# What is the volume of the larger sphere if the diameters of two spheres are in the ratio of 2:3 and the sum of their volumes is 1260 cu.m?

Apr 2, 2015

It is $972$ cu.m

The volume formula of spheres is:

$V = \left(\frac{4}{3}\right) \cdot \pi \cdot {r}^{3}$

We have sphere $A$ and sphere $B$.

${V}_{A} = \left(\frac{4}{3}\right) \cdot \pi \cdot {\left({r}_{A}\right)}^{3}$

${V}_{B} = \left(\frac{4}{3}\right) \cdot \pi \cdot {\left({r}_{B}\right)}^{3}$

As we know that ${r}_{A} / {r}_{B} = \frac{2}{3}$

$3 {r}_{A} = 2 {r}_{B}$
${r}_{B} = 3 {r}_{A} / 2$

Now plug ${r}_{B}$ to ${V}_{B}$

${V}_{B} = \left(\frac{4}{3}\right) \cdot \pi \cdot {\left(3 {r}_{A} / 2\right)}^{3}$

${V}_{B} = \left(\frac{4}{3}\right) \cdot \pi \cdot 27 {\left({r}_{A}\right)}^{3} / 8$

${V}_{B} = \left(\frac{9}{2}\right) \cdot \pi \cdot {\left({r}_{A}\right)}^{3}$

So we can now see that ${V}_{B}$ is $\left(\frac{3}{4}\right) \cdot \left(\frac{9}{2}\right)$ times bigger than ${V}_{A}$

So we can simplify things now:

${V}_{A} = k$
${V}_{B} = \left(\frac{27}{8}\right) k$

Also we know ${V}_{A} + {V}_{B} = 1260$

$k + \frac{27 k}{8} = 1260$

$\frac{8 k + 27 k}{8} = 1260$

$8 k + 27 k = 1260 \cdot 8$
$35 k = 10080$
$k = 288$

$k$ was the volume of $A$ and the total volume was $1260$. So the larger sphere's volume is $1260 - 288 = 972$