# Class 9 Mathematics Sample Paper

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.

## 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

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

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

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

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.

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

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

C

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

A

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

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

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.

∴ 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.

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 :

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:

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.

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]

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

∴ 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