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 2512 – 2502 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 cm2
(b) 181.5 cm2
(c) 100.5 cm2
(d) 150 cm2
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 r1 and r2 be the radii of two cylinder and h1 and h2 be their heights.
Now, required ratio = CSA of first cylinder / CSA of second cylinder
2πr1h1 / 2πr2h2 =(r1 / r2 ) = (h1 / h2 )= 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 = 1/3πr2h
= 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 x3 in polynomial 2x3 + 3x2 – 7x + 5.
Ans. The given polynomial can be written as
2x3 + 3x2 – 7x + 5
∴ Coefficient of x = –7
And coefficient of x3 = 2
OR
Factorise x2 + x – 20.
Ans. Let p(x) = x2 + x – 20
By splitting the middle term, we get
p(x) = x2+ 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 cm2 , find its diameter.
Ans. Let radius and slant height of the cone be 4x and 7x respectively.
Now, curved surface area = 792 cm2 [Given]
⇒ πrl = 792 ⇒ 22/7 × 4x X 7x = 792
⇒ x2 = 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 cm2 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 x4 + 1/x4 = 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 = 4πr2
New surface area = 4πR2 = 4π ( 7/10r)2 = 49π/25 r2
∴ Required percentage = 4πr2 – 49π/25 r2 / 4πr2 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 m3 = 4.368 m3
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) = 2x3 – 3x2 – 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, 2x3 – 3x2 – 17x + 30 = (x + 3) (x – 2) (2x – 5)
OR
If x + 1/x = 8 , find
(i) x2 + 1/x2
(ii) x4 + 1/x4
Ans. (i) We have, x + 1/x = 8
⇒ ( x + 1/x )2 = 82 [On squaring both sides]
⇒ x2+ 1/x2 + 2 X x X 1/x = 64 [ (a + b)2 = a2+ b2 + 2ab]
⇒ x2+ 1/x2 + 2 = 64
⇒ x2+ 1/x2 = 64 – 2 = 62
+ 1 + 2 = 64 ⇒ + 1 = 64 − 2 = 62
(ii) We have, x2+ 1/x2 = 62
⇒ ( x2+ 1/x2)2 = (62)2 [On squaring both sides]
⇒ ( x2)2 + (1/x2)2 + 2 X x2 X 1/x2 = 3844 [(a + b)2 = a2 + b2 + 2ab]
⇒ x4+ 1/x4 + 2 = 3844 ⇒ x4+ 1/x4 = 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
OP2 = OL2 + PL2 and OR2 = OM2 + RM2 [By Pythagoras theorem]
⇒ 100 = OL2 + 64 and 100 = OM2 + 36
⇒ OL2 = 36 and OM2 = 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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