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A boy is roaming on the roof of his house observes that the insects are there in bunches. He found '$p$' insects on day 1. On day two, the number of insects doubled and he takes out '$q$' insects and kills them. On the third day again, the insects left is doubled and he again takes out '$q$' insects and kill them. On the fourth day also he found the same, insects left is doubled and he takes out '$q$' insects again. He is surprised to find that there are no insects left on his roof.

Q.

Common Information Question: 2/3

The minimum number of insects that he kills on any of the last three days is:

 A.

8

 B.

7

 C.

0

 D.

1

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Solution:
Option(A) is correct

Day (1) -> Number of insects $-> p$
Day (2) -> Number of insects $-> 2p−q$
Day (3) -> Number of insects $-> 4p−3q$
Day (4) -> Number of insects $-> 8p−7q$

Thus,

\(8p-7q=0\)

\(\Rightarrow q=\dfrac{8p}{7}\)

$\Rightarrow P$ has to multiple of 7 so that $q$ is an integer
$\Rightarrow$ For $q$ is to minimum $p=7$
$\Rightarrow q$ = 8

 


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