Set-2 Sec-a

CBSE X Mathematics 2009 Solution (SET 2)

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

1.

Find the [HCF × LCM] for the numbers 105 and 120. Solution: For two numbers a and b, HCF × LCM = a × b For the given numbers 105 and 120: HCF × LCM = 105 × 120 HCF × LCM = 12600

2.

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 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.

3.

If

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

CBSE X Mathematics 2009 Solution (SET 2)

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

4.

7 . 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

5.

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

CBSE X Mathematics 2009 Solution (SET 2) Solution:

BR = BP

[Tangents drawn to a circle from a point outside the circle are equal]

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

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

6.

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

CBSE X Mathematics 2009 Solution (SET 2)

= 22 cm + 14 cm = 36 cm Thus, the perimeter of the protractor is 36 cm.

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.

In LMN,

L = 50° and

N = 60°. If ∆LMN

∆PQR, then find

Q.

Solution:

L + M + N = 180° Substituting L = 50° and 50 + M + 60° = 180° M = 70°

(Angle sum property) N = 60° in this equation:

It is given that LMN PQR. We know that corresponding angles in similar triangles are of equal measures. M = Q = 70° Thus, the measure of

Q is 70°.

CBSE X Mathematics 2009 Solution (SET 2)

9.

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

k k

[1 sin 2 sec

a 2 b2 ]

cos 2 ] 1 cos

k 1 Thus, the value of k is 1.

10.

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

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