Sec-3 Sec-a

CBSE X Mathematics 2009 Solution (SET 3)

CBSE X 2009 Mathematics Section A Question Number 1 to 10 carry 1 mark each.

1.

Find the discriminant of the quadratic equation 3 3x 2 10 x

3

0

Solution: For the quadratic equation ax2 + bx + c = 0, Discriminant = D = b2 – 4ac Hence, for the given equation, D = (10) 2 4(3 3)( 3) = 100 – 36 = 64 Thus, the discriminant of the given equation is 64.

2.

If

4 , a, and 2 are three consecutive terms of an A.P., then find the value of a. 5

Solution: If three terms a, b, and c are in A.P., then we have b – a = c – b 2b = a + c 4 If , a, and 2 are three consecutive terms of an A.P., then 5 4 2a 2 5 14 2a 5 7 a 5 Thus, the value of a is

3.

7 . 5

If the areas of two similar triangles are in the ratio 25:64, write the ratio of their corresponding sides.

CBSE X Mathematics 2009 Solution (SET 3)

Solution: We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. It is given that the areas of two similar triangles are in the ratio 25:64. 25 64

5 8

2

Thus, the ratio of the corresponding sides of the two similar triangles is 5:8.

4.

In figure 1, ∆ABC is circumscribing a circle. Find the length of BC.

Solution:

BR = BP However, BR = 3 cm BP = 3 cm AR = AQ

[Tangents drawn to a circle from a point outside the circle are equal] … (1) [Tangents drawn to a circle from a point outside the circle are equal]

However, AR = 4 cm AQ = 4 cm AQ + QC = AC QC = AC – AQ Using the values AQ = 4 cm and AC = 11 cm, QC = 11 cm – 4 cm QC = 7 cm CP = CQ CP = 7 cm

[Tangents drawn to a circle from a point outside the circle are equal] … (2)

BC = BP + CP On using equations (1) and (2), we obtain BC = 3 cm + 7 cm

CBSE X Mathematics 2009 Solution (SET 3)

BC = 10 cm Thus, the length of BC is 10 cm.

5.

Two coins are tossed simultaneously. Find the probability of getting exactly one head. Solution: If two coins are tossed simultaneously, then the possible outcomes are S = {HT, TH, TT, HH} Thus, the total number of possible outcomes is 4. Out of all the four outcomes, {HT} and {TH} are cases of exactly one head. Favourable outcomes 2 1 Required probability = Total possible outcomes 4 2

6.

Find the [HCF × LCM] for the numbers 100 and 190. Solution: For two numbers a and b, HCF × LCM = a × b For the given numbers 100 and 190: HCF × LCM = 100 × 190 HCF × LCM = 19000

7.

If 1 is a zero of the polynomial p(x) = ax2 – 3(a – 1) x – 1, then find the value of a. Solution: If 1 is a zero of polynomial p(x), then p(1) = 0. p(1) = a(1)2 – 3(a – 1) (1) – 1 = 0 a – 3a + 3 – 1 = 0 –2a = –2 a=1 Thus, the value of a is 1.

8.

If sec2θ (1 + sin θ) (1 – sin θ) = k, then find the value of k.

CBSE X Mathematics 2009 Solution (SET 3)

Solution: sec2 (1 + sin )(1 – sin ) = k sec 2 (1 sin 2 ) k [(a b)(a b)

sec 2 (cos 2 ) 1 cos 2 1 k

cos 2

[1 sin 2

k k

sec

a 2 b2 ]

cos 2 ] 1 cos

k 1 Thus, the value of k is 1.

9.

If the diameter of a semicircular protractor is 14 cm, then find its perimeter. Solution:

Diameter = 14 cm Diameter Radius = 2

14 cm 2

7 cm

Length of the semicircular part = πr

22 (7) 7

22 cm

Total perimeter = Length of semicircular part + Diameter = 22 cm + 14 cm = 36 cm Thus, the perimeter of the protractor is 36 cm.

10.

Find the number of solutions of the following pair of linear equations: x + 2y – 8 = 0 2x + 4y = 16 Solution: The given pair of linear equations is x + 2y – 8 = 0 2x + 4y – 16 = 0

CBSE X Mathematics 2009 Solution (SET 3)

On comparing with general equations a1 x b1 y c1 0 a2 x b2 y c2

0

we obtain a1 a2

1 b1 , 2 b2 a1 a2

b1 b2

2 4 c1 c2

1 c1 , 2 c2

8 16

1 2

1 2

Hence, the given pair of linear equations has infinitely many solutions.