CBSE Class 9 Mathematics Practice Paper 3 (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 18 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)

In which quadrant does the point (–5, –3) lie?(a) Quadrant I(b) Quadrant II(c) Quadrant III(d) Quadrant IV

The degree of the polynomial

5 x^{4} - 7 x^{2} + 2 x - 9

is:(a) 2(b) 4(c) 1(d) 9

Which of the following is a rational number?(a)

π

(b)

\sqrt{2}

(c)

\frac{3}{7}

(d)

0.1010010001...

The identity

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

is true for:(a) Integers only(b) Rational numbers only(c) Real numbers(d) Complex numbers only

The length of an arc of a circle with radius

r

and subtending an angle

θ

at the centre is:(a)

\frac{θ}{360^{\circ}} \times 2 π r^{2}

(b)

\frac{θ}{180^{\circ}} \times π r

(c)

\frac{θ}{360^{\circ}} \times π r^{2}

(d)

\frac{θ}{360^{\circ}} \times 2 π r

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

In an arithmetic progression, the

n^{t h}

term is given by

t_{n} = 5 n + 2

. The common difference is:(a) 5(b) 2(c) 7(d) 3

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 zero polynomial is:(a) 0(b) 1(c) Undefined(d) Indeterminate

The sum of two rational numbers is always:(a) Irrational(b) Rational(c) An integer(d) A natural number

The algebraic identity

a^{2} - b^{2} = ?

(a)

(a - b)^{2}

(b)

(a + b)^{2}

(c)

(a - b) (a + b)

(d)

(a - b) (a - b)

If the diameter of a circle is 14 cm, its circumference is: (use

π = \frac{22}{7}

)(a) 44 cm(b) 88 cm(c) 22 cm(d) 154 cm

A die is rolled. The probability of getting a number greater than 4 is:(a)

\frac{1}{6}

(b)

\frac{2}{6}

(c)

\frac{3}{6}

(d)

\frac{4}{6}

The sequence 2, 4, 8, 16, ... is an example of:(a) Arithmetic Progression(b) Geometric Progression(c) Constant Sequence(d) None of these

The point where the x-axis and y-axis intersect is called:(a) The origin(b) The vertex(c) The centre(d) The intersection point

The number of terms in the polynomial

7 y^{3} - 4 y^{2} + y - 10

is:(a) 1(b) 3(c) 4(d) 7

The decimal expansion of

\frac{5}{8}

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

The area of a triangle with base

b

and height

h

is:(a)

b \times h

(b)

\frac{1}{2} \times b \times h

(c)

2 \times b \times h

(d)

\frac{1}{2} \times (b + h)

Assertion (A): A circle can pass through any three non-collinear points.Reason (R): The perpendicular bisectors of the sides of a triangle intersect at a single point, which is the circumcentre.(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): In a geometric progression, the ratio of any term to its preceding term is constant.Reason (R): The formula for the

n^{t h}

term of a GP is

t_{n} = a r^{n - 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)

Locate the points P(–2, 3), Q(4, –1), R(–5, –4) and S(3, 5) on the Cartesian plane. In which quadrant does S lie?

f (x) = x^{3} - 6 x^{2} + 8 x - 2

, find

f (3)

Represent the rational numbers

\frac{2}{5}

and

- \frac{3}{4}

on the number line.

Expand

(2 a - 3 b)^{2}

using an appropriate identity.

A chord of length 12 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre.

C18 Marks

(Short Answer II Questions: 3 Marks Each)

A train travels at a uniform speed. If the train were 6 km/hr faster, it would have taken 4 hours less for a journey of 288 km. What is the original speed of the train?

Factorise:

16 x^{2} + 24 x y + 9 y^{2}

In Fig. 1, ABCD is a cyclic quadrilateral with AC and BD as diagonals. If AB = AD and AC = BD, prove that ABCD is a rectangle.[Fig. 1: A cyclic quadrilateral ABCD with diagonals AC and BD intersecting]ORProve that equal chords of a circle are equidistant from the centre.

A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting:(i) A king(ii) A red face card(iii) A spade

Find the sum of the first 15 terms of the AP: 10, 7, 4, ...ORFind the

25^{t h}

term of the AP whose first term is -8 and common difference is -3.

The area of a sector of a circle of radius 4 cm is

16 π

^{2}

. Find the angle subtended by the sector at the centre.

D20 Marks

(Long Answer Questions: 5 Marks Each)

A rectangular park is surrounded by a path of uniform width. The area of the park is 800 m

^{2}

. If the path width is 2m, the area of the path is 272 m

^{2}

. Find the dimensions of the park.ORA rectangular piece of paper has length 20 cm and width 12 cm. Four congruent squares are cut from the corners of the paper. The sides of the squares are x cm. The remaining paper is folded up to form an open box. Find the volume of the box in terms of x.

x + y + z = 15

and

x^{2} + y^{2} + z^{2} = 83

, find the value of

x y + y z + z x

.ORFactorise:

4 x^{2} - 9 y^{2}

In Fig. 2, a chord PQ is drawn in a circle of radius 13 cm such that its distance from the centre is 5 cm.[Fig. 2: A circle with centre O, radius 13 cm and a chord PQ at distance 5 cm from O](i) Find the length of the chord PQ. (2 marks)(ii) If another chord RS is drawn such that its distance from the centre is 12 cm, find the length of chord RS. (2 marks)(iii) Which chord is longer and why? 1mark

E12 Marks

(Case-based Questions: 4 Marks Each)

Case Study 1: Coordinate Geometry in Real Life
A city planner is designing a new neighbourhood. The main roads of the city form a coordinate system. The library is located at point L(–3, 2), the park is at P(4, 5), and the school is at S(2, –3).

(i) What is the distance between the library and the park? 1mark(ii) What is the distance between the park and the school? (1 mark)(iii) If the city planner wants to build a community centre at point C such that it is equidistant from the library and the school, which axis would the community centre lie on? Justify your answer. (2 marks)

Case Study 2: Polynomials in Cost Calculation
A shopkeeper sells two types of items. Item A costs $x$ rupees per piece, and item B costs $y$ rupees per piece. A customer buys 5 pieces of item A and 3 pieces of item B. Additionally, there is a service charge of ₹50.

(i) Write an algebraic expression representing the total cost for the customer. 1mark(ii) If item A costs ₹20 per piece and item B costs ₹30 per piece, calculate the total cost. (1 mark)(iii) If the shopkeeper decides to increase the price of item A by ₹5 and decrease the price of item B by ₹2, write the new expression for the total cost (excluding the service charge). (2 marks)

Case Study 3: Probability in a Game
Rohan is playing a game with a spinner that has 6 equal sectors numbered 1 to 6. He also has a bag containing 4 red marbles and 6 blue marbles. He spins the spinner and draws a marble from the bag.

(i) What is the probability of getting an even number on the spinner? 1mark(ii) What is the probability of drawing a blue marble from the bag? (1 mark)(iii) What is the probability that Rohan spins an odd number and draws a red marble? (2 marks)

Answer Key

SECTION A

(b) 4

(c)

\frac{3}{7}

(d)

\frac{θ}{360^{\circ}} \times 2 π r

(b) 1

(a) 5

(b) Quadrant II

(b) Rational

(c)

(a - b) (a + b)

(a) 44 cm

(b)

\frac{2}{6}

(b) Geometric Progression

(a) The origin

(a) Terminating

(b)

\frac{1}{2} \times b \times h

(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.

SECTION B

Q(4, –1) is in Quadrant IV. R(–5, –4) is in Quadrant III. S(3, 5) is in Quadrant I. P(–2, 3) lies in the Second Quadrant.

f (3) = 3^{3} - 6 (3^{2}) + 8 (3) - 2 = 27 - 54 + 24 - 2 = - 5

Represent

\frac{2}{5}

at 0.4 and

- \frac{3}{4}

at -0.75 on the number line.

\sqrt{7}

cannot be written as

\frac{p}{q}

where p, q are integers and

q \neq 0

(2 a - 3 b)^{2} = (2 a)^{2} - 2 (2 a) (3 b) + (3 b)^{2} = 4 a^{2} - 12 a b + 9 b^{2}

Let

r

be the radius and

d

be the distance from the centre. By Pythagoras theorem,

r^{2} = d^{2} + (\frac{chord}{2})^{2}

10^{2} = d^{2} + (\frac{12}{2})^{2} \Rightarrow 100 = d^{2} + 36 \Rightarrow d^{2} = 64 \Rightarrow d = 8

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} \Rightarrow \frac{15}{x y} = \frac{3}{10} \Rightarrow x y = 50

. The numbers are the roots of

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

, which 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:

r^{2} = d^{2} + (\frac{chord}{2})^{2} \Rightarrow 10^{2} = 12^{2} + (\frac{chord}{2})^{2} \Rightarrow 100 = 144 + (\frac{chord}{2})^{2}

, this is incorrect. Re-reading: distance is 12, radius is 10. This case is not possible as distance from centre cannot be greater than radius. Let's assume radius is 13 and distance is 12.

13^{2} = 12^{2} + (\frac{chord}{2})^{2} \Rightarrow 169 = 144 + (\frac{chord}{2})^{2} \Rightarrow (\frac{chord}{2})^{2} = 25 \Rightarrow \frac{chord}{2} = 5 \Rightarrow chord = 10

cm. Assuming the question meant radius 13, distance 12. If radius is 10 and chord is 16, this is also not possible. Let's assume radius is 10, chord is 12. Then

10^{2} = d^{2} + (12 / 2)^{2} \Rightarrow 100 = d^{2} + 36 \Rightarrow d = 8

cm. (Using values from Q25 for consistency).

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: For AP: 2, 7, 12, ...

a = 2, d = 5

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

.For AP: 10, 7, 4, ...

a = 10, d = - 3

S_{15} = \frac{15}{2} [2 a + (15 - 1) d] = \frac{15}{2} [2 (10) + 14 (- 3)] = \frac{15}{2} [20 - 42] = \frac{15}{2} [- 22] = 15 \times (- 11) = - 165

Radius

r = \frac{60}{2} = 30

cm. Distance covered in one revolution is the circumference.

C = 2 π r = 2 \times \frac{22}{7} \times 30 = \frac{1320}{7} \approx 188.57

cm.

SECTION D

Let dimensions of the park be

l

and

w

l w = 800

. New dimensions

(l + 4) (w + 4) = 800 + 272 = 1072

l w + 4 l + 4 w + 16 = 1072

800 + 4 (l + w) + 16 = 1072

4 (l + w) = 256 \Rightarrow l + w = 64

. Solve

l + w = 64

and

l w = 800

t^{2} - 64 t + 800 = 0

. Using quadratic formula

t = \frac{64 \pm \sqrt{64^{2} - 4 (800)}}{2} = \frac{64 \pm \sqrt{4096 - 3200}}{2} = \frac{64 \pm \sqrt{896}}{2} = \frac{64 \pm 8 \sqrt{14}}{2} = 32 \pm 4 \sqrt{14}

. Dimensions are

32 + 4 \sqrt{14}

m and

32 - 4 \sqrt{14}

m.OR: Let width be

w

. Length

l = 2 w + 3

. Perimeter

2 (l + w) = 48 \Rightarrow l + w = 24

(2 w + 3) + w = 24 \Rightarrow 3 w = 21 \Rightarrow w = 7

cm.

l = 2 (7) + 3 = 17

cm. Dimensions are 17 cm and 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)^{3} - 3 a b (a + b) = 10^{3} - 3 (21) (10) = 1000 - 630 = 370

.OR:

x^{2} + 11 x + 30

. Find two numbers whose product is 30 and sum is 11. These are 5 and 6. So,

(x + 5) (x + 6)

(i) Let O be the center. Let M be the midpoint of PQ. In right triangle OMP,

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

13^{2} = 5^{2} + M P^{2} \Rightarrow 169 = 25 + M P^{2} \Rightarrow M P^{2} = 144 \Rightarrow M P = 12

. Length of chord PQ =

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

cm. 2marks(ii) For chord RS, radius is 13 cm, distance is 12 cm.

13^{2} = 12^{2} + (\frac{R S}{2})^{2} \Rightarrow 169 = 144 + (\frac{R S}{2})^{2} \Rightarrow (\frac{R S}{2})^{2} = 25 \Rightarrow \frac{R S}{2} = 5 \Rightarrow R S = 10

cm. (2 marks)(iii) Chord PQ (24 cm) is longer than chord RS (10 cm) because it is closer to the centre (5 cm vs 12 cm). (1 mark)

E12 Marks

Case Study 1: Coordinate Geometry in Real Life

(i) Distance LP =

\sqrt{(4 - (- 3))^{2} + (5 - 2)^{2}} = \sqrt{7^{2} + 3^{2}} = \sqrt{49 + 9} = \sqrt{58}

units. 1mark(ii) Distance PS =

\sqrt{(2 - 4)^{2} + (- 3 - 5)^{2}} = \sqrt{(- 2)^{2} + (- 8)^{2}} = \sqrt{4 + 64} = \sqrt{68}

units. (1 mark)(iii) For the community centre C(x, y) to be equidistant from L(–3, 2) and S(2, –3), it must lie on the perpendicular bisector of LS. The midpoint of LS is

(\frac{- 3 + 2}{2}, \frac{2 - 3}{2}) = (- \frac{1}{2}, - \frac{1}{2})

. The slope of LS is

\frac{- 3 - 2}{2 - (- 3)} = \frac{- 5}{5} = - 1

. The slope of the perpendicular bisector is 1. Equation of perpendicular bisector:

y - (- \frac{1}{2}) = 1 (x - (- \frac{1}{2})) \Rightarrow y + \frac{1}{2} = x + \frac{1}{2} \Rightarrow y = x

. A point where

y = x

does not necessarily lie on any specific axis unless

x = y = 0

. However, if the community centre must be equidistant and lies on an axis, it would imply symmetry. If the y-coordinate of L and S were opposite and x-coordinates were same, it would be on x-axis. If x-coordinates were opposite and y-coordinates were same, it would be on y-axis. Here, it lies on the line

y = x

. If we need to pick an axis, it is likely that the question implies a situation where the perpendicular bisector is one of the axes. Re-interpreting: If the community centre C(x,y) is on the y-axis, then

x = 0

C L^{2} = (- 3 - 0)^{2} + (2 - y)^{2} = 9 + (2 - y)^{2}

C S^{2} = (2 - 0)^{2} + (- 3 - y)^{2} = 4 + (- 3 - y)^{2}

9 + 4 - 4 y + y^{2} = 4 + 9 + 6 y + y^{2} \Rightarrow 13 - 4 y = 13 + 6 y \Rightarrow 10 y = 0 \Rightarrow y = 0

. So C(0,0). If C is on the x-axis, then

y = 0

C L^{2} = (- 3 - x)^{2} + (2 - 0)^{2} = (- 3 - x)^{2} + 4

C S^{2} = (2 - x)^{2} + (- 3 - 0)^{2} = (2 - x)^{2} + 9

9 + 6 x + x^{2} + 4 = 4 - 4 x + x^{2} + 9 \Rightarrow 13 + 6 x = 13 - 4 x \Rightarrow 10 x = 0 \Rightarrow x = 0

. So C(0,0). The community centre is at the origin (0,0). The origin lies on both axes. The question asks which axis it would lie on. The perpendicular bisector is

y = x

. This line passes through the origin. The origin lies on both axes. Let's assume it asks which axis would be relevant IF it were on an axis. This seems to imply a case where the perpendicular bisector IS an axis. This happens when points are symmetric with respect to an axis. Here, points are not symmetric to either axis. The perpendicular bisector is

y = x

. The origin (0,0) is equidistant from L and S. So, the community centre can be at the origin. The origin lies on both axes. If forced to pick one, the context is missing. Let's state it lies on BOTH. If the question implies a single axis, then there might be an error or it's a trick. The perpendicular bisector

y = x

does not coincide with the x-axis (

y = 0

) or y-axis (

x = 0

) except at the origin. The most logical equidistant point is the origin. The origin lies on both axes.

Case Study 2: Polynomials in Cost Calculation

(i) Total cost =

5 x + 3 y + 50

. 1mark(ii) If

x = 20

and

y = 30

, Total cost =

5 (20) + 3 (30) + 50 = 100 + 90 + 50 = 240

rupees. (1 mark)(iii) New price of A =

x + 5

. New price of B =

y - 2

. New expression for total cost (excluding service charge) =

5 (x + 5) + 3 (y - 2) = 5 x + 25 + 3 y - 6 = 5 x + 3 y + 19

. (2 marks)

Case Study 3: Probability in a Game

(i) Spinner outcomes: {1, 2, 3, 4, 5, 6}. Even numbers: {2, 4, 6}. Total outcomes = 6. Favourable outcomes = 3. P(Even number) =

\frac{3}{6} = \frac{1}{2}

. 1mark(ii) Marble outcomes: {Red, Blue}. Total marbles = 4 + 6 = 10. Red = 4, Blue = 6. P(Blue marble) =

\frac{6}{10} = \frac{3}{5}

. (1 mark)(iii) P(Odd number on spinner) =

\frac{3}{6} = \frac{1}{2}

. P(Red marble) =

\frac{4}{10} = \frac{2}{5}

. Since the events are independent, P(Odd number and Red marble) = P(Odd number)

\times

P(Red marble) =

\frac{1}{2} \times \frac{2}{5} = \frac{2}{10} = \frac{1}{5}

. (2 marks)

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Class 9 Mathematics — Practice Paper 3

Answer Key