Class 9 Mathematics Sample Paper

Sample Papers Class 9

We have provided Class 9 Mathematics Sample Paper as per the latest CBSE examination pattern for the current academic year. The following CBSE Sample Papers for Class 9 Mathematics has been prepared based on the guess papers issued recently. Students will be able to practice these papers and get good marks in upcoming Mathematics exams for Class 9.

CBSE Sample Papers for Class 9 Mathematics

Term 2 Sample Papers
Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

PART – A
Section – I

1. Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, then CD =
(a) 4 cm
(b) 8 cm
(c) 2 cm
(d) None of these

Answer

A

2. In which of the following, it is not possible to construct the triangle?
(a) ΔABC, BC = 6 cm, ∠B = 60° and AB + AC = 4.7 cm
(b) ΔPQR, QR = 6 cm, ∠Q = 105° and PR + PQ = 8 cm
(c) ΔXYZ, XY + YZ + ZX = 8 cm, ∠Y = 40°, ∠Z = 120°
(d) None of these

Answer

A

3. The circumference of the base of a right circular cylinder is 235 cm. If the height of the cylinder is 1 m. Find the curved surface area of the cylinder.
(a) 235 m2
(b) 23500 m
(c) 23500 cm2
(d) 235 cm2

Answer

C

4. Hema dialed a phone number 100 times in a week, out of which she got the response 36 times. The probability that she will not get the response is
(a) 55/100
(b) 100/55
(c) 45/100
(d) 64/100

Answer

D

Section – II

Case study-based question is compulsory. Attempt any 4 sub parts. Each sub-part carries 1 mark.

5. Sandeep who is a postman has to deliver 3 letters in a society. The society is in circular shape. He delivers the letters to Rohan, Monu and Seema, the houses of them are situated at R, M and S respectively on a circular path of radius 5 km. He delivers the first letter to Rohan at R. After Rohan, he delivers letter to Seema at S and then to Monu at M. If the distance between Rohan and Seema and between Seema and Monu is 4 km each, then answer the following questions.

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

(i) If ∠1 = ∠ROS and ∠2 = ∠SOM, then
(a) ∠1 < ∠2
(b) ∠1 = ∠2
(c) ∠1 ≠ ∠2
(d) ∠1 > ∠2

Answer

B

(ii) If P is the point of intersection of OS and RM, then OP is perpendicular to
(a) SO
(b) SM
(c) RM
(d) SO

Answer

C

(iii) ∠RPS =
(a) ∠MPS
(b) ∠SOR
(c) ∠SOM
(d) ∠RSP

Answer

A

(iv) From which point, SP passes?
(a) Point R
(b) Point M
(c) Point O
(d) None of these

Answer

C

(v) If OP = 2.5 km, then find approximate distance between the houses of Rohan and Monu.
(a) 4.33 km
(b) 3 km
(c) 2.9 km
(d) 6.53 km

Answer

A

PART – B
Section – III

6. In a rhombus ABCD, if ∠A = 60°, then find the sum of ∠A and ∠C.
Ans. 
Since, a rhombus is a parallelogram.

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

∴ Its opposite angles are equal.
⇒ ∠A = ∠C
∴ ∠C = 60° [∠A = 60° (Given)]
Now, required sum = ∠A + ∠C
= 60° + 60° = 120°

7. The surface area of a sphere is 13.86 cm2. Then find the radius of the sphere.
Ans. 
Let r be the radius of sphere.
Since curved surface area of sphere = 13.86 cm2 [Given]
⇒ 4πr2 = 13.86
⇒ 4 × 22/7 × r2 = 13.86
⇒ r2 = 13.86 × 7/4 × 22 = 1.1025 ⇒ r = 1.05 cm

8. In figure, O is the centre of the circle. If chord AP = chord PB. Prove that OP is bisector of ∠AOB.

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

Ans. Given, AP = PB
We know that, equal chords of a circle subtend equal angles at the centre.
∴ ∠AOP = ∠BOP
Hence, OP is bisector of ∠AOB.

OR

ABCD is a cyclic quadrilateral such that ∠A = 110°, ∠B = 70°, ∠C = 80°, then ∠D = _____ .
Ans. We know that, sum of opposite angles of a cyclic quadrilateral is 180°.
∴ ∠B + ∠D = 180°
⇒ ∠D = 180° – ∠B ⇒ ∠D = 180° – 70° [∠B = 70°]
∴ ∠D = 110°

9. Find the zero of the polynomial q(x) = 3x + 2.
Ans. We have, q(x) = 3x + 2
Put q(x) = 0 ⇒ 3x + 2 = 0 ⇒ x = −2/3
∴ −2/3 is the zero of the polynomial q(x) = 3x + 2.

10. Factorize 2a7 – 128a.
Ans. We have, 2a7 – 128a = 2a(a6 – 64)
= 2a[(a3)2 – (8)2] = 2a(a3+ 8)(a3– 8) [ a2 – b2 = (a – b)(a + b)]
= 2a(a3+ 23)(a3– 23)
= 2a(a + 2)(a2 – 2a + 4)(a – 2)(a2+ 2a + 4) [Using (a3+ b3 ) = (a + b) (a2 – ab + b2 ) and (a3– b3 ) = (a – b) ( a2+ ab + b2 )]
= 2a(a + 2)(a – 2)(a2+ 2a + 4)(a2– 2a + 4)

Section – IV

11. In a parallelogram ABCD, diagonals AC and BD intersect at O and AC = 7.4 cm and BD = 6.2 cm. Find the length of AO and BO.
Ans. Since, ABCD is a parallelogram
⇒ AO = 1/2 AC = 1/2 × 7.4 = 3.7 cm
and BO = 1/2 BD = 1/2 × 6.2 = 3.1 cm 

12. If (x – b) is a factor of p(x) = x2 + ax – 20, then show that b = 20/a + b ; a + b ≠ 0.
Ans. Since zero of (x – b) is b, therefore by factor theorem,
p(b) = 0, if (x – b) is the factor of p(x).
∴ p(b) = b2 + a(b) – 20 = 0
⇒ b2 + ab = 20 ⇒ b(b + a) = 20 ⇒ b= 20/a + b

OR

If x – 1/x = 3 , then find the value of x2 + 1/x2 .
Ans. We have, x – 1/x = 3
⇒ (x – 1/x)2 = (3)2 [Squaring both sides]
⇒ x2 + 1/x2 – 2 X x X 1/x = 9
⇒ x2 + 1/x2 – 2 = 9 ⇒ x2 + 1/x2 = 9 + 2 = 11

13. Draw an angle of 80° with the help of a protractor and bisect it.
Ans. Steps of construction :

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

Step I : Draw an ∠AOB = 80° with the help of a protractor.
Step II : Taking O as centre and any suitable radius, draw an arc cutting OA at P and OB at Q.
Step III : Taking P as centre and with radius more than 1/2 PQ , draw an arc in the interior of ∠AOB.
Step IV : Taking Q as centre and with same radius as in step III, draw another arc intersecting the previous arc at R.
Step V : Join OR and produce it to C. 
Then, ray OR is the required angle bisector of ∠AOB.

14. Find the number of cubes of 2 cm side which can be made from cuboid of dimensions l = 22 cm,
b = 16 cm and h = 5 cm.
Ans. Volume of cuboid = 22 × 16 × 5 = 1760 cm3
Also, volume of 1 cube = (2)3 = 8 cm3
Now, number of cubes formed = Volume of cuboid / Volume of 1 cube
= 1760/8 =220

Section – V

15. For the polynomial, x3 + 2x+ 1/5 – 7/2 x2 – x6 , find
(i) the degree of the polynomial
(ii) the coefficient of x3
(iii) the coefficient of x6 
Ans.  Given polynomial can be written as
 x3/5 =2/5x + 1/5 – 7/2 x2 – x6
(i) Degree = 6
(ii) Coefficient of x3= 1/5
(iii) Coefficient of x6 = –1

16. A die is thrown 100 times and the outcomes are recorded as below:

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

If the die is thrown again, find the probability of getting
(i) the number 1
(ii) the number 6
(iii) an even number less than 6
Ans. Total number of trials = 100
(i) Let Ebe the event of getting 1 on throwing a die.
Then, P (getting 1) = P(E1) = 20/100 = 0.2
(ii) P (getting 6) = 15/100 = 0.15
(iii) P (getting an even number less than 6) = 35/100 = 0.35

OR

A box contains some balls whose total weight is 40 kg. There are 10 balls each of weight 1 kg, 5 balls each of weight 2 kg, 5 balls each of weight 2.5 kg and rest of the ball weighing 1.5 kg each. A ball is drawn at random from the box, find the probability that the ball drawn is of :
(i) weight 2.5 kg
(ii) weight neither 2.5 kg nor 1 kg
Ans. Weight of 10 balls of 1 kg each = 1 × 10 = 10 kg
Weight of 5 balls of 2 kg each = 2 × 5 = 10 kg
Weight of 5 balls of 2.5 kg each = 2.5 × 5 = 12.5 kg
∴ Weight of 20 balls = 10 + 10 + 12.5 = 32.5 kg
So, weight of remaining balls = 40 – 32.5 = 7.5 kg
Now, as weight of each remaining ball = 1.5 kg
∴ Number of remaining balls = 7.5/1.5 = 5
∴ Total number of balls = 10 + 5 + 5 + 5 = 25
(i) Number of balls having weight 2.5 kg = 5
∴ P(weight of ball is 2.5 kg) = 5/25 = 1/5 
(ii) Number of balls having weight neither 2.5 kg nor 1 kg = 5 + 5 = 10
∴ P(weight of ball is neither 2.5 kg nor 1 kg) = 10/25 = 2/5

17. Two circles of radii 10 cm and 8 cm intersect at two points and the length of the common chord is 12 cm. Find the distance between their centres.
Ans. Let O and O′ be the centres of the circles of radii 10 cm and 8 cm, respectively and PQ be their common chord of length 12 cm.

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

Then, we have OP = 10 cm,
O′P = 8 cm and PQ = 12 cm
∴ PL = 1/2 PQ = 6 cm [Since, when two circles intersect at two points, then their centres lie on the perpendicular bisector of the common chord]
In right angled ΔOLP,
OP2 = OL2 + LP2 [By Pythagoras theorem]

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

Now, in right angled ΔO′LP,
O′P2 = O′L2+ LP2 [By Pythagoras theorem]

Class 9 Mathematics Sample Paper Term 2 With Solutions Set A

∴ Distance between their centres, OO′ = OL + LO′
= (8 + 5.29) cm = 13.29 cm

Section – VI

18. Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm. 
Ans. Let the radius of the larger sphere be r1.
Weight of larger solid sphere (m1) = 5920 g
Weight of smaller solid sphere (m2) = 740 g
Diameter of the smaller sphere = 5 cm
⇒ Radius of the smaller sphere (r2 ) cm = 5/2 cm
Now, density (D) = Mass (M)/Volume (V)
or V = M/D ⇒ V1 = 5920/D cm3 …(i)
and V2 = 740/D cm3 …(ii)
On dividing (i) by (ii), we get V1/V2 = 592/74
⇒ 4/3πr1/ 4/3πr2= 592/74 ⇒ (r1/r2)3 = 592/74 ⇒ (r1/5/2)3 = 592/74
⇒ r1/125/8 = 592/74 ⇒ 8r13/125 = 592/74 ⇒ r13= 592/74 X 125/8 =125
∴ r1 = 5
Hence, the radius of the larger sphere is 5 cm.

OR

A cone of height 24 cm has a curved surface area 550 cm2. Find its volume.
Ans. Height of the cone(h) = 24 cm
Let r cm be the radius of the base and l cm be the slant height of the cone. Then
l = √r2 + h2 = √r+ 242 = √r2+ 576 …(i)
Now, curved surface area = 550 cm2 [Given]
⇒ 22/7 × r × √r2 + 576 = 550 [Using (i)]
⇒ r √r2 576 = 550 x 7/22 ⇒ r√r2+ 576 = 175
Squaring both the sides, we get
 r( r2 + 576) = 30625
⇒ ( r2)2 + 576 r2 – 30625 = 0
Let r2 = x. Then the above equation becomes
x2 + 576x – 30625 = 0
⇒ x2 + 625x – 49x – 30625 = 0 [By splitting the middle term]
⇒ x(x + 625) – 49(x + 625) = 0 ⇒ (x + 625)(x – 49) = 0
⇒ x + 625 = 0 or x – 49 = 0 ⇒ x = – 625 or x = 49
Now, x = –625 is not possible.
∴ x = 49 ⇒ r2= 49 ⇒ r = 7 cm
∴ Volume of the cone = 1/3πr2h
= 1/3 × 22/7 × 72 × 24 = 1232 cm3.

19. If x2 + 1/x2 = 79, then find the value of x3+ 1/x3
Ans.
We have, x2 + 1/x2 = 79
Consider, (x + 1/x)2 = x2 + 1/x2 + 2 X x X(1/x)
⇒ (x + 1/x)= x2 + 1/x2 + 2 = 79 + 2 = 81 ⇒ (x + 1/x) = ± 9 
Cubing both sides, we get (x + 1/x)3 = (±9)
⇒ x3 + 1/x+ 3 X x X 1/x (x + 1/x) = ±729 
⇒ x3 + 1/x+ 3(±9) = ±729 [Using (i)]
⇒ x3 + 1/x± 27 = ±729 ⇒ x3 + 1/x= ± 729+27
= (729 – 27) or (–729 + 27) = 702, – 702
∴ x3 + 1/x3 = ±702

OR

Find the value of the following polynomials at the number written against them.
(i) p(z) = 2z + 5z2 + 3z3 – 3 at z = –3
(ii) p(t) = t99 – t + 1 at t = –1
(iii) p(x) = 2x2 + 6x – 7 at x = 1/4
(iv) p(y) = 4y2 + 1/6 y + 2√3 at y = √3
Ans. (i) We have, p(z) = 2z + 5z2 + 3z3 – 3
∴ p (–3) = 2(–3) + 5(–3)2 + 3(–3)3 – 3
= –6 + 45 – 81 – 3 = –45
(ii) We have, p(t) = t99 – t + 1
∴ p (–1) = (–1)99 – (–1) + 1 = –1 + 1 + 1 = 1
(iii) We have, p(x) = 2x2 + 6x – 7
∴ p (1/4) = 2 (1/4)+6 (1/4) – 7 = 1/8 + 3/2 – 7 = -43/8
(iv) We have, p(y) = 4y+ 1/6 y + 2√3
∴ p (√3) = 4(√3)2 + 1/6 (√3) + 2√3
= 12 + √3/6 + 2√3 = 12 + 13 √3/6

Class 9 Mathematics Sample Paper