) 25&6
(p) 0&0015 (q) 0&0006 (r) 81&25 .
2. Tk. 2&50 3. Tk. 3&20 4. Tk. 98&50 5. 3&2
6. 22&05 7. 25&5 8. 0&05 9. 39&37 inches
10. Tk. 1&72 11. Tk. 15&75 12. 7 hours 13. Tk. 15
14. Tk. 38&75 15. 11 16. 37 55, 33&05

212 Answer
Exercise ă 11
1. (a)
20
1
(b)
20
3
(c)
4
1
(d)
20
7
(e)
5
3
(f)
2
1
(g)
4
3
(h)
20
17
(i)
12
1
(j)
6
1
2. (a) 25% (b) 40% (c) 62&5% (d) 10% (e) 41&67%
(f) 5% (g) 13&33% (h) 42&86% (i) 120% (j) 125%
(k) 130% (l) 135&38%
3. (1) Tk. 3 (2) Tk. 13
2
1
(3) Tk.
125
16
(4) 4 k.g. (5) 15 g.
(6) 6
3
2
k.m. (7) Tk.110 (8) Tk. 15 (9) Tk. 24
4. (a) 25% (b) 64% (c) 60% (d) 33
3
1
% (e) 62
2
1
%
(f) 87
2
1
% 5. 60 6. 18 persons 7. 3
3
1
% 8. 40 persons
9. 4% 10. 500 persons 11. 80% 12. 72%
13. 87
2
1
% 14. 405 persons 15. 14% 16. Tk. 2620
17. 20 Ruhi fish 18. 90% 19. Tk.100 20. Tk.800
21. 3% 22. 1200 persons 23. 3%
24. 10% profit 25. 20% loss 26. Tk.187&50 27. Tk.75&00
28. Tk.600&00 29. 12
2
1
% 30. Tk.20&00 31. 8%
32. 9% 33. 8% 34. Tk.70 35. Tk.252
36. 4 years 37. 6 years 38. Tk.1200.

Exercise– 12(a)
1. (a) 368 m.m. (b) 99305 m.m. (c) 8000952 m.m.
(d) 8245096 m.m. (e) 7250850 m.m.
2. (a) 0&588 metres (b) 43&257 metre (c) 90&366 metre
(d) 8010&25 metre (e) 6002&37 metre
3. (a) 22&03527 k.m. (b) 19&087090 k.m. (c) 125&075250 k.m.
(d) 78&040154 k.m. (e) 0&308096 k.m.
4. (a) 18156 m.g. (b) 4027452 m.g..(c) 9038100 m.g.
(d) 5065370 m.g. (e) 4002360 m.g.
5. (a) 29515 gram (b) 60182 gram (c) 82098&45 gram
(d) 9065&549 gram.
6. (a) 45&02559 k.g. (b) 12&098145 k.g. (c) 9&427505 k.g.
(d) 27&30697 k.g.
7. 3&85 quintals, 0&385 matric ton. 8. 85&321 matric ton
9. 169&74 k.g. 10. 39&59 seer (approx.)
11. 85&60 seer (approx.)
12. (a) 69 k.m. 9 h.m. 8 decametre 1 metre.
(b) 91 k.g. 8 h.g. 3 decagram 8 gram
13. (a) 69 k.m. 5 h.m. 7 decametre 8 metre
(b) 15 kg. 0 h.g. 4 decagram 4 gram
14. (a) 54 k.m. 4 h.m. 4 decam. 0 metre
(b) 56 k.m. 8 h.m. 1 decam. 2 metre
(c) 76 kg. 1 h.gram 3 decagram 6 gram
(d) 43 kg. 9 h.gram 1 decagram 2 gram
15. (a) 3 k.m. 2 h.metre 1 decametre 8 metre
(b) 5 k.m. 4 h. metre 5 decametre 3 metre
(c) 8 k.g. 7 h.gram 4 decagram 6 gram
16. 127 k.g. 6 h.gram 8 decagram 9 gram
17. 60 k.g. 500 gram
18. 318 k.g. 4 h.gram 1 decagram 1 gram
19. 15&60 k.m.
.

214 Answer
Exercise ă 12(b)
1. (a) 1080 square c.m. (b) 1&30 square metre (c) 1665 square metre
(d) 5186&50 square metre
2. (a) 729 square metre (b) 5776 square metre
(c) 41&4736 square metre
3. (a) 75 square metre (b) 225 square metre (c) 7&35 square metre
4. 270 metre 5. 53 c.m. 6. 28 metre
7. 2025 square metre 8. 168 square metre 9. 170&88 metre
10. 27 metre 11. 2&1 hector 12. 63 square c.m.
13. 35 square c.m. 14. 1125 square metre
Exercise ă13
1. (c) 1952 (d) 1964 (f) 1984
(g) 1996 (h) 2000 (j) 2008
2. (a) 18 years 8 monthes 18 days 2 hours 40 minutes
(b) 35 years 1 month 18 days 3 hours 1 minute 10 seconds
3. (a) 12 days 23 hours 19 minutes 40 seconds
(b) 6 years 7 months 2 days 20 hours 35 minutes 25 seconds
4. (a) 2 years 6 months 26 days 10 hours 20 minutes
(b) 3 years 7 months 20 days 2 hours 8 minutes 20 seconds
(c) 46 years 9 months 2 days 6 hours
5. (a) 1 year 6 months 4 days 16 hours
(b) 1 year 4 months 18 days 19 hours 41 minutes
6. 378600 seconds
7. 29352 hours
8. 1579680 minutes

9. (a) 1 month 28 days 20 hours 15 minutes
(b) 1 year 10 months 9 days 4 hours
(c) 2 years 4 months 23 days 8 hours
(d) 1 day 1 hour 59 minutes
10. 20 years 8 months 27 days 11. 22years 4 months 27 days
12. 20 years 2 months 19 days 13. 238 days
14. 20 February, 1989 15. 29 January, 1981
16. 8 June, 2000 17. 29 May, 2006
18. 19 February, 2028 19. 19
20. 17 hours 20 minutes 21. 10 : 55 p.m.
Exercise ă 18
1. (a) 15 (b) 27 (c) 20 (d) 210 (e) 295
2. (a) 8 (b) 52 (c) 200 (d) 100 (e) 100
3 (a) 24 (b) 3125 (c) 0 (d) 1776 (e) 4640
4. (a) 330 (b) 54 (c) 8 (d) 28
5 Tk. 40 6. Tk. 90.
Exercise ă 19 (a)
1. (1) (a) (2) (b) (3) (a)
2. Do yourself
Exercise ă 19 (b)
1. (1) (c) (2) (a) (3) (c) (4) (c)
2. 1665.2 persons 3. 85.8 persons 4. Do yourself
5. Do yourself 6. In Dhaka district 5887 people
** Population census-2001. National Report, Bangladesh Bureau of Statistics. Planning Division.

Mathematics mathematics elementary

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