Please refer to the Class 9 Mathematics Sample Paper below. These CBSE Sample Papers for Class 9 Mathematics have been prepared based on the latest guidelines and examination patterns issued for the current academic year. We have provided Term 1 and Term 2 sample papers with solutions. You can click on the links below and access the free latest CBSE Sample Papers for Mathematics for Standard 9. All guess papers for Mathematics Class 9 have been prepared by expert teachers. You will be able to understand the type of questions which are expected to come in the examinations and get better marks.

**CBSE Sample Papers for Class 9 Mathematics**

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

**PART – A****Section – I**

**1. The value of 251 ^{2} – 250^{2} is**

(a) 12

(b) 525

(c) 500

(d) 501

**Answer**

D

**2. In the given figure, ABCD and AEFG are two parallelograms. If ∠C = 50°, then ∠F = **

(a) 60°

(b) 50°

(c) 130°

(d) None of these

**Answer**

B

**3. In the given figure, O is the centre of the circle. ∠AOB = ∠COD = 55° and CD = 5.7 cm, then AB is equal to **

(a) 2.5 cm

(b) 10 cm

(c) 10/3 cm

(d) 5.7 cm

**Answer**

D

**4. The surface area of a cube whose side is 5.5 cm is**

(a) 125 cm^{2}

(b) 181.5 cm^{2}(c) 100.5 cm^{2}

(d) 150 cm^{2}

**Answer**

B

**Section – II**

**Case study-based question is compulsory. Attempt any 4 sub parts. Each question carries 1 mark.5**.

**Rohan has five coins, which he tossed simultaneously 250 times and record the outcomes as given in the table below.**

**If he wants to toss these 5 coins again, then answer the following questions.**

**(i) Probability of getting 3 tails is**

(a) 1/2

(b) 1/5

(c) 1/3

(d) 2/3

**Answer**

B

**(ii) Probability of getting 2 tails is**

(a) 3/50

(b) 7/25

(c) 7/50

(d) 3/25

**Answer**

C

**(iii) Probability of getting atleast 4 tails is**

(a) 7/25

(b) 3/25

(c) 3/50

(d) 7/50

**Answer**

A

**(iv) Probability of getting atmost 1 tail is**

(a) 23/50

(b) 19/50

(c) 7/50

(d) 16/23

**Answer**

B

**(v) Probability of getting no head is**

(a) 1/50

(b) 2/3

(c) 15/16

(d) 9/50

**Answer**

D

**PART – B****Section – III**

**6. Find the zero of the polynomial p(y) = y /64 −1.****Ans.** We have, p(y) = y/64 − 1 .

To find its zero, put p(y) = 0

∴ y/64 -1=0 ⇒ y/64 = 1 ⇒ y=64

∴ 64 is the zero of polynomial p(y) = y/64− 1 .

**7. In the given figure, if ∠ABC = 15°, then find ∠AOC. **

**Ans. **Since, the angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of circle.

∴ ∠AOC = 2∠ABC = 2 × 15° = 30°

**8. The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas.****Ans.** Let r_{1} and r_{2} be the radii of two cylinder and h_{1} and h_{2} be their heights.

Now, required ratio = CSA of first cylinder / CSA of second cylinder

2πr_{1}h_{1 }/ 2πr_{2}h_{2 }=(r_{1} / r_{2} ) = (h_{1} / h_{2} )= 2/3 x 5/3 = 10/9 or 10 : 9

**OR**

**A right triangle having sides 6 cm, 8 cm and 10 cm is revolved about the side of the length 8 cm. Find the volume of the solid so formed.****Ans.** Sides of the triangle are 6 cm, 8 cm and 10 cm. When this triangle is revolved about the side of 8 cm, we get a cone as shown in the figure.

Thus, radius of the base of the cone so formed (r) = 6 cm.

Height (h) = 8 cm.

∴ Volume of the cone so formed = 1/3πr^{2}h

= 1/3 × π ×(6)^{2} × 8 = 1/3 × 22/7 × 36 × 8

= 301.7 (approx.)

**9. In the given figure, M, N and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC. **

**Ans.** Since M, N and P are mid-points of AB, AC and BC respectively. Therefore by mid-point theorem, we have,

MN = (1/2)BC, MP = (1/2)AC and NP = (1/2)AB

⇒ BC = 6 cm, AC = 5 cm and AB = 7 cm

**10. Find the measure of each of the two angles formed by bisecting an angle of measure 150°. ****Ans. **The measure of each of the two angles formed by bisecting an angle of measure 150° = 1/2 × 150° = 75°.

**Section – IV**

**11. In the given figure, ABCD is a parallelogram. Find the value of (x + y). **

**Ans. **Since the sum of any two consecutive angles of a parallelogram is 180°.

∴ ∠ADC + ∠BCD = 180°

⇒ x + 80° = 180° ⇒ x = 100° …(i)

Also, ∠DAB = ∠BCD [Opposite angles of a parallelogram]

∴ ∠DAB = 80°

Now, ∠DAE + ∠DAB = 180° [Linear pair]

⇒ y + 80° = 180° ⇒ y = 100° …(ii)

∴ x + y = 100° + 100° = 200° [From (i) and (ii)]

**12. Write the coefficients of x and x ^{3 }in polynomial 2x^{3} + 3x^{2} – 7x + 5.**

**Ans.**The given polynomial can be written as

2x

^{3}+ 3x

^{2}– 7x + 5

∴ Coefficient of x = –7

And coefficient of x

^{3 }= 2

**OR**

**Factorise x ^{2 }+ x – 20.**

**Ans.**Let p(x) = x

^{2}+ x – 20

By splitting the middle term, we get

p(x) = x

^{2}+ 5x – 4x – 20 = x(x + 5) – 4(x + 5)

= (x – 4) (x + 5)

**13. The radius and slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm ^{2} , find its diameter.**

**Ans.**Let radius and slant height of the cone be 4x and 7x respectively.

Now, curved surface area = 792 cm

^{2}[Given]

⇒ πrl = 792 ⇒ 22/7 × 4x X 7x = 792

⇒ x

^{2 }= 792/88 = 9 ⇒ x = 3

∴ r = 4x = 12 cm

Hence, the diameter of the base of the cone is 2 × 12 = 24 cm.

**14. Find the height of a solid right circular cylinder, whose total surface area is equal to 314 cm ^{2} and the diameter of the base is 8 cm. (Use p = 3.14)**

**Ans.**Here, radius (r) = 8/2 cm = 4 cm

Let h be the height of the cylinder.

Now, TSA of cylinder = 2πr(r + h)

⇒ 314 = 2 × 3.14 × 4[4 + h]

⇒ 314/2 x 3.14 x 4 = 4 + h ⇒ 4 + h = 12.5

⇒ h = 12.5 – 4 = 8.5 cm

**Section – V**

**15. If x ^{4 }+ 1/x^{4 } = 194, then find the value of x + 1/x.**

**Ans.**

**16. The diameter of a sphere is decreased by 30%. What will be the percentage change in its total surface area?****Ans.** Let d be the diameter of the sphere

⇒ Radius (r) = d/2 ⇒ d = 2r

New diameter = d – 30% of d = d −3/10 d = 7/10 d

∴ New radius (R) = 1/2 (7/5 r) = 7/10r

Now, original surface area = 4πr^{2}

New surface area = 4πR^{2 }= 4π ( 7/10r)^{2 }= 49π/25 r^{2}

∴ Required percentage = 4πr^{2 }– 49π/25 r^{2} / 4πr^{2 } x 100 = 51%

**OR**

**A closed cuboidal tank can store 4368 litres of water. The external dimensions of the tank are 2.2 m × 1.7 m × 1.7 m. If the walls of the tank are 5 cm thick, then what is the thickness of the bottom (top) of the tank if they are same?****Ans. **Capacity of tank = 4368 litres = = 4368/1000 m^{3} = 4.368 m^{3}

Internal length of tank = (2.2 – 2 x 5/100)m = 2.1m

Internal breadth of tank = (1.7 – 2 x 5/100)m = 1.6m

Let the thickness of tank in the bottom be x m.

∴ Internal height of tank = (1.7 – 2x) m

Now, Internal volume of tank = Capacity of tank

⇒ 2.1 × 1.6 × (1.7 – 2x) = 4.368

⇒ 1.7 – 2x = 4.368 / 2.1 x 1.6 = 1.3

⇒ 2x = 1.7 – 1.3 = 0.4

⇒ x = 0.2 m = 0.2 × 100 cm = 20 cm

So, required thickness = 20 cm

**17. Draw a line segment AB = 8.5 cm and draw perpendiculars at A and B. Are these two perpendiculars parallel to each other?****Ans.** Steps of construction :

Step I : Draw a line segment AB = 8.5 cm.

Step II : With A as centre and any suitable radius, draw an arc cutting AB at P.

Step III : Keeping the radius same and starting from P, mark points C and D on the arc drawn in step II such that PC = CD.

Step IV : Draw AC and AD.

Step V : Draw AX bisector of ∠CAD. Then, ∠PAX = 90°.

Step VI : Similarly, draw angle 90° at B, as shown in figure. Then, we get the perpendiculars AX and BY.

Yes, the perpendiculars AX and BY are parallel to each other. [ Sum of co-interior angles is 180°]

**Section – VI**

**18. Show that (x + 3) is a factor of the polynomial f(x) = 2x ^{3} – 3x^{2} – 17x + 30 and hence factorise f(x).**

**Ans.**By factor theorem, (x + 3) will be a factor of f(x), f f(–3) = 0.

We have, f (–3) = 2 × (–3)

^{3}– 3 (–3)

^{2}– 17 × (–3) + 30

= –54 – 27 + 51 + 30 = 0

Hence, (x + 3) is a factor of f(x).

Now, let f(2) = 2(2)

^{3}– 3(2)

^{2}– 17(2) + 30

= 16 – 12 – 34 + 30 = 0

Therefore, 2 is the zero of f(x) i.e., (x – 2) is a factor of f(x).

Again, f (5/2) = 2(5/2)

^{3 }– 3(5/2)

^{2}– 17(5/2) + 30

= 125/4 – 75/4 – 85/2 + 30 = 0

Therefore, 5/2 is also a zero of f(x) i.e., (2x – 5) is factor of f(x).

Hence, 2x

^{3}– 3x

^{2}– 17x + 30 = (x + 3) (x – 2) (2x – 5)

**OR**

**If x + 1/x = 8 , find****(i) x ^{2 }+ 1/x^{2}**

**(ii) x**(i) We have, x + 1/x = 8

^{4 }+ 1/x^{4 }Ans.⇒ ( x + 1/x )

^{2}= 8

^{2 }[On squaring both sides]

⇒ x

^{2}+ 1/x

^{2}+ 2 X x X 1/x = 64 [ (a + b)

^{2}= a

^{2}+ b

^{2}+ 2ab]

⇒ x

^{2}+ 1/x

^{2}+ 2 = 64

⇒ x

^{2}+ 1/x

^{2}= 64 – 2 = 62

+ 1 + 2 = 64 ⇒ + 1 = 64 − 2 = 62

(ii) We have, x

^{2}+ 1/x

^{2 }= 62

⇒ ( x

^{2}+ 1/x

^{2})

^{2 }= (62)

^{2 }[On squaring both sides]

⇒ ( x

^{2})

^{2}+ (1/x

^{2})

^{2}+ 2 X x

^{2 }X 1/x

^{2}= 3844 [(a + b)

^{2}= a

^{2 }+ b

^{2}+ 2ab]

⇒ x

^{4}+ 1/x

^{4}+ 2 = 3844 ⇒ x

^{4}+ 1/x

^{4 }= 3844 – 2 = 3842

**19. During a practical activity in maths lab students were using circular geoboard. The angle subtended by an arc BC at the centre is (2a + 50°). Pallavi calculated ∠BAC as (a + 25°). **

**(i) Did she find correct answer? Justify it.****(ii) Find ∠BAC if a = 30°.****(iii) What will be the value of ∠BOC for a = 15°.****(iv) If a = 30°, then find the measure of reflex ∠BOC.****Ans.** Angle subtended by BC at centre = (2a + 50°)

We know that, angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of circle.

∴ ∠BOC = 2 × ∠BAC ⇒ 2 × ∠BAC = 2a + 50°

⇒ ∠BAC = 1/2 × 2 (a +25° )

∴ ∠BAC = a + 25°

(i) Yes, Pallavi calculated the right answer.

(ii) Given, a = 30°

∴ ∠BAC = a + 25° = 30° + 25° ⇒ ∠BAC = 55°

(iii) Given, ∠BOC = 2a + 50°

⇒ ∠BOC = 2 × 15° + 50° [a = 15°]

= 30° + 50° = 80°

(iv) Given, a = 30°, then ∠BOC = 2 × 30° + 50°

= 60° + 50° = 110°

Reflex ∠BOC = 360° – 110° = 250°

**OR**

**PQ and RS are two parallel chords of a circle whose centre is O and radius is 10 cm. If PQ = 16 cm and RS = 12 cm, find the distance between PQ and RS, if they lie****(i) on the same side of the centre O.****(ii) on opposite side of the centre O.****Ans. **We have, OP = OR = 10 cm, PQ = 16 cm and RS = 12 cm Now, draw a perpendicular from centre O to chords PQ and RS, meets at L and M respectively i.e., OL ⊥ PQ and OM ⊥ RS.

∴ PL = 8 cm and RM = 6 cm

[ Perpendicular drawn from the centre of a circle to a chord bisects the chord]

In right triangles OLP and OMR, we have

OP^{2} = OL^{2} + PL^{2} and OR^{2} = OM^{2} + RM^{2 }[By Pythagoras theorem]

⇒ 100 = OL^{2} + 64 and 100 = OM^{2} + 36

⇒ OL^{2 }= 36 and OM^{2} = 64

⇒ OL = 6 cm and OM = 8 cm

(i) In this case, from fig. I, we have Distance between PQ and RS

= LM = OM – OL = (8 – 6) cm = 2 cm

(ii) In this case, from fig. II, we have Distance between PQ and RS = LM

= OL + OM = (6 + 8) cm = 14 cm

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