Question

X
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(b) 32 +13-
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10
16
ज3 होल क्यूब आईफोन ​

Answer 1

Answer:

1. In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:

(a) 5, (b) 7, (c) 10, (d) 12, (e) 14

Solution:

Let the number of cows be x and their legs be 4x.

Let the number of chicken be y and their legs be 2x.

Total number of legs = 4x + 2y.

Total number of heads = x + y.

The number of legs was 14 more than twice the number of heads.

Therefore, 2 × (x + y) + 14 = 4x + 2y.

or, 2x + 2y + 14 = 4x + 2y.

or, 2x + 14 = 4x [subtracting 2y from both sides].

or, 14 = 4x – 2x [subtracting 2x from both sides].

or, 14 = 2x.

or, x = 7 [dividing by 2 on both sides].

Therefore, the number of cows = 7.

Answer: (b)

2. The roots of the equation ax2 + bx + c = 0 will be reciprocal if:

(a) a = b, (b) a = bc, (c) c = a, (d) c = b, (e) c = ab.

Solution:

Let k be one of the root of the given equation.

According to the problem,

1/k will be the other root of the given equation.

We know that, product of the roots of the equation = c/a.

Therefore, k × 1/k = c/a.

or, 1 = c/a.

or, a = c [multiplying a on both sides].

The roots of the equation ax2 + bx + c = 0 will be reciprocal if a = c.

Therefore, a = c or c = a.

Answer: (c)

3. If 8 ∙ 2x = 5(y+8), then when y = -8, x =

(a) -4, (b) -3, (c) 0, (d) 4, (e) 8

Solution:

8 ∙ 2x = 5(y+8).

or, 8 ∙ 2x = 5(-8+8).

or, 2x = 50.

or, 2x = 1 [anything to the power 0 is 1 then, 50 = 1].

Taking log on both sides,

log 2x = log 1.

or, x log 2 = 0 [since log 1 = 0].

or, x = 0/log 2 [dividing log 2 on both sides].

Therefore, x = 0.

Answer: (c)

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