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 $C=\frac{1}{2}$.

Answer 1

According to question let, $A=K\frac{\sqrt{B}}{C^3}$

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

$3=K\frac{\sqrt{256}}{2^3}=K\frac{16}{8}=2K$

$K=\frac{3}{2}$

So

$A=\frac{3\sqrt{B}}{2C^3}$

Now for $A=24,C=\frac{1}{2}$ we have,

$24=\frac{3\sqrt{B}}{2.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}}$

$\sqrt{B}=\frac{\mathrm{24.2.}\frac{1}{2}.\frac{1}{2}.\frac{1}{2}}{3}=\frac{48}{24}=2$

$B=2^2=4$