UNIT 4 PEA306.ppt

Mensuration is the branch of mathematics which deals with the study of
Geometric shapes, their area, volume and related parameters.
Through this Topic Candidates should be able to
• understand the basic concepts of mensuration
• observe the data given and interpret it from the given problem
• apply the Concepts to solve Company Specific Aptitude tests
• apply the Concepts to solve questions asked in various government exams
like SSC CGL, Banking Sector etc.
Learning Outcomes

 2D Figures: Area, Perimeter.
 3D Figures: Volume, Curved surface Area and Total surface Area.
Contents
2D Figures 3D Figures
Rectangle Cuboid
Square Cube
Triangle Cylinder
Circle Cone
Parallelogram Sphere
Trapezium Hemisphere

2D Figures:
Rectangle
• Area of Rectangle = Length × Breadth.
• Perimeter of a Rectangle = 2 × (Length + Breadth)
• Length of the Diagonal = √(Length2 + Breadth2 )

Square
• Area of a Square = Length × Length = (Length)2
• Perimeter of a square = 4 × Length
• Length of the Diagonal = √2 × Length

Isosceles Triangle
• Area of Isosceles Triangle:
(Where a = length of two equal side, b = length of
base of isosceles triangle.)

Equilateral Triangle
• Area of Equilateral Triangle

Circle & Semicircle
• Area of a circle = πr2 = 𝜋d2/4
• Circumference of a circle = 2πr = πd
• Circumference of a semicircle = πr + 2r
• Area of semicircle = 𝜋r2/2
(In the following formulae, r = radius and d =
diameter of the circle)

Parallelogram
• Area of a Parallelogram = Length × Height
• Perimeter of a Parallelogram = 2 × (Length + Breadth)

Trapezium
• Area of a trapezium =(1/2)× sum of parallel sides × distance between
parallel sides

1. The area of a square is 4096 sq. cm. Find the ratio of the breadth and
length of a rectangle whose length is twice the side of the square and
breadth is 24 cm less than the side of the square
A. 5:32
B. 7:16
C. 5:16
D. None of these

2. A wire in the form of a circle of radius 3.5 m is bent in the form of a rectangle
whose length and breadth in the ratio of 6:5. What is the area of rectangle.
A. 30
B. 60
C. 120
D. None of these

3. The circumference of two circles are 264 m and 352 m. Find the difference
between area of the larger and smaller circles.
A. 4123
B. 8642
C. 4312
D. 2612

4. What would be the cost of building 7 m wide garden around a circular field
with diameter equal to 280m, if the cost per sq. m for building the garden is Rs.
21
A. Rs. 156242
B. Rs. 248521
C. Rs. 132594
D. None of these

5. A cow is tied on the corner of a rectangular field of size 30m*20m by a 14 m
long rope. The area of the region that she can graze is
A. 350 sq. m
B. 196 sq. m
C. 154 sq. m
D. 22 sq. m

6. At each corner of a triangular field of side 26m, 28m and 30m a cow is
tethered by a rope of length 7m. The area un-grazed by the cow is
A. 336 sq. m
B. 259 sq. m
C. 154 sq. m
D. 77 sq. m

7. The radius of a circular field is equal to the side of a square field. If the
difference between the perimeter of the circular field and that of the square field
is 32m, what is the perimeter of the square field?
A. 84m
B. 95m
C. 56m
D. 28m

8. Two equal maximum sized circular plates are cut-off from a circular paper-
sheet of circumference 352 cm. The circumference of each circular plate is.
A. 176 cm
B. 180 cm
C. 165 cm
D. 150 cm

9. A wire, when bent in the form of a square, encloses a region having area 121
cm2. If the same wire is bent into the form of a circle, then the area of the circle
is?
A. 144 sq. cm
B. 180 sq. cm
C. 154 sq. cm
D. 176 sq. cm

10. Four circles having equal radii are drawn with centres at the four corners of
a square. Each circle touches the other two adjacent circle. If remaining
area of the square is 168 cm square, what is the size of the radius of the
circle?
A. 1.4 cm
B. 14 cm
C. 35 cm
D. 21 cm

11. The area of a triangle is 216 cm2 and its sides are in the ratio 3 : 4 : 5. The
perimeter of the triangle is
A. 6 cm
B. 12 cm
C. 36 cm
D. 72 cm

Q. No. Answer Q. No. Answer Q. No. Answer Q. No. Answer Q. No. Answer
1 C 2 A 3 C 4 C 5 C
6 B 7 C 8 A 9 C 10 B
11 D

3D Figures:
Cuboid
Let
l= length
b= breadth
h= height. Then,
• Volume = (l x b x h) cubic units.
• Lateral Surface area = 2( b + l) h sq. units.
• Surface area = 2(lb + bh + lh) sq. units.
• Diagonal = √(l2 + b2 + h2 ) units.

Cube
Let each edge of a cube be of length a. Then,
• Volume = a3 cubic units.
• Lateral Surface area = 4a2 sq. units
• Surface area = 6a2 sq. units.
• Diagonal = √3a units.
a
a
a

Cylinder
Let radius of base = r and Height (or length) = h. Then,
• Volume = (пr2h) cubic units.
• Curved surface area = (2пr h) sq. units.
• Total surface area = 2 п r(h + r) sq. units.

Cone
Let radius of base = r and Height = h. Then,
• Slant height, l = √h2 + r2 units.
• Volume = (1/3) п r2h cubic units.
• Curved surface area = (п rl) sq. units.
• Total surface area = (п rl + п r2) sq. units.

Sphere
Let the radius of the sphere be r. Then,
• Volume = (4/3) п r3 cubic units.
• Surface area = (4 п r2) sq. units.

Hemisphere
Let the radius of a hemisphere be r. Then,
• Volume = (2/3) п r3 cubic units.
• Curved surface area = (2 п r2) sq. units.
• Total surface area = (3 п r2) sq. units.

1. If the areas of the three adjacent faces of a cubical box are 120 cm square,
72 cm square and 60 cm square respectively, then the volume of the box is:
A. 800 cm cube
B. 680 cm cube
C. 700 cm cube
D. 720 cm cube

2. A wooden box measures 20 cm × 12 cm × 10 cm. the thickness of the wood
is 1 cm. The volume of the wood required to make the box is:
A. 960 cm cube
B. 900 cm cube
C. 1000 cm cube
D. 1100 cm cube

3. The size of a wooden block is (15 cm × 12 cm × 20 cm). How many such
blocks will be required to construct a solid wooden cube of minimum size?
A. 50
B. 40
C. 60
D. 55

4. Two solid cylinders of radii 4 cm and 5 cm and lengths 6 cm and 4 cm,
respectively are recast into cylindrical disc of thickness 1 cm. The radius of the
disc is
A. 7 cm
B. 14 cm
C. 21 cm
D. 28 cm

5. The radius of cross-section of a solid cylindrical rod of iron is 50 cm. The cylinder is
melted down and formed into 6 solid spherical balls of the same radius as that of the
cylinder. The length of the rod (in m) is
A. 0.8
B. 2
C. 3
D. 4

6. A cone, a hemisphere and a cylinder stand on equal bases of radius R and have
equal heights H. Their whole surfaces are in the ratio:
A. (√3+1) : 3 : 4
B. (√2+1) : 7 : 8
C. (√2+1) : 3 : 4
D. None of these

7. A large solid sphere of diameter 15 m is melted and recast into several small
spheres of diameter 3 m. What is the percentage increase in the surface area of the
smaller spheres over that of the large sphere?
A. 200%
B. 400%
C. 500%
D. can’t be determined

8. A hemispherical basin 150 cm in diameter holds water one hundred and twenty
times as much a cylindrical tube. If the height of the tube is 15 cm, then the
diameter of the tube (in cm) is:
A. 23
B. 24
C. 25
D. 26

9. A cylinder is circumscribed about a hemisphere and a cone is inscribed in the
cylinder so as to have its vertex at the Centre of one end and the other end as
its base. The volumes of the cylinder, hemisphere and the cone are
respectively in the ratio of:
A. 3 : √3 : 2
B. 3 : 2 : 1
C. 1 : 2 : 3
D. 2 : 3 : 1

10. The height of a cone is 40 cm. The cone is cut parallel to its base such that the
volume of the small cone is 1/64 of the cone. Find at which height from base the
cone is cut?
A. 20 cm
B. 30 cm
C. 25 cm
D. 22.5 cm

1 D 2 A 3 C 4 B 5 D
6 C 7 B 8 C 9 B 10 B

© Department of Analytical Skills
CLOCKS

CONCEPT BASE
DIAL:
• The face or dial of a clock is a circle whose circumference is divided into 60
equal parts, called minute spaces.
• They are marked by short lines in the face of the clock. However, the end of
every fifth minute space is marked longer than the others and the 60 minute
spaces are represented as 12 divisions or 12 hour spaces.

HANDS OF THE CLOCK:
• A clock has two hands; the smaller one is called the hour hand or short
hand while the larger one is called the minute hand or long hand as shown
in the figure above.
• The hands of the clock are the indicators of time.
• The time is read based on the positions of the hands with respect to minute
or hour spaces.

UNIT OF TIME:
• The unit of time in clocks is seconds.
• A second, the SI unit of time, is defined as the natural periodicity of the
radiation of a caesium-133 atom.
• A duration of 60 seconds is defined as a minute.
• In clocks, it is also defined as the time taken by the minute hand to move
across a minute space.
• A duration of 60 minutes is defined as an hour.
• In clocks, it is also defined as the time taken by the hour hand or short hand
to move across five minute spaces or one hour space.
• A day is also defined as the duration of 24 hours
• The time is measured owing to the movement of the hands of a clock. A
complete rotation of any of the hands would cover a measure of 360
degrees (circular displacement) .
• Hour hand covers an angle of 360° in 12 hours. So, the hour hand in one
hour will cover 360 /12 = 30°.
• So for every minute, the hour hand moves through 30 /60 = 0.5 °.

• Minute hand covers an angle of 360° in 1 hour. So for every minute, the
minute hand moves through 360/60 = 6 °.
• Both the hands of the clock move in same direction.
• So, their relative displacement for every minute is 5.5° or 11/2°.
• This 5.5° movement constitutes the movements of both the hands.
• So for every minute, both the hands give a relative displacement of 5.5°.

FORMULA TO FIND THE ANGLE BETWEEN THE HANDS OF THE CLOCK:
• Let us say in a race, Sam gives a head-start of 120 metres to Anand.
• Sam and Anand run at the speed of 10 m/s and 6 m/s respectively.
• Running in the same direction, the relative displacement is 10 − 6 = 4
metres for every second.
• Now, let us calculate the time taken by
(i) Sam to catch up with Anand
(ii) Sam to lead Anand by 20 metres
(i) Time taken by Sam to catch up with Anand
Once they begin to run, the initial gap of 120 metres starts decreasing at the
rate (relative speed) of 4 m/s.
Therefore, it takes
4 × t = 120 (Relative Speed × Time = Relative distance or gap)
t = 30 secs to catch up with Anand or to say that they are at the same point.

(ii) Time taken by Sam to lead Anand by 20 metres:
We know that Sam needs 30 secs to catch up with Anand and thereby
it can be considered that both of them are running, starting at the same point
after 30 secs.
Though Sam runs at 10 m/s, the gap between them increases at the
rate of their relative speed i.e. 4 m/s.
Hence, it takes 4 × t = 20, t = 5 secs to lead
Anand by 20 metres when they start from the same point.
Thus, from the starting point Sam takes
30 + 5 = 35 secs to lead Anand by 20 metres.
However, the answer can be directly found by taking the relative distance to be
120 + 20 = 140 metres
4 × t = 140
t = 35 secs.

From (i) and (ii), we can think of the hands of the clock as two runners running
in a circular track of 360°. Here, the angle between the hands of the clock
should be considered as the relative distance between them.
Therefore, we can formulate the following

𝜃=30°𝐻− 11/2𝑀
(when the minute hand is trailing the hour hand) 𝜃=11/2
𝑀− 30°𝐻
(when the minute hand is leading the hour hand)
 𝜃 = Angle between the hands of the clock
 30°𝐻 = Initial gap; H - the division pointed by the hour hand which indicates
the gap between the minute hand and hour hand initially when multiplied by
30°.
 11/2𝑀 = Relative displacement for a period of time; 11/2 - Relative speed of
the hands of the clock; M - no. of minute spaces moved by the minute hand.

COINCIDENCE OF THE HANDS OF THE CLOCK:
• At 12 o' clock, the hands of the clock overlap i.e. the angle formed between
them is 0°.
• To find the next time they overlap, the gap between them should be 360° so
that they are at the same point. Hence it takes,
11/2 × t = 360
t = 720/11 = 65 5/11 minutes
Thus, it can be easily concluded that the hands of the clock coincide every 65
5/11 minutes.

FREQUENCY OF FORMATION OF ANGLES BETWEEN THE HANDS OF
THE CLOCK:
• With the relative displacement, it is easier to find the frequency at which the
hands of a clock form a particular angle.
• Let's find the no. of times the hands of the clock form 0° in a day.
• We know that the hands of the clock coincide every 65 5/11 or 720/11
minutes. So, in a day there are 24 × 60 = 1440 minutes
• Therefore, in a day the hands of the clock coincide (1440 × 11) / 720 = 22
times.
• Thus, it can be inferred that the hands of the clock coincide once in every
hour except for two occasions i.e. from 11 to 1 o' clock they coincide only
once at 12 o' clock.
• So, it is 11 times in 12 hours and thus 22 times in 24 hours.
• The finding can be extended to other angles by applying the formula,
𝜃= 30°𝐻− 11/2𝑀 or
𝜃= 11/2𝑀− 30°𝐻
for every hour in the clock.

GAIN AND LOSS OF TIME IN AN INCORRECT CLOCK:
• A normal clock becomes incorrect when there is a change in the speeds of the hands of
the clock.
• As seen already, duration of a day is measured by the time taken by the hour hand to
move across the 12 divisions twice.
• So, when the speed of the hands of the clock increases they complete their rotation for
a day sooner.
• Thus, they result in loss of time when actually compared with the duration for a day in a
correct clock .
• Similarly, when the hands of the clock move slower than usual, they take more time to
complete the rotations for a day.
• Thus, they result in gaining more time for a day when compared with that in a correct
clock.

KEY POINTS:
• In 60 minutes, the minute hand gains 55 minutes on the hour hand
• The hands of the clock coincide every 65 5/11 minutes and for every hour,
both the hands coincide once
• The hands are in the same straight line when they are coincident or
opposite to each other
• When the two hands are at right angles, they are 15 minute spaces apart
• When the hands are in opposite directions, they are 30 minute spaces apart
• The hands of the clock form an angle of 0° or 180° − 22 times a day
• The hands of the clock form an angle of 1° to 179° − 44 times a day
• If a watch or a clock indicates 8:15, when the correct time is 8, it is said to
be 15 minutes faster than the correct time. On the other hand, if it indicates
7:45, when the correct time is 8, it is said to be 15 minutes slow.

1. Find the angle between the hands of a clock when the time is 5:40.
(a) 80⁰
(b) 160⁰
(c) 70⁰
(d) 120⁰

2. Find the angle between the minute hand and the hour hand of a clock when
the time is 7:20.
(a) 80⁰
(b) 90⁰
(c) 100⁰
(d) 110⁰

3. The reflex angle between the hands of a clock at 10:25 is
(a) 180⁰
(b) 162.5⁰
(c) 165⁰
(d) 197.5⁰

4. Find the angle between the hour hand and the minute hand of a clock when
the time is 15:25.
(a) 47.5 ⁰
(b) 45.5 ⁰
(c) 50 ⁰
(d) None of these

5. At what time between 5 p.m. and 6 p.m., do the hands of a clock coincide?
(a) 8 hour 29 3/11 min
(b) 9 hour 33 8/11 min
(c) 5 hour 27 3/11 min
(d) None of these

8. Find at what time (in minutes) past 8 o' clock but before 9 o' clock will the
hands of a clock be in the same straight line but not together.
(a) 100/11
(b) 110/11
(c) 120/11
(d) None of these

9. How many times in a day do the hands of a clock form 60⁰?
(a) 22
(b) 33
(c) 44
(d) 55

10. A few times per day, the minute hand of a clock is exactly above (or below)
the hour hand. How many times per day does this occur?
(a) 11 times
(b) 24 times
(c) 22 times
(d) 44 times

11. At how many times between 12 o'clock and 1 o' clock are the minute hand
and the hour hand of a clock at an angle of 90 degrees to each other?
(a) 4
(b) 6
(c) 3
(d) 2

12. How much does a watch gain per day, if its hands coincide every 64
minutes?
(a) 32 8/11 min
(b) 36 5/11 min
(c) 90 min
(d) 96 min

13. My watch was 8 minutes behind at 8 p.m. on Sunday but it was 7 minutes
ahead of time at 8 p.m. on Wednesday. During this period, at which time has
this watch shown the correct time?
(a) Tuesday 10.24 a.m.
(b) Wednesday 9.16 p.m.
(c) Tuesday 10.24 p.m.
(d) Wednesday 9.16 a.m.

14. The minute hand of a clock overtakes the hour hand at intervals of 65
minutes of the correct time. How much a day does the clock gain or lose in
minutes?
(a) 1440/143
(b) 1444/143
(c) 1400/143
(d) 4440/143

15. A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What
will be the true time when the clock indicates 1 p.m. on the following day?
(a) 48 min past 12
(b) 48 min past 11
(c) 45 min past 12
(d) 45 min past 11

16. At what time between 4 and 5 o'clock, will the hands of a watch point in
opposite directions?
(a) 54 min past 4
(b) 53 7/11 min past 4
(c) 54 7/11 min past 4
(d) 54 6/11 min past 4

17. A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the
afternoon of the same day, when the watch indicated quarter past 4 o'clock, the
true time is
(a) 4 p.m.
(b) 6 p.m.
(c) 5 p.m.
(d) 7 p.m.

18. At what time between 5.30 and 6 o' clock, will the hands of a clock be at
right angles?
(a) 43 5/11 min past 5
(b) 43 7/11 min past 5
(c) 5 p.m.
(d) 7 p.m.

19. Find the reflex angle between the hour hand and the minute hand of a clock
when the time is 11:25.
(a) 192.5 ⁰
(b) 230 ⁰
(c) 45 ⁰
(d) 72 ⁰

20. How many minutes is it until six o’clock if fifty minutes ago it was four times
as many minutes past three o’clock?
(a) 26 min
(b) 27 min
(c) 30 min
(d) None of these

Answer_Keys
1 C 2 C 3 D 4 A 5 C
6 C 7 B 8 C 9 C 10 C
11 D 12 A 13 A 14 A 15 A
16 D 17 A 18 B 19 A 20 A

Content
I. Introduction- Origin of calendar
II. Ordinary year and leap year
III. Odd day Concept
IV. Major types of questions:
1)Finding day for a date-
a)Conventional method
b)Shortcut
2) Finding a year having the same calendar as another year
V. Practice Questions

I. Introduction
The Calendar which we currently follow is called the Gregorian Calendar. It is
named after Pope Gregory XIII, who introduced it in October 1582.
The calendar was a refinement to the Julian calendar amounting to a. The
motivation for the reform was to stop the drift of the calendar and set the date for
Easter celebrations. Transition to the Gregorian calendar would restore the holiday
to the time of the year in which it was celebrated when introduced by the early
Church.
The reform was adopted initially by the Catholic countries of Europe. Protestants
and Eastern Orthodox countries continued to use the traditional Julian calendar and
adopted the Gregorian reform after a time, for the sake of convenience in
international trade.
The last European country to adopt the reform was Greece, in 1923.

The Gregorian reform modified the Julian calendar's scheme of leap
years as follows:
1. Every year that is exactly divisible by four is a leap year, except for years
that are exactly divisible by 100, but these centurial years are leap years if
they are exactly divisible by 400. For example, the years 1700, 1800, and
1900 are not leap years, but the years 1600 and 2000 are.
2. In addition to the change in the mean length of the calendar year from
365.25 days (365 days 6 hours) to 365.2425 days (365 days 5 hours 49
minutes 12 seconds

II. What are Ordinary and Leap years??
The time taken by the Earth to make one complete revolution is said to be a complete
Year, which is equivalent to exactly
365.2425 days.
But ignoring the .2425 we generalize an ordinary year to be of 365 days.
Further for approximate calculation we assume the 365.2425 days to be 365.25 days.
This extra 0.25 days in 4 years becomes 1 full day. Hence that extra day is added to a
year every 4 years. That particular year is called a leap year, which has one extra day
than the ordinary year, which means a leap year has 366 days in total.

365.2425
365.24
365.25
Ordinary year
365 days
Leap year
=Ordinary year+(0.25x4)
=365+1
=366 days
0.0025 when
ignored
0.24 approximated to 0.25 for
simpler calculation
Only the 365
days taken
The 0.25 becomes 1
extra day in 4 years

How to check for a leap year?
1. Non Century Years:
If any non-century year (not ending with “00”) is divisible by 4 (last two digits of
the year should be divisible by 4), then it is said to be a leap year.
Reason: As explained in the previous slide, the 0.25 extra days of 365.25 days
become one extra day every 4 years (0.25x4=1).
Hence every 4 years we have one leap year.
Ex: 1844, 1892, 1996, 2004, 2012, 2016 etc

2. Century Years:
But in case of a century year(Years ending with “00”,ex: 1400,1600,2000,etc),
the year will be a leap year only if it is divisible by 400.
Reason:
365.2425
365.25
Let us take an example. The 100th year is divisible by 4.
But it is not a leap year.
Why???
The approximation we make while assuming 365.24 days to be 365.25 days
(increasing the value by 0.01), leads to an extra 1 day which does not exist in
real. Hence this extra 1 day which is forming every 100 year should be removed
from the 100th year. So, although the hundredth year is divisible by 4 and should
be a leap year, we remove that extra one day of the leap year from the 100th year.
Hence it ends up as an ordinary year only.
But why is the 400th year a leap year then???

So, till now we learnt that the 100th year (Ex: 100,200,300 etc) is not a leap
although it is divisible by 4.
But why is 400th year a leap year then???
This answer also lies in the value 365.2425 days which is the time taken by the
earth for one revolution.
In 365.2425 we initially ignore 0.0025. which is a negligible value for one year.
But in 400 years, this 0.0025 becomes 1 complete day (0.0025x4=1) and we
add this extra day every 400th year.
Although like every 100th year, 400th year also should have been an ordinary
year, but because of the addition of this extra day(mentioned above), every 400th
year is a leap year.
And that is why every century leap year is divisible by 400.
Example: 400,1200,1600,2000 etc.

Let us take a quick quiz(Ask the students)
Q1. Is 1996 a leap year? Why?
Q2. Are all the years divisible by 4 leap years? Why?
Q3. How do you check for a century year, whether it is a leap year or not?
Why?

III. Odd day concept
If we are supposed to find the day of the week on a given date, we use the
concept of 'odd days'. In a given period, the number of days more than the
complete weeks are called odd days .
For example the number of odd days in 15 total days = 1, since remainder when
15 is divided by 7 is 1.
Similarly the number of odd days in the month of January is 3. Since January
has 31 days,
31/7= R(3).
Note: The odd day value will always lie from 0 to 6. If the odd day value is
more than 6, keep dividing it by 7 until the remainder value comes below
7. The remainder value smaller than 7 is the final odd day value.

Finding odd days
1. In a single year
Now that you know what are odd days, can you find out how many odd
days are there in an ordinary year and a leap year respectively???
1. Ordinary year:
Total no. of days=365 days
365/7= R(1)
Odd days = 1
2. Leap year:
Total number of days= 366
366/7= R(2)
Odd days= 2

Finding odd days
2. Through out a century/year span
Let us see through examples how to find the number of odd days through out a
century or year span.
Ex: Find the number of odd days in the first 35 years in a century.
Step1: Let us first find the number of leap years and ordinary years in these 35
years. 35 years have 8 leap years (4x8=32, closest to 35).
Therefore no. of ordinary years= 35-8 = 27 years.
Step2: Odd days in 8 leap years= 8x2=16 (Since 1 leap year has 2 odd days)
Odd days in 27 ordinary years= 27x1=27 (Since 1 ordinary year has 1 odd day)
Step3: Total number of odd days=(16+27)=43= 43/7=R(1)
Therefore, number of odd days in 35 years = 1

Following the same steps, find the number of odd days in the first
century(first 100 years).
100
24 76
48 76
Total no. of odd days=(48+76)=124
124/7=R(5)
Hence, the first 100 years(century) have 5 odd days in total.
Note: The first 100 years have 24 leap years, not 25.
Because 100th year is not a leap year as discussed in slide no.8.
(No. of leap years) (No. of ordinary years)
(Odd days in 76
ordinary years)
(Odd days in
24 leap years)

Odd days in the centuries
Since first 100 years is having 5 odd days, we can say:
First 200 years have 5x2=10, 10/7= R(3)
First 300 years have 5x3=15, 15/7= R(1)
First 400 years have 5x4=20, 20/7=R(6) + 1= 7= R(0)
And the cycle keeps repeating for the next centuries from here.
Note: 1 extra day has been repeated for 400 years, since 400th year is a leap
year as mentioned in earlier slides.
Until
Century
Up to Year Odd days Similar
centuries
1st 100 5 500 >
2nd 200 3 600 >
3rd 300 1 700 >
4th 400 0 800 >

Odd days in the months
Month No. of days Odd days
January 31 3
February 28,29 0,1
March 31 3
April 30 2
May 31 3
June 30 2
July 31 3
August 31 3
September 30 2
October 31 3
November 30 2
December 31 3

Concept Review Question
Q. If May 10, 1997 was a Monday, what will be the day on Oct 10, 2001?
A. Wednesday
B. Thursday
C. Friday
D. Saturday

Solution:
In this question the reference point is May 10, 1997 and we have to find the
number of odd days from May 10, 1997 up to Oct 10, 2001.
Now, from May 11, 1997 - May 10, 1998 = 1 odd day
May 11, 1998 - May 10, 1999 = 1 odd day
May 11, 1999 - May 10, 2000 = 2 odd days (2000 was leap year)
May 11, 2000 - May 10, 2001 = 1 odd day
Thus, the total number of odd days up to May 10, 2001 = 5.
Now, the remaining 21 days of May will give 0 odd days.
In June, we have 2 odd days; in July, 3 odd days; in August, 3 odd days; in
September,2 odd days and up to 10th October, we have 3 odd days. Hence,
total number of odd days = 18 i.e. 4 odd days.
Since, May 10, 1997 was a Monday, then 4 days after Monday will be Friday.
So, Oct 10, 2001 would be a Friday.

IV. 1. a) Finding the day for a particular date
1. Odd days in the year(Actual year-1)
2. Odd days in the month(Actual month-1)
3. Odd days in the date
Get the sum of the odd days and the final odd day value from it.
Ex: 15/07/2016
1. Odd days in 2015
2. Odd days in January to June
3. Odd days in the date(15/7= R(1))
Sum/7=Odd day value
Odd
day
value
1 2 3 4 5 6 7 or 0
Day of
the
week
Mon Tue Wed Thu Fri Sat Sun

Concept review Questions:
Using the above mentioned conventional method, find the day for the following
dates:
1. Independence day of India
2. Republic Day of India

IV. 2 b) Shortcut to find the day
1. Month Code:
1 4 4 0 2 5 0 3 6 1 4 6
Jan Feb Mar Apr May Jun July Aug Sep
t
Oct Nov Dec
2. Century Code
Years
between
Code/Odd
days
1600-1699 6
1700-1799 4
1800-1899 2
1900-1999 0
2000-2099 6
Steps:
Find the sum of:
1. Date
2. Last 2 digits of the year
3. Quotient of last two digits of the year when
divided by 4
4. Code of month
5. Odd days of the year
The odd days in the above sum value will give the
day
Note:
Incase of months of January and February in a leap year, subtract one odd day from
the total odd days.

Concept review Questions:
Using the Shortcut-method, find the day for the following dates:
1. Independence day of India
2. Republic Day of India

IV.3. Finding years with similar calendars
For two different years having the same calender, the following conditions
should be satisfied:
1. Both years must be of the same type. i.e., both years must be ordinary
years or both years must be leap years.
2. 1st January of both the years must be the same day of the week.
It simply means that for a year to have the same calendar with X year ,the total
odd days from X should be 0.
Let us understand with an example.

Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
Odd
Days
: 1 2 1 1 1 2 1 1 1 2 1
Example:
For a year to have the same calendar with 2007 ,the total odd days from 2007
should be 0.
As we know how to recognize an ordinary and leap year, and also we know the
number of odd days they have, we can make the below table:
Since up to 2017 the number of odd days become 0, hence the next year(2018)
will have the same calendar as 2007.
And also both the years are ordinary years, hence the first condition is also
satisfied.

Shortcut (Only when you are not crossing a an ordinary century year) :
Leap year calendar repeats every 28 years.
Here 28 is distributed as 6+11+11.
Rules:
a) If given year is at 1st position then next repeated calendar year is Given+6.
b) If given year is at 2nd position then next repeated calendar year is Given+11.
c) If given year is at 3rd position then next repeated calendar year is Given+11.
LY 1ST 2ND 3RD LY
28 6 11 11 28

Example:
Find the year which as same calendar as that of 2007 after it.
Sol:
Given year is 2007
According to the above Rule:
2007 is at the 3rd position. So, add 11 yrs.
2007+11=2018
so, the same Calendar after 2007 is 2018.
LY 1ST 2ND 3RD LY
28 6 11 11 28

Find the year after 2015 which has the same calendar as 2015.
A. 2019
B. 2023
C. 2026
D. 2029

The year after 1996 having the same calendar as of 1996 will be
A. 1999
B. 1998
C. 2001
D. None of these

Q. No. Answer Q. No. Answer
1 C 2 D

Q1. If today is Monday, what will be the day one year and 50 days from now?
(Tech Mahindra-2011)
A. Tuesday
B. Wednesday
C. Thursday
D. Can not be determined

Q3. If day before Yesterday it was Monday, What day will fall on day after
tomorrow ? (Tech Mahindra-2013)
a. Sunday
b. Tuesday
c. Saturday
d. Friday

Q4. Radha remembers that her father's birthday is after 16th but before 21st of
March, While her brother Mangesh remembers that his father's birthday is
before 22nd but after 19th of March. On which date is the birthday of their
father?
a) 19th
b) 20th
c) 21st
d) Cannot be determined
e) None of these

Q5. 1.12.91 is the first Sunday. Which is the fourth Tuesday of December ‘91?
(TCS-2015)
(a) 31.12.91
(b) 24.12.91
(c) 17.12.91
(d) 26.12.91

Q6. If the third day of the month is Monday, Which of the following will be the
fifth day from 21st of that month? (Tech Mahindra-2015)
a) Tuesday
b) Monday
c) Wednesday
d) Thursday
e) None of these

Q7. If 8th Dec 2007 was Saturday, then what day of the week was it on 8th Dec
2006? (Capegemini-2015)
(a) Monday
(b) Thursday
(c) Friday
(d) Sunday

Q8. Two brothers were expected to return here on the same day. Rajat returned
3 days earlier, but Rohit returned 4 days later. If Rajat returned on Thursday,
what was the expected day when both the brothers were to return home and
when did Rohit Return?(Infosys 2015)
a) Wednesday, Sunday
b) Thursday, Monday
c) Sunday, Thursday
d) Monday, Friday

Q9. 8th Dec 2009 was Tuesday, what day of the week was it on 8th Dec 2006?
(SSC,2016)
A. Sunday
B. Tuesday
C. Friday
D. Tuesday

Q11. What was the day of the week on January 1, 1998? (Tech Mahindra-2016)
a. Wednesday
b. Monday
c. Friday
d. Thursday

Q13. The day of week on July 1,2000 was
a) Monday
b) Friday
c) Saturday
d) Tuesday
e) None of the above

Q14. On what dates of August 1980 did Monday fall? (TCS-2014)
A. 4th, 11th, 18th, 25th
B. 3rd, 10th, 17th, 24th
C. 6th, 13th, 20th, 27th
D. 9th, 16th, 23rd, 30th

Q15. If we suppose the 60th independence day of India was on Thursday, then
the 85th independence day would have been on?
A. Monday
B. Wednesday
C. Friday
D. Sunday

Q16. If 1 Jan 2015 is Thursday, then probability of 22nd October to be
Wednesday will be? (Infosys-2015)
A. 0
B. 0.5
C. 1
D. Can not be determined

Q17. The calendar for 1992 is the same as for (Tech Mahindra-2012,2015)
a) 1997
b) 2020
c) 2016
d) 1994

Q18. The calendar of year 1982 is same as which year?
A. 1988
B. 1990
C. 1992
D. 1993

Q21. History Professor Nagarajan was talking to the students about a century
which has started with a Monday. What day India would be witnessing on the
last day of the century, the Professor was posing a question. Incidentally he
posed a question that the last day of the century cannot be:
(a)Monday (b) Tuesday
(c)Wednesday (d) Friday
Can you answer the Professor ‘s question?

Answer_Keys
1 D 2 A 3 D 4 B 5 B
6 C 7 C 8 C 9 C 10 D
11 D 12 B 13 C 14 A 15 A
16 A 17 B 18 D 19 C 20 B
21 B

UNIT 4 PEA306.ppt

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