ICSE 2025 EXAMINATION

Sample Question Paper - 1

Mathematics

Time Allowed: 2 hours and 30 minutes Maximum Marks: 80

General Instructions:

Answers to this Paper must be written on the paper provided separately.

You will not be allowed to write during the first 15 minutes.

This time is to be spent reading the question paper.

The time given at the head of this Paper is the time allowed for writing the answers.

Attempt all questions from Section A and any four questions from Section B.

All work, including rough work, must be clearly shown and must be done on the same sheet as the rest of the

answers.
Omission of essential work will result in a loss of marks.

The intended marks for questions or parts of questions are given in brackets [ ]

Mathematical tables are provided.

Section A
1. Question 1 Choose the correct answers to the questions from the given options: [15]
(a) A retailer purchases a fan for ₹1500 from a wholesaler and sells it to a consumer at 10% profit. If the [1]
sales are intra-state and the rate of GST is 12%, the cost of the fan to the consumer inclusive of tax is:

a) ₹1848 b) ₹1830

c) ₹1650 d) ₹1800
(b) A factory kept increasing its output by the same percentage every year. Then, the percentage, if it is [1]
known that the output is doubled in the last two years, will be

a) 44.4% b) 14.4%

c) 41.4% d) 44.1%
(c) When ax3 + 6x2 + 4x + 5 is divided by (x + 3), the remainder is -7. [1]
The value of constant a is

a) 2 b) -2

c) -3 d) 3
(d) If , then the value of matrix A5 is [1]

a) b)
 c) d)

(e) An AP starts with a positive fraction and every alternate term is an integer. If the sum of the first 11 [1]
terms is 33, then the fourth term is

a) 3 b) 6

c) 5 d) 2
(f) If (4, 3) and (-4, -3) are opposite two vertices of a rectangle, then other two vertices are [1]

a) (4, -3) and (-4, 3) b) (-4, -3) and (-4, -3)

c) (-4, 4) and (-3, 4) d) (4, -3) and (-3, 4)

(g) Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting [1]
AC at L and AD produced at E. The values of EL and ar ( AEL) are respectively

a) ar ( CBL) and BL b) 2BL and 4 ar ( CBL)

c) 4 ar ( CBL) and 2BL d) BL and ar ( CBL)

(h) A sphere of radius a units is immersed completely in water contained in a right circular cone of semi- [1]
vertical angle 30° and water is drained off from the cone till its surface touches the sphere. Then, the
volume of water remaining in the cone will be

a) b) a3

c) d) 5 a3

(i) Graph the range of the inequation on the number line. If the solution [1]
set is consider as a diagonal of a square on the number line, then the area of obtained figure, is

a) 11 sq units b) 14 sq units

c) 17 sq units d) 18 sq units
(j) The probability that the minute hand lies from 5 to 15 min in the wall clock, is [1]

a) b)

c) d)
(k) If , then An (where, n is a natural number) is equal to [1]

a) b)

c) d) l2 2

(l) The sum of the squares of the distances of a moving point (x, y) from two fixed points (a, 0) and (- a, [1]
0) is equal to a constant quantity 2b2. The value of x2 + y2 + a2 is equal to

a) b2 b) -a2

c) ab d) -b2

(m) If P, Q, S and R are points on the circumference of a circle of radius r, such that PQR is an equilateral [1]
triangle and PS is a diameter of the circle. Then, the perimeter of the quadrilateral PQSR will be
 a) 2( + 1)r b)

c) 2r d)
(n) Observe the data given in three sets [1]
P: 3, 5, 9, 12, x, 7, 2
Q: 8, 2, 1, 5, 7, 9, 3
R: 5, 9, 8, 3, 2, 7, 1
If the ratio between P's and Q's means is 7 : 5, then the ratio between P’s and R’s means is

a) 7 : 5 b) 5 : 7

c) 6 : 7 d) 7 : 6
(o) Assertion (A): Sum of first 10 terms of the arithmetic progression -0.5, -1.0, -1.5, ... is 27.5 [1]
Reason (R): Sum of n terms of an A.P. is given as Sn = where a = first term, d =
common difference.

a) Both A and R are true and R is the b) Both A and R are true but R is not the
correct explanation of A. correct explanation of A.

c) A is true but R is false. d) A is false but R is true.

2. Question 2 [12]
(a) Mrs. chopra deposits ₹1600 per month in a Recurring Deposit Account at 9% per annum simple [4]
interest. If she gets ₹65592 at the time of maturity, then find the total time for which the account was
held.
(b) Find the mean proportional of (a4 - b4)2 and [(a2 - b2)(a - b)]-2. [4]
(c) If cosec =x+ , then prove that cosec + cot = 2x or . [4]
3. Question 3 [13]
(a) The given solid figure is cylinder surmounted by a cone. The diameter of the base of the cylinder is 6 [4]
cm. The height of the cone is 4 cm and the total height of the solid is 25 cm. Take = .

Find the:
i. Volume of the solid
ii. Curved surface area of the solid
Give your answer correct to the nearest whole number.
(b) The equation of a line is y = 3x - 5. Write down the slope of this line and the intercept made by its on [4]
the Y-axis. Hence or otherwise, write down the equation of a line, which is parallel to the line and
which passes through the point (0, 5).
(c) Use graph paper for this question (Take 2 cm = 1 unit along both x and y axis). ABCD is a [5]
quadrilateral whose vertices are A(2, 2), B(2, - 2), C (0, -1) and D (0, 1)
 i. Reflect quadrilateral ABCD on the y-axis and name it as A'B'CD.
ii. Write down the coordinates of A' and B'
iii. Name two points which are invariant under the above reflection.
iv. Name the polygon A'B'CD.
Section B
Attempt any 4 questions
4. Question 4 [10]
(a) The price of a Barbie Doll is ₹ 3136 inclusive tax (under GST) at the rate of 12% on its listed price. A [3]
buyer asks for a discount on the listed price, so that after charging GST, the selling price becomes
equal to the listed price. Find the amount of discount which the seller has to allow for the deal.
(b) Find the values of k, for which the equation x2 + 5kx + 16 = 0 has no real roots. [3]
(c) The mean of the following distribution is 49. Find the missing frequency a. [4]

Class Interval 0-20 20-40 40-60 60-80 80 -100

Frequency 15 20 30 a 10

5. Question 5 [10]
(a) Find the values of x, y, a and b, when . [3]

(b) Two chords AB and CD of a circle intersect each other at a point E inside the circle. If AB = 9 cm, [3]
AE = 4 cm and ED = 6 cm, then find CE.
(c) Determine, whether the polynomial g(x) = x - 7 is a factor of f(x) = x3 - 6x2 - 19x + 84 or not. [4]
6. Question 6 [10]
(a) Find the points of trisection of the line segment joining the points (5, -6) and (-7, 5). [3]
(b) Prove the following identities. [3]

i. sin4 + cos4 = 1 - 2 sin2 cos2

ii. =
(c) 150 workers were engaged to finish a job in a certain number of days, 4 workers dropped out on [4]
second day, 4 more workers dropped out an third day and so on. It took 8 more days of finish the
work. Find the number of days in which the work was completed.
7. Question 7 [10]
(a) A two-digit positive number, such that the product of its digits is 6. If 9 is added to the number, then [5]
the digits interchange their places. Find the number.
(b) The marks obtained by 120 students in a test are given below: [5]

Marks Number of Students

0 - 10 5

10 - 20 9

20 - 30 16

30 - 40 22

40 - 50 26

50 - 60 18
 60 - 70 11

70 - 80 6

80 - 90 4

90 - 100 3

Draw an ogive for the given distribution on a graph sheet.

(Use suitable scale for ogive to estimate the following)
i. the median.
ii. the number of students who obtained more than 75% marks in the test.
iii. the number of students who did not pass the test, if minimum marks required to pass is 40.
8. Question 8 [10]
(a) Two players Niharika and Shreya play a tennis match. It is known that the probability of Niharika [3]
winning the match is 0.62. What is the probability of Shreya winning the match?
(b) A conical military tent is 5 m high and the diameter of the base is 24 m. Find the cost of canvas used [3]
in making this tent at the rate of ₹ 14 per sq m.
(c) In the given figure CE is a tangent to the circle at point C. ABCD is a cyclic quadrilateral. If ABC = [4]
93o and DCE = 35o

find:
i.
ii.
iii.
9. Question 9 [10]
(a) Given: A = {x : 3 < 2x - 1 < 9, x R}, B = {x : 11 3x + 2 23, x R} where R is the set of real [3]
number.
i. Represents A and B on number lines
ii. On the number line also mark A B.
(b) Find the missing frequency for the given frequency distribution table, if the mean, of the distribution [3]
is 18.

Class interval 11-13 13-15 15-17 17-19 19-21 21-23 23-25

Frequency 3 6 9 13 f 5 4

(c) In the given figure, M= N = 46o. Express x in terms of a, b and c, where a, b and c are the lengths [4]
of LM, MN and NK, respectively.
10. Question 10 [10]
(a) The ages of A and B are in the ratio 7 : 8. Six years ago, their ages were in the ratio 5 : 6. Find their [3]
present ages.
(b) Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents [3]
to the circle and measure their lengths.
(c) The angle of elevation from a point P of the top of a tower QR, 50 m high is 60o and that the tower [4]

PT from a point Q is 30o. Find the height to the tower PT, correct to the nearest metre.
 Solution

Section A
1. Question 1 Choose the correct answers to the questions from the given options:
(i) (a) ₹1848
Explanation: {
Here, selling price of fan = ₹1650
GST on fan = 12% of ₹ 1650
= 1650
= 198
Thus, cost of a fan to the consumer inclusive of tax
= ₹(1650 + 198) = ₹1848
(ii) (c) 41.4%
Explanation: {
Let P be the initial production (2 yr ago) and the increase in production every year be x%. Then, production at the end
of first year = P + =
Production at the end of second year
=P= +

=P =P
Since, the production is doubled in last two years.
P = 2P =2

(100 + x)2 = 2 (100)2 10000 + x2 + 200x = 20000

On comparing it with ax2 + bx + c = 0, we get
a = 1, b = 200 and c = -10000

By quadratic formula, x =

x=
= -100 100 = 100(-1 4 )
= 100 (-1 + 1.414) [ percentage cannot be negative]
= 100(0.414) = 41.4
Hence, the required percentage is 41.4%.
(iii) (a) 2
Explanation: {
Let f(x) = ax3 + 6x2 + 4x + 5
By remainder theorem, f(-3) = -7
a(-3)3 + 6(-3)2 + 4(-3) + 5 = -7
-27a + 54 - 12 + 5 = -7
-27a = -54
a=2
(iv)
(c)

Explanation: {
Given,

Now, A2 = A A =
 A4 = A2 A2 =

Now, A5 = A4 A =

(v) (d) 2
Explanation: {
Given, S11 = 33
(2a + 10d) = 33 [ Sn = [2a + (n - 1)d]
a + 5d = 3
i.e. a6 = 3 a4 = 2 [ alternate terms are integers and the given sum is possible]

(vi) (a) (4, -3) and (-4, 3)

Explanation: {
Since, the image of point (4, 3) under X-axis is (4, - 3) and the image of point (4, 3) under Y-axis is (-4, 3).

Other two vertices of the rectangle are (4, -3) and (-4, 3).
(vii) (b) 2BL and 4 ar ( CBL)
Explanation: {
In BMC and EMD, we have

BMC = EMD [vertically opposite angles]

MC = MD [ M is the mid-point of CD]
MCB = MDE [alternate angles]
So, by AAS congruence criterion, we have

BC = ED [ corresponding parts of congruent triangles are equal]

In AEL and CBL, we have
ALE = CLB [vertically opposite angles]
and EAL = BCL [alternate angles]
So, by AA criterion of similarity, we have

[ if two triangles are similar, then their corresponding sides are proportional]
 On taking first two terms, we get

= 2 [ AD = SC as sides opposite to parallelogram and DE = BC, proved above]

EL = 2BL ...(i)
Now, [ ratio of areas of two similar triangles is equal to the square of the ratio of their
corresponding sides]
= (2)2 [from Eq. (i)]

=4
ar ( AEL) = 4 ar ( CBL)
(viii) (b) a3
Explanation: {
Let radius of sphere be a, i.e. OK = OA = a.
Then, the centre O of a sphere will be centroid of the BCD

OA = AB AB = 3(OA)
In right angled OKB,
sin 30o = =
=
OB = 2a
Now, AB = OA + OB = a + 2a = 3a
Now, in right angled BAC,
= tan 30o =
AC = =
AC = units
Now, volume of a cone BCD = (AC)2 AB
= =
Volume of water remaining in the cone = Volume of the cone BCD - Volume of a sphere
= 3 a3 - a3 = a3 cu units
(ix) (d) 18 sq units
Explanation: {
Given,

[multiplying by 3 in each term]

-8 3x + 1 10
-8 - 1 3x + 1 - 1 10 - 1 [subtracting 1 from each term]
-9 3 x 9
[dividing by 3 each term]
-3 x 3
Since, x R.
Range of x is [-3, 3].
 Representation of range of x on the number line is given as

Now, consider the following figure

Here, AC = 6 units, which is a diagonal of square.

Let side of a square ABCD be a.
In right angled ABC,
AC2 = AB2 + BC2
62 = a2 + a2
36 = 2a2
a2 = 18
Now, area of a square ABCD= (Side)2 = a2 = 18 sq units.
(x) (a)
Explanation: {
In a wall clock, the minute hand cover the 60 min in on complete round.
Total number of possible outcomes = 60
The minute hand cover the time from 5 to 15 min,
Number of outcomes favourable to E
= Distance from 5 to 15 min = 10
Required probability =
(xi)
(b)

Explanation: {
We have, = 3l

An = (3l)n = 3nln = 3nl [ ln = l, for all natural numbers n]

(xii) (a) b2
Explanation: {
Let P (x, y) be the moving point.
Let given two fixed points be A (a, 0) and B (-a, 0).
According to the given condition,
PA2 + PB2 = 2b2
(x - a)2 + (y - 0)2 + (x + a)2 + (y - 0)2 = 2b2 [by distance formula]
x2 - 2ax + a2 + y2 + x2 + 2ax + a2 + y2 = 2b2
2x2 + 2y2 + 2a2 = 2b2
x2 + y2 + a2 = b2 [dividing both sides by 2]
(xiii) (a) 2( + 1)r
Explanation: {
 As PQR is an equilateral triangle, hence PS will be perpendicular to QP and will divide it into 2 equal parts.
Since, P and S will be supplementary, so
S = 120o and QSA = RSA = 60o
Now, PA = PQ cos 30o and OA = OQ sin 30o =

AS = OA = and PA = PO + OA = r +

Hence, PQ =

In QAS, AS = QS cos 60o QS = =r

Since, AQ = AR, AS is common and QAS = RAS = 90o

So, QS = RS.
Perimeter of PQSP = 2(PQ + QS) = 2( + 1)r
(xiv) (a) 7 : 5
Explanation: {
Mean of the observations of sets
P=
Q= =5
and R = =5
Ratio of means of sets P and Q = 7 : 5 [given]
Let P’s mean = 7y and Q’s mean = 5y.
5y = 5
y=1
Q's mean = 5
Now, P's mean = 7y
=7 1 38 + x = 48 x = 11
Mean of P : Mean of R = 7y : 5 = 7 1:5=7:5
(xv) (a) Both A and R are true and R is the correct explanation of A.
Explanation: {
Both are correct. Reason is the correct reasoning for Assertion.
Assertion, S10 = [2(-0.5) + (10 - 1)(-0.5)]
= 5[-1 - 4.5]
= 5(-5.5) = 27.5
2. Question 2
(i) p = ₹ 1600/month
r = 9% p.a.
m.v. = 65,592
n=?
m.v. =
65,592 =
65,592 = 1600n + 6n(n + 1)
65,592 = 1600n + 6n2 + 6n
6n2 + 1606n - 65,592 = 0
 n=

n=
n=
n=

n = 36 months
n = 3 years
or
or n =
or n = -303.66 months rejected.
As 'n' is no. of months here. So can't be -ve.
(ii) Let the mean proportional between (a4 - b4)2 and [(a2 - b2)(a - b)]-2 be x.

(a4 - b4)2, x and [(a2 - b2)(a - b)]-2 are in continued proportion.

(a4 - b4)2 : x = x : [(a2 - b2)(a - b)]-2
x2 = (a4 - b4)2 [(a2 - b2)(a - b)]-2

(iii)Given, cosec =x+ ...(i)

We know that, cot2 = cosec2 -1
cot2 = - 1 [from Eq. (i)]

cot2 = [ (a + b)2 = a2 + b2 + 2ab]

= =

= = [ a2 + b2 - 2ab = (a - b)2]
cot = ...(ii)
or cot = ...(iii)
On adding Eqs. (i) and (ii), we get
cosec + cot = 2x
Now, adding Eqs. (i) and (iii), we get
cosec + cot =
Hence, cosec + cot = 2x or .
3. Question 3
(i) Given total height of the solid = 25 cm
Height of the cone (h2) = 4 cm
Diameter of the cylinder = 6 cm
Height of the cylinder (h1) = 25 - 4 = 21 cm
Radius of the cone = Radius of the cylinder = (r)
= = 3 cm
Slant height of cone =
=
=
= 5 cm.
 i. Volume of the solid = Volume of cylinder + Volume of cone
=
=

=
=
= 631.71 cm3 632 cm3. (Approx.)
ii. Curved surface area of the solid = C.S.A of cylinder + C.S.A. of cone
= 2 rh1 + rl
= r(2h1 + l)
= 3(2 21 + 5)
= 3 47 = 443.14 cm2
Curved surface area = 443 cm2 (Approx.).
(ii) Given eqn of line y = 3x - 5
Comprare with y = mx + c we get.
Slope (m) = 3 and
y-intercept (c) = -5
Now slope of the line parellel to the given line will be 3 and it passes through (0, 5).
Thus eqn of line will be
y - y1 = m(x - x1)
y - 5 = 3(x - 0)
y - 5 = 3x
Y= 3x + 5
(iii)

i.

ii. Co-ordinates of A' (-2, 2)

Co-ordinates of B' (-2, - 2)
iii. Two invariant points are C (0, -1) and D (0,1).
iv. A'B'CD is an isosceles Trapezium polygon.
Section B
4. Question 4
(i) Let the List price of doll be ₹ x.
Total Amount = x + 12% of x
=x+
=
ATQ = 3136
x=
x = 2800
List price of doll ₹ 2800.
 Now the reduced price of the doll
= ₹(2800 - y)
amount of GST on ₹ (2800 - y)
12% of (2800 - y)
=₹ (2800 - y)
Now the selling price of doll
= (2800 - y) +
= (2800 - y)
= (2800 - y)
According to given condition, selling price of doll = list price of doll (i.e 2800)
i.e = 2800
2800 - y =
2800 - y = 2500
2800 - 2500 = y
y = 300
Hence, amount of discount is ₹ 300.
(ii) Given equation is x2 + 5kx + 16 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 1, b = 5k and c = 16
Now, discriminant, D = b2 - 4ac
= (5k)2 - 4 1 16 = 25k2 - 64
Since, the given equation has no real roots.
D<0
25k2 - 64 < 0
<0 <0

k2 < <k<
(iii) Class Frequency (f) x f

0-20 15 10 150

20-40 20 30 600

40-60 30 50 1500

60-80 a 70 70a

80-100 10 90 900

= 75 + a = 3150 + 70a
Given mean = 49
= 49
or, = 49
3150 + 70a = 3675 + 49a
70a - 49a = 3675 - 3150
21a = 525
a = 25
5. Question 5
(i) We know that two matrices are said to be equal if each matrix has the same number of rows and same number of
columns. Corresponding elements within each matrix are equal.
Given:

x + y = 5 ...(i),
2x - 3y = -5 ...(ii)
 a - b = 3 ...(iii)
a + b = -1 ...(iv)
Solving eqn (i) and (ii)

y=3
Putting the value of y in eqn (i)
x+3=5
x=2
Again solving eqn (iii) and (iv)

a=1
Putting the value of a in eqn (iv)
1 + b = -1
b = -2
x = 2, y = 3, a = 1, b = -2
(ii)

Given, two chords AB and CD are intersect each other at point E.

AB = 9 cm
AE = 4 cm
ED = 6 cm
So, BE = AB - AE = 9 - 4 = 5
So, AE EB = DE CE
4 5 = 6 CE
CE = = 3.34
(iii)g(x) = 0
x - 7 = 0, x = 7
By factor theorem, g(x) will be a factor of f(x)
if f(7) = 0
Now, f(7) =
= 343 - 294 - 133 + 84
= 427 - 427 = 0
f(7) = 0, So, g(x) is a factor of f(x).
6. Question 6
(i) Let P and Q be the points of trisection of AB.

Given points be A(5, -6) and B(-7, 5)

P divides AB in the ratio 1 : 2
[By section formula, ]
the coordinate P are
= =

Q divides AB in the ratio 2 : 1 then the coordinates of Q are

 =

Hence, the points of trisection of AB are and .

(ii) i. LHS = sin4 + cos4 = 1 - 2 sin2 cos2

= (sin2 + cos2 )2 - 2 sin2 cos2 [ a2 + b2 = (a + b)2 - 2ab]
= 12 - 2 sin2 cos2 = 1 - 2sin2 cos2 [ sin2 A + cos2 A = 1]
= RHS
Hence proved.
ii. = is true
if = is true
i.e. if = is true
i.e. if = 2 cosec is true

i.e. if = 2 cosec is true. [ cosec2 - cot2 = 1]

which is true.
Hence proved.
(iii)Let total work be 1 and let total work completed in days.
Total work
work dream in 1 day =
Number of days to complete work

This is the work done by 150 workers

work done by 1 worker in one day =
Number of workers work done per worker in 1 day Total work done in 1 day

150

146

142
Given that, In this manner, it took 8 more days to finish the work i.e. work finished in (n + 8) days.
+ ... + (n + 8) terms = 1
[150 + 146 + 142 + ... + (n + 8) terms] = 1
150 + 146 + 142 + ... +(n + 8) terms = 150n
Now a = 150 d = 146 - 150
= -4
diff. is equal It forms A.P.
Sn = [2a + (n - 1)d]
150 + 146 + 142 +.. (n = 8) terms = 150n becomes.
[2(150) + (n + 8 - 1)] = 150n
[150 - 2(n + 7)] = 150n
(n + 8)(150 - 2n - 14) = 150n
(n + 8)(136 - 2n) = 150n
136n - 2n2 + 1088 - 16n = 150n
2n2 - 120n - 1088 + 150n = 0
2n2 + 30n - 1088 = 0
n2 + 15n - 544 = 0
n2 + 32n - 17n - 544 = 0
n(n + 32) - 17(n + 32) = 0
(n + 32)(n - 17) = 0
n + 32 = 0 or n = 17
n = -32 n = 17
 Reject n = -32 as n should be natural no.
n = 17 work was complete in 17 + 8 = 25 days.
7. Question 7
(i) Let the two digit no. be 10x + y
product of their digits
i.e., xy = 6
y = ...(i)
According to the question
10x + y + 9 = 10y + x
9x - 9y + 9 = 0
x - y = -1 ...(ii)
Substituting the value of y from equal (i),
= -1
= -1
x2 - 6 = -x
x2 + x - 6 = 0
x2 + 3x - 2x - 6 = 0
x(x + 3) - 2(x + 3) = 0
(x - 2)(x + 3) = 0
x = 2, x = - 3 (according to question rejected as digits are never negative)
put the value of x in eqn. (i)
y= =3
Thus, x = 2 and y = 3
Hence, the required no. = 10 2 + 3 = 23
(ii) C.I. f c.f

0 - 10 5 5

10 - 20 9 14

20 - 30 16 30

30 - 40 22 52

40 - 50 26 78

50 - 60 18 96

60 - 70 11 107

70 - 80 6 113

80 - 90 4 117

90 - 100 3 120

N = 120
 i. Given, n = 120 which is even
Median = th term
= 60th term
Median = 43
ii. The number of students who obtained more than 75% marks in test = 120 - 110 = 10.
8. Question 8
(i) Let E and F denote the events that Niharika and Shreya win the match, respectively. It is clear that, if Niharika wins the
match, then Shreya losses the match and if Shreya wins the match, then Niharika losses the match. Thus, E and F are
complementary events.
P(E) + P(F) = 1
Since, probability of Niharika ’s winning the match, i.e. P(E) = 0.62
Probability of Shreya’s winning the match,
P(F) = P (Niharika losses the match)
= 1 - P(E)
= 1 - 0.62 = 038
(ii) Given:
h = 5m, d = 24 m, r = 12 m

l = 13 m
Convas Required = C.S.A of conical tent

Convas Required = 490.28 m2

Total cost = 490.28 14
= ₹6814
Hence, total cost of convas used = ₹6814
(iii) i. (Opposite angles of cyclic quadrilateral)

= 87o
ii. (Alternate segment theorem)
 iii. In = 180o (Sum of internal angles of a triangle = 180°)

9. Question 9
(i) i. A = {x : 3 < 2x - 1 < 9, x R}
3 < 2x - 1 < 9
3 + 1 < 2x - 1 + 1 < 9 + 1
4 < 2x < 10

2<x<5
A = (2, 5) R

B = {x : 11 3x + 2 23, x R}
11 3x + 2 23
11 - 2 3x + 2 - 2 23 - 2
9 3x 21

3 x 7
B = [3, 7] R

ii. A B = (2, 5) [3, 7]

= [3, 5)

(ii) fi xi
Class Interval Required mid value

11 - 13 3 12 36

13 - 15 6 14 84

15 - 17 9 16 144

17 - 19 13 18 234

19 - 21 f 20 20f

21 - 23 5 22 110

23 - 25 4 24 96

= 704 + 20f

mean
18 =
720 + 18f = 704 + 20f
f=8
(iii)

Given: In the given figure.

LMN = PNK = 46o
 LM || PN (as corresponding angles are equal)
Now consider LMK and PNK
LMK = PNK (corresponding angles are equal)
LKM = PKN (common)
LMK PNK (AA similarity)

x=
Hence we get the result x =
10. Question 10
(i) Let the present age of A and B are 7x and 8x respectively.
Then, 6 years ago their ages are 7x - 6 and 8x - 6
So,
6(7x - 6) = 5(8x - 6)
42x - 36 = 40x - 30
42x - 40x = -30 + 36
2x = 6
x= =3
Hence, the present age of A and B are 21 and 24.
(ii) i. Draw a circle with radius 6 cm and centre C.
ii. Take a point P at 10 cm from centre and join CP.
iii. Draw perpendicular bisector of CP which cuts CP at O.
iv. Take O as centre and OC as radius draw a circle which cuts the previous circle at A and B.
v. Join PA and PB.
vi. PA and PB are required tangents.

(iii)We have, RPQ = 60o and PQT = 30o

and QR = 50 m
Let PT = x m and PQ = y m
In PQR,
tan 60o =

or y = ...(i)
In PQT,
tan 30o =

or x = ...(ii)
From eq. (i) and (ii), we get
x= = 16.66
= 17 m (correct to the nearest meter)