Class 9 Question Paper 2025 Maths

ANNUAL EXAMINATION

Question Paper
2025
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SESSION ENDING EXAMINATION 2024-25
CLASS: IX ​ ​ ​ ​ ​ ​ ​ SUBJECT: MATHEMATICS
TIME: 3 HOURS​ ​ ​ ​ ​ ​ M. MARKS: 80

General Instructions:
1.​ All questions are compulsory.
2.​ This question paper contains 38 questions.
3.​ There is no overall choice. However internal choices are provided in some questions.
4.​ Question paper is divided into five sections: Section-A contains 18 Multiple Choice
Questions and 02 Assertion Reason based questions each carry 1 mark, Section-B
contains 05 questions each carry 2 marks, Section-C contains 06 questions each carry 3
marks, Section-D contains 4 question each carry 5 marks and Section-E contains 3
questions each carry 4 marks.
5.​ Draw neat and clean figures wherever required.

Q No Section-A (20x1 = 20 marks) Marks

1 Which of the following statements is true? 1
(a)​ Every irrational number can be represented as a fraction.
(b)​Every irrational number can be represented with the help of decimals.
(c)​ Every rational number can be represented as a fraction.
(d)​Every rational number can be represented as an integer.

2 If (33)2 = 9x then 4x =? 1
a) 1 b) 4 c) 16 d) 64

3 Which of the following expression is polynomial in one variable. 1
1
a) 4x2 – 3x + 7 b) 4x2 – 3y + 7 c) 2 𝑥 + 5 d) y + 𝑦

4 Find the value of: (–12)3 + (7)3 + (5)3 1
a) 1275 b) 1260 c) – 1240 d) – 1260

5 If P (x) = 2x2 – 4x + 3, then P (–1) =? 1
a) 1 b) 5 c) 9 d) – 3

6 Which one of the following options is true, the equation y = 3x + 5 has 1
a) A unique solution b) Two solutions
c) No solution d) Many solutions

7 Which one option is correct for the equation 3y – 2x = 5(x + y) – 4 expressed 1
in standard from of linear equation as ax + by + c = 0?
a) 7x + 2y – 4 = 0 b) – 7x + 2y + 4 = 0
c) – 7x – 2y – 4 = 0 d) 7x – 2y + 4 = 0

8 Name of each part of the plane formed by horizontal and vertical lines in a 1
cartesian plane.
a) Origin b) Abscissa c) Ordinate d) Quadrant

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9 In adjacent figure, according to Euclid’s 5th postulate, the pair of 1
angles, having the sum less than 180° is:
a) 1 and 2 b) 2 and 4
c) 1 and 3 d) 3 and 4

10 The supplement of an angle y is: 1
a) 90⁰ + y b) 90⁰ – y c) 180⁰ + y d) 180⁰ – y

11 In ∆ABC, BC = AB and ∠B = 80°. Then ∠A is equal to 1
a) 800 b) 500 c) 200 d) 100

12 If the diagonals of a quadrilateral are equal and bisect at right angles, then the 1
quadrilateral is
a) Square b) Rectangle c) Rhombus d) Parallelogram

13 A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The 1
acute angle between the diagonals is
a) 550 b) 500 c) 400 d) 250

14 Which of the following statement is incorrect? 1
a)​ Equal chords of a circle subtend equal angles at the centre.
b)​ The perpendicular from the centre of a circle to a chord trisects the chord.
c)​ Angles in the same segment of a circle are equal.
d)​ Equal chords of a circle are equidistant from the centre.

15 In adjacent figure, if ∠ABC = 20º, then ∠AOC is equal to: 1
a) 200 b) 500
c) 400 d) 700​

16 The edges of a triangular board are 6 m, 8 m and 10 m. what is the cost of 1
painting it at the rate of ₹5 per square metre
a) ₹70 b) ₹80 c) ₹120 d) ₹240

17 If volume and surface area of a sphere is numerically equal, then its diameter 1
is
a) 2 units b) 3 units c) 4 units d) 6 units

18 In the class intervals 15 – 25, 25 – 35, the number 25 is included in 1
a) 15 – 25 b) 25 – 35 c) both d) none

DIRECTION: In the question number 19 and 20, a statement of Assertion
(A) is followed by a statement of Reason (R). Choose the correct option from
the following.
a) Both Assertion and Reason are TRUE and Reason is the correct
explanation of Assertion.

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b) Both Assertion and Reason are TRUE but Reason is not the correct
explanation of Assertion.
c) Assertion is TRUE but Reason is FALSE.
d) Assertion is FALSE but Reason is TRUE.

19 Assertion (A): The polynomial p (x) = 4x3 – 3x2 + 5x – 6 when divided by (x 1
– 1) gives zero as the remainder.
Reason (R): (x – 1) is a factor of the polynomial p (x) = 4x3 – 3x2 + 5 x – 6

20 Assertion (A): If the angles of a quadrilateral are x, (x + 30), (x − 30) and 2x, 1
the measure of the smallest angle is 520.
Reason (R): The sum of all angles of a quadrilateral is 3600.

Section - B (5x2 = 10 marks)

21 Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the 2
following case:
p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1
OR
Find the value of k, if y + 3 is a factor of 3y2 + ky + 6

22 If a point C lies between two points A and B such that AC = BC, then prove 2
1
that AC = 2 AB. Explain by drawing the figure.

23 Prove that a diagonal of a parallelogram divides it into two congruent 2
triangles.
OR
In the rhombus PQRS, PQ = 5 cm, PR = 8 cm. Find the length of the diagonal
SQ.

24 The hollow sphere, in which the circus motorcyclist performs his stunts, has a 2
diameter of 14 m. Find the area available to the motorcyclist for riding.
OR
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity of pit in
kilolitres?

25 The air distances of four cities from Delhi (in km) are given 2
City Kolkata Mumbai Chennai Hyderabad
Distance from Delhi 1340 1100 1700 1250
(km)
Draw a bar graph to represent the above data.

Section - C (6x3 = 18 marks)

26 Find three different irrational numbers between the rational numbers
5
and 3
7
11
13
.
OR

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Express 0. 4 17 in the form of p/q , where p and q are integers and q ≠ 0.

27 1
Verify that: x3 + y3 + z3 − 3 xyz = 2 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2] 3
OR
3 2
Factorise : x – 3x – 9x – 5

28 Express the following linear equations in the form ax + by + c = 0 and indicate 3
the values of a, b and c in each case:
𝑦
(i) 3y + 2x = 9.35 (ii) x + 5 – 10 = 0 (iii) – 3x + 2 = 0

29 If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. 3

30 In the adjoining figure, LM is a line parallel to the y-axis at a 3
distance of 3 units. Find,

(i) What are the coordinates of the points P, R and Q?
(ii) What is the difference between the abscissa of the points L and
ordinate of M?

31 A random survey of the number of children 3
of various age groups playing in a park was
found as follows:

Draw a histogram to represent the data.

Section - D (4x5 = 20 marks)

32 Represent 9. 7 on the number line and justify your answer. 5
OR
Show how 6 can be represented on the number line by square root spiral.

33 Prove that the diagonal divides a parallelogram into two congruent triangles. 5
∆ABC in which E and F are mid point of AB and BC respectively, if AE = 4
cm find AB.

34 Three boys Ashutosh, Bharat and Mridul are playing a game by standing on a 5
circle of radius 10 m drawn in a park. Ashutosh throws a ball to Bharat, Bharat
to Mridul, Mridul to Ashutosh. If the distance between Ashutosh and Bharat
and between Bharat and Mridul is 12 m each, what is the distance between
Ashutosh and Mridul?

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OR
In the adjacent figure, ∠OAB = 30º and ∠OCB =
57º. Find
(i) ∠AOC
(ii) ∠ABC
(iii) If intersecting point of AB and OC is P, find
∠OPB

35 The surface area of a sphere of radius 5 cm is five times the area of the curved 5
surface of a cone of radius 4 cm. Find the height and the volume of the cone.
OR
A dome of a building is in the form of a hemisphere. From inside, it was
white-washed at the cost of Rs. 4989.60. If the cost of white-washing is Rs. 20
per square metre, find the
(i) Inside surface area of the dome, 2
(ii) Inner radius of the dome 1
(iii) Volume of the air inside the dome. 2

Section - E (4x3 = 12 marks)

36 Swimming Pool Construction
Rajan has a form house in Patiala. He has a swimming pool also inside his
form house, in the shape of a rectangular prism with a length of 5x + 10
meters, width 3x + 5 meters, and depth x + 2 meters.

On the basis of the above information, answer the following:
(a) Write a polynomial to represent the volume of the pool. 1
(b) If x = 1, calculate the volume of swimming pool. 1
(c) The cost of filling the pool with water is ₹8 per cubic meter. Determine the
total cost to fill the pool. 2
OR
If Rajan wants to paint the walls of the pool, Determine the area to be
painted.

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37 Two cars are moving on two parallel roads represented as AB and CD
respectively in the given figure. First car reaches at point E and takes a turn
towards its right at an angle of 50⁰. At the same time, second car reaches at
point F and takes a turn towards its left at an angle of 60⁰. They both meet at a
point G. Based on the above information and given figure, answer the
following question (without considering the width of the roads)

(a) What will be the measure of angle y marked in the figure?
(b) What will be the measure of ∠EGF marked as x?
(c) What will be the measure of reflex ∠EGF?
OR
What will be the measure of reflex ∠AEG? 1
2

1

38 Triangles are used in bridges because they evenly distribute weight without
changing their proportions. When force is applied on a shape like rectangle it
would flatten out. Before triangles were used in bridges, they were weak and
could not be very big. To solve that problem engineers would put a post in the
middle of a square and make it sturdier. Isosceles triangles were used to
construct a bridge in which the base and equal sides of an isosceles triangle are
in the ratio 2:3:3 and its perimeter is 40 m.

(a) What are the measurements of the sides of an isosceles triangle? 1
OR
Find the semi-perimeter of the above triangle.
(b) What is the area of the above isosceles triangle so formed? 2

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(c) Find the cost of painting the so formed triangle at the rate of ₹ 18.50 per 1
𝑚2.

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SESSION ENDING EXAMINATION: 2024-25
SUB: MATHEMATICS​ ​ ​ ​ ​ ​ ​ TIME: 3:00 Hours
CLASS: IX​ ​ ​ ​ ​ ​ ​ ​ ​ Max. M.: 80 Marks

MARKING SCHEME
SECTION – A

1.​ c) Every rational number can be represented as a fraction. 1

2.​ d) 64 1

3.​ a) 4x2 – 3x + 7 1

4.​ d) – 1260 1

5.​ c) 9 1

6.​ d) Many solutions 1

7.​ a) 7x + 2y – 4 = 0 1

8.​ d) Quadrant 1

9.​ c) 1 and 3 1

10.​ d) 180⁰ – y 1

11.​ b) 500 1

12.​ a) Square 1

13.​ b) 500 1

14.​ b) The perpendicular from the centre of a circle to a chord trisects the chord. 1

15.​ c) 400​ 1

16.​ c) ₹120 1

17.​ d) 6 units 1

18.​ b) 25 – 35 1

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19.​ a) Both Assertion and Reason are TRUE and Reason is the correct explanation of Assertion. 1

20.​ d) Assertion is FALSE but Reason is TRUE. 1

SECTION – B

21.​ According to the factor theorem (x – a) is a factor of p(x), if p(a) = 0
Now g(x) = x + 1 = 0 ​ x=-1 ½
⇒​ P(x) = 2x3 + x2 – 2x – 1
⇒​ P(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1 = – 2 + 1 + 2 – 1 = 0 1
⇒​ So, we conclude that g(x) is a factor of p(x) ½

OR
½
⇒​ If y + 3 is a factor then p (- 3) = 0 [y + 3 = 0, so that y = - 3]
⇒​ P(x) = 3y2 + ky + 6 = 0
½
⇒​ P(-3) = 3(-3)2 + k(-3) + 6 = 0
1
⇒​ k = 11

22.​ Since it is given AC = BC ½
By figure we can say that AC + BC = AB
AC + AC = AB (given BC = AC) ½
2 AC = AB
​ AC = ½ AB 1

23.​ Given: ABCD be a parallelogram and AC be a diagonal ½
To Prove: ∆ABC and ∆CDA are congruent
Proof: Correct figure and proof 1½
OR
⇒​ Figure: PQRS is a rhombus,
⇒
​ Since diagonals bisect each other at 900
⇒​ So, O is the mid-point of PR and PO = RO ½
⇒​ Also, PO = PR/2 = 8/2 = 4 cm and PQ = 5 cm
½
⇒​ POQ is right angled triangle so by Pythagoras Theorem
⇒​ OQ = 3 cm
⇒​ Hence SQ = 2 × OQ = 2 × 3 = 6 𝑐𝑚
1
24.​ Given that: dimeter = 14 m

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⇒​ Radius = 14/2 = 7 m ½

⇒​ Surface area of sphere = 4 π r2 = 4×
22
×7 × 7 1
7

⇒​ 4 × 22 × 7 = 616
½
⇒​ Area available to the motorcyclist = 616 m2

OR
3.5 ½
Given that: diameter = 3.5 m, so r = 2
and h = 12 m
1 1 22 3.5 3.5 1
Volume of cone = 3 π r2 h = 3 × 7
× 2
× 2
× 12 = 38.5 m3
½
Capacity of pit in kilolitres = 38.5 ×1000/1000 = 38.5 kilolitres
25.​ Correct scale
Horizontal line and vertical line ½

Correct bar graph 1½

SECTION – C

26.​ Rational numbers 5
= 0.714… 3
7

11
and 13 .= 0.846…

Correct three irrational numbers
OR
Given that Express 0. 4 17
⇒​ Let x = 0.4171717… (1) ½

Multiply both side by 100
½
⇒​ 100 x = 41.71717… (2)

Subtract (1) from (2) 1
⇒​ 99 x = 41.3
1
⇒​ x = 41.3/99 = 413/990

27.​ RHS = 1 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
2

1
1
⇒​ 2 (x + y + z) [(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)]

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1
⇒​ 2 (x + y + z) [2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx]
1
⇒​ (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
⇒​ x2 + xy2 + xz2 – x2y – xyz – zx2 + yx2 + y3 + yz2 – xy2 – y2z – xyz + zx2 + zy2 + z3 –
xyz – yz2 – xz2
1
⇒​ x3 + y3 + z3 − 3 xyz = LHS

OR
P(x) = x3 – 3x2 – 9x – 5
According to the factor theorem, if p(a) = 0 then (x – a) is a factor of p(x)
∴ Factors of 5 = ± 1, ± 5 ½

⇒​ If P(1) = 0, then (x – 1) is a factor
P(1) = 13 – 3.12 – 9.1 – 5 = 1 – 3 – 9 – 5 = – 16 ≠ 0, so (x – 1) is not a factor
½
P(–1) = (–1)3 – 3.( –1)2 – 9.( –1) – 5 = – 1 – 3 + 9 – 5 = 0, so (x + 1) is a factor
⇒​ P(x) = x3 – 3x2 – 9x – 5
1
⇒​ = x2 (x + 1) – 4x (x + 1) – 5 (x + 1) = (x + 1) (x2 – 4x – 5)
1
⇒​ = (x + 1) ((x2 – 5x + x – 5) = (x + 1) (x + 1) (x – 5)

28.​ (i) 3y + 2x = 9.35 𝑦
(ii) x + 5 – 10 = 0 (iii) – 3x + 2 = 0 1

⇒​ 2x + 3y – 9.35 = 0 1
⇒​ – 3x + 0.y + 2 = 0 for
⇒​ x + 5 y – 10 = 0
⇒​ a = – 3, b = 0, c = + 2 each
⇒​ a = 2, b = 3,
1
⇒​ a = 1, b = 5 , c = –
c = – 9.35
10

29.​ It is given that Point (3, 4) lies on the 3y = ax + 7, ½
⇒​ So it should satisfy the linear equation 3y = ax + 7,
⇒​ Substitute x = 3 and y = 4 in the given equation 3y = ax + 7 ½
⇒​ 3y = ax + 7
⇒​ 3×4 = a×3 + 7 1
⇒​ 12 – 7 = 3a
⇒​ A = 5/3 1

30.​ (i) Coordinates of the points P, R and Q ½
⇒​ P = (3, 2) for
⇒​ Q = (3, – 1) each
⇒​ R = (3, 0)

(ii) Abscissa of the point L = 3 and ordinate of point M = – 3

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⇒​ Difference = 3 – (– 3) = 3 + 3 = 6

31.​ Correct
calculations 1½

Correct histogram 1½

SECTION – D

32.​ Correct representaion/ construction of 9. 7 on number 4

line 1

Justification

OR
​ 5

Show how 6 can be represented on the
number line by square root spiral.

33.​ Correct statement 1
Given, to prove, contruction 1
Correct proof 2

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AB = 8 cm 1

34.​ Now onward Ashutosh = A, Bharat = B and Mridul = M
Given: A circle of centre O and radius 10 m
It means OA = OB = OM = 10 m
and AB = BM = 12 m
Construction: Join OA, OB, OM, AM and OL ⊥ AB 1
Let intersection point of OB and AM be K
Solution: In ∆OAK & ∆OMK
⇒​ OA = OM (Radii of same circle)
⇒​ OK = OK (Common side)
⇒​ ∠𝐴𝑂𝐾 = ∠𝑀𝑂𝐾 (Angle subtended by equal chords at centre are equal)
⇒​ So ∆OAK ≅ ∆OMK (By ASA Congruence rule)
⇒​ Therefore AK = MK & ∠𝑂𝐾𝐴 = ∠𝑂𝐾𝑀 = 900 (CPCT)
⇒​ Hence AM ⊥ OB or AK ⊥ OB 1

Since OL ⊥ AB, so L is the mid-point of AB,
⇒​ BL = 6 m
⇒​ Therefore ∆OLB is right angled triangle, so by Pythagoras theorem OL = 8 m
1
⇒​ Now, Area of triangle OAB = ½ × AB × OL = ½ ×12 × 8 = 48 m2 (1)
⇒​ Again Area of triangle OAB = ½ × AK × OB = ½ × 𝐴𝐾 × 10 = 5×AK m2 (2)
⇒​ By eq. (1) and eq. (2) AK = 9.6 m 1
⇒​ Hence AM = AK + KM = 9.6 + 9.6 = 19.2 m 1

OR
In the adjacent figure, ∠OAB = 30º and ∠OCB = 57º
In ∆OAB,
⇒​ ∠OAB = ∠OBA = 300 (Angle opp. to equal side i.e. radii)
⇒​ ∠AOB = 180 – 2 x 300 = 1200
½

In ∆OCB,
⇒​ ∠OCB = ∠OBC = 570 (Angle opp. to equal side i.e. radii)
⇒​ ∠BOC = 180 – 2 x 570 = 1200 = 660 ½

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(i) ∠AOC = ∠AOB – ∠BOC = 1200 – 660 = 540 1
1
(ii) ∠ABC = ∠OBC – ∠OBA = 570 – 300 = 270
(iii) Intersecting point of AB and OC is P
∠OPB = ∠AOP + ∠OAP (Exterior angle property)
= ∠AOC + ∠OAB = 540 + 300 = 840 2

35.​ Radius of sphere = 5 cm
⇒​ Surface area of sphere = 4 π r2 = 100 π ½

Radius of cone = 4 cm and height be ‘h’ & slant height be ‘l’
½
⇒​ Curved surface of cone = π r l = 4 π l

As per given situation
⇒​ Surface area of square = 5 × CSA of cone 1
⇒​ 100 π = 5 × 4 π l = 20 π l l = 5 cm
1
2 2
Also h = 𝑙 −𝑟 , l = 5 cm, r = 4 cm
2
Now height ‘h’ = 3 cm (by pythagoras theorem)
1 1
⇒​ Volume of cone = 3 π r2 h = 3 × π × 4 ×4 ×3 = 16 π = 16 × 3.14 = 50.24 cm2

OR
We know that Area white washed × Cost of white wash = Total cost ½

⇒​ Area white washed × 20 = 4989.60
½
⇒​ Area white washed = 4989.60 / 20 = 249.48 m2

Now, area white washed is curved surface area of hemisphere
⇒​ ∴ Inner surface area of dome = 249.48 m2

Let the radius of dome be ‘r’ m
Surface area of dome = 249.48 m2
⇒​ 2πr2 = 249.48 m2 2
2 2
⇒​ 2 × 22/7 × r = 249.48 m
⇒​ r2 = 249.48 × 72 × 22 = 39.69 m2
⇒​ r = √39.69 = 6.3 m
2
Volume of the air inside the dome = Volume of hemisphere
⇒​ 2/3 π r3
⇒​ 2/3 × 22/7 × 6.3 × 6.3 × 6.3 = 523.908 m3

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SECTION – E (Competency Based Questions)

36.​ Swimming Pool Construction
(a) Polynomial to represent the volume of the pool = Volume of cuboid
= l.b.h = (5x+10) . (3x+5) . (x+2) = 5.(3x3 + 17x2 + 32x +20)
(c) If x = 1, Volume of pool = 360 m3
(iii) The cost of filling the pool = 360 ×8 = ₹2880 1
OR 1
Area of the walls of the pool = 2 (l+b).h = 2 (5x+10+3x+5).(x+2) = (16x2 + 62x + 60) m2 2
37.​ (a) Angle y = 1200 1
(b) Angle x = 1100 2

(c) Reflex ∠EGF = 3600 – 1100 = 2500 1
OR
Reflex ∠AEG = 3600 – 500 = 3100

38.​ (a) Sides of an isosceles triangle are 10 m, 15 m, 15 m 1
OR
Semi-perimeter of the triangle = 40/2 = 20 m
(b) Area of isosceles triangle = ½ base . height

= ½ . 10 . 10 2 = 50 ×1. 41
= 70.50 m2 2

(c) Cost of painting = ₹18.50 × 70.50 = ₹1304.25 1

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