15. Two vases are cylindrical in shape. The larger vase has diameter $20 \mathrm{~cm}$. The smaller vase has diameter $10 \mathrm{~cm}$ and height $16 \mathrm{~cm}$. The larger vase is partially filled with water. Then the empty smaller vase, with the open end at the top, is slowly pushed down into the water, which flows over its rim. When the smaller vase is pushed right down, it is half full of water.
What was the original depth of the water in the larger vase?
A $10 \mathrm{~cm}$
B $12 \mathrm{~cm}$
C $14 \mathrm{~cm}$
D $16 \mathrm{~cm}$
E $18 \mathrm{~cm}$
Solution
C
We use the formula $V=\pi r^{2} h$ for the volume $V$ of a cylinder of radius $r$ and height $h$. Expressed in terms of the diameter $d$ of the cylinder, where $r=\frac{1}{2} d$, this becomes $V=\frac{1}{4} \pi d^{2} h$.

When the smaller vase is pushed right down and the water finishes flowing over the rim, the water in the larger vase comes to the top of the smaller cylinder and so has the same depth as height of the smaller cylinder, that is, $16 \mathrm{~cm}$.

So the volume of water in the larger vase is that of a cylinder with diameter $20 \mathrm{~cm}$ and height $16 \mathrm{~cm}$, less the volume of the empty half of the smaller vase.
The volume of a cylinder with diameter $20 \mathrm{~cm}$ and height $16 \mathrm{~cm}$ is, $\frac{1}{4} \pi \times 20^{2} \times 16 \mathrm{~cm}^{3}=$ $1600 \pi \mathrm{cm}^{3}$. The volume of the empty half of the smaller cylinder is $\frac{1}{4} \pi \times 10^{2} \times 8 \mathrm{~cm}^{3}=200 \pi \mathrm{cm}^{3}$.
It follows that the volume of the water in the cylinder is, in $\mathrm{cm}^{3}$,
$$
1600 \pi-200 \pi=1400 \pi \text {. }
$$

Now suppose that the original depth of the water in the cylinder is $x \mathrm{~cm}$. It follows that the volume of the water in the cylinder is, in $\mathrm{cm}^{3}$,
$$
\frac{1}{4} \pi \times 20^{2} \times x=100 \pi x .
$$

Since these two expressions for the volume of the water must be the same,
$$
100 \pi x=1400 \pi,
$$
from which it follows that $x=14$.
Therefore the original depth of the water in the cylinder was $14 \mathrm{~cm}$.