Please solve q 42?

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Answer 1 · 2018-04-18T12:14:22

The answer is $"option (B)"$

Let the number of diamonds robbed be $=x$

To the first watchman, he gave

$(x/2+2)$ and the number remaining is $(x/2-2)$

To the second watchman, he gave

$(1/2(x/2-2)+2=x/4+1)$ and the number remaining is $(1/2(x/2-2)-2=x/4-3)$

To the third watchman, he gave

$(1/2(x/4-3)+2=x/8+1/2)$ and the number remaining is $(1/2(x/4-3)-2=x/8-7/2)$

But,

$x/8-7/2=1$

$x/8=1+7/2=9/2$

$x=9/2*8=36$ diamonds

The answer is $"option (B)"$