Please refer to Class 10 Mathematics Sample Paper with solutions provided below. All sample papers for Mathematics Class 10 have been designed as per the latest paper pattern issued by CBSE for the current academic year. Students should practice these guess papers for Class 10 Mathematics as it will help them to gain more understanding of the type of questions that are expected to be asked in upcoming Class 10 Mathematics exams. Please click on the links below to access free CBSE Sample Papers for Class 10 Mathematics.

## Class 10 Mathematics Sample Paper Term 2 With Solutions Set A

#### SECTION – A

**1. In the given figure, the radii of two concentric circles are 7 cm and 8 cm. If PA = 15 cm then find PB.**

**Answer : **Here, OA ⊥ PA and OB ⊥ PB

[∵ Tangent at any point of a circle is perpendicular to the radius through the point of contact.]

In ΔPAO, OP^{2} = AP^{2} + OA^{2} = 15^{2} + 8^{2} = 225 + 64 = 289

⇒ OP = 17 cm

In ΔPBO, PB^{2} = OP^{2} – OB^{2} = 17^{2} – 7^{2} = 289 – 49 = 240

⇒ PB = √240 cm = 4√15 cm

**2. Find the roots of the equation x2 + 7x + 10 = 0 by using quadratic formula.****Answer :** Given, x^{2} + 7x + 10 = 0

Comparing with ax2 + bx + c = 0, we get a = 1, b = 7 and c = 10

Hence, the roots of the given equation are – 2 and – 5.

**3. Which term of the A.P. 27, 24, 21, …… is zero?****Answer :** Given A.P. = 27, 24, 21, ……….. .

Here, a = 27 and d = 24 – 27 = – 3 and, an = 0

∴ a_{n} = a + (n – 1)d ⇒ 0 = 27 + (n – 1)(– 3)

⇒ – 3n + 3 = – 27

⇒ – 3n = – 27 – 3 = – 30 ⇒ n = 10.

**OR**

**In an Arithmetic Progression, if d = – 4, n = 7, a _{n} = 4, then find a.**

**Answer**

**:**We have, d = – 4, n = 7 and a

_{n}= 4

∴ a

_{n}= a + (n – 1)d 1 ⇒ 4 = a + (7 – 1)(– 4)

⇒ 4 = a + 6 (– 4)

⇒ = a – 24 ⇒ a = 4 + 24 ⇒ a = 28.

**4. A toy is in the form of a cone mounted on a hemisphere. The diameter of the base of the cone and that of hemisphere is 18 cm and the height of cone is 12 cm. Calculate the surface area of the toy. [Take π = 3.14]****Answer :** Radius of the base of the cone and hemisphere (r) = 18/2 = 9 cm

Height of cone (h) = 12 cm

Slant height of cone, l = √(r^{2} + h^{2})

⇒ l = √(9^{2} +12^{2}) = √(81+144) = √225 =15cm

Total surface area of toy = Curved surface area of hemisphere + Curved surface area of cone

= 2πr^{2} + πrl = πr(2r + l) = 3.14 × 9(2 × 9 + 15)

= 3.14 × 9 × 33 = 932.58 cm^{2}

**5. Find the value of mode, using an empirical relation, when it is given that mean and median are 10.5 and 9.6 respectively.****Answer :** We know, the empirical relationship is

Mode = 3 Median – 2 Mean

= 3(9.6) – 2(10.5) [ Median = 9.6 and Mean = 10.5 ]

= 28.8 – 21.0 = 7.8

**6. Find the value(s) of k so that the quadratic equation 3x ^{2} − 2kx + 12 = 0 has equal roots.**

**Answer :**Comparing with ax

^{2}+ bx + c = 0, we have a = 3, b = −2k, c = 12

For equal roots, D = 0 ⇒ b

^{2}– 4ac = 0

⇒ (−2k)

^{2}− 4 × 3 × 12 = 0

⇒ 4k

^{2}– 144 = 0 ⇒ k

^{2}= 36 ⇒ k = ±6

**OR**

**Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.****Answer :** Let the three consecutive natural numbers be x, x + 1 and x + 2.

∴ (x + 1)^{2} = (x + 2)^{2} – (x)^{2} + 60 ⇒ x^{2} + 2x + 1 = x^{2} + 4x + 4 – x^{2} + 60

⇒ x^{2} – 2x – 63 = 0 ⇒ x^{2} – 9x + 7x – 63 = 0 ⇒ x(x – 9) + 7(x – 9) = 0

⇒ (x – 9)(x + 7) = 0

Thus, x = 9 or x = – 7

Rejecting – 7, we get x = 9

Hence, three numbers are 9, 10 and 11.

#### SECTION – B

**7. The two palm trees are of equal heights and are standing opposite to each other on either side of the river, which is 80 m wide. From a point O between them on the river the angles of elevation of the top of the trees are 60° and 30°, respectively. Find the height of the trees and the distances of the point O from the trees. (use √3 = 1.73)****Answer :** Let BD = width of river = 80 m

AB = CD = height of both trees = h

BO = x, OD = 80 – x

In ΔABO, ∠B = 90°, tan 60° = h/x ⇒ √3 = h/x ⇒ h = √3x ….. (i)

In ΔCDO, ∠D = 90°, tan 30° = h/(80 – x) ⇒ 1/√3 = h/(80 – x) …..(ii)

Solving (i) and (ii), we get x = 20 m

h = √3x [From eqn. (i)]

= 1.73 × 20 m = 34.6 m

The height of the trees = h = 34.6 m

BO = x = 20 m and

DO = 80 – x = 80 – 20 = 60 m

**OR**

**Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of two ships as observed from the top of light house are 60° and 45°. If the height of the light house is 200 m, find the distance between the two ships. [Use √3 = 1.73]****Answer :** Let d m be the distance between the two ships. Suppose the distance of one of the ships from the light house is x m, then the distance of the other ship from the light house is (d – x) m.

Thus, the distance between two ships is approximately 315.60 m.

**8. If the mean of the following distribution is 54, find the missing frequency x.**

**Answer : **

Mean = (∑f_{1}x_{1}) / ∑f_{1} ⇒ 54 = (3600+90x) / (80+x) ⇒ 54(80+x) = 3600 + 9x

⇒ 4320 + 54x = 3600 + 90x

⇒ 36x = 720 ⇒ x = 20

**9. Draw a line segment AB of length 9 cm. With A and B as centres, draw circles of radius 5 cm and 3 cm respectively. Construct tangents to each circle from the centre of the other circle.****Answer :**

**Steps of Construction:**

(i) Draw a line segment AB of 9 cm.

(ii) Taking A and B as centres draw two circles of radii 5 cm and 3 cm respectively.

(iii) Perpendicular bisect the line AB. Let midpoint of AB be C.

(iv) Taking C as centre draw a circle of radius AC which intersects the two circles at point P, Q,R and S.

(v) Join BP, BQ, AS and AR.

Hence, BP, BQ and AR, AS are the required tangents.

**10. Find the value of p from the following data, if its mode is 48.**

**Answer :** Here, modal class=40 – 50

#### SECTION – C

**11. A hemispherical depression is cut out from one face of a cubical block of side 7 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Find the surface area of the remaining solid. [Use π =22/7]****Answer :** Diameter of hemisphere = Edge of cube = 7 cm Radius of hemisphere(r) = 7/2 cm

Required surface area = surface area of cube – area of top of hemisphere + curved surface area of hemisphere

= 6a^{2} – πr^{2} + 2πr^{2} = 6a^{2} + πr^{2}

= 6(7)^{2} + π(7/2)^{2} = 6 × 49 + 22/7 × 7/2 × 7/2

= 294 + 38.5 = 332.5 cm^{2}

**12. PQ is a tangent to a circle with centre O at point P. If ΔOPQ is an isosceles triangle, then find ∠OQP.****Answer :** In ΔOPQ, ∠P + ∠Q + ∠O =180° (∠O = ∠Q, isosceles triangle)

⇒ 2∠Q + ∠P = 180°

⇒ 2∠Q + 90° = 180°

⇒ 2∠Q = 90°

⇒ ∠Q = 45°

**OR**

**The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger circle at P on producing. Find the length of AP.****Answer :** ∠APB = 90° (angle in semi-circle)

and ∠ODB = 90° (radius is perpendicular to tangent)

∴ ΔABP ~ ΔOBD (by AA similarity)

⇒ AB/OB = AP/OD ⇒ 26/13 = AP/8 ⇒ AP = 16cm

**Case Study- 1**

**13. There are two windows in a house. First window is at the height of 2 m above the ground and other window is 4 m vertically above the lower window. Ankit and Radha are sitting inside the two windows at points G and F respectively. At an instant, the angles of elevation of a balloon from these windows are observed to be 60° and 30° as shown below.**

**Based on the above information, answer the following questions.**

(i) Find the value of h.

(ii) What is the height of the balloon from the ground?**Answer :**

(ii) Height of the balloon from the ground = BE

= BC + CD + DE = 2 + 4 + 2 = 8 m

**Case Study- 2**

**14. Anita’s mother start a new shoe shop. To display the shoes, she put 3 pairs of shoes in 1st row, 5 pairs in 2nd row, 7 pairs in 3rd row and so on.**

**On the basis of above information, answer the following questions.**

(i) If she puts a total of 120 pairs of shoes, then find the number of rows required.

(ii) What is the difference of pairs of shoes in 17th row and 10th row.**Answer :** Number of pairs of shoes in 1st, 2nd, 3rd row, … are 3, 5, 7, …

So, it forms an A.P. with first term a = 3, d = 5 – 3 = 2

(i) Let n be the number of rows required.

∴ Sn = 120 ⇒ (n/2) [2(3) + (n − 1)2] = 120

⇒ n2 + 2n – 120 = 0 ⇒ n2 + 12n – 10n – 120 = 0

⇒ (n + 12) (n – 10) = 0 ⇒ n = 10

So, 10 rows required to put 120 pairs.

(ii) No. of pairs in 17th row = a17 = 3 + 16(2) = 35

No. of pairs in 10th row = a10 = 3 + 9(2) = 21

∴ Required difference = 35 – 21 = 14