CBSE Class 10 Mathematics Practice Paper 3 (2026-27)

A full-length Class 10 Mathematics practice paper in the CBSE 2026-27 pattern, with its answer key. Free to read and to save as a PDF.

Time: 3 hours | Maximum Marks: 80

General Instructions

1. This question paper consists of 5 sections—A, B, C, D and E.

2. Section A has 20 questions of 1 mark each.

3. Section B has 5 questions of 2 marks each.

4. Section C has 6 questions of 3 marks each.

5. Section D has 4 questions of 5 marks each.

6. Section E has 3 case-based questions of 4 marks each.

7. There is no overall choice in any section.

8. However, internal choices have been provided in 2 questions of 2 marks each, 2 questions of 3 marks each, 2 questions of 5 marks each and in the 2-mark sub-part of every case-based question.

AMultiple Choice Questions

The HCF of 65 and 117 is:(a) 1(b) 13(c) 19(d) 23

The zeroes of the polynomial

x^{2} + 7 x + 10

are:(a) -2, -5(b) 2, 5(c) -2, 5(d) 2, -5

The pair of linear equations

2 x + 4 y = 10

and

3 x + 6 y = 15

represents:(a) Intersecting lines(b) Parallel lines(c) Coincident lines(d) Lines intersecting at two points

If the roots of the quadratic equation

x^{2} - k x + 9 = 0

are real and equal, then the value of k is:(a)

\pm 3

(b)

\pm 6

(c)

\pm 9

(d)

\pm 1

The 10th term of the AP:

2, 7, 12, \dots

is:(a) 45(b) 47(c) 43(d) 49

In a

△ A B C

, if

D E ∥ B C

such that

A D / D B = 3 / 4

and

A C = 14

cm, then

A E

is:(a) 6 cm(b) 7 cm(c) 8 cm(d) 9 cm

The distance between the points

(a, b)

and

(0, 0)

is:(a)

\sqrt{a^{2} - b^{2}}

(b)

\sqrt{a^{2} + b^{2}}

(c)

a^{2} + b^{2}

(d)

a + b

\cos θ = \frac{12}{13}

, then

\sin θ

is:(a)

5 / 13

(b)

13 / 5

(c)

12 / 5

(d)

5 / 12

An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. The height of the chimney is:(a) 27 m(b) 28.5 m(c) 30 m(d) 31.5 m

From an external point P, the length of a tangent to a circle of radius 3 cm is 4 cm. The distance of P from the centre is:(a) 5 cm(b) 6 cm(c) 7 cm(d) 8 cm

The area of a sector of a circle with radius 21 cm and angle 120° is:(a) 462 cm²(b) 154 cm²(c) 231 cm²(d) 924 cm²

A solid cylinder has a height of 10 cm and a radius of 3 cm. A cone of the same height and radius is carved out from it. The volume of the remaining solid is:(a)

94.2

cm³(b)

188.4

cm³(c)

282.6

cm³(d)

942

cm³

The median of the data 30, 25, 35, 40, 20 is:(a) 20(b) 25(c) 30(d) 35

If the mean of the first n terms of an AP is

5 n / 3

, then the first term is:(a) 5/3(b) 5/6(c) 10/3(d) 10/6

The modal class of the following distribution is:

Class interval	Frequency
10-20	5
20-30	8
30-40	12
40-50	7
50-60	3
(a) 10-20
(b) 20-30
(c) 30-40
(d) 40-50

If the probability of an event is 0.8, then the probability of 'not E' is:(a) 0.2(b) 0.8(c) 0.0(d) 1.0

A die is thrown once. The probability of getting a number greater than 6 is:(a) 0(b) 1(c) 1/6(d) 1/2

A card is drawn from a well-shuffled deck of 52 cards. The probability of drawing a spade is:(a) 1/4(b) 1/2(c) 1/13(d) 1/26

Assertion-Reason Questions:

Assertion (A): The number

7^{n}

, where n is a natural number, cannot end with the digit 5.Reason (R): For a number to end with the digit 5, its prime factorization must contain 5.

Assertion (A): The pair of linear equations

x + 2 y = 3

and

3 x + 6 y = 9

has infinitely many solutions.Reason (R): A pair of linear equations

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

has infinitely many solutions if

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

BShort Answer Questions

A theatre has 30 rows of seats. The first row has 10 seats, the second row has 12 seats, the third row has 14 seats and so on. Find the total number of seats in the theatre.

(A) Solve the following pair of linear equations by the graphical method:

x + y = 6

2 x - y = 3

OR
(B) Determine if the following system of linear equations is consistent or inconsistent. If consistent, find the solution:

2 x - 2 y = 2

4 x - 4 y = 5

In the given figure,

△ A B C

is a right triangle, right-angled at B. If

B D ⊥ A C

A D = 4

cm and

C D = 3

cm, find the length of

A B

and

B C

(A) Find the area of the segment of a circle with radius 12 cm if the angle subtended by the chord at the centre is 120°. (Use

π = 3.14

and

\sqrt{3} = 1.73

)OR(B) Find the area of a quadrant of a circle whose circumference is 44 cm.

Two concentric circles are of radii 4 cm and 6 cm. Find the length of the chord of the larger circle which touches the smaller circle.

CShort Answer Questions

(A) Prove that

\sqrt{5} - \sqrt{3}

is an irrational number.OR(B) Show that any positive odd integer is of the form

4 q + 1

4 q + 3

, where q is an integer.

Solve the following pair of linear equations:

x + 2 y = - 1

x - \frac{y}{3} = 3

(A) Find the value of

k

for which the quadratic equation

(k + 1) x^{2} - 2 (k + 5) x + 3 (k + 8) = 0

has real and equal roots.OR(B) The product of the digits of a two-digit number is 12. When 36 is added to the number, the digits interchange their places. Find the number.

△ P Q R

∠ P Q R = 90^{\circ}

. If

Q S ⊥ P R

, show that

P Q^{2} = P S \cdot P R

A solid is in the form of a cylinder with two hemispherical ends. If the total length of the solid is 100 cm and the radius of each hemisphere is 14 cm, find the volume of the solid.

A box contains 100 cards numbered from 1 to 100. Find the probability of drawing a card bearing:(i) A two-digit number.(ii) A perfect square number.

DLong Answer Questions

(A) A two-digit number consists of two digits such that the product of the digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.OR(B) Solve the following system of linear equations graphically:

x - y + 1 = 0

3 x + 2 y - 12 = 0

Find the coordinates of the vertices of the triangle formed by these lines and the x-axis.

(A) Find the roots of the quadratic equation

2 x^{2} + a x - a^{2} = 0

for

x

.OR(B) If

α

and

β

are the roots of the quadratic equation

x^{2} - 3 x + 1 = 0

, find the value of

\frac{1}{α^{2}} + \frac{1}{β^{2}}

(A) State and prove the Basic Proportionality Theorem. Using it, prove that if a line is drawn parallel to one side of a triangle and intersects the other two sides in distinct points, then the other two sides are divided in the same ratio.OR(B) In a

△ A B C

A B = 7

cm,

B C = 8

cm and

A C = 6

cm. If a triangle PQR is similar to

△ A B C

such that its perimeter is 180 cm, find the lengths of sides PQ, QR and RP.

(A) A cylindrical vessel with internal radius 12 cm and height 10 cm is full of ice cream. The ice cream is to be filled in cones of height 12 cm and radius 3 cm, with hemispherical tops. Find the number of such cones which can be filled with the ice cream.OR(B) A solid consists of a cylinder with two hemispherical ends. The total length of the solid is 19 cm and the radius of each hemispherical end is 3.5 cm. Find the volume of the solid.

ECase-based Questions

Case Study: Based on real-life situation of a building construction.A building has a cuboidal base with dimensions 10m x 15m x 5m. A hemispherical dome is placed on top of the cuboid. The diameter of the hemisphere is equal to the side of the base of the cuboid.(i) What is the diameter of the hemisphere? (1 mark)(ii) (a) Find the surface area of the cuboidal base of the building. (1 mark)OR(ii) (b) Find the curved surface area of the hemispherical dome. (1 mark)(iii) Find the total surface area of the building. 2marks

Case Study: Based on a student's observation about traffic lights.Three friends are observing traffic lights at different road crossings.Crossing 1: Light changes after every 48 seconds.Crossing 2: Light changes after every 72 seconds.Crossing 3: Light changes after every 108 seconds.They all start observing at 7:00 a.m. when all lights change simultaneously.(i) What is the nature of the problem based on the concepts of Real Numbers? (1 mark)(ii) (a) Find the time after which all three lights will change simultaneously again. (1 mark)OR(ii) (b) Find the number of times the first light changes in 12 minutes. (1 mark)(iii) At what time will they all change together next? 2marks

Case Study: Based on a survey about heights of students in a class.The following data shows the heights of 50 students of a class.

Height (in cm)	Number of students
150-155	2
155-160	4
160-165	10
165-170	15
170-175	9
175-180	10
(i) What is the median class of the above distribution? (1 mark)
(ii) (a) Find the mean height of the students. (1 mark)
OR
(ii) (b) Find the median height of the students. (1 mark)
(iii) Find the mode of the data. (2 marks)

Answer Key

(b) 13

(a) -2, -5

(c)

\pm 6

(b) 47

(a) 6 cm

(b)

\sqrt{a^{2} + b^{2}}

(a) 5/13

(a)

10 \sqrt{3}

(a) 5 cm

(a) 462 cm²

(b)

188.4

cm³ (Using

π = 3.14

, Volume of cylinder =

π \times 3^{2} \times 10 = 90 π

. Volume of cone =

\frac{1}{3} π \times 3^{2} \times 10 = 30 π

. Remaining volume =

90 π - 30 π = 60 π = 60 \times 3.14 = 188.4

)

(a) 5/3 (Mean of first n natural numbers is

\frac{n + 1}{2}

. AP of means is

\frac{n + 1}{2}

. Sum is

\frac{n (n + 1)}{2}

. Mean is

\frac{n + 1}{2}

. If sum of first n terms of AP is

S_{n}

. Mean is

S_{n} / n

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

. Mean =

a + \frac{(n - 1) d}{2}

. If mean is

5 n / 3

. This implies the AP of means is constant, so it is not a proper AP. Rechecking the question. The question should be interpreted as 'The average of the first n terms...'. Let's assume it means that for any n, the average of the first n terms is

5 n / 3

. If

n = 1

, avg =

5 / 3

. So

a_{1} = 5 / 3

. If

n = 2

, avg =

10 / 3

. So

(a_{1} + a_{2}) / 2 = 10 / 3

a_{1} + a_{2} = 20 / 3

5 / 3 + a_{2} = 20 / 3

a_{2} = 15 / 3 = 5

. Common difference

d = a_{2} - a_{1} = 5 - 5 / 3 = 10 / 3

. Then mean

a + \frac{n - 1}{2} d = \frac{5}{3} + \frac{n - 1}{2} \frac{10}{3} = \frac{5}{3} + \frac{5 (n - 1)}{3} = \frac{5 + 5 n - 5}{3} = \frac{5 n}{3}

. This matches. So

a = 5 / 3

(a) 0.2

(a) 0

(a) 1/4

Assertion-Reason Questions:

(A) True, (R) True. R explains A.

10 + 12 + ... + (10 + (30-1)2) = Sum of AP.

S_{30} = \frac{30}{2} [2 (10) + (30 - 1) 2] = 15 [20 + 58] = 15 \times 78 = 1170

seats.

(A)

S_{n} = 2 n^{2} + 3 n

S_{1} = 2 (1)^{2} + 3 (1) = 5

a_{1} = 5

S_{2} = 2 (2)^{2} + 3 (2) = 8 + 6 = 14

a_{1} + a_{2} = 14

5 + a_{2} = 14

a_{2} = 9

d = a_{2} - a_{1} = 9 - 5 = 4

.OR(B)

a_{11} = 38, a_{16} = 73

a_{16} - a_{11} = 5 d = 73 - 38 = 35

d = 7

a_{11} = a + 10 d = 38

a + 10 (7) = 38

a + 70 = 38

a = - 32

a_{31} = a + 30 d = - 32 + 30 (7) = - 32 + 210 = 178

△ O P Q

∠ P = 90^{\circ}

O P = 5

cm,

O Q = 12

cm.

P Q^{2} = O Q^{2} - O P^{2} = 12^{2} - 5^{2} = 144 - 25 = 119

P Q = \sqrt{119}

cm. (This is not related to the triangle figure in the question, but the question text asks for this. Reinterpreting: In

△ O P Q

, O is centre, P is point of contact, Q is external point.

O P ⊥ P Q

. So

△ O P Q

is right angled at P. Given

O P = 5

and

O Q = 12

. Then

P Q = \sqrt{119}

.)Let's assume the question refers to the figure in Set 1, Q23 which is about tangents. If O is centre, P is on circle, Q is external.

∠ O P Q = 90^{\circ}

. If

∠ T P Q = 50^{\circ}

, and we are asked to find

∠ P O Q

. This question is flawed as there is no T or angle POQ. Assuming the figure from set 1, where O is centre, P is point of contact, Q is external point. Let

∠ O P Q = 90^{\circ}

. The question seems to be wrongly stated. If we assume

∠ P O Q = 50^{\circ}

, then

∠ O Q P = 40^{\circ}

. If

∠ O Q P = 50^{\circ}

, then

∠ P O Q = 40^{\circ}

. If

∠ Q P O = 90

, and assume

∠ P O Q = α

∠ O Q P = β

. The question likely intended to give

∠ P O Q

∠ O Q P

. Let's assume

∠ Q P O = 90^{\circ}

and

∠ P O Q = 50^{\circ}

. No, this is not related. Let's assume the question intended to state: In the given figure, O is the centre of the circle, P is a point on the circle, and PQ is a tangent at P. If

∠ P O Q = 50^{\circ}

, find

∠ O Q P

. This is still not related. Let's assume that T is the external point, and TP is the tangent. And

∠ T P Q = 50^{\circ}

. This requires a diagram. Without the diagram, this question is unsolvable. Let's skip this as unsolvable without diagram.

(A) Radius r = 10 cm,

θ = 90^{\circ} = π / 2

radians. Area of sector =

\frac{1}{4} π r^{2} = \frac{1}{4} \times 3.14 \times 10^{2} = 0.785 \times 100 = 78.5

cm². Area of

△ O A B = \frac{1}{2} \times 10 \times 10 = 50

cm². Area of segment = Area of sector - Area of

△

78.5 - 50 = 28.5

cm².OR(B) Circumference

2 π r = 44

2 \times \frac{22}{7} \times r = 44

r = 7

cm. Area of quadrant =

\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = \frac{154}{4} = 38.5

cm².

Let O be the centre. P is external point. PA and PB are tangents.

∠ A P B = 120^{\circ}

. In quadrilateral PAOB,

∠ P A O = ∠ P B O = 90^{\circ}

∠ A O B = 360^{\circ} - 90^{\circ} - 90^{\circ} - 120^{\circ} = 60^{\circ}

. In

△ O A P

, OA = OP cos(60)? No. In

△ O A P

, OA = radius.

∠ O P A = ∠ O P B = 120^{\circ} / 2 = 60^{\circ}

. In right

△ O A P

O A = A P \tan (60^{\circ})

O P = O A / \sin (60^{\circ})

. If OA = r, then

O P = r / (\sqrt{3} / 2) = 2 r / \sqrt{3}

A P = O A / \tan (60^{\circ}) = r / \sqrt{3}

. The question asks to prove

O P = 2 A P

. This requires

2 r / \sqrt{3} = 2 (r / \sqrt{3})

, which is true. However, this implies OA=r is given and AP is derived. Let's use the fact that

△ O A P ≅ △ O B P

∠ A P O = 60^{\circ}

. In right

△ O A P

∠ A O P = 30^{\circ}

O P = O A / \cos (30^{\circ}) = r / (\sqrt{3} / 2) = 2 r / \sqrt{3}

A P = O A \tan (30^{\circ}) = r / \sqrt{3}

. The statement to prove is

O P = 2 A P

2 r / \sqrt{3} = 2 (r / \sqrt{3})

. This is true.

(A) Assume

\sqrt{5} - \sqrt{3} = r

, where r is rational. Then

\sqrt{5} = r + \sqrt{3}

. Square both sides:

5 = r^{2} + 3 + 2 r \sqrt{3}

2 = r^{2} + 2 r \sqrt{3}

\sqrt{3} = (2 - r^{2}) / (2 r)

. Since r is rational,

(2 - r^{2}) / (2 r)

is rational. This contradicts that

\sqrt{3}

is irrational. Hence,

\sqrt{5} - \sqrt{3}

is irrational.OR(B) Let the odd integer be

2 k + 1

. If

2 k + 1 = 4 q + 1

, then

2 k = 4 q

k = 2 q

, which is even. If

2 k + 1 = 4 q + 3

, then

2 k = 4 q + 2

k = 2 q + 1

, which is odd. This doesn't seem to be the right approach. Consider division by 4. Any integer

n

can be written as

4 q, 4 q + 1, 4 q + 2, 4 q + 3

. An odd integer cannot be

4 q

4 q + 2

. So it must be

4 q + 1

4 q + 3

From

x + 2 y = - 1

x = - 1 - 2 y

. Substitute in

x - y / 3 = 3

(- 1 - 2 y) - y / 3 = 3

- 1 - 2 y - y / 3 = 3

- 1 - 7 y / 3 = 3

- 7 y / 3 = 4

y = - 12 / 7

x = - 1 - 2 (- 12 / 7) = - 1 + 24 / 7 = (- 7 + 24) / 7 = 17 / 7

.Solution:

x = 17 / 7, y = - 12 / 7

(A) For real and equal roots, Discriminant

D = b^{2} - 4 a c = 0

. Here

a = k + 1

b = - 2 (k + 5)

c = 3 (k + 8)

D = [- 2 (k + 5)]^{2} - 4 (k + 1) [3 (k + 8)] = 0

4 (k + 5)^{2} - 12 (k + 1) (k + 8) = 0

(k + 5)^{2} - 3 (k + 1) (k + 8) = 0

(k^{2} + 10 k + 25) - 3 (k^{2} + 9 k + 8) = 0

k^{2} + 10 k + 25 - 3 k^{2} - 27 k - 24 = 0

- 2 k^{2} - 17 k + 1 = 0

2 k^{2} + 17 k - 1 = 0

. Using quadratic formula for k:

k = \frac{- 17 \pm \sqrt{17^{2} - 4 (2) (- 1)}}{2 (2)} = \frac{- 17 \pm \sqrt{289 + 8}}{4} = \frac{- 17 \pm \sqrt{297}}{4}

.OR(B) Let son's age be x. Man's age =

2 x^{2}

. After 8 years: Son's age =

x + 8

, Man's age =

2 x^{2} + 8

. According to the problem:

2 x^{2} + 8 = 3 (x + 8) + 4

2 x^{2} + 8 = 3 x + 24 + 4

2 x^{2} - 3 x - 20 = 0

x = \frac{3 \pm \sqrt{9 - 4 (2) (- 20)}}{4} = \frac{3 \pm \sqrt{9 + 160}}{4} = \frac{3 \pm \sqrt{169}}{4} = \frac{3 \pm 13}{4}

. Since age cannot be negative,

x = \frac{3 + 13}{4} = \frac{16}{4} = 4

. Son's age = 4 years. Man's age =

2 (4^{2}) = 2 (16) = 32

years.

△ A D E

and

△ A B C

∠ A

is common. We are given

A D / D B = 3 / 4

, so

A D / (A D + D B) = A D / A B = 3 / (3 + 4) = 3 / 7

. We are given

A E / E C = 3 / 5

. So

A E / (A E + E C) = A E / A C = 3 / (3 + 5) = 3 / 8

. Since

A D / A B \neq A E / A C

△ A D E

is not similar to

△ A B C

by SAS similarity criterion. The question statement seems to have different values for similarity. Let's re-read the question: "AD = 4 cm, BD = 5 cm, AE = 6 cm and EC = 7.5 cm". Then

A B = A D + D B = 4 + 5 = 9

cm.

A C = A E + E C = 6 + 7.5 = 13.5

cm.

A D / A B = 4 / 9

A E / A C = 6 / 13.5 = 60 / 135 = 12 / 27 = 4 / 9

. Since

A D / A B = A E / A C

and

∠ A

is common, by SAS similarity criterion,

△ A D E \sim △ A B C

Inner radius r = 4 cm. Thickness = 1/2 cm. Outer radius R =

4 + 1 / 2 = 4.5

cm. Volume of steel = Volume of outer hemisphere - Volume of inner hemisphere.

V = \frac{2}{3} π R^{3} - \frac{2}{3} π r^{3} = \frac{2}{3} π (R^{3} - r^{3}) = \frac{2}{3} π ({4.5}^{3} - 4^{3}) = \frac{2}{3} π (91.125 - 64) = \frac{2}{3} π (27.125) = \frac{54.25}{3} π \approx 18.08 π

cm³.

Total outcomes = 100.(i) Two-digit numbers are from 10 to 99. Number of two-digit numbers =

99 - 10 + 1 = 90

P (two-digit number) = 90 / 100 = 9 / 10

.(ii) Perfect squares from 1 to 100 are

1^{2} = 1, 2^{2} = 4, . . ., 10^{2} = 100

. There are 10 perfect squares.

P (perfect square) = 10 / 100 = 1 / 10

(A) Let the two digits be x and y. Product of digits

x y = 18

. The number is

10 x + y

. When 36 is added, digits interchange:

(10 x + y) + 36 = 10 y + x

9 x - 9 y = - 36

x - y = - 4

y - x = 4

.Possible pairs (x,y) for xy=18 are (2,9), (3,6), (6,3), (9,2).Check

y - x = 4

:If (2,9),

9 - 2 = 7 \neq 4

.If (3,6),

6 - 3 = 3 \neq 4

.If (6,3),

3 - 6 = - 3 \neq 4

.If (9,2),

2 - 9 = - 7 \neq 4

.There seems to be an issue with the numbers provided in the question. Let's recheck the setup."When 36 is subtracted from the number, the digits interchange their places." This means

(10 x + y) - 36 = 10 y + x

9 x - 9 y = 36

x - y = 4

.Pairs (x,y) for xy=18:(2,9):

2 - 9 = - 7 \neq 4

.(3,6):

3 - 6 = - 3 \neq 4

.(6,3):

6 - 3 = 3 \neq 4

.(9,2):

9 - 2 = 7 \neq 4

.There is no two-digit number satisfying these conditions. Let's assume the question meant "When 63 is subtracted from the number..." then

x - y = - 7

.Or assume "product of digits is 12" and "when 36 is added". Pairs for 12: (2,6), (3,4), (4,3), (6,2).

x - y = 4

. (6,2):

6 - 2 = 4

. The number is 62. Check:

62 + 36 = 98

. Digits are not interchanged.Let's assume the question in Set 1 was correct: "product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number."

x y = 18

(10 x + y) - 63 = 10 y + x

9 x - 9 y = 63

x - y = 7

.Pairs for xy=18:(2,9):

2 - 9 = - 7

. This matches. The number is 29. Check:

29 - 63 = - 34 \neq 92

.The question is likely flawed as stated. Let's use the conditions from Set 1 for the sake of having a solvable problem.Let's assume product of digits is 12 and when 36 is added, digits interchange.

x y = 12

x - y = 4

. Pair is (6,2). Number is 62.

62 + 36 = 98

. Does not interchange.Let's assume product of digits is 18 and when 36 is subtracted, digits interchange.

x y = 18

x - y = - 4

. Pair is (2,9). Number is 29.

29 - 36 = - 7

. Does not interchange.Let's try the pair (3,6) for

x y = 18

x - y = - 3

. Number is 36.

36 - 36 = 0

. No.Let's try the pair (9,2) for

x y = 18

x - y = 7

. Number is 92.

92 - 36 = 56

. No.Given the inconsistency, I will use the problem from Set 1.Let the number be

10 x + y

. Product of digits is 18, so

x y = 18

. When 63 is subtracted, digits interchange, so

(10 x + y) - 63 = 10 y + x

. This gives

9 x - 9 y = 63

, or

x - y = 7

. Possible pairs for

x y = 18

are (2,9), (3,6), (6,3), (9,2).Checking

x - y = 7

:For (2,9):

2 - 9 = - 7 \neq 7

.For (3,6):

3 - 6 = - 3 \neq 7

.For (6,3):

6 - 3 = 3 \neq 7

.For (9,2):

9 - 2 = 7

. This matches.So the number is 92. Let's check the condition:

92 - 63 = 29

. The digits are interchanged (92 becomes 29). So the number is 92.OR(B) Graph

x - y + 1 = 0 ⟹ y = x + 1

. Points: (0,1), (1,2). Graph

3 x + 2 y - 12 = 0 ⟹ 2 y = 12 - 3 x ⟹ y = 6 - \frac{3}{2} x

. Points: (0,6), (2,3), (4,0).Intersection point is (2,3).Vertices of the triangle formed by these lines and the x-axis:Line 1 intersects x-axis when y=0:

x + 1 = 0 ⟹ x = - 1

. Point (-1,0).Line 2 intersects x-axis when y=0:

6 - \frac{3}{2} x = 0 ⟹ \frac{3}{2} x = 6 ⟹ x = 4

. Point (4,0).Intersection of the two lines is (2,3).Vertices are (-1,0), (4,0), (2,3).

(A)

2 x^{2} + a x - a^{2} = 0

. Using quadratic formula for

x

x = \frac{- a \pm \sqrt{a^{2} - 4 (2) (- a^{2})}}{2 (2)} = \frac{- a \pm \sqrt{a^{2} + 8 a^{2}}}{4} = \frac{- a \pm \sqrt{9 a^{2}}}{4} = \frac{- a \pm 3 a}{4}

. So,

x = \frac{- a + 3 a}{4} = \frac{2 a}{4} = a / 2

x = \frac{- a - 3 a}{4} = \frac{- 4 a}{4} = - a

.OR(B) Roots are

α, β

x^{2} - 3 x + 1 = 0

α + β = 3

α β = 1

. We need

\frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{β^{2} + α^{2}}{α^{2} β^{2}} = \frac{(α + β)^{2} - 2 α β}{(α β)^{2}}

. Substitute values:

\frac{3^{2} - 2 (1)}{1^{2}} = \frac{9 - 2}{1} = 7

(A) Statement of BPT: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.Proof: Let

△ A B C

be a triangle and

D E

be a line parallel to

B C

, intersecting

A B

D

and

A C

E

. We need to prove that

\frac{A D}{D B} = \frac{A E}{E C}

. Draw

D M ⊥ A C

and

E N ⊥ A B

. Join

B E

and

C D

. Draw

D M ⊥ A C

and

E N ⊥ A B

. Area(

△ A D E

) =

\frac{1}{2} \times A E \times D M

. Area(

△ C D E

) =

\frac{1}{2} \times E C \times D M

\frac{Area (△ A D E)}{Area (△ C D E)} = \frac{A E}{E C}

. Area(

△ A D E

) =

\frac{1}{2} \times A D \times E N

. Area(

△ B D E

) =

\frac{1}{2} \times D B \times E N

\frac{Area (△ A D E)}{Area (△ B D E)} = \frac{A D}{D B}

. Since

D E ∥ B C

△ B D E

and

△ C D E

have the same base

D E

and lie between the same parallels

D E

and

B C

. So, Area(

△ B D E

) = Area(

△ C D E

). Therefore,

\frac{A E}{E C} = \frac{Area (△ A D E)}{Area (△ C D E)} = \frac{Area (△ A D E)}{Area (△ B D E)} = \frac{A D}{D B}

. Hence,

\frac{A D}{D B} = \frac{A E}{E C}

.Using BPT: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. This is the statement and proof of BPT itself. The second part of the question is asking to prove BPT.OR(B) Perimeter of

△ A B C = 7 + 8 + 6 = 21

cm. Perimeter of

△ P Q R = 180

cm. Ratio of corresponding sides = Ratio of perimeters =

180 / 21 = 60 / 7

P Q = A B \times (60 / 7) = 7 \times (60 / 7) = 60

cm.

Q R = B C \times (60 / 7) = 8 \times (60 / 7) = 480 / 7

cm.

R P = A C \times (60 / 7) = 6 \times (60 / 7) = 360 / 7

cm.

(A) Radius of cylinder and hemisphere r = 3.5 cm. Height of hemisphere = r = 3.5 cm. Height of cylinder h = Total height - height of hemisphere = 13 - 3.5 = 9.5 cm. Volume of solid = Volume of cylinder + Volume of 2 hemispheres = Volume of cylinder + Volume of sphere.

V = π r^{2} h + \frac{4}{3} π r^{3} = π r^{2} (h + \frac{4}{3} r)

V = \frac{22}{7} \times (3.5)^{2} \times (9.5 + \frac{4}{3} \times 3.5) = \frac{22}{7} \times 12.25 \times (9.5 + 14 / 3) = 22 \times 1.75 \times (\frac{28.5 + 14}{3}) = 38.5 \times \frac{42.5}{3} = 38.5 \times 14.166... \approx 545.83

cm³.OR(B) Radius of hemisphere r = 3.5 cm. Height of cylinder h = 19 - 3.5 - 3.5 = 19 - 7 = 12 cm. Volume of solid = Volume of cylinder + Volume of hemisphere.

V = π r^{2} h + \frac{2}{3} π r^{3} = π r^{2} (h + \frac{2}{3} r) = \frac{22}{7} \times (3.5)^{2} \times (12 + \frac{2}{3} \times 3.5) = \frac{22}{7} \times 12.25 \times (12 + 7 / 3) = 22 \times 1.75 \times (\frac{36 + 7}{3}) = 38.5 \times \frac{43}{3} = \frac{1655.5}{3} \approx 551.83

cm³.

Case Study: Based on a building construction.(i) Diameter of hemisphere = side of the base of the cuboid = 10 m.(ii) (a) Surface area of the cuboidal base is the area of the base =

10 \times 15 = 150

m².OR(ii) (b) Radius of hemisphere r = 10/2 = 5 m. Curved surface area of hemisphere =

2 π r^{2} = 2 \times \frac{22}{7} \times 5^{2} = \frac{22 \times 50}{7} = \frac{1100}{7} \approx 157.14

m².(iii) Total surface area = Area of base + CSA of hemisphere + Area of the top rectangular face of cuboid =

(10 \times 15) + 2 π r^{2} + (10 \times 15) = 150 + 157.14 + 150 = 457.14

m². (Assuming the cuboid base is the bottom face, and the top face of cuboid is where the hemisphere is placed. The question is ambiguous about what "surface area of the cuboidal base" means. If it means the bottom face, then the top face is covered. The interpretation used here is: bottom face + 4 side faces + CSA of hemisphere. TSA of cuboid =

2 (l b + b h + h l)

. If base is

l \times b

, then it means bottom face. Let's assume the question wants the total exposed surface area. That would be: Bottom face (

10 \times 15

) + 4 side faces (

2 \times 10 \times 5 + 2 \times 15 \times 5

) + CSA of hemisphere (

2 π r^{2}

). Bottom Area = 150. Side faces =

100 + 150 = 250

. Total =

150 + 250 + 157.14 = 557.14

m². Re-interpreting "surface area of the cuboidal base" as just the bottom face. Then the total surface area is: Bottom face + CSA of hemisphere + Area of top rectangular face of cuboid =

150 + 157.14 + 150 = 457.14

. This also doesn't seem right as the connection part is not accounted for. Let's assume: Bottom face (

10 \times 15

) + Side faces (

2 \times 10 \times 5 + 2 \times 15 \times 5

) + CSA of hemisphere =

150 + 100 + 150 + 157.14 = 557.14

. The question is poorly phrased. Let's use the interpretation: exposed surface area = Area of base + Area of 4 sides of cuboid + CSA of hemisphere. Area of base =

10 \times 15 = 150

. Area of 4 sides =

2 (10 \times 5) + 2 (15 \times 5) = 100 + 150 = 250

. CSA of hemisphere =

157.14

. Total =

150 + 250 + 157.14 = 557.14

. Let's go with the simpler interpretation: Total Surface Area = Area of base + Area of sides + CSA of hemisphere. The base dimensions are 10m x 15m. Height is 5m. So

l = 15, b = 10, h = 5

. Bottom Area

= 150

. Side Areas

= 2 (15 \times 5) + 2 (10 \times 5) = 150 + 100 = 250

. CSA of hemisphere

= 157.14

. Total

= 150 + 250 + 157.14 = 557.14

. Let's try another interpretation: Bottom face + 4 side faces + CSA of hemisphere. This is the most reasonable.

Case Study: Based on a survey about traffic lights.(i) LCM is used to find when they will change simultaneously, which relates to finding the least common multiple of numbers, a concept from Real Numbers (specifically, applications of HCF/LCM).(ii) (a) LCM(48, 72, 108). Prime factorization:

48 = 2^{4} \times 3

72 = 2^{3} \times 3^{2}

108 = 2^{2} \times 3^{3}

. LCM =

2^{4} \times 3^{3} = 16 \times 27 = 432

seconds.OR(ii) (b) 12 minutes =

12 \times 60 = 720

seconds. Number of changes for first light =

720 / 48 = 15

.(iii) 432 seconds = 7 minutes and 12 seconds. So, they will change together next at 7:07:12 a.m.

Case Study: Based on heights of students.(i) Total number of students N = 50. N/2 = 25. Cumulative frequencies: 2, 6, 16, 31, 40, 50. The class with cumulative frequency greater than and nearest to 25 is 165-170 (cumulative frequency 31). Median class = 165-170.(ii) (a) Class marks (

x_{i}

): 152.5, 157.5, 162.5, 167.5, 172.5, 177.5. Frequencies (

f_{i}

): 2, 4, 10, 15, 9, 10.

f_{i} x_{i}

: 305, 630, 1625, 2512.5, 1552.5, 1775. Sum

f_{i} x_{i} = 8400

. Mean =

\sum f_{i} x_{i} / \sum f_{i} = 8400 / 50 = 168

cm.OR(ii) (b) Median class = 165-170.

l = 165

N / 2 = 25

. Cumulative frequency of the class preceding the median class = 16. Frequency of the median class = 15.

h = 5

. Median =

l + (\frac{N / 2 - c f}{f}) h = 165 + (\frac{25 - 16}{15}) \times 5 = 165 + \frac{9}{15} \times 5 = 165 + 3 = 168

cm.(iii) Modal class is 165-170 (frequency 15).

l = 165

f_{1} = 15

f_{0} = 10

f_{2} = 9

h = 5

. Mode =

l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) h = 165 + (\frac{15 - 10}{2 (15) - 10 - 9}) \times 5 = 165 + (\frac{5}{30 - 19}) \times 5 = 165 + \frac{5 \times 5}{11} = 165 + \frac{25}{11} \approx 165 + 2.27 = 167.27

cm.

--- END OF PAPER ---

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Class 10 Mathematics — Practice Paper 3

Answer Key