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Question

If A varies directly as the square root of B and inversely as the cube of C, and if A=3 when B=256 and C=2, find B when A=24 and C=12.

Solution
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According to question let, A=KBC3

where K is a constant and A=3,B=256,C=2 satisfies the equation, So

3=K25623=K168=2K

K=32

So

A=3B2C3

Now for A=24,C=12 we have,

24=3B2.12.12.12

B=24.2.12.12.123=4824=2

B=22=4


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