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.
According to question let, A=K√BC3
where K is a constant and A=3,B=256,C=2 satisfies the equation, So
3=K√25623=K168=2K
K=32
So
A=3√B2C3
Now for A=24,C=12 we have,
24=3√B2.12.12.12
√B=24.2.12.12.123=4824=2
B=22=4