Series HRS Code-30/2 Summative Assessment –II Subject Mathematics class 10 CBSE Board 2014

SECTION-A

1. The probability that a number selected at random from the number 1,2,3,…..15. is a multiple of 4 is

(A)4/15 (B) 2/15 (C)1/5 (D)1/3

Ans: 1/5

2. The angle of depression of car parked on the road from the top of a 150m high tower is 300 . The distance of the car from the tower in m meter is

(A)50Ö3 (B)150Ö3 (C) 50Ö2 (D)75

Ans: (B)150Ö3

3. Two circle touches externally at P . AB is common tangent to the circle touching them at A and B . The value of <APB is

(A)300 (B)450 (C)600 (D) 900

Ans: (B)450

4. If k,2k -1 and 2k + 1 are three consecutive term of AP then value of k is

(A)2 (B)3 (C)-3 (D)5

Ans: (B)3

5. A chord of a circle of radius 10 cm subtends a right angle at its centre. The length of chord is

(A) 5Ö2 (B)10Ö2 (C)5/Ö2 (D)10Ö3

Ans: (B)10Ö2

6. ABCD is a rectangle whose three vertices are B(4,0), C(4,3) and D(0,3) . The length of one of its diagonal is

(A) 5 (B)4 (C)3 (D)25

Ans: (D)25

7. In a right triangle ABC , right angled at B ,BC = 12cm and AB = 5cm . The radius of circle inscribe in the triangle (in cm) is

(A) 4 (B)3 (C)2 (D) 1

Ans: (C)2

8. In a family of 3 children, the probability of having at least one boy is

(A) 7/8 (B)1/8 (C) 5/8 (D) ¾

Ans: (A) 7/8

SECTION-B

9.In fig-01,common tangent AB and CD to the circles with centers O1 and O2 intersect at E. Prove that AB=CD

Solution:

AE= CE [tangent from E] and CE = ED [tangent from E]

Adding them we get AE+ CE = CE + ED

AB = CD

10. The incircle of an isosceles triangle ABC , in which AB=AC, touches side BC , CA and AB at D,E and F respectively. Prove that BD = CD

Solution:

AB = AC

BF + AF = AE + CE --------(i)

BF = BD and CE = CD [tangent from B and C] -----------------(ii)

using (i)and (ii)

BD + AE = AE + CD

BD = CD

11. Two different dice are tossed together. Find the probability

(i) that the number on each side is even

(ii)The sum of the number appear on two dice is 5

Solution:

(i) Total favorable outcomes are 22,42,62,24,44,64,26,46,66 = 9

Number of all possible outcomes = 36

P[E] = 9/36= ¼

(ii) Total favorable outcomes are 41,32,23,14 = 4

Number of all possible outcomes = 36

P[E] = 4/36= 1/9

12. The total surface area of a solid hemisphere is 462cm2 , find the volume?

Ans: Total surface area of solid hemisphere= 3pr2

462 = 3 x 22/7 x r2

r = 7cm

Volume of solid hemisphere = 2/3 x pr3 = 2/3 x ( 22/7) x 7 x 7 x 7 = 2156/3 = 718.66cm3

13. Find the number of natural number between 101 and 199 which are by both 2 and 5.

Solution:

a1 = 110 and d = 10 an = 190

an = a + (n-1)d

190 = 110 + (n-1)10

(190 -110 )/10 = n-1

8+1= n

n = 9

Hence, there are 9 natural number between 101 and 199 which are by both 2 and 5.

14. Find the value of k for which the quadratic equation 9x2 – 3kx + k hs equal root.

Solution:

For equal root: D = 0

b2 – 4ac = 0

(- 3k)2 – 4 x 9 x k = 0

9k2 = 36k

k = 4 Hence, Value of k for which the quadratic equation 9x2 – 3kx + k ha equal root = 4

SECTION-A

1. The probability that a number selected at random from the number 1,2,3,…..15. is a multiple of 4 is

(A)4/15 (B) 2/15 (C)1/5 (D)1/3

Ans: 1/5

2. The angle of depression of car parked on the road from the top of a 150m high tower is 300 . The distance of the car from the tower in m meter is

(A)50Ö3 (B)150Ö3 (C) 50Ö2 (D)75

Ans: (B)150Ö3

3. Two circle touches externally at P . AB is common tangent to the circle touching them at A and B . The value of <APB is

(A)300 (B)450 (C)600 (D) 900

Ans: (B)450

4. If k,2k -1 and 2k + 1 are three consecutive term of AP then value of k is

(A)2 (B)3 (C)-3 (D)5

Ans: (B)3

5. A chord of a circle of radius 10 cm subtends a right angle at its centre. The length of chord is

(A) 5Ö2 (B)10Ö2 (C)5/Ö2 (D)10Ö3

Ans: (B)10Ö2

6. ABCD is a rectangle whose three vertices are B(4,0), C(4,3) and D(0,3) . The length of one of its diagonal is

(A) 5 (B)4 (C)3 (D)25

Ans: (D)25

7. In a right triangle ABC , right angled at B ,BC = 12cm and AB = 5cm . The radius of circle inscribe in the triangle (in cm) is

(A) 4 (B)3 (C)2 (D) 1

Ans: (C)2

8. In a family of 3 children, the probability of having at least one boy is

(A) 7/8 (B)1/8 (C) 5/8 (D) ¾

Ans: (A) 7/8

SECTION-B

9.In fig-01,common tangent AB and CD to the circles with centers O1 and O2 intersect at E. Prove that AB=CD

Solution:

AE= CE [tangent from E] and CE = ED [tangent from E]

Adding them we get AE+ CE = CE + ED

AB = CD

10. The incircle of an isosceles triangle ABC , in which AB=AC, touches side BC , CA and AB at D,E and F respectively. Prove that BD = CD

Solution:

AB = AC

BF + AF = AE + CE --------(i)

BF = BD and CE = CD [tangent from B and C] -----------------(ii)

using (i)and (ii)

BD + AE = AE + CD

BD = CD

11. Two different dice are tossed together. Find the probability

(i) that the number on each side is even

(ii)The sum of the number appear on two dice is 5

Solution:

(i) Total favorable outcomes are 22,42,62,24,44,64,26,46,66 = 9

Number of all possible outcomes = 36

P[E] = 9/36= ¼

(ii) Total favorable outcomes are 41,32,23,14 = 4

Number of all possible outcomes = 36

P[E] = 4/36= 1/9

12. The total surface area of a solid hemisphere is 462cm2 , find the volume?

Ans: Total surface area of solid hemisphere= 3pr2

462 = 3 x 22/7 x r2

r = 7cm

Volume of solid hemisphere = 2/3 x pr3 = 2/3 x ( 22/7) x 7 x 7 x 7 = 2156/3 = 718.66cm3

13. Find the number of natural number between 101 and 199 which are by both 2 and 5.

Solution:

a1 = 110 and d = 10 an = 190

an = a + (n-1)d

190 = 110 + (n-1)10

(190 -110 )/10 = n-1

8+1= n

n = 9

Hence, there are 9 natural number between 101 and 199 which are by both 2 and 5.

14. Find the value of k for which the quadratic equation 9x2 – 3kx + k hs equal root.

Solution:

For equal root: D = 0

b2 – 4ac = 0

(- 3k)2 – 4 x 9 x k = 0

9k2 = 36k

k = 4 Hence, Value of k for which the quadratic equation 9x2 – 3kx + k ha equal root = 4

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