# Class 9 Mathematics Sample Paper

Please refer to Class 9 Mathematics Sample Paper with solutions provided below. All sample papers for Mathematics Class 9 have been designed as per the latest paper pattern issued by CBSE for the current academic year. Students should practice these guess papers for Class 9 Mathematics as it will help them to gain more understanding of the type of questions that are expected to be asked in upcoming Class 9 Mathematics exams. Please click on the links below to access free CBSE Sample Papers for Class 9 Mathematics.

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

Part – I

1. If x2 + kx + 6 = (x + 2)(x + 3) for all x, the value of k is :
(a) 1
(b) – 1
(c) 5
(d) 3

C

2. D, E and F are mid-points of sides BC, CA and AB of ΔABC. If perimeter of ΔABC is 12.8 cm, then perimeter of ΔDEF is :
(a) 17 cm
(b) 38.4 cm
(c) 25.6 cm
(d) 6.4 cm

D

3. If PQRS is a parallelogram, then ∠Q – ∠S is equal to :
(a) 90°
(b) 120°
(c) 180°
(d) 0°

D

4. If in a quadrilateral ABCD, ∠A = 90° and AB = BC = CD = DA, then ABCD is a :
(a) parallelogram
(b) rectangle
(c) square
(d) rhombus

C

5. Three chords AB, CD and EF of a circle are respectively 3 cm, 3.5 cm, and 3.8 cm away from the centre. Then which of the following is correct ?
(a) AB > CD > EF
(b) AB < CD < EF
(c) AB = CD = EF
(d) AB = CD < EF

A

6. AD is a diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm, the distance of AB from the centre of the circle is :
(a) 17 cm
(b) 15 cm
(c) 4 cm
(d) 8 cm

D

7. The region between an arc and two radii, joining the centre to the end points of the are is called :
(a) Sector
(b) Segment
(c) Semicircle
(d) None of the above

A

8. The radii of two right circular cylinders are in the ratio of 4 : 5 and their heights are in the ratio 2 : 3. The ratio of their curved surface area is equal to :
(a) 8 : 15
(b) 36 : 81
(c) 2 : 3
(d) 16 : 25

A

9. The radius of a cylinderical wire is decreased to one-third. If its volume remains the same, its length will increase to :
(a) 2 times
(b) 3 times
(c) 6 times
(d) 9 times

D

10. Which of the following cannot be experimental probability of an event ?
(a) 2/3
(b) 1
(c) 0
(d) 3/2

D

Case Study Based Questions

CASE STUDY-1 :
In a newly constructed park which is situated in the heart of city Hyderabad, an architect has form a structure in the given shape. The shape has a cuboid, which is standing on the two cylindrical beams. The dimensions of the cuboid are 1.5 m, 3 m and 0.5 m. The dimensions of the cylinders are of height 2 m and diameter 0.6 m.

11. As the structure is made from the concrete, how much volume of concrete is required to make the cuboidal shape?
(a) 1.75 m3
(b) 2.20 m3
(c) 2.25 m3
(d) 1.25 m3

C

12. What is formula for calculating the lateral surface area of the cylinder ?
(a) r2
(b) 2 rh
(c) r2h
(d) 2r3

B

13. What is the volume of two cylinders ?
(a) 1.20 m3
(b) 1.134 m3
(c) 3 m3
(d) 2.2 m3

B

14. If the cuboid needs to be painted red, how much area need to be painted ?
(a) 5.2 m2
(b) 13 m2
(c) 6.75 m2
(d) 5.7 m2

B

15. If a cloth is needed to cover the cylindrical part, how much cloth is needed ?
(a) 8.25 m2
(b) 1.25 m2
(c) 4.50 m2
(d) 7.536 m2

D

CASE STUDY-2 :
Diwali Fest is an annual South Asian arts and culture festival, produced by the Diwali Celebration Society. In the Diwali fest, a game is played which is like that, there is a spinner on which some numbers are written. The numbers on spinner are 2, 5, 7, 9, 12, 16. Depending on the condition of stall owner, if a particular number comes, than a die will be thrown.

16. What is the probability, that spinner stops on an even number?
(a) 1/8
(b) 1/4
(c) 1/2
(d) 1/16

C

17. A player will get a special prize, if spinner stops on a perfect square:
(a) 1/3
(b) 1/2
(c) 1/4
(d) 1/8

A

18. If the player gets a chance to throw, a dice, what is the probability of getting a multiple of 2 on dice?
(a) 1/3
(b) 1/2
(c) 1/4
(d) 1/8

B

19. If a dice is thrown, what is the probability of getting a number less than 4?
(a) 1/2
(b) 1/4
(c) 1/8
(d) 2/7

A

20. An event whose probability to occur is 1. Then this type of event is called :
(a) Impossible event
(b) Possible event
(c) Inependent event
(d) Sure event

D

Part – II
Section – A

21. Find the value of the polynomial p(z = 3z2 – 4z + 17 , when z = 3.
Ans.
p(z)= 3z2 – 4z + √17
So, p(3) = 3(3)– (4 x 3) + √17
= (3 x 9) – 12 + √17  = 27 – 12 + √17 = 15 + √17

22. In the given figure, O is the centre of the circle, OM ⊥ BC, OL ⊥ AB, ON ⊥ AC and OM = ON = OL. Is ΔABC equilateral ? Give reasons.

Ans. Given : OL ⊥ AB, OM ⊥ BC and ON ⊥ AC
Also,      OM = ON = OL = Perpendicular distance of chords from the centre of a circle
∴          AB = BC = AC [Chords equidistant from the centre of a circle are equal.]
∴  ΔABC is an equilateral triangle.

23. The height of the cone is 15 cm. If its volume is 1570 cm3, find the radius of the base (use π = 3.14).
Ans.
Let the radius of the base of the cone be r cm.
Height, h = 15 cm            (Given)
Volume of cone = 1570 cm3        (Given)
So,                 1/3πr2h = 1570
or 1/3 x 3.14 x rx 15 = 1570
⇒                            r2 = 1570 x 3 / 3.14 x 15 = 100
⇒                             r = 10 cm.

Section – B

24. If x = –1/3 x = is a zero of the polynomial p(x) = 27x3 – ax2 – x + 3, then find the value of a.
Ans.
If x = –1/3 is a zero of the polynomial
27x3 – ax2 – x + 3,

25. If the polynomial p(x) = x4 – 2x3 + 3x2– ax + 8 is divided by (x – 2), it leaves a remainder 10. Find the value of a.
Ans.
p(x)= x4 – 2x3 + 3x2 – ax + 8
Using Remainder Theorem
10 = p(2) = 24 – 2 x 23 + 3 x 22 – a x 2 + 8
or     10 = 16 – 16 + 12 – 2a + 8
or    – 2 = – 2a + 8
or   – 2a = – 10
i.e.,      a = 5

26. 1.1 cu.cm of copper is to be drawn into a cylindrical wire 0.5 cm in diameter. Calculate the length of the wire.
Ans.
Let the length of wire = l cm
Here ‘r’ = 0.5/2  = 0.25 cm
∴ Volume of wire = πr2h
= π (0.25)2 h
But, volume of the copper = 1.1 cm3
Therefore, 22/7 x 1/4 x 1/4 x l = 1.1       [Here h = l]
⇒       l = 1.1 x 4 x 4 x 7 / 22 = 28/5 cm = 5.6 cm

Section – C

27. The polynomials x3 + 2x2 – 5ax – 8 and x3 + ax2 – 12x – 6 when divided by (x – 2) and (x – 3) leave remainders p and q respectively. If q – p = 10, find the value of a.
Ans.
Let   p(x) = x3 + 2x2 – 5ax – 8 and
g(x) = x+ ax2 – 12x – 6
When divided by (x – 2) and (x – 3) leave remainders p and q respectively
p(x) = x+ 2x2 – 5ax – 8
Now,      p(2) = 23 + 2 × 22 – 5a × 2 – 8
= 8 + 8 – 10a – 8
So,             p = 8 – 10a                        …(1)
Also,       g(3) = 33 + a × 32 – 12 × 3 – 6
So,              q = 27 + 9a – 36 – 6
or                q = –15 + 9a
Now, it is given that q – p = 10
Therefore,  – 15 + 9a – 8 + 10a = 10
⇒                               19a – 23 = 10
⇒                                       19a = 33
⇒                                           a = 33/19

28. If X, Y and Z are the mid-points of sides BC, CA and AB of ΔABC respectively, prove that AZXY is a parallelogram.
Ans.
In ΔABC
X and Z are mid-points of sides BC and AB respectively.
⇒     XZ || AC (By mid point-theorem)
or    XZ || AY 1
X and Y are mid-points of sides BC and AC respectively
∴      XY || AB (By mid-point theorem)
⇒     XY || AZ
Since in quad AZXY, both pair of opposite sides are parallel
∴ AZXY is parallallelogram.

29. In the given alongside figure, ΔABC is a triangle inscribed in a circle, with centre O, such that AB = AC and ∠BEC = 100°. Find the value of x and y.

Ans. ABCD is a cyclic quadrilateral.
∴    ∠A + 100° = 180°     (Opposite angles of cyclic quadrilateral)
⇒               ∠A = 80°
In    ΔABC, AB = AC (Given)
⇒           ∠ABC = ∠ACB = x
∴     ∠A + 2 ∠x = 180°        (Angle sum property of a triangle)
⇒             2∠x = 180° – 80° = 100°
⇒                 x = 50°
BDCE is also a cyclic quadrilateral
∴ y + 100° = 180° ⇒ y = 180° – 100° = 80°