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 m^{2} (b) 23500 m^{2 }(c) 23500 cm^{2}

(d) 235 cm^{2 }

**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. **

**(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.

∴ 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 cm ^{2}. Then find the radius of the sphere.Ans. **Let r be the radius of sphere.

Since curved surface area of sphere = 13.86 cm

^{2}[Given]

⇒ 4πr

^{2}= 13.86

⇒ 4 × 22/7 × r

^{2}= 13.86

⇒ r

^{2}= 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 2a ^{7} – 128a.**

**Ans.**We have, 2a

^{7}– 128a = 2a(a

^{6}– 64)

= 2a[(a

^{3})

^{2}– (8)

^{2}] = 2a(a

^{3}+ 8)(a

^{3}– 8) [ a

^{2}– b

^{2}= (a – b)(a + b)]

= 2a(a

^{3}+ 2

^{3})(a

^{3}– 2

^{3})

= 2a(a + 2)(a

^{2}– 2a + 4)(a – 2)(a

^{2}+ 2a + 4) [Using (a

^{3}+ b

^{3}) = (a + b) (a

^{2}– ab + b

^{2}) and (a

^{3}– b

^{3}) = (a – b) ( a

^{2}+ ab + b

^{2})]

= 2a(a + 2)(a – 2)(a

^{2}+ 2a + 4)(a

^{2}– 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) = x ^{2} + 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) = b

^{2}+ a(b) – 20 = 0

⇒ b

^{2}+ ab = 20 ⇒ b(b + a) = 20 ⇒ b= 20/a + b

**OR**

**If x – 1/x = 3 , then find the value of x ^{2} + 1/x^{2} .**

**Ans.**We have, x – 1/x = 3

⇒ (x – 1/x)

^{2}= (3)

^{2 }[Squaring both sides]

⇒ x

^{2}+ 1/x

^{2}– 2 X x X 1/x = 9

⇒ x

^{2}+ 1/x

^{2}– 2 = 9 ⇒ x

^{2}+ 1/x

^{2}= 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 cm^{3}

Also, volume of 1 cube = (2)^{3} = 8 cm^{3}

Now, number of cubes formed = Volume of cuboid / Volume of 1 cube

= 1760/8 =220

**Section – V**

**15. For the polynomial, x ^{3} + 2x+ 1/5 – 7/2 x^{2} – x^{6} , find**

**(i) the degree of the polynomial**

**(ii) the coefficient of x**

^{3}**(iii) the coefficient of x**

^{6}**Ans.**Given polynomial can be written as

x

^{3}/5 =2/5x + 1/5 – 7/2 x

^{2}– x

^{6}

(i) Degree = 6

(ii) Coefficient of x

^{3}= 1/5

(iii) Coefficient of x

^{6}= –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 E_{1 }be the event of getting 1 on throwing a die.

Then, P (getting 1) = P(E_{1}) = 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,

OP^{2} = OL^{2} + LP^{2} [By Pythagoras theorem]

Now, in right angled ΔO′LP,

O′P^{2} = O′L^{2}+ LP^{2} [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 r_{1}.

Weight of larger solid sphere (m_{1}) = 5920 g

Weight of smaller solid sphere (m_{2}) = 740 g

Diameter of the smaller sphere = 5 cm

⇒ Radius of the smaller sphere (r_{2} ) cm = 5/2 cm

Now, density (D) = Mass (M)/Volume (V)

or V = M/D ⇒ V_{1} = 5920/D cm^{3} …(i)

and V_{2} = 740/D cm^{3} …(ii)

On dividing (i) by (ii), we get V_{1}/V_{2} = 592/74

⇒ 4/3πr_{1}^{3 }/ 4/3πr_{2}^{3 }= 592/74 ⇒ (r_{1}/r_{2})^{3} = 592/74 ⇒ (r_{1}/5/2)^{3} = 592/74

⇒ r_{1}^{3 }/125/8 = 592/74 ⇒ 8r_{1}^{3}/125 = 592/74 ⇒ r_{1}^{3}= 592/74 X 125/8 =125

∴ r_{1} = 5

Hence, the radius of the larger sphere is 5 cm.

**OR**

**A cone of height 24 cm has a curved surface area 550 cm ^{2}. 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 = √r

^{2}+ h

^{2}= √r

^{2 }+ 24

^{2}= √r

^{2}+ 576 …(i)

Now, curved surface area = 550 cm

^{2}[Given]

⇒ 22/7 × r × √r

^{2}+ 576 = 550 [Using (i)]

⇒ r √r

^{2}576 = 550 x 7/22 ⇒ r√r

^{2}+ 576 = 175

Squaring both the sides, we get

r

^{2 }( r

^{2}+ 576) = 30625

⇒ ( r

^{2})

^{2}+ 576 r

^{2}– 30625 = 0

Let r

^{2}= x. Then the above equation becomes

x

^{2}+ 576x – 30625 = 0

⇒ x

^{2}+ 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 ⇒ r

^{2}= 49 ⇒ r = 7 cm

∴ Volume of the cone = 1/3πr

^{2}h

= 1/3 × 22/7 × 7

^{2}× 24 = 1232 cm

^{3}.

**19. If x ^{2} + 1/x^{2} = 79, then find the value of x^{3}+ 1/x^{3}Ans.** We have, x

^{2}+ 1/x

^{2}= 79

Consider, (x + 1/x)

^{2}= x

^{2}+ 1/x

^{2}+ 2 X x X(1/x)

⇒ (x + 1/x)

^{2 }= x

^{2}+ 1/x

^{2}+ 2 = 79 + 2 = 81 ⇒ (x + 1/x) = ± 9

Cubing both sides, we get (x + 1/x)

^{3}= (±9)

^{3 }

⇒ x

^{3}+ 1/x

^{3 }+ 3 X x X 1/x (x + 1/x) = ±729

⇒ x

^{3}+ 1/x

^{3 }+ 3(±9) = ±729 [Using (i)]

⇒ x

^{3}+ 1/x

^{3 }± 27 = ±729 ⇒ x

^{3}+ 1/x

^{3 }= ± 729+27

= (729 – 27) or (–729 + 27) = 702, – 702

∴ x

^{3}+ 1/x

^{3}= ±702

**OR**

**Find the value of the following polynomials at the number written against them.****(i) p(z) = 2z + 5z ^{2} + 3z^{3} – 3 at z = –3**

**(ii) p(t) = t**

^{99}– t + 1 at t = –1**(iii) p(x) = 2x**

^{2}+ 6x – 7 at x = 1/4**(iv) p(y) = 4y**

^{2}+ 1/6 y + 2√3 at y = √3**Ans.**(i) We have, p(z) = 2z + 5z

^{2}+ 3z

^{3}– 3

∴ p (–3) = 2(–3) + 5(–3)2 + 3(–3)

^{3}– 3

= –6 + 45 – 81 – 3 = –45

(ii) We have, p(t) = t

^{99}– t + 1

∴ p (–1) = (–1)

^{99}– (–1) + 1 = –1 + 1 + 1 = 1

(iii) We have, p(x) = 2x

^{2}+ 6x – 7

∴ p (1/4) = 2 (1/4)

^{2 }+6 (1/4) – 7 = 1/8 + 3/2 – 7 = -43/8

(iv) We have, p(y) = 4y

^{2 }+ 1/6 y + 2√3

∴ p (√3) = 4(√3)

^{2}+ 1/6 (√3) + 2√3

= 12 + √3/6 + 2√3 = 12 + 13 √3/6