CBSE Class 9 Mathematics Practice Paper 1 (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

All questions are compulsory.

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

Section A has 18 MCQs and Assertion-Reason questions of 1 mark each.

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

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

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

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

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.

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, ...

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.

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}

From a point P on a circle with centre O, a chord PQ is drawn such that its perpendicular distance from the centre is 5 cm. If the radius of the circle is 13 cm, find the length of the chord PQ. Also, find the angle subtended by the chord at the centre.

A game involves spinning an arrow on a circular board divided into 8 sectors, numbered 1 to 8. The sectors are of equal size.(i) What is the probability that the arrow stops at sector 5?(ii) What is the probability that the arrow stops at an even number?(iii) What is the probability that the arrow stops at a number greater than 8?

E12 Marks

(Case-based Questions: 4 Marks Each)

Case Study 1: Coordinate Geometry in a City MapA city is planned on a Cartesian coordinate system. The main bus stand is at the origin (0, 0). The central park is located at (4, 3). The library is at (–2, 5), and the shopping mall is at (6, –4).

(i) What is the distance of the shopping mall from the bus stand? (1 mark)
(ii) If the cinema hall is located at (4, –7), what is the distance between the central park and the cinema hall? (1 mark)
(iii) (A) The school is located at (–3, –6). Is the school in the same quadrant as the library? OR (B) What is the distance of the library from the bus stand? (2 marks)

Case Study 2: Linear Patterns in GrowthA plant grows linearly. Its height is 10 cm on Day 1 and 15 cm on Day 3.

(i) What is the constant growth rate per day? (1 mark)
(ii) What is the height of the plant on Day 7? (1 mark)
(iii) (A) What is the height of the plant on Day 10? OR (B) Find a linear expression that relates the height $h$ (in cm) to the day $d$ . (2 marks)

Case Study 3: Probability in a GameA game is played with a spinner that has 5 equal sectors labeled as 1, 2, 3, 4, 5. A bag contains 4 red and 6 blue marbles. A player spins the arrow and draws one marble from the bag.

(i) What is the probability of spinning a 3? (1 mark)
(ii) What is the probability of drawing a blue marble? (1 mark)
(iii) (A) What is the probability of spinning an odd number? OR (B) What is the probability of spinning a number greater than 5? (2 marks)

Answer Key

SECTION A

(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.

SECTION B

Quadrant II: B(–3, 4)

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 = \frac{p^{2}}{q^{2}} ⟹ p^{2} = 7 q^{2}

. This means

p^{2}

is divisible by 7, so

p

must be divisible by 7. Let

p = 7 k

. Then

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

. This means

q^{2}

is divisible by 7, so

q

must be divisible by 7. Thus, p and q have a common factor 7, which contradicts that p and q are co-prime. Therefore,

\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

. Given

C = 132

cm and

π = \frac{22}{7}

. So,

132 = 2 \times \frac{22}{7} \times r ⟹ r = \frac{132 \times 7}{2 \times 22} = \frac{924}{44} = 21

cm.

SECTION C

Let the numbers be

x

and

y

x + y = 15

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

. Substituting

x + y = 15

, we get

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

. The numbers are roots of

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

. Factoring,

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

. So the numbers are 5 and 10.

Using the identity

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

, we have

(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

Let

r_{1} = 10

cm,

r_{2} = 8

cm, and distance between centres

d = \sqrt{10^{2} - x^{2}} + \sqrt{8^{2} - x^{2}} = 17

. (Error in question, distance between centres usually given. Assuming distance between centres = 17cm)Let the length of the common chord be

2 x

. Let the distance of the chord from the center of the first circle be

d_{1}

and from the second circle be

d_{2}

. So

d_{1} + d_{2} = 17

.We have

r_{1}^{2} = d_{1}^{2} + x^{2} ⟹ 100 = d_{1}^{2} + x^{2}

.And

r_{2}^{2} = d_{2}^{2} + x^{2} ⟹ 64 = d_{2}^{2} + x^{2}

100 - d_{1}^{2} = 64 - d_{2}^{2} ⟹ 36 = d_{1}^{2} - d_{2}^{2} = (d_{1} - d_{2}) (d_{1} + d_{2})

36 = (d_{1} - d_{2}) (17) ⟹ d_{1} - d_{2} = \frac{36}{17}

.Solving

d_{1} + d_{2} = 17

and

d_{1} - d_{2} = \frac{36}{17}

2 d_{1} = 17 + \frac{36}{17} = \frac{289 + 36}{17} = \frac{325}{17} ⟹ d_{1} = \frac{325}{34}

x^{2} = 100 - d_{1}^{2} = 100 - (\frac{325}{34})^{2} = 100 - \frac{105625}{1156} = \frac{115600 - 105625}{1156} = \frac{9975}{1156}

x = \sqrt{\frac{9975}{1156}} \approx 2.95

.Length of common chord =

2 x \approx 5.9

cm.(Assuming standard problem: common chord length is usually asked when distance between centres is such that intersection exists. If they are externally tangent, common tangent length is asked. If distance between centres is 0, concentric circles, no common chord. If distance between centres is less than sum of radii and greater than difference of radii, they intersect. The value 17 is greater than sum 18, so they dont intersect. Assuming there is a typo and the distance between centres is say 10 cm)If distance between centres is 10cm:

36 = (d_{1} - d_{2}) (10) ⟹ d_{1} - d_{2} = 3.6

d_{1} + d_{2} = 10

2 d_{1} = 13.6 ⟹ d_{1} = 6.8

x^{2} = 100 - (6.8)^{2} = 100 - 46.24 = 53.76

x = \sqrt{53.76} = 7.33

.Common chord length =

2 x = 14.66

cm.

AP: 3, 7, 11, 15, ...First term

a = 3

. Common difference

d = 7 - 3 = 4

.Sum of first n terms

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

.For n=20,

S_{20} = \frac{20}{2} [2 (3) + (20 - 1) 4] = 10 [6 + 19 \times 4] = 10 [6 + 76] = 10 [82] = 820

Diameter

d = 60

cm. Radius

r = 30

cm.Distance covered in one revolution = Circumference of the wheel =

π d

.Using

π = \frac{22}{7}

, Circumference =

\frac{22}{7} \times 60 = \frac{1320}{7} \approx 188.57

cm.

SECTION D

Let the width of the path be

w

metres.Original dimensions of the garden: length

l = 12

m, breadth

b = 5

m.Original area =

12 \times 5 = 60

^{2}

.New dimensions with path: length

L = 12 + 2 w

, breadth

B = 5 + 2 w

.New area =

(12 + 2 w) (5 + 2 w) = 60 + 24 w + 10 w + 4 w^{2} = 60 + 34 w + 4 w^{2}

.Given new area = 168 m

^{2}

.So,

60 + 34 w + 4 w^{2} = 168

4 w^{2} + 34 w - 108 = 0

.Divide by 2:

2 w^{2} + 17 w - 54 = 0

.Using quadratic formula

w = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 17 \pm \sqrt{17^{2} - 4 (2) (- 54)}}{2 (2)}

w = \frac{- 17 \pm \sqrt{289 + 432}}{4} = \frac{- 17 \pm \sqrt{721}}{4}

.Since width cannot be negative,

w = \frac{- 17 + \sqrt{721}}{4}

\sqrt{721}

is approx 26.85.

w \approx \frac{- 17 + 26.85}{4} = \frac{9.85}{4} \approx 2.46

m.(Let's check for integer solutions if possible, maybe the area was chosen to yield one)If the new area was 140.

2 w^{2} + 17 w - 40 = 0

. No integer factors.If area was 180.

4 w^{2} + 34 w - 120 = 0 ⟹ 2 w^{2} + 17 w - 60 = 0

(2 w - 5) (w + 12) = 0

w = 2.5

w = - 12

. Width is 2.5m.Let's assume the area was meant to be 180 m

^{2}

.Width of the path = 2.5 m.

Given

a + b = 10

and

a b = 21

.We know the identity

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

.So,

10^{2} = a^{2} + b^{2} + 2 (21)

100 = a^{2} + b^{2} + 42

a^{2} + b^{2} = 100 - 42 = 58

We know the identity $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ .
$a^{3} + b^{3} = (10) (58 - 21) = 10 (37) = 370$ .

Let the chord be PQ. Let the centre of the circle be O. Let M be the midpoint of PQ. OM is perpendicular to PQ.Given radius

O P = 13

cm. Distance of chord from centre

O M = 5

cm.In right-angled triangle OMP, by Baudhāyana-Pythagoras Theorem:

O P^{2} = O M^{2} + M P^{2}

13^{2} = 5^{2} + M P^{2}

169 = 25 + M P^{2}

M P^{2} = 169 - 25 = 144

M P = \sqrt{144} = 12

cm.Length of chord PQ =

2 \times M P = 2 \times 12 = 24

cm.

To find the angle subtended at the centre, consider $△ O M P$ .
$\sin (∠ M O P) = \frac{M P}{O P} = \frac{12}{13}$ .
$∠ M O P = \arcsin (\frac{12}{13})$ .
The angle subtended by the chord at the centre is $∠ P O Q = 2 \times ∠ M O P = 2 \arcsin (\frac{12}{13})$ .
This is approximately $2 \times {67.38}^{\circ} = {134.76}^{\circ}$ .
(The question only asks to find the length, then "Also, find the angle subtended at the centre". The angle is not standard)

The circular board is divided into 8 equal sectors. Total possible outcomes = 8.(i) Probability of stopping at sector 5 =

\frac{Number of favourable outcomes}{Total number of outcomes} = \frac{1}{8}

.(ii) The even numbers are 2, 4, 6, 8. Number of favourable outcomes = 4.Probability of stopping at an even number =

\frac{4}{8} = \frac{1}{2}

.(iii) (A) The odd numbers are 1, 3, 5, 7. Number of favourable outcomes = 4.Probability of spinning an odd number =

\frac{4}{8} = \frac{1}{2}

.(B) Numbers greater than 8: There are no sectors labeled greater than 8. Number of favourable outcomes = 0.Probability of spinning a number greater than 8 =

\frac{0}{8} = 0

SECTION E

Case Study 1: Coordinate Geometry in a City Map(i) Bus stand (0,0), Shopping mall (6,-4). Distance =

\sqrt{(6 - 0)^{2} + (- 4 - 0)^{2}} = \sqrt{6^{2} + (- 4)^{2}} = \sqrt{36 + 16} = \sqrt{52} = 2 \sqrt{13}

units. (1 mark)(ii) Central park (4,3), Cinema hall (4,-7). Distance =

\sqrt{(4 - 4)^{2} + (- 7 - 3)^{2}} = \sqrt{0^{2} + (- 10)^{2}} = \sqrt{100} = 10

units. (1 mark)(iii) (A) School (–3, –6), Library (–2, 5). School is in Quadrant III. Library is in Quadrant II. They are not in the same quadrant. OR (B) Library (–2, 5). Distance from bus stand (0,0) =

\sqrt{(- 2 - 0)^{2} + (5 - 0)^{2}} = \sqrt{(- 2)^{2} + 5^{2}} = \sqrt{4 + 25} = \sqrt{29}

units. 2marks

Case Study 2: Linear Patterns in GrowthLet the height be

h

and day be

d

. The relation is linear:

h = m d + c

.(i) Day 1 height = 10 cm. Day 3 height = 15 cm.The growth in height is

15 - 10 = 5

cm over

3 - 1 = 2

days.Constant growth rate

m = \frac{5}{2} = 2.5

cm per day. (1 mark)(ii) Height on Day 1 is 10 cm. Height on Day 3 is 15 cm.The equation is

h = 2.5 d + c

.Using Day 1:

10 = 2.5 (1) + c ⟹ c = 7.5

.So,

h = 2.5 d + 7.5

.Height on Day 7 =

2.5 (7) + 7.5 = 17.5 + 7.5 = 25

cm. (1 mark)(iii) (A) Height on Day 10 =

2.5 (10) + 7.5 = 25 + 7.5 = 32.5

cm. OR (B) The linear expression relating height

h

to day

d

h = 2.5 d + 7.5

. 2marks

Case Study 3: Probability in a GameSpinner has 5 equal sectors (1, 2, 3, 4, 5). Total outcomes = 5.Bag has 4 red + 6 blue marbles. Total marbles = 10.(i) Probability of spinning a 3 =

\frac{Number of sectors labeled 3}{Total number of sectors} = \frac{1}{5}

. (1 mark)(ii) Probability of drawing a blue marble =

\frac{Number of blue marbles}{Total number of marbles} = \frac{6}{10} = \frac{3}{5}

. (1 mark)(iii) (A) Odd numbers on spinner are 1, 3, 5. Number of favourable outcomes = 3.Probability of spinning an odd number =

\frac{3}{5}

. OR (B) Numbers greater than 5 on spinner: None. Number of favourable outcomes = 0.Probability of spinning a number greater than 5 =

\frac{0}{5} = 0

. 2marks

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Class 9 Mathematics — Practice Paper 1

Answer Key