CBSE Class 9 Mathematics Practice Paper 2 (2026-27)

A full-length Class 9 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. All questions are compulsory.

2. The question paper consists of five sections: A, B, C, D and E.

3. Section A has 20 MCQs and Assertion-Reason questions of 1 mark each.

4. Section B has 5 Short Answer I questions of 2 marks each.

5. Section C has 6 Short Answer II questions of 3 marks each.

6. Section D has 4 Long Answer questions of 5 marks each.

7. Section E has 3 Case-based questions of 4 marks each.

8. There is no overall choice. However, internal choice is provided in 2 Questions of section B, 2 Questions of section C, 2 Questions of section D and in the 2-mark sub-part of every case study question.

A20 Marks

(MCQ/Assertion-Reason Questions: 1 Mark Each)

A point with coordinates (–7, 5) lies in which quadrant?(a) Quadrant I(b) Quadrant II(c) Quadrant III(d) Quadrant IV

The degree of the polynomial

2 x^{3} + 7 x^{2} - 5 x + 10

is:(a) 1(b) 2(c) 3(d) 10

Which of the following is an irrational number?(a)

\frac{22}{7}

(b) 3.14(c)

\sqrt{5}

(d) 0.333...

The identity

(a - b)^{2} = a^{2} - 2 a b + b^{2}

is true for:(a) All real values of a and b(b) Only positive values of a and b(c) Only specific values of a and b(d) Only integer values of a and b

The circumference of a circle with radius

r

is given by:(a)

π r^{2}

(b)

2 π r

(c)

π d^{2}

(d)

\frac{2 π r}{360^{\circ}}

The probability of an event that is impossible to occur is:(a) 1(b) 0(c) 0.5(d) Indeterminate

The explicit formula for the nth term of an arithmetic progression is

t_{n} = a + (n - 1) d

. Here, 'd' represents:(a) The first term(b) The nth term(c) The common difference(d) The number of terms

A point

(x, y)

is plotted on the Cartesian plane. If

x > 0

and

y < 0

, the point lies in:(a) Quadrant I(b) Quadrant II(c) Quadrant III(d) Quadrant IV

The degree of the constant polynomial 5 is:(a) 0(b) 1(c) 5(d) Undefined

Between any two distinct rational numbers, there lies:(a) Exactly one rational number(b) No rational number(c) Infinitely many rational numbers(d) At most one irrational number

The algebraic identity

(a + b) (a - b) = ?

(a)

a^{2} + 2 a b + b^{2}

(b)

a^{2} - 2 a b + b^{2}

(c)

a^{2} - b^{2}

(d)

a^{2} + b^{2}

If the radius of a circle is 7 cm, its area is: (use

π = \frac{22}{7}

)(a) 154 cm

^{2}

(b) 44 cm

^{2}

^{2}

(d) 77 cm

^{2}

A die is rolled. The probability of getting a number less than 3 is:(a)

\frac{1}{6}

(b)

\frac{2}{6}

(c)

\frac{3}{6}

(d)

\frac{4}{6}

The sequence 3, 6, 9, 12, ... is an example of:(a) Geometric Progression(b) Arithmetic Progression(c) Fibonacci Sequence(d) Prime Number Sequence

The coordinates of the origin are:(a) (1, 1)(b) (0, 1)(c) (1, 0)(d) (0, 0)

The number of terms in the polynomial

4 x^{2} - 2 x + 1

is:(a) 1(b) 2(c) 3(d) 4

The decimal expansion of

\frac{1}{3}

is:(a) Terminating(b) Non-terminating, repeating(c) Non-terminating, non-repeating(d) Terminating and repeating

The area of a rectangle with length

l

and breadth

b

is:(a)

l + b

(b)

2 (l + b)

(c)

l \times b

(d)

\frac{l}{b}

Assertion (A): The point (2, 3) and (3, 2) represent the same point in the Cartesian plane.Reason (R): The order of coordinates matters in representing a point in the Cartesian plane.(a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true.

Assertion (A): The sum of probabilities of all possible outcomes of an experiment is 1.Reason (R): Probability is a measure of the likelihood of an event, ranging from 0 to 1.(a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true.

B10 Marks

(Short Answer I Questions: 2 Marks Each)

Plot the points A(2, 3), B(–3, 4), C(–4, –2), and D(1, –5) on graph paper. Which of these points lie in the second quadrant?

Find the value of the polynomial

P (x) = 3 x^{2} - 5 x + 2

when

x = - 1

Show that

\sqrt{7}

is an irrational number by using the method of contradiction.

Factorise:

4 x^{2} + 12 x y + 9 y^{2}

The circumference of a circle is 132 cm. Find its radius. (Use

π = \frac{22}{7}

)

C18 Marks

(Short Answer II Questions: 3 Marks Each)

The sum of two numbers is 15. The sum of their reciprocals is

\frac{3}{10}

. Find the numbers.

Expand

(3 x - 2 y + z)^{2}

using an appropriate identity.

Two circles of radii 10 cm and 8 cm have their centres distance apart. Find the length of the common chord.ORA chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre.

A bag contains 3 red balls, 5 green balls and 4 blue balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is:(i) Red(ii) Green(iii) Blue

Find the sum of the first 20 terms of the arithmetic progression: 3, 7, 11, 15, ...ORFind the 10th term of the AP: 2, 7, 12, ...

A bicycle wheel has a diameter of 60 cm. Calculate the distance covered by the wheel in one revolution.

D20 Marks

(Long Answer Questions: 5 Marks Each)

A rectangular garden has dimensions 12m × 5m. A path of uniform width is to be constructed all around the garden, increasing the total area to 168 m

^{2}

. Find the width of the path.ORThe length of a rectangle is 3 cm more than twice its width. If the perimeter of the rectangle is 48 cm, find its dimensions.

a

and

b

are two numbers such that

a + b = 10

and

a b = 21

, find the value of

a^{2} + b^{2}

. Also, find the value of

a^{3} + b^{3}

.ORFactorise:

x^{2} + 11 x + 30

A circular disc of radius 14 cm has two circles of radius 3.5 cm each drawn on it. The centres of the two smaller circles are on the radius of the larger circle such that they touch each other at the centre of the larger circle. Find the area of the remaining portion of the disc. (Use

π = \frac{22}{7}

)ORFind the area of the shaded region in the figure, where ABCD is a square of side 14 cm and APB and CPD are semicircles. (Use

π = \frac{22}{7}

)[Diagram showing a square with two semicircles drawn on opposite sides as diameters, meeting at the center. The shaded region is the area within the square but outside the two semicircles.]

A pack of playing cards consists of 52 cards, of which 16 are colour cards (4 Aces, 4 Kings, 4 Queens, 4 Jacks). A card is drawn at random from the pack.(i) Find the probability that the card drawn is a colour card.(ii) Find the probability that the card drawn is an Ace.(iii) Find the probability that the card drawn is neither a colour card nor an Ace.

E12 Marks

(Case-based Questions: 4 Marks Each)

Case Study: A school decided to promote physical fitness among its students. A circular running track of radius 21 metres has been constructed in the school playground. A rectangular area for other sports is located at the centre of the circle.(i) What is the perimeter of the running track? (2 marks)(ii) What is the area of the running track? 2marks

Case Study: A tailor uses a linear polynomial to estimate the cost of stitching a shirt. The fixed cost is

150 a n d t h e c o s t p e r s h i r t i s

50. The tailor observes that for 10 shirts, the bill is

650. (i) I f

i s t h e t o t a l c o s t a n d

i s t h e n u m b e r o f s h i r t s, w r i t e t h e l i n e a r p o l y n o m i a l r e p r e s e n t i n g t h e t o t a l c o s t . (1 m a r k) (i i) F i n d t h e c o s t o f s t i t c h i n g 25 s h i r t s . (1 m a r k) (i i i) I f t h e t o t a l c o s t w a s

1150, how many shirts were stitched? 2marks

Case Study: In a class of 50 students, the marks obtained in Mathematics are given below. The students are grouped based on their marks.

Marks Range	Number of Students
0-20	5
20-40	10
40-60	15
60-80	12
80-100	8
(i) If a student is chosen at random, what is the probability that the student scored marks between 40 and 60? (1 mark)
(ii) What is the probability that a student scored less than 40 marks? (1 mark)
(iii) What is the probability that a student scored 80 marks or more? (2 marks)

Answer Key

(b) Quadrant II

(c)

\sqrt{5}

(a) All real values of a and b

(b)

2 π r

(b) 0

(d) Quadrant IV

(a) 0

(c)

a^{2} - b^{2}

(a) 154 cm

^{2}

(b)

\frac{2}{6}

(b) Arithmetic Progression

(d) (0, 0)

(b) Non-terminating, repeating

(c)

l \times b

(d) A is false but R is true.

(a) Both A and R are true and R is the correct explanation of A.

Point C(–4, –2) lies in the second quadrant. (Mistake in question: C is in Quadrant III. B is in Quadrant II. Correction: Point B(–3, 4) lies in the second quadrant.)

P (- 1) = 3 (- 1)^{2} - 5 (- 1) + 2 = 3 (1) + 5 + 2 = 3 + 5 + 2 = 10

Assume

\sqrt{7} = \frac{p}{q}

where p, q are co-prime integers,

q \neq 0

. Then

7 q^{2} = p^{2}

. This implies

p^{2}

is divisible by 7, so p must be divisible by 7. Let

p = 7 k

. Then

7 q^{2} = (7 k)^{2} = 49 k^{2}

, so

q^{2} = 7 k^{2}

. This implies

q^{2}

is divisible by 7, so q must be divisible by 7. Since both p and q are divisible by 7, they are not co-prime, which contradicts our assumption. Hence,

\sqrt{7}

is irrational.

4 x^{2} + 12 x y + 9 y^{2} = (2 x)^{2} + 2 (2 x) (3 y) + (3 y)^{2} = (2 x + 3 y)^{2}

Circumference

C = 2 π r

132 = 2 \times \frac{22}{7} \times r ⟹ r = \frac{132 \times 7}{44} = 3 \times 7 = 21

cm.

Let the numbers be x and y.

x + y = 15

and

\frac{1}{x} + \frac{1}{y} = \frac{3}{10}

\frac{x + y}{x y} = \frac{3}{10} ⟹ \frac{15}{x y} = \frac{3}{10} ⟹ x y = 50

. The numbers are roots of

t^{2} - 15 t + 50 = 0

, which gives

(t - 5) (t - 10) = 0

. The numbers are 5 and 10.

(3 x - 2 y + z)^{2} = (3 x)^{2} + (- 2 y)^{2} + z^{2} + 2 (3 x) (- 2 y) + 2 (- 2 y) (z) + 2 (z) (3 x) = 9 x^{2} + 4 y^{2} + z^{2} - 12 x y - 4 y z + 6 z x

OR: Let the distance be d.

10^{2} = d^{2} + 8^{2} ⟹ 100 = d^{2} + 64 ⟹ d^{2} = 36 ⟹ d = 6

cm.

Total balls = 3 + 5 + 4 = 12.(i) P(Red) =

\frac{3}{12} = \frac{1}{4}

.(ii) P(Green) =

\frac{5}{12}

.(iii) P(Blue) =

\frac{4}{12} = \frac{1}{3}

OR:

a = 2, d = 5

t_{10} = a + (10 - 1) d = 2 + 9 (5) = 2 + 45 = 47

Diameter

d = 60

cm. Circumference =

π d = \frac{22}{7} \times 60 = \frac{1320}{7}

cm or approx 188.57 cm.

OR: Let width be

w

. Length

l = 2 w + 3

. Perimeter

2 (l + w) = 48 ⟹ 2 (2 w + 3 + w) = 48 ⟹ 2 (3 w + 3) = 48 ⟹ 6 w + 6 = 48 ⟹ 6 w = 42 ⟹ w = 7

cm. Length

l = 2 (7) + 3 = 17

cm. Dimensions are 17 cm × 7 cm.

a^{2} + b^{2} = (a + b)^{2} - 2 a b = 10^{2} - 2 (21) = 100 - 42 = 58

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) = (10) (58 - 21) = 10 (37) = 370

.OR:

x^{2} + 11 x + 30 = x^{2} + 5 x + 6 x + 30 = x (x + 5) + 6 (x + 5) = (x + 5) (x + 6)

OR: Square side = 14 cm. Radius of semicircle = 7 cm. Area of square =

14^{2} = 196

^{2}

. Area of two semicircles = Area of one circle =

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

^{2}

. Shaded Area = Area of square - Area of two semicircles =

196 - 154 = 42

^{2}

Total cards = 52. Colour cards = 16 (4 Kings, 4 Queens, 4 Jacks, 4 Aces).(i) P(Colour card) =

\frac{16}{52} = \frac{4}{13}

.(ii) P(Ace) =

\frac{4}{52} = \frac{1}{13}

.(iii) P(Neither colour card nor Ace) = P(Not colour card AND Not Ace). Colour cards include Aces. So P(Not colour card) =

1 - \frac{16}{52} = \frac{36}{52} = \frac{9}{13}

. P(Not Ace) =

1 - \frac{4}{52} = \frac{48}{52} = \frac{12}{13}

. Cards that are neither colour nor Ace are number cards (2 to 10). There are 9 number cards in each of the 4 suits, so

9 \times 4 = 36

. P(Neither colour card nor Ace) =

\frac{36}{52} = \frac{9}{13}

Radius

r = 21

m.(i) Perimeter (Circumference) =

2 π r = 2 \times \frac{22}{7} \times 21 = 2 \times 22 \times 3 = 132

m. (2 marks)(ii) Area =

π r^{2} = \frac{22}{7} \times 21^{2} = 22 \times 21 \times 3 = 1386

^{2}

. 2marks

Linear polynomial:

y = 50 x + 150

.(i)

y = 50 x + 150

. (1 mark)(ii) For 25 shirts,

x = 25

. Cost =

50 (25) + 150 = 1250 + 150 = 1400

. (1 mark)(iii)

1150 = 50 x + 150 ⟹ 50 x = 1000 ⟹ x = 20

shirts. 2marks

Total students = 50.(i) Number of students scoring between 40 and 60 = 15. P(40-60) =

\frac{15}{50} = \frac{3}{10}

. (1 mark)(ii) Number of students scoring less than 40 = 5 + 10 = 15. P(less than 40) =

\frac{15}{50} = \frac{3}{10}

. (1 mark)(iii) Number of students scoring 80 or more = 8. P(80 or more) =

\frac{8}{50} = \frac{4}{25}

. 2marks

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Class 9 Mathematics — Practice Paper 2

Answer Key