142 MATHEMATICS

8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the

angle of elevation of the top of the statue is 60° and from the same point the angle of
elevation of the top of the pedestal is 45°. Find the height of the pedestal.

9. The angle of elevation of the top of a building from the foot of the tower is 30° and the
angle of elevation of the top of the tower from the foot of the building is 60°. If the tower
is 50 m high, find the height of the building.

10. Two poles of equal heights are standing opposite each other on either side of the road,
which is 80 m wide. From a point between them on the road, the angles of elevation of
the top of the poles are 60° and 30°, respectively. Find the height of the poles and the
distances of the point from the poles. OA

11. A TV tower stands vertically on a bank Gs

: /foa
of a canal. From a point on the other peony
bank directly opposite the tower, the
angle of elevation of the top of the
tower is 60°. From another point 20 m ,
away from this point on the line joing .
this point to the foot of the tower, the A / Pa
angle of elevation of the top of the' _ D 20m C u
tower is 30° (see Fig. 9.12), Find the . B
height of the tower and the width of : Fig. 9.12
the canal. .

12. From the top of a7 m high building, the angle of elevation of the top of a cable tower is
60° and the angle of depression of its foot is 45°. Determine the height of the tower.

13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of
depression of two ships are 30° and 45°. If one ship is exactly behind the other on the
same side of the lighthouse, find the distance between the two ships.

14. A1.2 mtall girl spots a balloon moving wo @)
with the wind in a horizontal line at a ® wa
height of 88.2 m from the ground. The C a
angle of elevation of the balloon from oot a
the eyes of the girl at any instant is oS be 88.2 m
60°. After some time, the angle of < 600."
elevation reduces to 30° (see Fig. 9.13). i ANB bead : Y
Find the distance travelled by the
balloon during the interval. Fig. 9.13

15. Astraight highway leads to the foot of a tower. A man standing at the top of the tower
observes a car at an angle of depression of 30°, which is approaching the foot of the

Reprint 2025-26


SOME APPLICATIONS OF TRIGONOMETRY 143
tower with a uniform speed. Six seconds later, the angle of depression of the car is found
to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

9.2 Summary

In this chapter, you have studied the following points :

1. @ The line of sight is the line drawn from the eye of an observer to the point in the
object viewed by the observer.

(i) The angle of elevation of an object viewed, is the angle formed by the line of sight
with the horizontal when it is above the horizontal level, i.e., the case when we raise
our head to look at the object.

(iii) The angle of depression of an object viewed, is the angle formed by the line of sight
with the horizontal when it is below the horizontal level, i.e., the case when we lower
our head to look at the object. °
2. The height or length of an object or the distance between two distant objects can be
determined with the help of trigonometric ratios. oe
CJ
Reprint 2025-26


[a] an [a]
aa
rhetaene
[en] sor
1062CH10
10.1 Introduction -
You have studied in Class IX that a circle is a collection of all points ina plane
which are at a constant distance (radius) from a fixed point (centre). You have
also studied various terms related to a circle like chord, segment, sector, arc etc.
Let us now examine the different situations that can arise when a circle and a line
are given in a plane. , :
So, let us consider a circle and a line PQ. There can be three possibilities given
in Fig. 10.1 below:
P. .
P S) - P
AS
. B
Q : Q Q
(i) (ii) (ili)
: Fig. 10.1
In Fig. 10.1 (4), the line PQ and the circle have no common point. In this case,
PQ is called a non-intersecting line with respect to the circle. In Fig. 10.1 (i), there
are two common points A and B that the line PQ and the circle have. In this case, we
call the line PQ a secant of the circle. In Fig. 10.1 (iii), there is only one point A which
is common to the line PQ and the circle. In this case, the line is called a tangent to the
circle.
Reprint 2025-26


CIRCLES 145

You might have seen a pulley fitted over a well whichis used Zax
in taking out water from the well. Look at Fig. 10.2. Here the rope @
on both sides of the pulley, if considered as a ray, is like a tangent |
to the circle representing the pulley. (6)

Is there any position of the line with respect to the circle
other than the types given above? You can see that there cannot
be any other type of position of the line with respect to the circle. Fie. 10.2
In this chapter, we will study about the existence of the tangents 18. iN
to a circle and also study some of their properties.

10.2 Tangent to a Circle .
In the previous section, you have seen that a tangent* to a circle is a line that
intersects the circle at only one point. -

To understand the existence of the tangent to a circle at a point, let us perform
the following activities: . ou
Activity 1 : Take a circular wire and attach a straight wire AB at a point P of the
circular wire so that it can rotate about the point P in a plane. Put the system on a table
and gently rotate the wire AB about the point P to get different positions of the straight
wire [see Fig. 10.3(i)]. . ,

In various positions, the wire intersects the AN A
circular wire at P and at another point Q, or Q, or 6
Q,, etc. In one position, you will see that it will KL '
intersect the circle at the point P only (see position An | G
A'B' of AB). This shows that a tangent exists at . ys
the point P of the circle. On rotating further, you X -
can observe that in all other positions of AB, it will R, Yrs
intersect the circle at P and at another point, say R, B BY RS
or R, or R,, etc. So, you can observe that there is
only one tangent at a point of the circle. Fig. 10.3 (i)

While doing activity above, you must have observed that as the position AB
moves towards the position A' B', the common point, say Q,, of the line AB and the
circle gradually comes nearer and nearer to the common point P. Ultimately, it coincides
with the point P in the position A'B' of A'"B"'. Again note, what happens if 'AB' is
rotated rightwards about P? The common point R, gradually comes nearer and nearer
to P and ultimately coincides with P. So, what we see is:

The tangent to a circle is a special case of the secant, when the two end
points of its corresponding chord coincide.

*The word 'tangent' comes from the Latin word 'tangere', which means to touch and was
introduced by the Danish mathematician Thomas Fineke in 1583.
Reprint 2025-26


146 MATHEMATICS
Activity 2 : On a paper, draw a circle and a p"

secant PQ of the circle. Draw various lines

parallel to the secant on both sides of it. You P

will find that after some steps, the length of p'

the chord cut by the lines will gradually

decrease, i.e., the two points of intersection of

the line and the circle are coming closer and

closer [see Fig. 10.3(41)]. In one case, it

becomes zero on one side of the secant and in \ \Q""
another case, it becomes zero on the other side Q

of the secant. See the positions P'Q' and P'"'Q" \
of the secant in Fig. 10.3 (ii). These are the Q
tangents to the circle parallel to the given secant °

PQ. This also helps you to see that therecannot = * Fig. 10.3 (i)

be more than two tangents parallel to a given - ou

secant, - _

This activity also establishes, what you must have observed, while doing
Activity 1, namely, a tangent is the secant when both of the end points of the
corresponding chord coincide.

The common point of the tangent and the circle is called the point of contact
[the point A in Fig. 10.1 (iii)]and the tangent is said to touch the circle at the
common point. so : ,

Now look around you. Have you seen a bicycle
or acart moving? Look at its wheels. All the spokes
of a wheel are along its radii. Now note the position
of the wheel with respect to its movement on the
ground. Do you see any tangent anywhere?

(See Fig. 10.4). In fact, the wheel moves along a line
which is a tangent to the circle representing the wheel.
Also, notice that in all positions, the radius through
the point of contact with the ground appears to be at
right angles to the tangent (see Fig. 10.4). We shall Fig. 10.4
now prove this property of the tangent.
Theorem 10.1 : The tangent at any point of a circle is perpendicular to the
radius through the point of contact.
Proof : We are given a circle with centre O and a tangent XY to the circle at a
point P. We need to prove that OP is perpendicular to XY.
Reprint 2025-26


CIRCLES 147
Take a point Q on XY other than P and join OQ (see Fig. 10.5).
The point Q must lie outside the circle.
(Why? Note that if Q lies inside the circle, XY
will become a secant and not a tangent to the
circle). Therefore, OQ is longer than the radius
OP of the circle. That is,
OQ > OP.
Since this happens for every point on the
line XY except the point P, OP is the y ~~. xX
shortest of all the distances of the point O to the P Q
points of XY. So OP is perpendicular to XY. ; ,
(as shown in Theorem A1.7.) Fig. 10.5 .
Remarks . ",
1. By theorem above, we can also conclude that at any point on a circle there can be
one and only one tangent. tO °
2. The line containing the radius through the point of contact is also sometimes called
the 'normal' to the circle at the point.
EXERCISE 10.1
1. How many tangents can acirclehave?
2. Fillinthe blanks: _./
(i) A tangent to a circle intersects it in _____ point (s).
(i) A line intersecting a circle in two points is calleda_-____--
(ii) A circle can have _________ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called _________.
3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at
a point Q so that OQ = 12 cm. Length PQ is:
(A) 12cm (B) 13cm (C) 85cm (D) J119 cm.
4, Draw a circle and two lines parallel to a given line such that one is a tangent and the
other, a secant to the circle.
10.3 Number of Tangents from a Point on a Circle
To get an idea of the number of tangents from a point on a circle, let us perform the
following activity:
Reprint 2025-26


148 MATHEMATICS
Activity 3 : Draw a circle on a paper. Take a

point P inside it. Can you draw a tangent to the A"

circle through this point? You will find that all

the lines through this point intersect the circle in SX)

two points. So, it is not possible to draw any

tangent to a circle through a point inside it

[see Fig. 10.6 (i)].

Next take a point P on the circle and draw
tangents through this point. You have already (i) °
observed that there is only one tangent to the
circle at such a point [see Fig. 10.6 (ii)].

Finally, take a point P outside the circle and
try to draw tangents to the circle from this point. ~
What do you observe? You will find that you P
can draw exactly two tangents to the circle "
through this point [see Fig. 10.6 (iii)]. tO

We can summarise these facts as follows: -

Case 1: There is no tangent to a circle passing .

through a point lying inside the circle. (ul)

Case 2 : There is one and only one tangent toa

circle passing through a point lying on the circle. T,
% /

Case 3 : There are exactly two tangents to a

circle through a point lying outside the circle.

In Fig. 10.6 (iii), T,and T, are the points of P
contact of the tangents PT, and PT,
respectively. Hs

2 T

The length of the segment of the tangent ,
from the external point P and the point of contact (iii)
with the circle is called the length of the tangent Fig. 10.6
from the point P to the circle.

Note that in Fig. 10.6 (iii), PT, and PT, are the lengths of the tangents from P to
the circle. The lengths PT, and PT, have a common property. Can you find this?
Measure PT, and PT,. Are these equal? In fact, this is always so. Let us give a proof
of this fact in the following theorem.

Reprint 2025-26


CIRCLES 149

Theorem 10.2 : The lengths of tangents drawn

from an external point to a circle are equal. Q

Proof : We are given a circle with centre O, a

point P lying outside the circle and two tangents

PQ, PR on the circle from P (see Fig. 10.7). We P

are required to prove that PQ = PR.

For this, we join OP, OQ and OR. Then

Z OQP and Z ORP are right angles, because R

these are angles between the radii and tangents, '

and according to Theorem 10.1 they are right Fig. 10.7

angles. Now in right triangles OQP and ORP, ,
OQ = OR (Radii of the same circle)
OP=OP _. . (Common)

Therefore, A OQP = A ORP -- . (RHS)

This gives PQ = PR _ . 7 (CPCT)

Remarks : oe

1. The theorem can also be proved by using the Pythagoras Theorem as follows:

PQ' = OP? - OQ' = OP? - OR' = PR? (As OQ = OR)

which gives PQ = PR. 7 "

2. Note also that Z OPQ = Z OPR. Therefore, OP is the angle bisector of Z QPR,

i.e., the centre lies on the bisector of the angle between the two tangents.

Let us take some examples.

Example 1 : Prove that in two concentric circles,

the chord of the larger circle, which touches the

smaller circle, is bisected at the point of contact. c

Solution : We are given two concentric circles

C, and C, with centre O and a chord AB of the A

larger circle C, which touches the smaller circle

C, at the point P (see Fig. 10.8). We need to prove

that AP = BP. B

Let us join OP. Then, AB is a tangent to C, at P Fig. 10.8

and OP is its radius. Therefore, by Theorem 10.1,

OP L AB

Reprint 2025-26


150 MATHEMATICS
Now AB is a chord of the circle C, and OP L AB. Therefore, OP is the bisector of the
chord AB, as the perpendicular from the centre bisects the chord,
1.¢., AP = BP
Example 2 : Two tangents TP and TQ are drawn
to a circle with centre O from an external point T. p
Prove that Z PTQ = 2 Z OPQ. P
Solution : We are given a circle with centre O, TS
an external point T and two tangents TP and TQ
to the circle, where P, Q are the points of contact
(see Fig. 10.9). We need to prove that Oo
Z PTQ= 2 Z OPQ Fig. 10.9 -
Let ZPTQ=9 : OO
Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle."
Loe. , od
Therefore, Z TPQ = Z TQP= 5 (180° - 8) = 90° - 5°
Also, by Theorem 10.1, ZOPT=90° |
. . 1
So, Z OPQ = Z OPT - Z TPQ = 50° [90° 50)
f° ) :
NL ~] 1
. = -§9=- ZPT
2 2 Q
This gives Z PTQ= 2 Z OPQ
Example 3 : PQ is a chord of length 8 cm of a P
circle of radius 5 cm. The tangents at P and Q
intersect at a point T (see Fig. 10.10). Find the Sem
length TP. 8 cm
Solution : Join OT. Let it intersect PQ at the T RTO
point R. Then A TPQ is isosceles and TO is the
angle bisector of Z PTQ. So, OT L PQ
and therefore, OT bisects PQ which gives O
PR =RQ=4cm.
5 5 DD Fig. 10.10
Also, OR = .OP* - PR? =,/5°- 4 cm =3cm.
Reprint 2025-26


CIRCLES 151
Now, 2ZTPR+ Z RPO = 90° = Z TPR + Z PTR (Why?)
So, ZRPO=Z2Z PTR
Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.
i, TP RP. TP 4 20
This gives PO = Ro' Le. = 3 or TP = 3 cm.
Note : TP can also be found by using the Pythagoras Theorem, as follows:
Let TP =x and TR=y. Then .
P= y+16 (Taking right A PRT) d)
V?4+52= (y +3) (Taking right A OPT) © (2)
Subtracting (1) from (2), we get "
32 16 . _
25= 6y-7 or y= G3 - |
2 .
Therefore, ee () +16= 206+) =" irom (1)
2002 |
or x= 3 :
Co' EXERCISE 10.2
In Q.1 to 3, choose the correct option and give justification.
1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from
the centre is 25 cm. The radius of the circle is T
(A) Tom (B) 12cm : <\
(C) 15cm . > (D) 24.5cm
2. In Fig. 10.1 1, if TP and TQ are the two tangents \ ww \
to acircle with centre O so that Z POQ = 110°, Q
then Z PTQ is equal to
(A) 60° - (B) 70°
©) 80° (D) 90° Fig. 10.11
3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other
at angle of 80°, then Z POA is equal to
(A) 50° (B) 60°
(C) 70° (D) 80°
Reprint 2025-26


152 MATHEMATICS
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the centre.
6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4
cm. Find the radius of the circle.
7, Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.
8. Aquadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that
AB+CD=AD+BC °
R Cc
D x P Y ,
0 Op
A | xX] y
P Bo - QB
Fig. 10.12 Fig. 10.13
9, In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with centre O and
another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove
that Z AOB = 90°. .) °
10. Prove that the angle between the two tangents drawn from an external point to a circle
is supplementary to the angle subtended by the line-segment joining the points of
contact at the centre.
11. Prove that the parallelogram circumscribing a A
circle is arhombus.
12. Atriangle ABC is drawn to circumscribe a circle
of radius 4 cm such that the segments BD and
DC into which BC is divided by the point of
contact D are of lengths 8 cm and 6 cm
respectively (see Fig. 10.14). Find the sides AB
and AC.
13. Prove that opposite sides of a quadrilateral ¢ B
. wpe . <-6 cm->D<-38 cm -->
circumscribing a circle subtend supplementary
angles at the centre of the circle. Fig. 10.14
Reprint 2025-26


CIRCLES 153
10.4 Summary
In this chapter, you have studied the following points :
1. The meaning of a tangent to a circle.
2. The tangent to a circle is perpendicular to the radius through the point of contact.
3. The lengths of the two tangents from an external point to a circle are equal.

CO

Reprint 2025-26


ia

[al s5 jG

'a F

t Sh

meee Tai

1062CH12
11.1 Areas of Sector and Segment of a Circle . Q
You have already come across the terms sector and i. qialor
segment of a circle in your earlier classes. Recall __ "
that the portion (or part) of the circular regionenclosed | :
by two radii and the corresponding arc is called a
sector of the circle and the portion (or part) of the
circular region enclosed between a chord and the
: . : A B
corresponding arc is called a segment of the circle. P
Thus, in Fig. 11.1, shaded region OAPB is a sector Fig. 11.1
of the circle with centre O. Z AOB is called the
angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of
the circle. For obvious reasons, OAPB is called the minor sector and
OAQB is called the major sector. You can also see that angle of the major sector is
360° - Z AOB. ,
Now, look at Fig. 11.2 in which AB is a chord Q
of the circle with centre O. So, shaded region APB is Mai
a segment of the circle. You can also note that Segm ont
unshaded region AQB is another segment of the circle 0 B
formed by the chord AB. For obvious reasons, APB °
is called the minor segment and AQB is called the Ss
major segment. SX,
&
Remark : When we write 'segment' and 'sector' x $
we will mean the 'minor segment' and the 'minor ;
sector' respectively, unless stated otherwise. Fig. 11.2
Reprint 2025-26


AREAS RELATED TO CIRCLES 155
Now with this knowledge, let us try to find some O
relations (or formulae) to calculate their areas.
Let OAPB be a sector of a circle with centre
O and radius r (see Fig. 11.3). Let the degree
measure of Z AOB be 0. A
You know that area of a circle (in fact of a
circular region or disc) is nr'.
In a way, we can consider this circular region to A B
be a sector forming an angle of 360° (i.e., of degree P "
measure 360) at the centre O. Now by applying the Fig. 11.3
Unitary Method, we can arrive at the area of the
sector OAPB as follows: ; ,
When degree measure of the angle at the centre is 360, area of the
sector = mr? . oe ,
So, when the degree measure of the angle at the centre is 1, area of the
nw a oe "
sector = ~~: oe "
360 . , .
Therefore, when the degree measure of the angle at the centre is 0, area of the
;
Tr 0 2 :
tor = -x6=-=-xmMr', .
sector 360 360 / :
Thus, we obtain the following relation (or formula) for area of a sector of a
circle: QO ) ;
Area of the sector of angle @=--- x nr?
rea of the sector o angle 360 ,
where r is the radius of the circle and 0 the angle of the sector in degrees.
Now, a natural question arises : Can we find Q
the length of the arc APB corresponding to this
sector? Yes. Again, by applying the Unitary
Method and taking the whole length of the circle
(of angle 360°) as 2mr, we can obtain the required a
. 0
length of the arc APB as 360 x 2nr, yn
9 A™__.- B
P
=. --x2
So, length of an are of a sector of angle 0 360 xX 20 , Fig, 11.4
Reprint 2025-26


156 MATHEMATICS
Now let us take the case of the area of the
segment APB of a circle with centre O and radius r
(see Fig. 11.4). You can see that :
Area of the segment APB = Area of the sector OAPB - Area of A OAB
Q 2
= --~- Xr -areaof AOAB
360
Note : From Fig. 11.3 and Fig. 11.4 respectively, you can observe that:
Area of the major sector OAQB = ar - Area of the minor sector OAPB
and Area of major segment AQB = n° - Area of the minor segment APB
Let us now take some examples to understand these concepts (or results).
Example 1 : Find the area of the sector of a circle
with radius 4 cm and of angle 30°. Also, find the area
of the corresponding major sector (Use m = 3.14).
Solution : Given sector is OAPB (see Fig. 11.5). - a \
Area of the sector = o x mr? a v
~ 360
. A p B
=o 3 14x 4x 4 om? P
360 . . Fig. 11.5
: 12.56 |
C) = 3 cm? = 4.19cm? (approx.)
\ = XQ
Area of the corresponding major sector
= mr - area of sector OAPB
8 = (3.14 x 16 - 4.19) cm?
4, "= 46.05 cm? = 46.1 cm?' (approx.)
oe 360 - 6
Alternatively, area of the major sector = st x Tr?
: = [P05 |x 3.14% 16 cm?
360
= 0 3.14 16cm? = 46.05 cm?
360
= 46.1 cm? (approx.)
Reprint 2025-26


AREAS RELATED TO CIRCLES 157
Example 2 : Find the area of the segment AYB x
shown in Fig. 11.6, if radius of the circle is 21 cm and A J B
22 120°.
Z AOB = 120°. (Use t= =) Cee
Fig. 11.6 '
Solution : Area of the segment AYB
= Area of sector OAYB - Area of AOQAB (1)
120. 22 | .
Now, area of the sector OAYB = 360 x = x 21x 21cm? = 462 cm'. (2)
For finding the area of A OAB, draw OM _ AB as shown in Fig. 11.7.
Note that OA = OB. Therefore, by RHS congruence, A AMO = A BMO.
. » . : ; 1 ° oO
So, M is the mid-point of AB and Z AOM = Z BOM = 5 x 120° = 60°,
L = Ls
et OM =x cm A B
"om .
So, fromAOMA, \ -- =cos 60° a oy
> > NS -_ OA . 7 <> a
. 1 1 °
x
-=- | cos 60° =- i
Or, 172 ( ;] Fig. 11.7
- . 21
or, . XG
. . 21
So, OM = - cm
2
. AM J3
Al . -- =si ° = -
SO, OA sin 60 5
21V3
So, AM = aN cm
2x21
Therefore, AB = 2 AM = DONS cm = 21V3cm
Reprint 2025-26


158 MATHEMATICS
1 1 21
So, area of A OAB = 5 ABXOM = > x 21V3 x > em'
441
= -3cm? (3)
4
441 2
Therefore, area of the segment AYB = [460 - 3 cm" [From (1), (2) and (3)]
21
= 7 (88 - 21V3)em? :
EXERCISE 11.1
22 .
Unless stated otherwise, use 7 = 7: .
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
2. Find the area of a quadrant of a circle whose circumference is 22 cm.
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute
hand in 5 minutes.
4. Achord of acircle of radius 10 cm subtends a right angle at the centre. Find the area of
the corresponding: (i) minorsegment (ii) major sector. (Use m=3.14)
5. Inacircle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
A \ .
(i) the length of the arc (ii) area of the sector formed by the arc
~M " 5
(iii) area of the segment formed by the corresponding chord
6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find
the areas of the corresponding minor and major
i <«¢ *.¢ ~
segments of the circle. vs 0 "3 vie ?, ' mn
(Use m=3.14 and ./3 = 1.73) Sfet ti fe% 208 23 +:
So ere Pose tos uy et
7. Acchord of a circle of radius 12 cm subtends an SESS St Je ye
angle of 120° at the centre. Find the area of the 3-240 ce 23 233 Pas
corresponding segment of the circle. +o BX wa + ot 3
+e ATR o
(Usen=3.14 and 3 = 1.73) 93S" : Ds
i. + d/ete ' +
8. A horse is tied to a peg at one corner of a square * "+, oe% pawn
shaped grass field of side 15 m by means of a5 m 7 247
long rope (see Fig. 11.8). Find Fig. 11.8
Reprint 2025-26


AREAS RELATED TO CIRCLES 159
(i) the area of that part of the field in which the
horse can graze.
(ii) the increase in the grazing area if the rope
were 10 m long instead of 5 m. (Use x =3.14) lok\ [se
9. A brooch is made with silver wire in the form
of a circle with diameter 35 mm. The wire is also
used in making 5 diameters which divide the
circle into 10 equal sectors as shown in
Fig. 11.9. Find: °
. Fig. 11.9
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch. .

10. Anumbrella has 8 ribs which are equally spaced
(see Fig. 11.10). Assuming umbrella to be a flat
circle of radius 45 cm, find the area between the
two consecutive ribs of the umbrella. _ oo |

11. Acar has two wipers which do not overlap. Each : 1
wiper has a blade of length 25 cm sweeping - WY
through an angle of 115°. Find the total area Fig. 11.10
cleaned at each sweep of the blades.

12. To warn ships for underwater rocks, a lighthouse" ' gr ve ON
spreads a red coloured light over a sector of I SA
angle 80° to a distance of 16.5 km. Find the area A h
of the sea over which the ships are warned. SJ a
(Use n= 3.14) $ Ps

' s

13. A round table cover has six equal designs as Ye 4
shown in Fig. 11.11. If the radius of the cover is an LE
28 cm, find the cost of making the designs at the -
rate of 0.35 per cm'. (Use ./3 = 1.7) Fig. 11.11

14. Tick the correct answer in the following :

Area of a sector of angle p (in degrees) of a circle
with radius R is
P P 2 P P 2
--xX2nR --xmR --x2nR --x2nR
() 180 (8) 180 © 360 ©) 720
Reprint 2025-26


160 MATHEMATICS
11.2 Summary
In this chapter, you have studied the following points :
1. Length of an arc of a sector of a circle with radius r and angle with degree measure 0 is
8 x 2mr.
360
2. Area of a sector of a circle with radius r and angle with degree measure 9 is a x Tr,
3. Area of segment of a circle "
= Area of the corresponding sector - Area of the corresponding triangle.
OO.
Reprint 2025-26


it a A

any _

(a) thr

1062CH13
12.1 Introduction .
From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and
sphere (see Fig. 12.1). You have also learnt how to find their surface areas and volumes.
(i) Cc) (ii) . (iii) (iv)
. Fig, 12.1
In our day-to-day life, we come across a number of solids made up of combinations
of two or more of the basic solids as shown above.
You must have seen a truck with a
container fitted on its back (see Fig. 12.2), 8)
carrying oil or water from one place to RY
another. Is it in the shape of any of the four |.
basic solids mentioned above? You may NK WK A
guess that it is made of a cylinder withtwo [J - TT Fr
hemispheres as its ends. \e) - 0
Fig. 12.2
Reprint 2025-26


162 MATHEMATICS

Again, you may have seen an object like the
one in Fig. 12.3. Can you name it? A test tube, right! 3
You would have used one in your science laboratory. _0°
This tube is also a combination of a cylinder and a
hemisphere. Similarly, while travelling, you may have
seen some big and beautiful buildings or monuments
made up of a combination of solids mentioned above.

If for some reason you wanted to find the
surface areas, or volumes, or capacities of such ;
objects, how would you do it? We cannot classify .
these under any of the solids you have already studied. Fig. 12.3

In this chapter, you will see how to find surface areas and volumes of such
objects. .

12.2 Surface Area of a Combination of Solids a .
Let us consider the container seen in Fig. 12.2. How do we find the surface area of
such a solid? Now, whenever we come across a new problem, we first try to see, if
we can break it down into smaller problems, we have earlier solved. We can see that
this solid is made up of a cylinder with two hemispheres stuck at either end. It would
look like what we have in Fig. 12.4, after we put the pieces all together.
SX
--_>(f N ) --
@0-- CO--
ML .
: Fig. 12.4

If we consider the surface of the newly formed object, we would be able to see
only the curved surfaces of the two hemispheres and the curved surface of the cylinder.

So, the total surface area of the new solid is the sum of the curved surface
areas of each of the individual parts. This gives,

TSA of new solid = CSA of one hemisphere + CSA of cylinder

: + CSA of other hemisphere
where TSA; CSA stand for "Total Surface Area' and 'Curved Surface Area'
respectively.

Let us now consider another situation. Suppose we are making a toy by putting
together a hemisphere and a cone. Let us see the steps that we would be going
through.

Reprint 2025-26


SURFACE AREAS AND VOLUMES 163
First, we would take a cone and a hemisphere and bring their flat faces together.
Here, of course, we would take the base radius of the cone equal to the radius of the
hemisphere, for the toy is to have a smooth surface. So, the steps would be as shown
in Fig. 12.5.
fa
oa

"

& -_ OS
Fig. 12.5 -

At the end of our trial, we have got ourselves a nice round-bottomed toy. Now if
we want to find how much paint we would require to colour the surface of this toy,
what would we need to know? We would need to know the surface area of the toy,
which consists of the CSA of the hemisphere and the CSA of the cone.

So, we can say: . oO _
Total surface area of the toy = CSA of hemisphere + CSA of cone

Now, let us consider some examples. :

Example I : Rasheed got a playing top (lattu) as his (seen \
birthday present, which surprisingly had no colour on k \
it. He wanted to colour it with his crayons. The top is 5cem
shaped like a cone surmounted by a hemisphere
(see Fig 12.6). The entire top is 5 cm in height and
the diameter of the top is 3.5 cm. Find the area he io.
22 ;
has to colour, (Take m = 7 ) Fig. 12.6
Solution : This top is exactly like the object we have discussed in Fig. 12.5. So, we
can conveniently use the result we have arrived at there. That is :
. _ TSA of the toy = CSA of hemisphere + CSA of cone
1
Now, the curved surface area of the hemisphere = 3 Anr °) = 2nr?
_|2~x 22 35 33 cm?
7 2 2
Reprint 2025-26


164 MATHEMATICS
Also, the height of the cone = height of the top - height (radius) of the hemispherical part
3.5
= |5-- |em =3.25 cm
2
3.5)
So, the slant height of the cone (J) = yr? +h? = 2 + (3.25)? cm =3.7 cm (approx.)
22 3.5 2
Therefore, CSA of cone = mrl = 7 x > x 3.7 | cm "
This gives the surface area of the top as ,
22.3.5. 3. 22. 3. |
a2 2 4 32 39) om? +f 2? x 32 3.7 Jem?
7 2 2 7 2° yO
22 3. ll , co
=> x "2 (3.5 + 3.7) cm' = 3X (3.5 + 3.7) cm? = 39.6 cm" (approx.)
You may note that 'total surface area of the top' is not the sum of the total
surface areas of the cone and hemisphere. ° 4.2 em
Example 2 : The decorative block shown : (....y >
in Fig. 12.7 is made of two solids - a cube . Pa
and a hemisphere. The base of the block isa .
cube with edge 5 cm, and the hemisphere 5 om,
fixed on the top has a diameter of 4.2 cm.
Find the total surface area of the block.
22
(Take n= -) Sem
7 oo Fig. 12.7
Solution : The total surface area of the cube = 6 x (edge) =6 x5 x5 cm' = 150 cm'.
Note that the part of the cube where the hemisphere is attached is not included in the
surface area.
So, ~ the surface area of the block = TSA of cube - base area of hemisphere
° + CSA of hemisphere
= 150--nr +2 ar = (150 + ar') cm
= 150 cm? + 22 42 42 cm?
7 2 2
= (150 + 13.86) cm? = 163.86 cm?
Reprint 2025-26


SURFACE AREAS AND VOLUMES 165
Example 3 : A wooden toy rocket is in the
shape of a cone mounted on a cylinder, as
shown in Fig. 12.8. The height of the entire 6m
rocket is 26 cm, while the height of the conical
: . . 26 cm
part is 6 cm. The base of the conical portion
has a diameter of 5 cm, while the base
diameter of the cylindrical portion is 3 cm. If
the conical portion is to be painted orange
and the cylindrical portion yellow, find the Scm
area of the rocket painted with each of these
colours. (Take m = 3.14) .
base of cylinder
Solution : Denote radius of cone by r, slant
. . . base of cone
height of cone by /, height of cone by A, radius >
of cylinder by 7' and height of cylinder by hr'. - oo
Then r = 2.5 cm, h = 6 cm, r = 1.5 cm, - ow
h' = 26 -6 = 20 cmand oe Fig. 12.8
l= Jr th? = J2.57 +6? om =6.5 cm
Here, the conical portion has its circular base resting on the base of the cylinder, but
the base of the cone is larger than the base of the cylinder. So, a part of the base of the
cone (a ring) is to be painted. ,
: \ .
So, the area to be painted orange = CSA of the cone + base area of the cone
So - base area of the cylinder
= mrl+ar-n(ry
= m[(2.5 x 6.5) + (2.5 - (1.5)] cm?
SO = 7[20.25] cm? = 3.14 x 20.25 cm?
SO = 63.585 cm?
Now, _ the area to be painted yellow = CSA of the cylinder
+ area of one base of the cylinder
a = 2nrh' + n(r'y
= nr (2h' +r)
= (3.14 x 1.5) (2 x 20 + 1.5) cm?
= 4.71 x 41.5 cm?
= 195.465 cm?
Reprint 2025-26


166 MATHEMATICS
Example 4 : Mayank made a bird-bath for his garden 30cm
in the shape of a cylinder with a hemispherical ee
depression at one end (see Fig. 12.9). The height of SEES)
the cylinder is 1.45 m and its radius is 30 cm. Find the 145m
22
total surface area of the bird-bath. (Take 1 => )
Solution : Let h be height of the cylinder, and r the | | |
common radius of the cylinder and hemisphere. Then, Fig. 12.9 ;
the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere
= 2arh + 2nr = 2nr(h+r)
22 '
= 2x a 30(145 +30) cm?
= 33000cm? =3.3m?"
EXERCISE 12.1.
22 .
Unless stated otherwise, take m = 7

1. 2 cubes each of volume 64 cm? are joined end to end. Find the surface area of the

resulting cuboid. - _
sa™ .

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The
diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the
inner surface area of the vessel.

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius.
The total height of the toy is 15.5 cm. Find the total surface area of the toy.

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest
diameter the hemisphere can have? Find the surface area of the solid.

5. A hemispherical depression is cut out from one face of a cubical wooden block such
that the diameter / of the hemisphere is equal to the edge of the cube. Determine the
surface area of the remaining solid.

6. A medicine capsule is in the shape of a
cylinder with two hemispheres stuck to each 5mm
of its ends (see Fig. 12.10). The length of <--_-_-_
the entire capsule is 14 mm and the diameter 14mm
of the capsule is 5 mm. Find its surface area. Fig. 12.10

Reprint 2025-26


SURFACE AREAS AND VOLUMES 167
7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and
diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the
top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of
the canvas of the tent at the rate of 500 per m2. (Note that the base of the tent will not

be covered with canvas.)

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm',

9. A wooden article was made by scooping «Kg '
out a hemisphere from each end of a solid AI
cylinder, as shown in Fig. 12.11. If the
height of the cylinder is 10 cm, and its :
base is of radius 3.5 cm, find the total . MU
surface area of the article. . QUT

12.3. Volume of a Combination of Solids oO Fig. 12.11
In the previous section, we have discussed how to find the surface area of solids made
up of a combination of two basic solids. Here, we shall see how to calculate their
volumes. It may be noted that in calculating the surface area, we have not added the
surface areas of the two constituents, because some part of the surface area disappeared
in the process of joining them. However, this will not be the case when we calculate
the volume. The volume of the solid formed by joining two basic solids will actually be
the sum of the volumes of the constituents, as we see in the examples below.
Se" .
Example 5 : Shanta runs an industry in
a shed which is in the shape of a cuboid
surmounted by a half cylinder (see Fig.
12.12). If the base of the shed is of
dimension 7 m x 15 m, and the height of (oa,
the cuboidal portion is 8 m, find the volume A
of air that the shed can hold. Further, ;
suppose the machinery in the shed 8m:
occupies a total space of 300 m°, and '
there are 20 workers, each of whom
occupy about 0.08 m? space on an v
average. Then, how much air is in the Fig. 12.12
22
shed? (Take m = 7?
Reprint 2025-26


168 MATHEMATICS
Solution : The volume of air inside the shed (when there are no people or machinery)
is given by the volume of air inside the cuboid and inside the half cylinder, taken
together.
Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively.
Also, the diameter of the half cylinder is 7 m and its height is 15 m.
; ., 1 ,
So, the required volume = volume of the cuboid + 3 volume of the cylinder
1 22
= 115K 7x84 bx ec 15 |? = 1128.75 m?-
2 7 2 2 ;
Next, the total space occupied by the machinery = 300 m? °
And the total space occupied by the workers = 20 x 0.08 m3 = 1.6 m?-
Therefore, the volume of the air, when there are machinery and workers
= 1128.75 - (300.00 + 1.60) = 827.15 m3
Example 6 : A juice seller was serving his ; :
customers using glasses as shown in Fig. 12.13. ,
The inner diameter of the cylindrical glass was
5 cm, but the bottom of the glass had a'.
hemispherical raised portion which reduced the
capacity of the glass. If the height of a glass eects,
was 10 cm, find the apparent capacity of the Dates'
glass and its actual capacity. (Use a = 3.14.) Fig. 12.13
Solution : Since the inner diameter of the glass = 5 cm and height = 10 cm,
the apparent capacity of the glass = urh
. . = 3.14x2.5 x 2.5 x 10 cm? = 196.25 cm?
But the actual capacity of the glass is less by the volume of the hemisphere at the
base of the glass.
. i 2 2 3
i.e., * itis less by 3 tr = 3 x 3.14 2.52.5 x 2.5cm™ = 32.71 cm
So, the actual capacity of the glass = apparent capacity of glass - volume of the
hemisphere
= (196.25 - 32.71) cm'
= 163.54 cm?
Reprint 2025-26


SURFACE AREAS AND VOLUMES 169
: i is i oo OAS
Example 7: A solid toy is in the form of a EC - FF
hemisphere surmounted by a right circular cone. The ZA
height of the cone is 2 cm and the diameter of the pao Nc
; , oo
base is 4 cm. Determine the volume of the toy. Ifa
right circular cylinder circumscribes the toy, find the \ /
difference of the volumes of the cylinder and the toy. HYP
(Take x = 3.14) Fig. 12.14
Solution : Let BPC be the hemisphere and ABC be the cone standing on the base
of the hemisphere (see Fig. 12.14). The radius BO of the hemisphere (as well as
1 ;
of the cone) = > x4cm=2cm.
2 3, 1_, "
So, volume of the toy = 3 + 3 h ; .
2 3,1 2 - 3 3
= 3 <8 l4x@) +3 X34 (Q2)° x2} cm = 25.12 cm
Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of
the base of the right circular cylinder = HP = BO = 2 cm, and its height is
EH = AO + OP=(24+2)cm=4cm
So, the volume required = volume of the right circular cylinder - volume of the toy
fe (3.14 x 2? x 4 ~ 25.12) cm?
. = 25.12 cm? :
NOY 5
Hence, the required difference of the two volumes = 25.12 cm*.
EXERCISE 12.2
s 22
Unless stated otherwise, take T= 7 .

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being
equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid
in terms of 7.

2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with
two cones attached at its two ends by using a thin aluminium sheet. The diameter of the
model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume
of air contained in the model that Rachel made. (Assume the outer and inner dimensions
of the model to be nearly the same.)

Reprint 2025-26


170 MATHEMATICS

3. A gulab jamun, contains sugar syrup up to about KD
30% of its volume. Find approximately how much
syrup would be found in 45 gulab jamuns, each
shaped like a cylinder with two hemispherical ends
with length 5 cm and diameter 2.8 cm (see Fig. 12.15). eee

fee

4. A pen stand made of wood is in the shape of a Fig. 12.15
cuboid with four conical depressions to hold pens. -
The dimensions of the cuboid are 15 cm by 10 cm by
3.5 cm. The radius of each of the depressions is 0.5
cm and the depth is 1.4 cm. Find the volume of
wood in the entire stand (see Fig. 12.16). |

5. A vessel is in the form of an inverted cone. Its .
height is 8 cm and the radius of its top, which is
open, is 5 cm. It is filled with water up to the brim. . ME VV VV
When lead shots, each of which is a sphere of radius ce
0.5 cm are dropped into the vessel, one-fourth of | |
the water flows out. Find the number of lead shots . .

d di . Fig. 12.16
ropped in the vessel.

6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which
is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the
pole, given that 1 cm? of iron has approximately 8g mass. (Use m = 3.14)

fs.'

7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on
a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water
such that it touches the bottom. Find the volume of water left in the cylinder, if the radius
of the cylinder is 60 cm and its height is 180 cm.

8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter
of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its
volume to be 345 cm'. Check whether she is correct, taking the above as the inside
measurements, and 7 = 3.14.

12.4 Summary
In this chapter, you have studied the following points:

1. To determine the surface area of an object formed by combining any two of the basic
solids, namely, cuboid, cone, cylinder, sphere and hemisphere.

2. To find the volume of objects formed by combining any two of a cuboid, cone, cylinder,
sphere and hemisphere.

Reprint 2025-26


[a]
eee
PY ae
[a] eer,
1062CH14
13.1 Introduction °
In Class IX, you have studied the classification of given data into ungrouped as well as
grouped frequency distributions. You have also learnt to represent the data pictorially
in the form of various graphs such as bar graphs, histograms (including those of varying
widths) and frequency polygons. In fact, you went a step further by studying certain
numerical representatives of the ungrouped data, also called measures of central
tendency, namely, mean, median and mode. In this chapter, we shall extend the study
of these three measures, i.e., mean, median and mode from ungrouped data to that of
grouped data. We shall also discuss the concept of cumulative frequency, the
cumulative frequency distribution and how to draw cumulative frequency curves, called
ogives. ( ) °
ML .
13.2 Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the
observations divided by the total number of observations. From Class IX, recall that if
Xp Xye oo X, are observations with respective frequencies fof, ..6f, then this
means observation x, occurs f, times, x, occurs f, times, and so on.
Now, the sum of the values of all the observations = Sx, +fx,+...+fx,, and
the number of observations =f, +f,+...+f..
So, the mean x of the data is given by
. t= AX t+ fary tot A %n
fithteoth,
Recall that we can write this in short form by using the Greek letter 2 (capital
sigma) which means summation. That is,
Reprint 2025-26


172 MATHEMATICS
DA
x=
Ls
i=l
; ; oo - 2x 2... vo
which, more briefly, is written as x = Ta , if it is understood that i varies from
1 ton.
Let us apply this formula to find the mean in the following example.
Example 1 : The marks obtained by 30 students of Class X of a certain school in a
Mathematics paper consisting of 100 marks are presented in table below. Find the
mean of the marks obtained by the students.
Marks obtained | 10 | 20 | 36 50 | 56 70 | 72 88 | 92 | 95
(x,)
Number of 1) 1) 374 73]2)4)] 47172727371
students (/,)
Solution: Recall that to find the mean marks, we require the product of each x, with
the corresponding frequency f,. So, let us put them in a column as shown in Table 13.1.
- Table 13.1 ©
Marks obtained (x,) Number of students (f,)
10 1 10
20 1 20
36 3 108
40 4 160
50 3 150
56 2 112
60 4 240
70 4 280
72 1 72
80 1 80
88 2 176
92 3 276
95 1 95
Reprint 2025-26


STATISTICS 173
_ Lfx, 1779
Now, x= = - = 59.3
xf, 30
Therefore, the mean marks obtained is 59.3.

In most of our real life situations, data is usually so large that to make a meaningful
study it needs to be condensed as grouped data. So, we need to convert given ungrouped
data into grouped data and devise some method to find its mean.

Let us convert the ungrouped data of Example 1 into grouped data by forming
class-intervals of width, say 15. Remember that, while allocating frequencies to each
class-interval, students falling in any upper class-limit would be considered in the next
class, ¢.g., 4 students who have obtained 40 marks would be considered in the classinterval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped
frequency distribution table (see Table 13.2). . -

Table 13.2. o
Class interval 10-25 | 25-40] 40-55 | 55-70 | 70-85] 85-100
wmwrormon] 2 > 7 «lols

Now, for each class-interval, we require a point which would serve as the

1 a
representative of the whole class. /t is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each
class can be chosen to represent the observations falling in the class. Recall that we
find the mid-point of a class (or its class mark) by finding the average of its upper and
lower limits. That is, : .

Class saark = Upper class limit + Lower class limit
2
7 ° _ 10425 .

With reference to Table 13.2, for the class 10-25, the class mark is > 1.€.,
17.5. Similarly, we can find the class marks of the remaining class intervals. We put
them in Table 13.3. These class marks serve as our x,'s. Now, in general, for the ith
class interval, we have the frequency f, corresponding to the class mark x, We can
now proceed to compute the mean in the same manner as in Example 1.

Reprint 2025-26


174 MATHEMATICS
Table 13.3
Class interval | Number of students (/,) Class mark fe
10 - 25 2 17.5 35.0
25 - 40 3 32.5 97.5
40 - 55 7 47.5 332.5
55 - 70 6 62.5 375.0
70 - 85 6 715 465.0
85 - 100 6 92.5 555.0
rhe | Pini 18600
The sum of the values in the last column gives us X f.x,. So, the mean x of the
given data is given by Co
_ "fx, . ee
xr TAX, = 1860.0 = 62 .
Xf, . 30 . ; -

This new method of finding the mean is known as the Direct Method.

We observe that Tables 13.1 and 13.3 are using the same data and employing the
same formula for the calculation of the mean but the results obtained are different.
Can you think why this is so, and which one is more accurate? The difference in the
two values is because of the mid-point assumption in Table 13.3, 59.3 being the exact
mean, while 62 an approximate mean. -

y

Sometimes when the numerical values of x, and f, are large, finding the product
of x, and f, becomes tedious and time consuming. So, for such situations, let us think of
a method of reducing these calculations.

We can do nothing with the f's, but we can change each x, to a smaller number
so that our calculations become easy. How do we do this? What about subtracting a
fixed number from each of these x,'s? Let us try this method.

The first step is to choose one among the x; s as the assumed mean, and denote
it by 'a'. Also, to further reduce our calculation work, we may take 'a' to be that x,
which lies in the centre of x,, x,,...,x,. So, we can choose a = 47.5 or a = 62.5. Let
us choose a = 47.5.

The next step is to find the difference d, between a and each of the x,'s, that is,
the deviation of 'a' from each of the x,'s.
1e., d,=x,-a=x,-475

Reprint 2025-26


STATISTICS 175
The third step is to find the product of d, with the corresponding f,, and take the sum
of all the f,d,'s. The calculations are shown in Table 13.4.
Table 13.4
Class interval Number of Class mark d,= x,- 47.5 Sd,
students (f,) (x,)
10 - 25 2 17.5 -30 -60
25 - 40 3 32.5 -15 45
40 - 55 7 47.5 0 0
55 - 70 6 62.5 15 90
70 - 85 6 7715 30 180
85 - 100 6 92.5 45 270
ye | |=
dy
So, from Table 13.4, the mean of the deviations, d = yf
Now, let us find the relation between d and x.
Since in obtaining d,, we subtracted 'a' from each x,, so, in order to get the mean
¥ , we need to add 'a' to d . This can be explained mathematically as:
f- ) ;
y - Lfd,
M f - oe - d =: ii
ean of deviations : ye. f
> Uf (x, -a
So, d= BA-@
- Lf,
yy 2Fix; - Ufa
SO Of OE
yf,
a a acti
. Xf;
. = xX-a
So, X=at+d
"fd,
i.e., X= at Hid
Xi
Reprint 2025-26


176 MATHEMATICS

Substituting the values of a, Lfd, and Lf, from Table 13.4, we get

xX = 475 4 9° L475 +14.5= 62,
30
Therefore, the mean of the marks obtained by the students is 62.

The method discussed above is called the Assumed Mean Method.
Activity 1 : From the Table 13.3 find the mean by taking each of x, (i.e., 17.5, 32.5,
and so on) as 'a'. What do you observe? You will find that the mean determined in
each case is the same, i.e., 62. (Why 7?)

So, we can say that the value of the mean obtained does not depend on the
choice of 'a'. .

Observe that in Table 13.4, the values in Column 4 are all multiples of 15. So, if
we divide the values in the entire Column 4 by 15, we would get smaller numbers to
multiply with f. (Here, 15 is the class size of each class interval.) .

x; - 4a . - . : _ :

So, let u, = hl where a is the assumed mean and s is the class size.

Now, we calculate u, in this way and continue as before (i.e., find f, u,and
then Yfu.). Taking h = 15, let us form Table 13.5.

-~ "Table 13.5
x; -a

10-25 2 17.5 -30 -2 -4

25 - 40 3 32.5 -15 -1 -3

40-55 7 47.5 0 0 0

55 - 70 6 62.5 15 1 6

70 - 85 6 715 30 2 12

85 - 100 6 92.5 45 3 18

- <Lfu,
L uo
et x f
Here, again let us find the relation between yw and x.
Reprint 2025-26


STATISTICS 177
xX; --a
We have, "= ---
i h
(x; - a)
xf, > _
Therefore, ee 1 2X; - O2fi
Xf; h Lf;
12a Bh
hl Sf Xf, °
] _ .
= 7/* -a :
"[¥ - a] |
So, hu=xX-a : . .
ie., Foathi 9° |
So X= at h Phils
oi
Now, substituting the values of a, h, Sf, and Xf, from Table 14.5, we get
- 5
- X= 475+4+15x (*]
~) / 30
NL = 47.5+ 14.5 =62
So, the mean marks obtained by a student is 62.
The method discussed above is called the Step-deviation method.
We note that: - >
@ the step-deviation method will be convenient to apply if all the ds have a
common factor.
e The mean obtained by all the three methods is the same.
e@ The assumed mean method and step-deviation method are just simplified
forms of the direct method,
@ The formula ¥ =a+ Au still holds if a and / are not as given above, but are
any non-zero numbers such that u, = a
Let us apply these methods in another example.
Reprint 2025-26


178 MATHEMATICS
Example 2 : The table below gives the percentage distribution of female teachers in
the primary schools of rural areas of various states and union territories (U.T.) of
India. Find the mean percentage of female teachers by all the three methods discussed
in this section.
Percentage of | 15-25] 25-35 | 35-45] 45-55 | 55-65] 65-75] 75-85
female teachers
Number of 11 7 4 4 2 1
States/U.T.
Source : Seventh All India School Education Survey conducted by NCERT
Solution : Let us find the class marks, x,, of each class, and put them in a column
(see Table 13.6): . oe
Table 13.6 9 -
Percentage of female Number of Xx,
teachers States /U.T. (f,)
15 - 25 6 20
25 - 35 11 30
35 - 45 7 40
45-55 4 50
55 - 65 4 60
65 - 75 2 70
75 - 85 1 80
x, -50
Here we take a = 50, h = 10, then d, = x,- 50 and 4, =o
We now find d, and u, and put them in Table 13.7.
Reprint 2025-26


STATISTICS 179
Table 13.7
Percentage of | Number of | x, | d,=x,-50 yet Ix, | fd, fu,
female states/U.T. 10
teachers (f,)
15 - 25 6 20 -30 -3 120 | -180 | -18
25 - 35 ll 30 -20 -2 330 | -220 | -22
35 - 45 7 40 -10 -1 280} -70 | -7
45-55 4 50 0 0 200 0 0
55 - 65 4 60 10 1 240 40 4
65 - 75 2 70 20 2 140 40 4
75 - 85 1 80 30 3 80 30 3
Pw | ss | | | _[nm
From the table above, we obtain Xf = 35, Lfx,= 1390, ~ |
Sid. =- 360, Sfu, = -36.
Using the direct method, x = 2% - 139 _ 39 74
of 35s
Using the assumed mean method, - .
NK ) Yfd. - _
x= a + id = 50 + 2) _ 39.71
Lf 35
Using the step-deviation method,
_ fy .
ge at| MH |p = 504/22) 10=30.71
" Xf; 35
Therefore, the mean percentage of female teachers in the primary schools of
rural areas is 39.71.
Remark : The result obtained by all the three methods is the same. So the choice of
method to be used depends on the numerical values of x, and f. If x, and f, are
sufficiently small, then the direct method is an appropriate choice. If x, and f, are
numerically large numbers, then we can go for the assumed mean method or
step-deviation method. If the class sizes are unequal, and x, are large numerically, we
can still apply the step-deviation method by taking / to be a suitable divisor of all the d,'s.
Reprint 2025-26


180 MATHEMATICS
Example 3 : The distribution below shows the number of wickets taken by bowlers in
one-day cricket matches. Find the mean number of wickets by choosing a suitable
method. What does the mean signify?
Number of 20-60 | 60-100] 100-150) 150-250] 250 - 350 | 350 - 450
wickets
Number of 7 5 16 12 2 3
bowlers
Solution : Here, the class size varies, and the x,s are large. Let us still apply the stepdeviation method with a = 200 and h = 20. Then, we obtain the data as in Table 13.8.
Table 13.8 . oo
d,
Number of Number of x. d=x.-- 200 u; => wf.
. i i i 20 ivi
wickets bowlers
taken (f)
20 - 60 7 40 -160 -8 -56
60 - 100 5 80 -120 -6 -30
100 - 150 16 125 -75 -3.75 -60
150 - 250 12 200 0 0 0
250 - 350 2 300 100 5 10
350 - 450 3 400 200 10 30
es
_ -106
So, 7 = 106 | Therefore, ¥ = 200+ 20( I") = 200 - 47.11 = 152.89.
45 45
This tells us that, on an average, the number of wickets taken by these 45 bowlers
in one-day cricket is 152.89.
Now, let us see how well you can apply the concepts discussed in this section!
Reprint 2025-26


STATISTICS 181
Activity 2 :

Divide the students of your class into three groups and ask each group to do one of the
following activities.

1. Collect the marks obtained by all the students of your class in Mathematics in the
latest examination conducted by your school. Form a grouped frequency distribution
of the data obtained.

2. Collect the daily maximum temperatures recorded for a period of 30 days in your
city. Present this data as a grouped frequency table.

3, Measure the heights of all the students of your class (in cm) and form a grouped
frequency distribution table of this data.
After all the groups have collected the data and formed grouped frequency

distribution tables, the groups should find the mean in each case by the method which
they find appropriate. , me
EXERCISE 13.1. _
1. Asurvey was conducted by a group of students as a part of their environment awareness
programme, in which they collected the following data regarding the number of plants in
20 houses in a locality. Find the mean number of plants per house.
sO
Which method did you use for finding the mean, and why?

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in 2) 500-520 | 520-540 | 540-560 | 560-580 | 580-600
Find the mean daily wages of the workers of the factory by using an appropriate method.

3. The following distribution shows the daily pocket allowance of children of a locality.

The mean pocket allowance is Rs 18. Find the missing frequency /.

Daily pocket 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25

allowance (in 2)

sonar | 7 Le | Lets p ete

Reprint 2025-26


182 MATHEMATICS
4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per
minute were recorded and summarised as follows. Find the mean heartbeats per minute
for these women, choosing a suitable method.
Number of heartbeats | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86
per minute
ornsome [fe pe Pe pepe p
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These
boxes contained varying number of mangoes. The following was the distribution of
mangoes according to the number of boxes.
Number of mangoes | 50 - 52 | 53-55 | 56-58 | 59 - 61 62 - 64
Find the mean number of mangoes kept in a packing box. Which method of finding
the mean did you choose? . ' " _
6. The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure | 100-150 | 150-200 | 200-250 | 250-300 | 300-350
(in %)
Number of 4 5 2 2 2
households
Find the mean daily expenditure on food by a suitable method.
7. To find out the concentration of SO, in the air (in parts per million, i.e., ppm), the data
was collected for 30 localities in a certain city and is presented below:
Concentration of SO, (in ppm)
0.00 - 0.04 4
0.04 - 0.08 9
0.08 - 0.12 9
0.12 - 0.16 2
0.16 - 0.20 4
0.20 - 0.24 2
Find the mean concentration of SO, in the air.
Reprint 2025-26


STATISTICS 183
8. Aclass teacher has the following absentee record of 40 students of a class for the whole
term. Find the mean number of days a student was absent.
Number of 6 - 10] 10 - 14] 14 - 20] 20 -28 | 28 - 38)38 - 40
days
Number of 11 10 7 4 4 3 1
students
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate. .
Literacy rate (in %) 45-55 | 55-65 | 65 - 75 | 75 - 85 | 85 - 95
ES
13.3 Mode of Grouped Data - : 2
Recall from Class IX, a mode is that value among the observations which occurs most
often, that is, the value of the observation having the maximum frequency. Further, we
discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining
a mode of grouped data. It is possible that more than one value may have the same
maximum frequency. In such situations, the data is said to be multimodal. Though
grouped data can also be multimodal, we shall restrict ourselves to problems having a
single mode only. ae ) ;
Let us first recall how we found the mode for ungrouped data through the following
example. .
Example 4 : The wickets taken by a bowler in 10 cricket matches are as follows:
2. .6 '4 5 0 2 1 3 2 3
Find the mode of the data.
Solution : Let us form the frequency distribution table of the given data as follows:
Number of 1 2 3 4 5
wickets
Number of 1 1 3 2 1 1 1
matches
Reprint 2025-26


184 MATHEMATICS

Clearly, 2 is the number of wickets taken by the bowler in the maximum number
(i.e., 3) of matches. So, the mode of this data is 2.

In a grouped frequency distribution, it is not possible to determine the mode by
looking at the frequencies. Here, we can only locate a class with the maximum
frequency, called the modal class. The mode is a value inside the modal class, and is
given by the formula:

Mode = (ae lsh
2h fo- hr
where /= lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
J, = frequency of the modal class, °
J, = frequency of the class preceding the modal class, '
f, = frequency of the class succeeding the modal class.

Let us consider the following examples to illustrate the use of this formula.
Example 5: A survey conducted on 20 households in a locality by a group of students
resulted in the following frequency table for the number of family members in a
household:

Number of 7 2 2 1
families

Find the mode of this data.

Solution : Here the maximum class frequency is 8, and the class corresponding to this
frequency is 3 - 5. So, the modal class is 3 - 5.
Now

modal class = 3 -5, lower limit (/) of modal class = 3, class size (h) = 2

frequency (f,) of the modal class = 8,

frequency (f,) of class preceding the modal class = 7,

frequency (f,) of class succeeding the modal class = 2.

Now, let us substitute these values in the formula :
Reprint 2025-26


STATISTICS 185
Mode = (ae lsh
2h ~ fi 07 f 2
= 3+ _ 8-7 x2=347=3,286
2x8-7-2 7

Therefore, the mode of the data above is 3.286.
Example 6 : The marks distribution of 30 students in a mathematics examination are
given in Table 13.3 of Example |. Find the mode of this data. Also compare and
interpret the mode and the mean.
Solution : Refer to Table 13.3 of Example 1. Since the maximum number of students
(i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 - 55. Therefore,

the lower limit (/) of the modal class=40,

the class size (h) = 15, . - _

the frequency (f,) of modal class = 7, _ °

the frequency (f,) of the class preceding the modal class = 3,

the frequency (f,) of the class succeeding the modal class = 6.
Now, using the formula: - _ ,

Cc) Mode = 1 + [she )m
a ° 2fi- fo -h

t Mode = 40+ -=3- |x 15 252
we ge ode = 4-6-3 =
So, the mode marks is 52. .
Now, from Example 1, you know that the mean marks is 62.

So, the maximum number of students obtained 52 marks, while on an average a
student obtained 62 marks.
Remarks : °
1. In Example 6, the mode is less than the mean. But for some other problems it may
be equal or more than the mean also.
2. It depends upon the demand of the situation whether we are interested in finding the
average marks obtained by the students or the average of the marks obtained by most

Reprint 2025-26


186 MATHEMATICS
of the students. In the first situation, the mean is required and in the second situation,
the mode is required.
Activity 3 : Continuing with the same groups as formed in Activity 2 and the situations
assigned to the groups. Ask each group to find the mode of the data. They should also
compare this with the mean, and interpret the meaning of both.
Remark : The mode can also be calculated for grouped data with unequal class sizes.
However, we shall not be discussing it.

EXERCISE 13.2 :

1. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years) 15-25 | 25-35 | 35-45 | 45-55 | 55-65
namberafpatenss | 6 | [| a | ow | s |

Find the mode and the mean of the data given above. Compare and interpret the two
measures of central tendency. - . °

2. The following data gives the information on the observed lifetimes (in hours) of 225
electrical components : . ,

Lifetimes (in hours) 20-40 | 40-60 | 60-80 | 80-100) 100-120
\ "
Determine the modal lifetimes of the components.

3. The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure :

Expenditure (in) Number of families
1000 - 1500 24
1500 - 2000 40
2000 - 2500 33
2500 - 3000 28
3000 - 3500 30
3500 - 4000 22
4000 - 4500 16
4500 - 5000 7
Reprint 2025-26


STATISTICS 187
4. The following distribution gives the state-wise teacher-student ratio in higher
secondary schools of India. Find the mode and mean of this data. Interpret the two
measures.
Number of students per teacher Number of states /U.T.
15 - 20 3
20 - 25 8
25 - 30 9
30-35 10
35-40 3
40 - 45 0
45 - 50 0
50-55 2
5. The given distribution shows the number of runs scored by some top batsmen of the
world in one-day international cricket matches. © ° :
3000 - 4000 4
4000 - 5000 18
5000 - 6000 9
6000 - 7000 7
7000 - 8000 6
8000 - 9000 3
9000 - 10000 1
10000 - 11000 1
Find the mode of the data.

6. A student noted the number of cars passing through a spot on a road for 100
periods each of 3 minutes and summarised it in the table given below. Find the mode
of the data :

10-20] 20-30 | 30-40] 40-50 | 50-60] 60-70] 70 - 80
wos [7 t*letel[~[ofsfs
Reprint 2025-26


188 MATHEMATICS
13.4 Median of Grouped Data
As you have studied in Class IX, the median is a measure of central tendency which
gives the value of the middle-most observation in the data. Recall that for finding the
median of ungrouped data, we first arrange the data values of the observations in
n+l ; ;
ascending order. Then, if n is odd, the median is the 3 th observation. And, ifn
is even, then the median will be the average of the stn and the (3 + 7 th observations.
Suppose, we have to find the median of the following data, which gives the
marks, out of 50, obtained by 100 students in a test : . :
First, we arrange the marks in ascending order and prepare a frequency table as
follows : ;
Table 13.9.
Marks obtained Number of students
(Frequency)
20 6
25 20
28 24
29 28
33 15
38 4
42 2
43 1
Reprint 2025-26


STATISTICS 189
Here n = 100, which is even. The median will be the average of the " th and the
2
G + i observations, i.e., the 50th and 51st observations. To find these
observations, we proceed as follows:
Table 13.10
Marks obtained Number of students
20 6 .
upto 25 6+20= 26 ;
upto 28 26 + 24=50
upto 29 50 + 28 =78
upto 33 78 +15=93
upto 38 93+4=97
upto 42 97+2=99
upto 43 99+ 1=100
Now we add another column depicting this information to the frequency table
above and name it as cumulative frequency column.
\ ) Table 13.11
Marks obtained Number of students Cumulative frequency
20 6 6
25 20 26
28 24 50
29 28 78
33 15 93
38 4 97
42 2 99
43 1 100
Reprint 2025-26


190 MATHEMATICS
From the table above, we see that:
50th observaton is 28 (Why?)
51st observation is 29
28 + 29
So, Median = > = 28.5
Remark : The part of Table 13.11 consisting Column | and Column 3 is known as
Cumulative Frequency Table. The median marks 28.5 conveys the information that
about 50% students obtained marks less than 28.5 and another 50% students obtained
marks more than 28.5.
Now, let us see how to obtain the median of grouped data, through the following
situation. . : .
Consider a grouped frequency distribution of marks obtained, out of 100, by 53
students, in a certain examination, as follows: - " .
Table 13.12 -
0-10 5
10 - 20 3
20 - 30 4
30 - 40 3
40 - 50 3
50 - 60 4
60 - 70 7
70 - 80 9
, 80 - 90 7
90 - 100 8
From the table above, try to answer the following questions:
How many students have scored marks less than 10? The answer is clearly 5.
Reprint 2025-26


STATISTICS 191

How many students have scored less than 20 marks? Observe that the number
of students who have scored less than 20 include the number of students who have
scored marks from 0 - 10 as well as the number of students who have scored marks
from 10 - 20. So, the total number of students with marks less than 20 is 5 + 3, ie., 8.
We say that the cumulative frequency of the class 10-20 is 8.

Similarly, we can compute the cumulative frequencies of the other classes, i.e.,
the number of students with marks less than 30, less than 40, . . ., less than 100. We
give them in Table 13.13 given below:

Table 13.13 :
Marks obtained Number of students
(Cumulative frequency)

Less than 10 5

Less than 20 5+3=8

Less than 30 8+4=12

Less than 40 124+3=15

Less than 50 15+3=18

Less than 60 18+4=22

Less than 70 22+7=29

Less than 80 29+9=38

Less than 90 384+7=45

Less than 100 45+8=53

The distribution given above is called the cumulative frequency distribution of
the less than type. Here 10, 20, 30, .. . 100, are the upper limits of the respective
class intervals. ~ .

We can similarly make the table for the number of students with scores, more
than or equal to 0, more than or equal to 10, more than or equal to 20, and so on. From
Table 13.12, we observe that all 53 students have scored marks more than or equal to
0. Since there are 5 students scoring marks in the interval 0 - 10, this means that there
are 53 - 5 = 48 students getting more than or equal to 10 marks. Continuing in the
same manner, we get the number of students scoring 20 or above as 48 - 3 = 45, 30 or
above as 45 - 4 = 41, and so on, as shown in Table 13.14.

Reprint 2025-26


192 MATHEMATICS
Table 13.14
Marks obtained Number of students
(Cumulative frequency)
More than or equal to 0 53
More than or equal to 10 53-5 =48
More than or equal to 20 48 --3 =45
More than or equal to 30 45-4=41
More than or equal to 40 41-3 =38
More than or equal to 50 38 --3 =35 .
More than or equal to 60 35-4=31
More than or equal to 70 31-7=24
More than or equal to 80 24-9=15
More than or equal to 90 15-7= 8
The table above is called a cumulative frequency distribution of the more
than type. Here 0, 10, 20, .. ., 90 give the lower limits of the respective class intervals.
Now, to find the median of grouped data, we can make use of any of these
cumulative frequency distributions.
Let us combine Tables 13.12 and 13.13 to get Table 13.15 given below:
so Table 13.15
Number of students (/) Cumulative frequency (cf)
0-10 5 5
10 - 20 3 8
20 - 30 4 12
30 - 40 3 15
40 - 50 3 18
50 - 60 4 22
60 - 70 7 29
70 - 80 9 38
80 - 90 7 45
90 - 100 8 53
Now in a grouped data, we may not be able to find the middle observation by
looking at the cumulative frequencies as the middle observation will be some value in
Reprint 2025-26


STATISTICS 193
a class interval. It is, therefore, necessary to find the value inside a class that divides
the whole distribution into two halves. But which class should this be?
n
To find this class, we find the cumulative frequencies of all the classes and 2°
We now locate the class whose cumulative frequency is greater than (and nearest to)
n n
> This is called the median class. In the distribution above, n = 53. So, 7 = 26.5.
Now 60 - 70 is the class whose cumulative frequency 29 is greater than (and nearest
to) ~ i.e, 26.5 |
0) 5» Le., 26.5.
Therefore, 60 - 70 is the median class. .
After finding the median class, we use the following formula for calculating the
median. . a, .
* cf oe 2
Median= / + | 2-- xh, -
where != lower limit of median class, .
n= number of observations,
f°? \
cf = cumulative frequency of class preceding the median class,
\ a
f= frequency of median class,
h= class size (assuming class size to be equal).
. . n
Substituting the values 5 26.5, 1= 60, cf =22,f=7,h=10
in the formula above, we get
26.5 - 22
Median= 60 + (25-22) x10
: 60 45
= + -_
7
= 66.4
So, about half the students have scored marks less than 66.4, and the other half have
scored marks more than 66.4.
Reprint 2025-26


194 MATHEMATICS
Example 7 : A survey regarding the heights (in cm) of 51 girls of Class X of a school
was conducted and the following data was obtained:
Height (in cm) Number of girls

Less than 140 4

Less than 145 11

Less than 150 29

Less than 155 40 :

Less than 160 46 ,

Less than 165 51 ,
Find the median height. . oe
Solution : To calculate the median height, we need to find the class intervals and their
corresponding frequencies. sO . .

The given distribution being of the less than type, 140, 145, 150,..., 165 give the
upper limits of the corresponding class intervals. So, the classes should be below 140,
140 - 145, 145 - 150, ..., 160 - 165. Observe that from the given distribution, we find
that there are 4 girls with height less than 140, i.e., the frequency of class interval
below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with
height less than 140. Therefore, the number of girls with height in the interval
140 - 145 is 11 - 4 = 7. Similarly, the frequency of 145 - 150 is 29 - 11 = 18, for
150 - 155, it is 40 - 29 = 11, and so on. So, our frequency distribution table with the
given cumulative frequencies becomes:

ao Table 13.16

Class intervals Cumulative frequency

Below 140 4 4
140 - 145 7 11
145 - 150 18 29
150 - 155 11 40
155 - 160 6 46
160 - 165 5 51

Reprint 2025-26


STATISTICS 195
n 51 . . oa,
Now n=5l. So, 5 25.5 . This observation lies in the class 145 - 150. Then,
l (the lower limit) = 145,
cf (the cumulative frequency of the class preceding 145 - 150) = 11,
f (the frequency of the median class 145 - 150) = 18,
h (the class size) = 5.
5 - cf :
Using the formula, Median = / + Ff x h | we have
Median = 145+ (ase SoU) a .
18 - .
145 725 149.03 . .
TOG TN
So, the median height of the girls is 149.03 cm. -
This means that the height of about 50% of the girls is less than this height, and
50% are taller than this height... - °
Example 8 : The median of the following data is 525. Find the values of x and y, if the
total frequency is 100. | J .
0- 100 2
100 - 200 5
200 - 300 x
300 - 400 12
400 - 500 17
500 - 600 20
600 - 700 y
700 - 800 9
800 - 900 7
900 - 1000 4
Reprint 2025-26


STATISTICS 197

Now, that you have studied about all the three measures of central tendency, let
us discuss which measure would be best suited for a particular requirement.

The mean is the most frequently used measure of central tendency because it
takes into account all the observations, and lies between the extremes, i.c., the largest
and the smallest observations of the entire data. It also enables us to compare two or
more distributions. For example, by comparing the average (mean) results of students
of different schools of a particular examination, we can conclude which school has a
better performance. "

However, extreme values in the data affect the mean. For example, the mean of
classes having frequencies more or less the same is a good representative of the data.
But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20,
21, 18, then the mean will certainly not reflect the way the data behaves. So, in such
cases, the mean is not a good representative of the data. _ .

In problems where individual observations are not important, and we wish to find
out a 'typical' observation, the median is more appropriate, e.g., finding the typical
productivity rate of workers, average wage in a country, etc. These are situations
where extreme values may be there. So, rather than the mean, we take the median as
a better measure of central tendency. . _

aa
In situations which require establishing the most frequent value or most popular
~~
item, the mode is the best choice, e.g., to find the most popular T. V. programme being
watched, the consumer item in greatest demand, the colour of the vehicle used by
most of the people, etc.
Remarks : . .
1. There is a empirical relationship between the three measures of central tendency :
3 Median = Mode + 2 Mean
2. The median of grouped data with unequal class sizes can also be calculated. However,
we shall not discuss it here.
Reprint 2025-26


198 MATHEMATICS
EXERCISE 13.3
1. The following frequency distribution gives the monthly consumption of electricity of
68 consumers of a locality. Find the median, mean and mode of the data and compare
them.
65 - 85 4
85 - 105 5
105 - 125 13 |
125 - 145 20
145 - 165 14
165 - 185 8
185 - 205 4
2. Ifthe median of the distribution given below is 28.5, find the values of x and y.
0- 10 5
10- 20 x
20- 30 20
30 - 40 15
40 - 50 y
. 50- 60 5
ee
3. A life insurance agent found the following data for distribution of ages of 100 policy
holders. Calculate the median age, if policies are given only to persons having age 18
years onwards but less than 60 year.
Reprint 2025-26


STATISTICS
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89 ,
Below 50 92
Below 55 98
Below 60 100
4, The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and
the data obtained is represented in the following table :
118 - 126 3
127-135 5
136 - 144 9
145 - 153 12
154 - 162 5
163-171 4
. 172-180 2
Find the median length of the leaves.
(Hint : The data needs to be converted to continuous classes for finding the median,
since the formula assumes continuous classes. The classes then change to
117.5 - 126.5, 126.5 - 135.5, ..., 171.5 - 180.5.)
Reprint 2025-26


200 MATHEMATICS

5. The following table gives the distribution of the life time of 400 neon lamps :

Life time (in hours) Number of lamps

1500 - 2000 14

2000 - 2500 56

2500 - 3000 60

3000 - 3500 86
3500 - 4000 74

4000 - 4500 62
4500 - 5000 48

Find the median life time of a lamp. . ", ,

6. 100 surnames were randomly picked up from a local telephone directory and the
frequency distribution of the number of letters in the English alphabets in the surnames
was obtained as follows: se :

Number of letters 10-13 | 13-16 | 16-19
female |» |= fw ls].
Determine the median number of letters in the surnames. Find the mean number of
letters in the surnames? Also, find the modal size of the surnames.

~~ ae .

7. The distribution below gives the weights of 30 students of a class. Find the median
weight of the students.

Weight (in kg) 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75

woman] 2 [> [ele lols le

13.5 Summary
In this chapter, you have studied the following points:
1. The mean for grouped data can be found by :
Dfex,
(i) the direct method: x = Xi
Yi
= Lid;
(ii) the assumed mean method : x = a + Sp
Reprint 2025-26


STATISTICS 201
Zhu,
(iii) the step deviation method: X =a+ (2) xh,
with the assumption that the frequency of a class is centred at its mid-point, called its
class mark.
2. The mode for grouped data can be found by using the formula:
Mode = / (fhe
2h -fo- hr
where symbols have their usual meanings.
3. The cumulative frequency of a class is the frequency obtained by adding the frequencies
of all the classes preceding the given class. .
4. The median for grouped data is formed by using the formula: .
7 _ of Be
Median = 7 + 2 xh, _
'| Ff
where symbols have their usual meanings.
. A NOTETO THE READER
For calculating mode and median for grouped data, it should be
ensured that the class intervals are continuous before applying the
formulae. Same condition also apply for construction of an ogive.
Further, in case of ogives, the scale may not be the same on both the axes.
Reprint 2025-26


at
Se
eat
ant be
oll
[] ape a
1062CH15
The theory of probabilities and the theory of errors now constitute
a formidable body of great mathematical interest and of great
practical importance. - , °
Bo ~ -R.S. Woodward
14.1 Probability - A Theoretical Approach
Let us consider the following situation: - .
Suppose a coin is tossed at random,
When we speak of a coin, we assume it to be 'fair', that is, it is symmetrical so
that there is no reason for it to come down more often on one side than the other.
We call this property of the coin as being 'unbiased'. By the phrase 'random toss',
we mean that the coin is allowed to fall freely without any bias or interference.

We know, in advance, that the coin can only land in one of two possible ways -
either head up or tail up (we dismiss the possibility of its 'landing' on its edge, which
may be possible, for example, if it falls on sand). We can reasonably assume that each
outcome, head or tail, is as likely to occur as the other. We refer to this by saying that
the outcomes head and tail, are equally likely.

For another example of equally likely outcomes, suppose we throw a die
once. For us, a die will always mean a fair die. What are the possible outcomes?
They are 1, 2, 3, 4, 5, 6. Each number has the same possibility of showing up. So
the equally likely outcomes of throwing a die are 1, 2, 3, 4, 5 and 6.

Reprint 2025-26


PROBABILITY 203

Are the outcomes of every experiment equally likely? Let us see.

Suppose that a bag contains 4 red balls and 1 blue ball, and you draw a ball
without looking into the bag. What are the outcomes? Are the outcomes - a red ball
and a blue ball equally likely? Since there are 4 red balls and only one blue ball, you
would agree that you are more likely to get a red ball than a blue ball. So, the outcomes
(a red ball or a blue ball) are not equally likely. However, the outcome of drawing a
ball of any colour from the bag is equally likely. So, all experiments do not necessarily
have equally likely outcomes.

However, in this chapter, from now on, we will assume that all the experiments
have equally likely outcomes.

In Class IX, we defined the experimental or empirical probability P(E) of an
event E as ; :

P(E) Number of trials in which the event happened
7 Total number of trials = -

The empirical interpretation of probability can be applied to every event associated
with an experiment which can be repeated a large number of times. The requirement
of repeating an experiment has some limitations, as it may be very expensive or
unfeasible in many situations. Of course, it worked well in coin tossing or die throwing
experiments. But how about repeating the experiment of launching a satellite in order
to compute the empirical probability of its failure during launching, or the repetition of
the phenomenon of an earthquake to compute the empirical probability of a multistoreyed building getting destroyed in an earthquake?

% wo .

In experiments where we are prepared to make certain assumptions, the repetition
of an experiment can be avoided, as the assumptions help in directly calculating the
exact (theoretical) probability. The assumption of equally likely outcomes (which is
valid in many experiments, as in the two examples above, of a coin and of a die) is one
such assumption that leads us to the following definition of probability of an event.

The theoretical probability (also called classical probability) of an event E,
written as P(E), is defined as

P(E) Number of outcomes favourable to E
Se on - 0000000000
Number of all possible outcomes of the experiment
where we assume that the outcomes of the experiment are equally likely.
We will briefly refer to theoretical probability as probability.
This definition of probability was given by Pierre Simon Laplace in 1795.
Reprint 2025-26


204 MATHEMATICS

Probability theory had its origin in the 16th century when

an Italian physician and mathematician J.Cardan wrote the

first book on the subject, The Book on Games of Chance. 2 prt !

Since its inception, the study of probability has attracted 2 'f 7

the attention of great mathematicians. James Bernoulli ee

(1654 - 1705), A. de Moivre (1667 - 1754), and . { , |

Pierre Simon Laplace are among those who made significant y 1, y

contributions to this field. Laplace's Theorie Analytique clef Me

des Probabilités, 1812, is considered to be the greatest ee

contribution by a single person to the theory of probability. Pierre Simon Laplace

In recent years, probability has been used extensively in (1749 - 1827)

many areas such as biology, economics, genetics, physics, os

sociology etc. . °

Let us find the probability for some of the events associated with experiments
where the equally likely assumption holds. oe -
Example 1 : Find the probability of getting a head when a coin is tossed once. Also
find the probability of getting a tail. . .
Solution : In the experiment of tossing a coin once, the number of possible outcomes
is two - Head (H) and Tail (T). Let E be the event 'getting a head'. The number of
outcomes favourable to E, (.e., of getting a head) is 1. Therefore,
fo :
P(E) =P th ; _ Number of outcomes favourable toE 1
- a Number of all possible outcomes 2
Similarly, if Fis the event 'getting a tail', then
1
- P(F) = P(tail) = > (Why ?)

Example 2: A bag contains a red ball, a blue ball and a yellow ball, all the balls being
of the same size. Kritika takes out a ball from the bag without looking into it. What is
the probability that she takes out the

(i) yellow ball? (ii) red ball? (iti) blue ball?
Solution : Kritika takes out a ball from the bag without looking into it. So, it is equally
likely that she takes out any one of them.

Reprint 2025-26


PROBABILITY 205

Let Y be the event 'the ball taken out is yellow', B be the event 'the ball taken
out is blue', and R be the event 'the ball taken out is red'.
Now, the number of possible outcomes = 3,
(i) The number of outcomes favourable to the event Y = 1.
Ss PCY) *

0, =3
a . 1 - 1

Similarly, (ii) PCR) = 3 and (iii) P(B) = 3
Remarks :
1. An event having only one outcome of the experiment is called an elementary
event. In Example 1, both the events E and F are elementary events. Similarly, in
Example 2, all the three events, Y, B and R are elementary events. |
2. In Example 1, we note that : P(E) + P(F) = 1 a, .
In Example 2, we note that : P(Y) + PCR) + P(B) = 1 . .

Observe that the sum of the probabilities of all the elementary events of
an experiment is |. This is true in general also. -
Example 3 : Suppose we throw a die once. (i) What is the probability of getting a
number greater than 4 ? (ii) What is the probability of getting a number less than or
equal to 4? °
Solution : (i) Here, let E be the event 'getting a number greater than 4'. The number
of possible outcomes is six : 1,2, 3, 4, 5 and 6, and the outcomes favourable to E are 5
and 6, Therefore, the number of outcomes favourable to E is 2. So,

2 1
. P(E) = P(number greater than 4) = 673
(ii) Let F be the event ' getting a number less than or equal to 4'.
Number of possible outcomes = 6
Outcomes favourable to the event F are 1, 2, 3, 4.
So, the number of outcomes favourable to F is 4.
Therefe P(F 42
erefore, = 6-3
Reprint 2025-26


206 MATHEMATICS
Are the events E and F in the example above elementary events? No, they are
not because the event E has 2 outcomes and the event F has 4 outcomes.
Remarks : From Example 1, we note that
P(E) + P(F . + . 1 (1)
+ = 7tT=
(E) +P) = 5 +5
where E is the event 'getting a head' and F is the event 'getting a tail'.
From (i) and (ii) of Example 3, we also get
1 2 :
P(E) + P(F) = ~ +> =1 (2)
3 3
where E is the event 'getting a number >4' and F is the event 'getting a number < 4'.
Note that getting a number not greater than 4 is same as getting a number less
than or equal to 4, and vice versa. . a.
In (1) and (2) above, is F not the same as 'not E'? Yes, it is. We denote the event
'not E' by E. oe . .
So, P(E) + P(not E) = 1 :
1.¢., P(E) + P(E) = 1, which givesus P(E) =1-- P(E).
In general, it is true that for an event E,
PE)=1-PE)
_ | ']
The event E, representing 'not E', is called the complement of the event E.
We also say that E and E are complementary events.
Before proceeding further, let us try to find the answers to the following questions:
(i) What is the probability of getting a number 8 in a single throw of a die?
(ti) What is the probability of getting a number less than 7 in a single throw of a die?
Let us answer (i) : ,
We know that there are only six possible outcomes in a single throw of a die. These
outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8, so there is no
outcome favourable to 8, i.e., the number of such outcomes is zero. In other words,
getting 8 in a single throw of a die, is impossible.
: 0
So, P(getting 8) = 67 0
Reprint 2025-26


PROBABILITY 207

That is, the probability of an event which is impossible to occur is 0. Such an
event is called an impossible event.

Let us answer (ii) :

Since every face of a die is marked with a number less than 7, it is sure that we
will always get a number less than 7 when it is thrown once. So, the number of
favourable outcomes is the same as the number of all possible outcomes, which is 6.

6

Therefore, P(E) = P(getting a number less than 7) = 6 =1
So, the probability of an event which is sure (or certain) to occur is 1. Such an event
is called a sure event or a certain event. ,
Note : From the definition of the probability P(E), we see that the numerator (number
of outcomes favourable to the event E) is always less than or equal to the denominator
(the number of all possible outcomes). Therefore, oe

OSP(E)S1 -

Now, let us take an example related to playing cards. Have you seen a deck of
playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards eachspades (4), hearts (¥), diamonds (#) and clubs (#). Clubs and spades are of black
colour, while hearts and diamonds are of red colour. The cards in each suit are ace,
king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face
cards. "

oN
Example 4 : One card is drawn from a well-shuffled deck of 52 cards. Calculate the
probability that the card will _ .
(i) be an ace,
(ii) not be an ace.
Solution : Well-shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck. Let E be the event 'the card is an ace'.

The number of outcomes favourable to E = 4

The number of possible outcomes = 52 (Why ?)

Therek pe = 4-4

erefore, (E) = 52 13
(ii) Let F be the event 'card drawn is not an ace'.
The number of outcomes favourable to the event F = 52-4 =48 (Why?)
Reprint 2025-26


208 MATHEMATICS
The number of possible outcomes = 52
Theref P(F 8
erefore, (F) = 52. 13
Remark : Note that F is nothing but E. Therefore, we can also calculate P(F) as
follows: P(F) = P(E) = 1 -- P(E) = 1 1."
ollows: P(F) = P(E) = 1- P(E) = BB
Example 5 : Two players, Sangeeta and Reshma, play a tennis match. It is known
that the probability of Sangeeta winning the match is 0.62. What is the probability of
Reshma winning the match?
Solution ;: Let S and R denote the events that Sangeeta wins the match and Reshma
wins the match, respectively. , OO
The probability of Sangeeta's winning = P(S) = 0.62 (given) .
The probability of Reshma's winning = P(R) = 1 - P(S) -
~ [As the events R and S are complementary]
= 1-0.62 = 0.38
Example 6 : Savita and Hamida are friends, What is the probability that both will
have (i) different birthdays? (Gi) the same birthday? (ignoring a leap year).
pos
Solution ; Out of the two friends, one girl, say, Savita's birthday can be any day of the
year. Now, Hamida's birthday can also be any day of 365 days in the year.
We assume that these 365 outcomes are equally likely.
Gi) If Hamida's birthday is different from Savita's, the number of favourable outcomes
for her birthday is 365 - 1 = 364
tea -- 364
So, P (Hamida's birthday is different from Savita's birthday) = 365
(ii) P(Savita and Hamida have the same birthday)
: = 1--P (both have different birthdays)
: 364 _
= |--- [Using P(E) = 1 - P(E)]
365 g P(E)
I
365
Reprint 2025-26


PROBABILITY 209
Example 7 : There are 40 students in Class X of a school of whom 25 are girls and 15
are boys. The class teacher has to select one student as a class representative. She
writes the name of each student on a separate card, the cards being identical. Then
she puts cards in a bag and stirs them thoroughly. She then draws one card from the
bag. What is the probability that the name written on the card is the name of (i) a girl?
(ii) a boy?
Solution : There are 40 students, and only one name card has to be chosen.
Gi) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25 (Why?)
; ; ; a)
Therefore, P (card with name of a girl) = P(Girl) = 40 = 8
(i) The number of outcomes favourable for a card with the name of a boy = 15 (Why?)
15.3 |
Therefore, P(card with name of a boy) = P(Boy) = 40 = 8
Note ; We can also determine P(Boy), by taking : "
. : 5 3
P(Boy) = 1 - P(not Boy) = 1 - P(Girl) = 1- 8 = 8
Example 8: A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn
at random from the box, what is the probability that it will be
& * \
(i) white? (ii) blue? (iii) red?
Se" .
Solution : Saying that a marble is drawn at random is a short way of saying that all the
marbles are equally likely to be drawn. Therefore, the
number of possible outcomes = 3 +2+4=9 (Why?)
Let W denote the event 'the marble is white', B denote the event 'the marble is blue'
and R denote the event 'marble is red'.
(i) The number of outcomes favourable to the event W = 2
Ss ° PCW) 2
0, / = 9
a . 31 - 4
Similarly, Gi) P(B)= - ==> and (iii) P(R)= -
9 3 9
Note that P(W) + P(B) + P(R) = 1.
Reprint 2025-26


210 MATHEMATICS
Example 9 : Harpreet tosses two different coins simultaneously (say, one is of 71
and other of ~ 2), What is the probability that she gets at least one head?

Solution : We write H for 'head' and T for 'tail'. When two coins are tossed
simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all
equally likely. Here (H, H) means head up on the first coin (say on ® 1) and head up
on the second coin (¥ 2), Similarly (H, T) means head up on the first coin and tail up on
the second coin and so on.

The outcomes favourable to the event E, 'at least one head' are (H, H), (H, T)
and (T, H). (Why?) °
So, the number of outcomes favourable to E is 3.

3 a
Therefore, P(E) = 4 .
. 3 ;
i.e., the probability that Harpreet gets at least one head is 4°.
Note : You can also find P(E) as follows: _ -
_ . =. 1
P(E)= 1-P()=1- I = 3 (sinc P(E) = Pino head) = i)
4 4 | 4

Did you observe that in all the examples discussed so far, the number of possible
outcomes in each experiment was finite? If not, check it now.

There are many experiments in which the outcome is any number between two
given numbers, or in which the outcome is every point within a circle or rectangle, etc.
Can you now count the number of all possible outcomes? As you know, this is not
possible since there are infinitely many numbers between two given numbers, or there
are infinitely many points within a circle. So, the definition of (theoretical) probability
which you have learnt so far cannot be applied in the present form. What is the way
out? To answer this, let us consider the following example :

Example 10* : In a musical chair game, the person playing the music has been
advised to stop playing the music at any time within 2 minutes after she starts
playing. What is the probability that the music will stop within the first half-minute
after starting?
Solution : Here the possible outcomes are all the numbers between 0 and 2. This is
the portion of the number line from 0 to 2 (see Fig. 14.1).
tt
0 4 1 2
Fig. 14.1
* Not from the examination point of view.
Reprint 2025-26


PROBABILITY 211
Let E be the event that "the music is stopped within the first half-minute'.
The outcomes favourable to E are points on the number line from 0 to 5 .
1 1
The distance from 0 to 2 is 2, while the distance from 0 to 3 is 2°
Since all the outcomes are equally likely, we can argue that, of the total distance
1
of 2, the distance favourable to the event E is 2°
1
So, P(E) = Distance favourable to the event E _ 2_ lo. ;
Total distance in which outcomes canlie 2 4
Can we now extend the idea of Example 10 for finding the probability as the ratio of
the favourable area to the total area? . . ,
Example 11* : A missing helicopter is reported to have crashed somewhere in the
rectangular region shown in Fig. 14.2. What is the probability that it crashed inside the
lake shown in the figure? .
-__----- 6km ----->
i a a ee
& we 2 " | cya Ne ee Nee See =
Ww ° e ° Coreen =
+ a 2S, rt ye aN t
LMS AD NES
ET Nas Nhe NUN ES &
ite OL I
MEL ies SMS OLN: IN . |
A. 9g .- >
: Fig. 14.2
Solution : The helicopter is equally likely to crash anywhere in the region.
Area of the entire region where the helicopter can crash
= (4.5 x 9) km? = 40.5 km?
* Not from the examination point of view.
Reprint 2025-26


212 MATHEMATICS
Area of the lake = (2.5 x 3) km? = 7.5 km?
Therefore, P (helicopt hed in the lake) 1 2B 3
erefore, elicopter crashed in the lake) = 70> = 705 = 53
Example 12 : A carton consists of 100 shirts of which 88 are good, 8 have minor
defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which
are good, but Sujatha, another trader, will only reject the shirts which have major
defects. One shirt is drawn at random from the carton. What is the probability that
(i) itis acceptable to Jimmy?
(ii) it is acceptable to Sujatha? "
Solution : One shirt is drawn at random from the carton of 100 shirts. Therefore,
there are 100 equally likely outcomes. _ _ -
(i) The number of outcomes favourable (ie, acceptable) to Jimmy = 88 (Why?)
. '. 88
Therefore, P (shirt is acceptable to Jimmy) = 00 > 0.88
(i) The number of outcomes favourable to Sujatha =88+8=96 (Why?)
fo . 96
So, P (shirt is acceptable to Sujatha) = --~ = 0.96
ae ~ 100
Example 13 : Two dice, one blue and one grey, are thrown at the same time. Write
down all the possible outcomes. What is the probability that the sum of the two numbers
appearing on the top of the dice is
(i) 8? - Gi) 13? , (iii) less than or equal to 12?
Solution : When the blue die shows '1', the grey die could show any one of the
numbers 1, 2, 3, 4, 5, 6. The same is true when the blue die shows '2', '3', '4', 'S' or
'6'. The possible outcomes of the experiment are listed in the table below; the first
number in each ordered pair is the number appearing on the blue die and the second
number is that on the grey die.
Reprint 2025-26


PROBABILITY 213
1 2 3 4 5 6
1 G,I) (4,2) (13) (4) (5) (1,6
<4> 2 2,1) @2) (@,3) @4) (2D (@,6
3 B,D G2) G3) G.4-"G3,5), 6)
4 (4,1) (4,2) (4,374,4)-4,5) (4,6)
6 (6,1) ((6,2)6,3) (6,4) (6,5) (6,6)
Fig. 14.3 a
Note that the pair (1, 4) is different from (4, 1). (Why?) _
So, the number of possible outcomes = 6 x 6 = 36, °
(i) The outcomes favourable to the event 'the sum of the two numbers is 8' denoted
by E, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (see Fig. 14.3)
i.e., the number of outcomes favourable to E = 5.
H C) P(E) = -
ence, Ke (E) =. 36
(ii) As you can see from Fig. 14.3, there is no outcome favourable to the event F,
'the sum of two numbers is 13',
" 0
8 , P(F) = -=0
So, : (P= 36
(iii) As you can see from Fig. 14.3, all the outcomes are favourable to the event G,
'sum of two numbers < 12',
S . P(G) = 26. 1
0, ( ) ~ 36
Reprint 2025-26


214 MATHEMATICS
EXERCISE 14.1
1. Complete the following statements:
(i) Probability of an event E + Probability of the event 'notE' =.
(i) The probability of an event that cannot happen is ________ . Such an event is
called.
(iii) The probability of an event that is certain to happen is ___ . Such an event
is called .
(iv) The sum of the probabilities of all the elementary events of an experiment is
(v) The probability of an event is greater than or equal to _____ and less than or
equalto__ .
2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start. ~
(i) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(ui) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. Itis a boy ora girl. _

3. Why is tossing a coin considered to be a fair way of deciding which team should get the
ball at the beginning of a football game?

4. Which of the following cannot be the probability of an event?

» LN |
(A) 3 (B) -1.5 (C) 15% (D) 0.7

5. If P(E) = 0.05, what is the probability of 'not E'?

6. A bag contains lemon flavoured candies only. Malini takes out one candy without
looking into the bag. What is the probability that she takes out

(ij) an orange flavoured candy?
(i) a lemon flavoured candy?

7. It is given that in a group of 3 students, the probability of 2 students not having the
same birthday is 0.992. What is the probability that the 2 students have the same
birthday?

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag.
What is the probability that the ball drawn is (1) red? (ii) not red?

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken
out of the box at random. What is the probability that the marble taken out will be
(i)red? (ii) white? (iii) not green?

Reprint 2025-26