Quant Boosters - Kamal Lohia - Set 1
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Q27) Two circles with radii 15cm and 20cm intersect each other at two points A and B such that AB is 25cm. Find the difference of non-common area of two circles
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It is simply the difference of areas of the two circles i.e. pi * 20^2 - pi * 15^2 = 175 * pi = 550 sq cm.
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Q28) A point P lies inside unit square ABCD such that angle(APB) is acute while angle(BPC) is obtuse. Find the area of the locus of point P.
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If we draw two semicircles on AB and BC as diameters, then P must lie inside the semicircle on BC but outside the semicircle on AB.
So the required area comes out to be 1/4
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Q29) Arc AB subtends an angle of 120 degrees on the center of the circle, O.
Q is a point in the major sector of arc AB. Which of the following may be true?
I. angle(AQB) > 120
II. angle(AQB) = 120
III. angle(AQB) < 120
a. III only
b. both II and III
c. both I and II
d. I, II and III
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Answer : D i.e. I, II and II all may be true.
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Q30) S is a set of first 100 natural numbers. 13 numbers are selected randomly. At least 'n' of 13 numbers give same remainder when divided by 5. What is the least value of n ?
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There are 5 possible different remainders i.e. 0, 1, 2, 3, 4 and to get the least number with same remainders, we should distribute the 13 numbers as widely as possible. In WORST scenario, we'll have at least ONE REMAINDER which is being shared by three numbers. Hence the answer will be 3.
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Just trying to generalize @kamal_lohia sir's solution
In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n
[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)
Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)
So the number of distinct integers would be [n/2] + [n/4] + 1if n = 100,
number of distinct integers would be [100/2] + [100/4] + 1 = 76if n = 2014,
number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511if n = 13
number of distinct integers would be [13/2] + [13/4] + 1 = 10
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