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Answers
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)
4. A circle of radius 10 inches has its center at the vertex C of an equilateral triangle ABC and passes through the other two vertices. The side AC extended through C intersects the circle at D. The number of degrees of angle ADB is:
(a) 15, (b) 30, (c) 60, (d) 90, (e) 120