# Class 10 Mathematics Sample Paper

We have provided Class 10 Mathematics Sample Paper as per the latest CBSE examination pattern for the current academic year. The following CBSE Sample Papers for Class 10 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 10.

## CBSE Sample Papers for Class 10 Mathematics

Class 10 Mathematics Sample Paper Term 2 With Solutions Set A

SECTION – A

1. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then find ∠ POA
Answer : Since OA ⊥ PA and OB ⊥ PB
Then ∠OAP=90⁰ and ∠OBP=90⁰
Now we know that Centre lies on the bisector of the angle between the two tangents.
∴ ∠OPA = ∠OPB
So, ∠OPA= 1/2 ∠APB = 1/2 × 80 = 40⁰
In ΔOPA, ∠POA + ∠OPA + ∠OAP = 180⁰
∠POA + 40⁰ + 90⁰ = 180⁰
⇒ ∠POA + 130⁰ = 180⁰ ⇒ ∠POA = 180⁰− 130⁰ ⇒ ∠POA = 50⁰

2. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
Answer : The given series can be written in the reverse way as
253, 248, 243, …, 13, 8, 5
Now for the new AP,
first term, a = 253 and common difference, d = 248 − 253 = −5, n = 20
Therefore, using nth term formula, we get, a20 = a + (20 − 1)d
⇒ a20 = 253 + (19)(−5)
⇒ a20 = 253 − 95
⇒ a = 158
Therefore, 20th term from the last term of the AP 3, 8, 13, …, 253.is 158.

3. Find the roots of the quadratic equation 6×2 – x – 2 = 0.
Answer : 6x2 – x – 2 = 0.
⇒ 6×2 – 4x + 3x – 2 = 0
⇒ 2x(3x – 2) +1(3x – 2) = 0
⇒ (3x – 2) (2x + 1) = 0
⇒ x = 2/3 and x = -1/2 are the roots of the given quadratic equation.

OR

Find the values of k for quadratic equation kx (x – 2) + 6 = 0, so that they have two equal roots.
Answer : Given, kx(x − 2) + 6 = 0
⇒ kx2 − 2kx + 6 = 0
Since the roots are equal,
⇒ b2 = 4ac ⇒ (−2k)2 = 4(k)(6) ⇒ 4k2 = 4k(6) ⇒ k = 0 or k = 6

∴ k = 6 (∵ k = 0 is not possible)

4. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer : Let the base of the right triangle be x cm.
Its altitude = (x − 7) cm
From Pythagoras theorem, Base² + Altitude² = Hypotenuse²
⇒ x² + (x – 7)² = 13² ⇒ x² + x² + 49 – 14x = 169
⇒ 2x² – 14x – 120 = 0 ⇒ x² – 7x – 60 = 0
⇒ x² – 12x + 5x – 60 = 0 ⇒ x(x – 12) + 5(x – 12) = 0 ⇒ (x – 12)(x + 5) = 0
Either x − 12 = 0 or x + 5 = 0, i.e., x = 12 or x = −5
Since sides are positive, x can only be 12.
Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 − 7)
cm = 5 cm.

5. The mode of the following frequency distribution is 34.5. Find the value of x.

Here, mode = 34.5. So, modal class is 30 – 40, f1 = x, f0 = 10 and f2 = 8

6. A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Answer : Volume of cone = 1/3×π×r2×h = 1/3×π×6×6×24 cm3
If r is the radius of the sphere, then its volume is 4/3 πr3.
Since, the volume of clay in the form of the cone and the sphere remains the same, we have
4/3 ×π×r3 = 1/3×π×6×6×24 ⇒ r3 = 3 × 3 × 24 = 33 × 23
So r = 3 × 2 = 6
Therefore, the radius of the sphere is 6 cm.

OR

A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.
Answer : The volume of the rod = π×(1/2)2 × 8 cm3 = 2π cm3
The length of the new wire of the same volume = 18 m = 1800 cm
If r is the radius (in cm) of cross-section of the wire, its volume = π × r2 ×1800 cm3
Therefore, π × r2 × 1800 = 2π
⇒ r2 = 1/900 ⇒ r = 1/30
So, the diameter of the cross section, i.e., the thickness of the wire is 1/15 cm, i.e., 0.67mm (approx.).

SECTION – B

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

We have N = 30 and N/2 = 15.
Median class = 55 − 60, so l = 55, f = 6, cf = 13 and h = 60 − 55 = 5

8. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Step I: Draw a line segment AB = 8 cm.
Step II: With A as centre, draw a circle of radius 4 cm and let it intersect the line segment AB in M.
Step III: With B as centre, draw a circle of radius 3 cm.
Step IV: With M as centre, draw a circle of radius AM and let it intersect the given two circles in P, Q and R, S.
Step V: Join AP, AQ, BR and BS.
These are the required tangents.

9. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.
Answer : In the below figure, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river.
P is a point on the bridge at a height of 3 m, i.e., DP = 3 m.
We are interested to determine the width of the river, which is the length of the side AB of the Δ APB.
Now, AB = AD + DB
In right Δ APD, ∠ A = 30°. So, tan300 = PD/AD ⇒ 1/√3 = 3/AD ⇒ AD = 3√3
Also, in right Δ PBD, ∠ B = 45°. So, BD = PD = 3 m.
Now, AB = BD + AD = 3 + 3√3 = 3 (1 + √3 ) m.
Therefore, the width of the river is 3 (√3 + 1)m .

OR

The angles of depression of the top and the bottom of an 8 m tall building from the top of a multistoreyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.
Answer : In the below figure, PC denotes the multistoreyed building and AB denotes the 8 m tall building.
PB is a transversal to the parallel lines PQ and BD. Therefore, ∠ QPB and ∠ PBD are alternate angles, and so are equal.
So ∠ PBD = 30°. Similarly, ∠ PAC = 45°.
In right Δ PBD, we have PD/BD = tan 300 = 1/√3 ⇒BD = √3

In right Δ PAC, we have PD/AC = tan450 =1 ⇒ PC =AC
Also, PC = PD + DC, therefore, PD + DC = AC.
Since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD = PD √3

So, the height of the multi-storeyed building is {4(√3 + 1) + 8}m = 4(3 + √3)m and the distance
between the two buildings is also 4(3 + √3)m.

10. If the mean of the distribution given below is 16.5, then find the values of a and b.

Here, Σf = a + b + 25 = 40 ⇒ a + b = 15, Σfu = b – 14, h = 5, A = 17.5

SECTION – C

11. If the ratio of the sum of the first n terms of two APs is (7n + 1) : (4n + 27), then find the ratio of their 9th terms.
Answer : Let the first terms be a and a′ and d and d′ be their respective common differences.

12. Prove that the parallelogram circumscribing a circle is a rhombus.
Answer : We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle)
with centre O.
Since tangents to a circle from an external point are equal in length,
∴ AP = AS, BP = BQ, CR = CQ and DR = DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
But AB = CD [opposite sides of ABCD]
∴ AB + CD = AD + BC ⇒ 2 AB = 2 BC
⇒ AB = BC
Similarly AB = DA and DA = CD
Thus, AB = BC = CD = AD
Hence ABCD is a rhombus.

OR

In the below figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

Answer : Since, the tangents drawn to a circle from an external point are equal.
∴ AP = AC
In Δ PAO and Δ AOC, we have:
AO = AO [Common]
OP = OC [Radii of the same circle]
AP = AC
⇒ Δ PAO ≅ Δ AOC [SSS Congruency]
∴ ∠PAO = ∠CAO
∠PAC = 2 ∠CAO …(1)
Similarly ∠CBQ = 2 ∠CBO …(2)
Again, we know that sum of internal angles on the same side of a transversal is 180°.
∴ ∠PAC + ∠CBQ = 180°
⇒ 2 ∠CAO + 2 ∠CBO = 180° [From (1) and (2)]
⇒ ∠CAO + ∠CBO = 180°/2 = 90° …(3)
Also ∠CAO + ∠CBO + ∠AOB = 180° [Sum of angles of a triangle]
⇒ 90° + ∠AOB = 180°
⇒ ∠AOB = 180° − 90°
⇒ ∠AOB = 90°

CASE STUDY QUESTION – 1

13. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens.
The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm.

Based on the above information, answer the following questions.
(i) Find the volume of four conical depressions in the entire stand
(ii) Find the volume of wood in the entire stand
Answer : (i) Dimensions of the cuboid are 15 cm, 10 cm and 3.5 cm.
∴ Volume of the cuboid = 15 × 10 × 35/10 cm3
= 15 × 35 cm3
= 525 cm3
Since each depression is conical with base radius (r) = 0.5 cm and depth (h) = 1.4 cm,
∴ Volume of each depression (cone)

Since there are 4 depressions,
∴ Total volume of 4 depressions

(ii) Volume of the wood in entire stand
= [Volume of the wooden cuboid] − [Volume of 4 depressions]

CASE STUDY QUESTION – 2

14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After 30 seconds, the angle of elevation reduces to 30° (see the below figure).

Based on the above information, answer the following questions. (Take √3 =1.732)
(i) Find the distance travelled by the balloon during the interval.
(ii) Find the speed of the balloon.
Answer : (i) In the figure, let C be the position of the observer (the girl).

A and P are two positions of the balloon.
CD is the horizontal line from the eyes of the (observer) girl.
Here PD = AB = 88.2 m − 1.2 m = 87 m

Thus, the required distance between the two positions of the balloon = 58 √3 m
= 58 x 1.73 = 100.46 m (approx.)
(ii) Speed of the balloon = Distance/time = 100.46/30 = 3.35 m/s (approx.)