Class X Session 2025-26

Subject - Mathematics (Standard)

Sample Question Paper - 09

Time Allowed: 3 hours Maximum Marks: 80

General Instructions:

1. This question paper contains 38 questions.

2. This Question Paper is divided into 5 Sections A, B, C, D and E.

3. In Section A, Questions no. 1-18 are multiple choice questions (MCQs) and questions no. 19 and 20 are

AssertionReason based questions of 1 mark each.

4. In Section B, Questions no. 21-25 are very short answer (VSA) type questions, carrying 02 marks each.

5. In Section C, Questions no. 26-31 are short answer (SA) type questions, carrying 03 marks each.

6. In Section D, Questions no. 32-35 are long answer (LA) type questions, carrying 05 marks each.

7. In Section E, Questions no. 36-38 are case study based questions carrying 4 marks each with sub parts of the
values of 1, 1 and 2 marks each respectively.

8. All Questions are compulsory. However, an internal choice in 2 Question of Section B, 2 Questions of Section C

and 2 Questions of Section D has been provided. An internal choice has been provided in all the 2 marks questions

of Section E .

9. Draw neat and clean figures wherever required.

10. Take wherever required if not stated.

11. Use of calculators is not allowed.

Section A
1. The HCF of 95 and 152, is [1]

a) 57 b) 19

c) 1 d) 38
2. If a is rational and √b is irrational, then a + √b is: [1]

a) a rational number b) a natural number

c) an irrational number d) an integer

–
3. If the sum and the product of zeroes of a quadratic polynomial are 2√3 and 3 respectively, then a quadratic [1]
polynomial is:
– – 2
a) x 2
− 2√3x − 3 b) (x − √3)

– –
c) x 2
+ 2√3x + 3 d) x
2
+ 2√3x − 3

4. If α and β are the zeroes of the polynomial ax2 - 5x + c and α + β = αβ = 10, then: [1]

Page 1 of 20
 a) a = 5, c = 1

2
b) a = 1

2
,c=5

c) a = 5

2
,c=1 d) a = 1, c = 5

5. In △ABC, if ∠ C = 3∠ B = 2(∠ A + ∠ B), then ∠ C = [1]

a) 120o b) 60o

c) 150o d) 90o

6. The value of a so that the point (3, a) lies on the line represented by 2x - 3y = 5 is [1]
−1
a) b)
1

3 3

c) – 1 d) 1

7. The roots of the quadratic equation 9a2b2x2 - 16abcdx - 25c2d2 = 0 are [1]

−25cd −cd
a) and b) and
25cd cd

9ab ab 9ab ab

−25cd −cd
c) and d) and
cd 25cd

9ab ab 9ab ab

8. The discriminant of the quadratic equation 2x2 + x - 1 = 0 is: [1]

a) 7 b) 9

c) -9 d) -7

9. If 9th term of an A.P. is zero, then its 29th term is ________ its 19th term. [1]

a) Equal to b) Half of

c) Twice of d) Thrice of
10. What is the common difference of an AP in which a18 - a14 = 32? [1]

a) -8 b) 4

c) -4 d) 8
11. In △ABC, D and E are points on side AB and AC respectively such that DE || BC and AD : DB = 3 : 1. If EA = [1]
3.3 cm, then AC =

a) 5.5 cm b) 4.4 cm

c) 4 cm d) 1.1 cm
12. The diameter of a circle is of length 6 cm. If one end of the diameter is (-4, 0), the other end on x-axis is at: [1]

a) (2, 0) b) (6, 0)

c) (4, 0) d) (0, 2)

13. cot2θ -
1
is equal to: [1]
2
sin θ

a) -1 b) 2

c) 1 d) -2
14. If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to [1]
–
a) √3 b) 1

√2

c) 1 d) 1

15. A circle is inscribed in a quadrilateral ABCD in which ∠B = 90

o
, if AD = 23 cm, AB = 29 cm and DS = 5 cm, [1]

Page 2 of 20
 then radius of circle is :

a) 13 cm b) 12 cm

c) 11 cm d) 14 cm
16. A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into the water until it [1]
is completely immersed. The water level in the vessel will rise by

a) 4

9
cm b) 9

4
cm

c) 2

9
cm d) 9

2
cm

17. Consider the following frequency distribution: [1]

Class 0-5 6-11 12-17 18-23 24-29

Frequency 13 10 15 8 11

The upper limit of the median class is

a) 17.5 b) 17

c) 18 d) 18.5
18. Which of the following is not probability of an event? [1]

a) 1

13
% b) 0.89
1

c) 0.89 d) 52%
19. Assertion (A): D and E are points on the sides AB and AC respectively of a △ABC such that DE||BC then the [1]
value of x is 11, when AD = 4cm, DB = (x - 4)cm, AE = 8cm and EC = (3x - 19)cm.
Reason (R): If a line divides any two sides of a triangle in the same ratio then it is parallel to the third side.

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

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

20. Assertion (A): The point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in the ratio 2 : 7 [1]
internally.
Reason (R): Three points A, B and C are collinear if area of △ABC = 0.

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

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

Section B
21. Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468. [2]
22. The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator [2]
and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.
23. If the mid-point of the line segment joining the points A (3, 4) and B (k, 6) is P (x, y) and x + y - 10 = 0, find the [2]

Page 3 of 20
 value of k.
OR
Write the ratio in which the line segment doming the points A (3, - 6) and B (5, 3) is divided by X-axis.
2

24. Prove that:

sec θ+tan θ
= (
1+sin θ
)
[2]
sec θ−tan θ cos θ

OR
1+cos θ
Prove the trigonometric identity : sin θ

1+cos θ
+
sin θ
= 2 cosec θ .
25. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice [2]
is 7?
Section C
26. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages [3]
to save ₹ 2000 per month, then find their monthly incomes.
OR
Draw the graph of the pair of equations 2x + y = 4 and 2x – y = 4. Write the vertices of the triangle formed by these
lines and the y-axis. Also find the area of this triangle.
27. Find the value of p for which the points (-1, 3), (2, p) and (5, -1) are collinear. [3]

28. If
cos α
= m and
cos α
= n show that (m2 + n2) cos2β= n2 [3]
cos β sin β

29. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and [3]
median PM of △PQR (see figure). Show that △ABC ∼ △P QR .

OR
In Fig. QA and PB are perpendiculars to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm. Find AQ.

30. A square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. [3]
[Use π = 3.14]

31. Prove that the tangents drawn from an external point to a circle are equal in length. [3]
Section D
32. A journey of 192 km from a town A to town B takes 2 hours more by an ordinary passenger train than a super [5]
fast train. If the speed of the faster train is 16 km/h more, find the speed of the faster and the passenger train.
OR

Page 4 of 20
 Solve: x−1

2x+1
+
2x+1

x−1
= 2, x ≠ −
1

2
,1

33. In an A.P., the nth term is 1

and the mth term is 1
. Find (i) (mn)th term, (ii) sum of first (mn) terms. [5]
m n

34. From the top of a building 15 m high, the angle of elevation of the top of a tower is found to be 30o. Form the [5]

bottom of the same building, the angle of elevation of the top of the tower is found to be 45o. Determine the
height of the tower and the distance between the tower and the building.
OR
A parachutist is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100 m
apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of
the point where he falls on the ground from the just observation point.
35. The table below gives the percentage distribution of female teachers in the primary schools of rural areas of [5]
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 female teachers 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 75 - 85

Number of states/U.T. 6 11 7 4 4 2 1

Section E
36. Read the following text carefully and answer the questions that follow: [4]
Two friends Govind and Pawan decided to go for a trekking. During summer vacation, they went to Panchmarhi.
While trekking they observed that the trekking path is in the shape of a parabola. The mathematical
representation of the track is shown in the graph.

i. What are the zeroes of the polynomial whose graph is given? (1)
ii. What will be the expression of the given polynomial p(x)? (1)
iii. What is the product of the zeroes of the polynomial which represents the parabola? (2)
OR
In the standard form of quadratic polynomial, ax2 + bx + c, whar are a, b, and c? (2)
37. Read the following text carefully and answer the questions that follow: [4]
Singing bowls (hemispherical in shape) are commonly used in sound healing practices. Mallet (cylindrical in
shape) is used to strike the bowl in a sequence to produce sound and vibration.

Page 5 of 20
 One such bowl is shown here whose dimensions are:
Hemispherical bowl has outer radius 6 cm and inner radius 5 cm.
Mallet has height of 10 cm and radius 2 cm.
i. What is the volume of the material used in making the mallet? (1)
ii. The bowl is to be polished from inside. Find the inner surface area of the bowl. (1)
iii. Find the volume of metal used to make the bowl. (2)
OR
Find total surface area of the mallet. (Use π = 3.14) (2)
38. Read the following text carefully and answer the questions that follow: [4]
To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a
class library. There are two sections- section A and section B of grade X. There are 32 students in section A and
36 students in section B.

i. What is the minimum number of books you will acquire for the class library, so that they can be distributed
equally among students of Section A or Section B? (1)
ii. If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF
(32, 36) is (1)
iii. 7 × 11 × 13 × 15 + 15 is a (2)
OR

If p and q are positive integers such that p = ab2 and q = a2b, where a, b are prime numbers, then the LCM
(p, q) is (2)

Page 6 of 20
 Solution

Section A

1.
(b) 19
Explanation:
Using the factor tree for 95, we have:

Using the factor tree for 152, we have:

Therefore,
95 = 5 × 19

3
152 = 2 × 19

HCF (95, 152) = 19

2.
(c) an irrational number
Explanation:
Let a be rational and √b is irrational.
If possible let a + √b be rational.
Then a + √b is rational and a is rational.
⇒ [(a + √b) − a] is rational [Difference of two rationals is rational]

⇒ √b is rational.

This contradicts the fact that √b is irrational.

The contradiction arises by assuming that a + √b is rational.
Therefore, a + √b is irrational.

3.
–
(b) (x − √3) 2

Explanation:
–
α + β = 2√3

αβ = 3

required quad. poly is

2
x − (α + β)x + αβ = 0

2 –
x − (2√3)x + 3 = 0

2 –
x − 2√3x + 3 = 0
– 2
(x − √3) = 0

4.
(b) a = 1

2
,c=5

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 Explanation:
P(x) = x2 - (sum of roots)x + (product of roots)
x2 - (10)x + 10 = 0
x2 - 5x + 5 = 0 ...(1)
1

Given, ax2 - 5x + c = 0 ...(2)

comparing (1) and (2), we get,
a= 1

2
,c=5

5. (a) 120o
Explanation:
Since ∠ A + ∠ B + ∠ C = 180o ... (i)
∠ C = 3∠ B = 2(∠ A+∠ B)

3∠ B = 2(∠ A+∠ B)
3∠ B - 2∠ B = 2∠ A
∠ B = 2∠ A

∠A = ∠
B

from (i),
∠
B

2
+ ∠ B + 3∠ B = 180o
9∠ = 180o
B

∠ B = 40o
∠ C = 3∠ B

∠ C = 3 × 40 = 120o
6.
(b) 1

Explanation:
2x - 3y = 5
⇒ 2 × 3 - 3 × a = 5

⇒ 6 - 3a = 5
⇒ a =
1

7.
25cd −cd
(d) 9ab
and ab

Explanation:
Using factorisation method
9a2b2x2 - 16abcdx - 25c2d2 = 0
⇒ 9a2b2x2 - 25abcdx + 9abcdx - 25c2d2 = 0
⇒ abx(9abx - 25cd) + cd(9abx - 25cd) = 0

⇒ (abx + cd)(9abx - 25cd) = 0

⇒ abx + cd = 0 and 9abx - 25cd = 0

−cd
⇒ x= ab
and x = 25cd

9ab

8.
(b) 9
Explanation:
D = b2 - 4ac
Here, a = 2, b = 1 and c = -1
b2 - 4ac = (1)2 - 4(2)(-1)
=1+8
=9

Page 8 of 20
 9.
(c) Twice of
Explanation:
Let 1st term of A.P. be a and common difference be d.
Now, a9 = 0 ⇒ a + 8d = 0 ⇒ a = -8d ...(i)
Now, a29 = a + 28d = -8d + 28d ...(ii)
⇒ a29 = 20d
Also, a19 = a + 18d = -8d + 18d = 10d
⇒ 2 × a19 = 2 × 10d = 20d ...(iii)
From (ii) and (iii), we have
a29 = 2 a19

10.
(d) 8
Explanation:
a18 - a14 = 32
(18 - 14)d = 32
⇒ 4d = 32

⇒ d = 8.

11.
(b) 4.4 cm
Explanation:
In ΔABC, DE || BC
AD : DB = 3 : 1, EA = 3.3 cm

Let EC = x
∵ In △ABC, DE || BC
AD AE 3 3.3
= ⇒ =
DB EC 1 x

3.3
⇒ x = = 1.1cm
3

∴ AC = AE + EC = 3.3 + 1.1 = 4.4 cm

12. (a) (2, 0)

Explanation:
Let the other point on x axis is (x, 0)
By distance formula
−−−−−−−−−−−−−− −
√(x + 4)2 + (0 − 0)2 = (6)

x+4=6
x=2
Hence the point is (2, 0)
13. (a) -1
Explanation:

Page 9 of 20
 Given: cot2θ - 1

2
sin θ
2

= cos

2
θ
−
1

2
sin θ sin θ
2
cos θ−1
= 2
sin θ
2
− sin θ
= 2
= -1
sin θ

[∵ 1 - cos2θ = sin2θ]
14.
(c) 1
Explanation:
We have, x tan 45° cos 60° = sin 60° cot 60°
1 √3 1 x 1
⇒ x × 1 × = × ⇒ =
2 2 √3 2 2

1
⇒ x = × 2 = 1
2

15.
(c) 11 cm
Explanation:

Here DS = DR = 5 cm
⇒ AS = 23 – 5 = 18 cm

And AS = AP = 18 cm
⇒ BP = 29 – 18 = 11 cm

∵ OP⊥ AB and OQ ⊥ BC

∴∠ OQB = ∠ OPB = 90 and ∠ B = 90 ∘ ∘

Also ∠ POQ = 90 ∘

Therefore, OPBQ is a square.

∴ BQ = OQ = 11 cm

Therefore Radius of circle = 11 cm

16.
(b) 9

4
cm

Explanation:
Radius of sphere (r1) = 3 cm
∴ Volume = 4

3
πr
3
=
4

3
π(3) cm
3 3

= 36π cm3
∴ Volume of water in the cylinder = 36π cm3
Radius of cylindrical vessel (r2) = 4 cm
Let h be its height, then
2 2
πr h = 36π ⇒ π(4) h = 36π
2
36π 9
⇒ 16πh = 36π ⇒ h = = cm
16π 4

17. (a) 17.5

Explanation:
Given, classes are not continuous, so we make continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of
each class.

Page 10 of 20
 Class Frequency Cumulative frequency

-0.5-5.5 13 13

5.5-11.5 10 23

11.5-17.5 15 38

17.5-23.5 8 46

23.5-29.5 11 57
Here, N

2
=
57

2
= 28.5 , which lies in the interval 11.5 - 17.5.
Hence, the upper limit is 17.5.
18.
(b) 0.89
1

Explanation:
1

0.89

Probability of an event can never be greater than 1.

19.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:
If a line divides any two sides of a triangle in the same ratio then it is parallel to the third side. This is the Converse of the Basic
Proportionality theorem.
So, the Reason is correct.
8
By Basic Proportionality theorem, we have AD

DB
=
AE

EC
⇒
4

x−4
=
3x−19

⇒ 4(3x - 19) = 8(x - 4)

⇒ 12x - 76 = 8x - 32
⇒ 4x = 44 ⇒ x = 11 cm

So, Assertion is correct.

But reason (R) is not the correct explanation of assertion (A).

20.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:
Using section formula, we have
k×6+1×(−3)
-1 = k+1

-k - 1 = 6k - 3
7k = 2
k= 2

Ratio is 2 : 7 internally.
Also, if ar(△ABC) = 0
A, B and C all these points are collinear.

Section B
21. The smallest number divisible by 520 and 468 = LCM(520,468)
Prime factors of 520 and 468 are :
520 = 23 × 5 × 13
468 = 2 × 2 × 3 × 3 × 13
Hence LCM(520,468) = 23 × 32 × 5 × 13 = 8 × 9× 5× 13 = 4680

Now the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
=LCM(520,468)-17
=4680-17

Page 11 of 20
 =4663
22. Let the numerator and denominator of fraction be x and y respectively.
Then, the fraction is . x

As per first condition

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator.
x + y = 2x + 4
⇒ -x + y = 4........(i)
According to the second condition,
If the numerator and denominator are increased by 3, they are in the ratio 2 : 3.
x+3 2
=
y+3 3

⇒ 3x + 9 = 2y + 6
⇒ 3x - 2y = -3.......(ii)
Multiply (i) by -2, we get
-2x + 2y = 8 ......... (iii)
Adding (ii) and (iii) , we get
and 3x - 2x = -3 + 8
⇒ x=5
Substituting x = 5 in (i), we get
5-y=4
y=9
Hence, the required fraction is 5

9
3+k 4+6
23. Mid-point of the line segment joining A(3, 4) and B(k, 6) = 2
,
2
3+k
= 2
,5

3+k
Then, 2
,5 = (x, y)
3+k
Therefore, 2
= x and 5 = y
Since x + y - 10 = 0, we have
3+k

2
+ 5 - 10 = 0
i.e., 3 + k = 10
Therefore, k = 7
OR
The point lies on x-axis
Its ordinate will be = 0
Let the point P(x, 0) divides the line-segment joining the points A(3, -6) and B(5, 3) in the ratio m:n.
m y2 +n y1 m×3+n(−6)
∴ 0 = ⇒ 0 =
m+n m+n

⇒ 3m − 6n = 0 ⇒ 3m = 6n
m 6 2
⇒ = =
n 3 1

∴ Ratio = 2:1
24. L. H. S. = sec θ+tan θ

sec θ−tan θ

Multiplying and dividing by (sec θ + tan θ)

sec θ+tan θ sec θ+tan θ
= ×
sec θ−tan θ sec θ+tan θ
2
(sec θ+tan θ)
2 2
= [∵ (a + b)(a − b) = a − b ]
2 2
sec θ− tan θ
2
(sec θ+tan θ)
2 2
= [∵ sec θ − tan θ = 1]
1
2 2
1 sin θ 1+sin θ
= ( + ) = ( )
cos θ cos θ cos θ

∴ L.H.S. = R. H. S.

OR
1+cos θ
LHS = sin θ

1+cos θ
+
sin θ

(1+cos θ)×(1+cos θ)
= sin θ×sin θ
+
(1+cos θ)×sin θ sin θ×(1+cos θ)
2 2
sin θ+1+2 cos θ+ cos θ
= sin θ×(1+cos θ)

Using the Pythagorean identity cos2 θ + sin2 θ = 1

Page 12 of 20
 2+2 cos θ
=
sin θ×(1+cos θ)

2(1+cos θ)
= sin θ×(1+cos θ)

= 2

sinθ

Now 1

sinθ
= cosec θ
= 2 cosec θ.
25. Two dice are thrown simultaneously. [given]
So, that number of possible outcomes = 36
Sum of the numbers appearing on the dice is 7.
So, the possible ways are (1, 6), (2,5), (3, 4), (4, 3), (5, 2) and (6, 1).
Number of possible ways = 6
∴ Required probability =
6 1
=
36 6

Section C
26. Let us denote the incomes of the two-person by ₹ 9x and ₹ 7x and their expenditures by ₹ 4y and ₹ 3y respectively.
Then the equations formed in the situation is given by :
9x – 4y = 2000 ...(i)
and 7x – 3y = 2000 ...(2)
Step 1: Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of y equal. Then, we get the equations:
27x – 12y = 6000 ...(3)
28x – 12y = 8000 ...(4)
Step 2: Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients of y are the same. So, we get
(28x – 27x) – (12y – 12y) = 8000 – 6000
i.e., x = 2000
Step 3: Substituting this value of x in (1), we get
9(2000) – 4y = 2000
i.e., y = 4000
So, the solution of the equations is x = 2000, y = 4000. Therefore, the monthly incomes of the persons are ₹18,000 and ₹14,000
respectively.
OR
The given pair of linear equations
2x + y = 4
and 2x-y = 4
Table for line 2x + y = 4
x 0 2

y 4 0

Points A B
and table for line 2x - y = 4
x 0 2

y -4 0

Points C B

Page 13 of 20
 So the Graphical representation of both lines is as above.
Here, both lines and Y - axis from a △ABC .
Hence, the vertices of a △ABC are A (0,4), B(2,0) and C(0,- 4) where A and C are obtained by putting x = 0 in the given
equations abd B is obtained by solving them together.
∴ Required area of △ABC = 2 × Area of △AOB
△ABC = 2 × ( × 4 × 2) = 8 sq. units.
1

Hence, the required area of the triangle is 8 sq units.

27. Let A → (–1, 3) B → (2, p) and C → (5, –1)

If the points A, B and C are collinear then let B divide AC in the ratio K : 1 internally.
(K)(5)+(1)(−1) (K)(−1)+(1)(3)
Then, B → { K+1
,
K+1
}

5K−1 −K+3
⇒ B → ( , )
K+1 K+1

But, B is given to be (2,p)

5K−1
∴ = 2
K+1

⇒ 5K - 1 = 2( K + 1)
⇒ 5K - 1 = 2K + 2
⇒ 5K - 2K = 2 + 1

⇒ 3K = 3
3 −K+3
⇒ K =
3
= 1 and K+1
= p

−1+3
⇒ = p
1+1

⇒ p=1
Hence, the required value of p is 1.
cos α cos α
28. Given, m = ......(1) and, n =
cos β sin β
.......(2)

LHS = (m2 +n2) cos2β

2 2

⇒ LHS = (
cos

2
α
+
cos

2
α
) cos
2
β [ from (1) & (2) ]
cos β sin β
2 2 2 2
cos α sin β + cos α cos β
2
⇒ LHS = ( ) cos β
2 2
cos β sin β
2 2
sin β + cos β
2 2
⇒ LHS = cos α( ) cos β
2
cos 2 β sin β

⇒ LHS = cos
2
α(
2
1

2
) cos
2
β [ Since, sin2A + cos2A =1]
cos β sin β

2 2
cos α cos α
⇒ L. H. S. = = ( )
2 sin β
sin β

∴ L. H. S. = n [ from (2) ]
2

= R.H.S . Hence, Proved.

Page 14 of 20
29.

It is given that:
AB BC AD
= =
PQ QR PM

1
BC
AB AD BC BD
....................................(i)
2
⇒ = = = =
PQ PM QR 1 QM
QR
2

In △ABD and △P QM , we have

AB

PQ
= = [from(i)]
AD

PM
BD

QM

∴ △ABD ∼ △P QM [by SSS-similarity criteria].

And also, ∠B = ∠Q [corresponding angles of similar triangles are equal].

Now, in △ABC and △P QR , we have
∠B = ∠Q [proved above]

and AB
=
PQ
[from(i)]. BD

QM

∴ △ABC ∼ △P QR [by SAS-similarity criteria].

OR
In triangles AOQ and BOP, we have

∠OAQ = ∠ OBP [Each equal to 90°]

∠AOQ = ∠ BOP [Vertically opposite angles]
Therefore, by AA- criterion of similarity, we obtain
ΔAOQ ∼ ΔBOP

AO OQ AQ
⇒ = =
BO OP BP

AO AQ
⇒ =
BO BP

10 AQ
⇒ =
6 9
10×9
⇒ AQ = = 15cm
6

30.

−−−− − −−−−−
2 2
OB = √O A + AB
−−− −−−−−
2 2
= √20 + 20
−−−−−− −−
= √400 + 400
−−−
= √800
−−−−−−
= √400 × 2
– –
OB = 20√2 cm or, radius = 20√2
Area of shaded region = Area of sector OQBPO - Area of square OABC
90
∘ – – 2
= ∘
× 3.14 × 20√2 ⋅ 20√2 − (20)
360
1
= × 3 ⋅ 14 × 800 − 400
4

= 2(314) − 400

Page 15 of 20
 = 628 − 400

= 228

Required Area = 228 cm2.

31. The attached figure shows two tangents, SK and SR drawn to circle with center O from an external point K.

To prove that: SK = RK
Proof:
Normal and tangent at a point on the circle are perpendicular to each other.
∠ OSK = ∠ ORK = 90o
Using Pythagoras Theorem,
OK2 = OS2 + SK2 ...(i)
OK2 = OR2 + RK2 ...(ii)
Subtracting (ii) from (i),
OK2 - OK2 = OS2 + SK2 - OR2 - RK2
⇒ SK2 = RK2 ∵ OS = OR
SK = RK
Section D
32.

Let speed of passenger train be x km/h

∴ speed of superfast train = (x + 16) km/h

By question, T passenger = and T superfast 192

x
=
192

(x+16)

or, 192

x
−
192

x+16
= 2

or, 192(x + 16) − 192x = 2 (x 2

+ 16x)

or, 192x + 192 × 16 − 192x = 2 (x + 16x) 2

192x + 3072 - 192x = 2 (x + 16x) ( divide throughout by 2, we get,

2

96x + 1536 - 96x = (x 2

+ 16x)

or x(x + 48) - 32(x + 48) = 0

or, (x - 32) (x + 48) = 0
or, x = 32 or - 48
Since speed can't be negative, therefore - 48 is not possible.
∴ Speed of passenger train = 32 km/h and Speed of fast train = 48 km/h

OR
Given
x−1 2x+1
+ = 0
2x+1 x−1
x−1 2x+1
Let 2x+1
be y so x−1
=
1

∴ Substituting this value

2
y +1
y +
1

y
= 2 or y
= 2

or y 2
+ 1 = 2y

or y − 2y + 1 = 0
2

or (y − 1) = 0 2

Putting y = ,
x−1

2x+1

x−1

2x+1
= 1 or x − 1 = 2x + 1
or x = -2

Page 16 of 20
33. an = 1

a + (n - 1)d = 1

1
am =
n

a + (m - 1)d = 1

On solving,
a= 1

mn

d= 1

mn

i. amn = 1

mn
+ (mn − 1) ×
1

mn

1+mn−1
= mn
=1
mn 1
ii. Smn = 2
(
mn
+ 1)

1+mn
= 2

34. According to question it is given that a building AB of height 15 m and tower CD of h meter respectively.

Angle of elevation ∠ DAE = 30o

Angle of elevation ∠ DBC = 45o
To find: BC and CD
Proof: In right △DEA, using Pythagoras theorem
DE

x
= tan 30
∘
( ∵ AE = BC = x)
h−15 1
⇒ =
x √3

⇒ h − 15 =
x
.......(i)
√3

In right △DCB, by Pythagoras theorem

h ∘
= tan 45
x

⇒ h = x .....(ii)

Putting the value of h in equation (i)

x
x-15 = [from (i)]
√3

x
⇒ 15 = x −
√3

√3
⇒ 15 = (1 − )x
3

3− √3
⇒ 15 = ( )x
3

⇒ 45 = (3 − 1.732)x
45
⇒ = x
1.268

x = 35.49
Thus h = 35.49 m
OR

Page 17 of 20
 Let PC be the height h of the parachutist and makes an angle of elevations between 45° and 60°respectively at two observing
points 100m apart from each other
Let AB = 100
Let distance (BC) = x m
In ΔP BC
o PC
tan 60 =
BC
– h
⇒ √3 =
x

⇒ x =
h
...(i)
√3

In ΔP AC
o PC
tan 45 =
AC
h
⇒ 1 =
100+x

⇒ h = 100 + x

⇒ h = 100 +
h
[From (1)]
√3

– – –
⇒ √3h = 100√3 + h [Multiply by √3]
– –
⇒ √3h − h = 100√3
– –
⇒ h (√3 − 1) = 100√3

100√3 √3+1
⇒ h = × ...(2)
√3−1 √3+1

100(3+ √3)
⇒ h =
3−1

⇒ h = 50 (3 + 1.732)

⇒ h = 50 × 4.732 = 236.6m

Put value of h from equation (2) in equation (1)

1 100√3
x = ×
√3 √3−1

100 √3+1
⇒ x = ×
√3−1 √3+1

1000(1.732+1)
⇒ x =
3−1
1000×2.732
⇒ x = = 100 × 1.366 = 136.6m
2

∴ Height of parachutist = 236.6m

Distance of first point from Where he falls = 136.6 m
35. Let, a = 50
C.I. Number of states/ U.T. (fi) xi di = xi - 50 fidi

15 - 25 6 20 -30 -180

25 - 35 11 30 -20 -220

35 - 45 7 40 -10 -70

45 - 55 4 50 0 0

55 - 65 4 60 10 40

65 - 75 2 70 20 40

75 - 85 1 80 30 30
From table, Σf i di = −360, Σfi = 36
Σfi di
we know that, mean=x̄ = a + ¯
¯

Σfi

Page 18 of 20
 −360
= 50 +
35

= 39.71

Section E
36. i. Point of intersection of graph of polynomial, gives the zeroes of the polynomial.
∴ zeroes = -4 and 7
ii. Since, zero's are α = -4, β = 7
α + β = -4 + 7 = 3

αβ = -4 × 7 = -28

P(x) = x2 - (Sum of zeroes)x + product of zeroes

P(x) = x2 - 3x + (-28)
P(x) = x2 - 3x - 28
iii. Product of zeroes = -4 × 7
= -28
OR
a is a non-zero real number, b and c are any real numbers c.
37. i. mallet is cylindrical in shape.
Volume = π × r2 × h
=π × (2 cm)2 × 10cm
=π × 4 cm2 × 10 cm
= 40π cm3
ii. Inner Surface Area = 2π × (5 cm)2
Inner Surface Area = 2π × 25 cm2
Inner Surface Area = 50π cm2
iii. Outer Hemisphere Volume = ( 2

3
)π (6 cm)3
=( 2

3
)π (216 cm3)
≈ 144π cm3
Inner Hemisphere Volume = ( 2

3
)π (5 cm)3
=( 2

3
)π (125 cm3)
cm3
250π
≈
3

Volume of Metal Used = Outer Hemisphere Volume - Inner Hemisphere Volume

= (144π cm3) - ( cm3)
250π

) cm3
432π 250π
=( 3
- 3

cm3
182π
= 3

OR
Lateral Surface Area = 2π rh
= 2 × 3.14 × 2 cm × 10 cm
= 125.6 cm2
For the top circular end:
Area = 3.14 × (2 cm)2 = 12.56 cm2
For the bottom circular end (which is the same as the top):
Area = 12.56 cm2
Total Surface Area = Lateral Surface Area + 2 × Circular End Areas
= 125.6 cm2 + 2 × 12.56 cm2
= 125.6 cm2 + 25.12 cm2
= 150.72 cm2
So, the total surface area of the mallet is 150.72 square centimeters.

Page 19 of 20
38. i. The number of students in Section A is 32, and the number of students in Section B is 36.
Step 1: Find the prime factors of each number:
32 = 2 × 2 × 2 × 2 × 2
36 = 2 × 2 × 3 × 3
Step 2: Identify the common and uncommon prime factors. The common ones are 2 × 2.
Step 3: Multiply the common and uncommon prime factors together to get the LCM:
LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
So, the minimum number of books needed to be acquired for the class library is 288, so they can be distributed equally among
students of Section A or Section B.
ii. Step 1: Find the prime factors of each number:
32 = 2 × 2 × 2 × 2
36 = 2 × 2 × 3 × 3
Step 2: Identify the common prime factors and their minimum exponent:
The common prime factors are 2 × 2.
Step 3: Calculate the HCF by multiplying the common prime factors:
HCF = 2 × 2 = 4
So, the HCF of 32 and 36 is 4.
iii. Given number (7 × 11 × 13 × 15 + 15)
It can also be written as 15 (7 × 11 × 13 +1).
As it is a product of two composite numbers
hence it is a composite number.
OR
Given:
p = ab2
q = a2b
Take the highest power of each prime factor:
LCM = a2 × b2
So, the LCM of p and q is a2b2.

Page 20 of 20