WAEC 2008 · Paper 2 · Q3

(a)
A bucket is $12\text{ cm}$ in diameter at the bottom, $20\text{ cm}$ in diameter at the open end and $16\text{ cm}$ deep. If the bucket is filled with water and emptied into a cylindrical tin of diameter $28\text{ cm}$ , calculate the depth of water in the tin. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

The bucket is a frustum: a cone with its tip cut off.
The radii are $R = 10\text{ cm}$ at the top and $r = 6\text{ cm}$ at the bottom, and the depth is $h = 16\text{ cm}$ .
Use the frustum volume $V = \frac13\pi h(R^2 + Rr + r^2)$ .
Put in the numbers: $V = \frac13 \times \frac{22}{7} \times 16 \times (100 + 60 + 36)$
$= \frac13 \times \frac{22}{7} \times 16 \times 196$ .
Simplify: $V = \frac{68\,992}{21}$
$= 3285\frac13\text{ cm}^3$ .
The tin has radius $14\text{ cm}$ , so its base area is $\frac{22}{7} \times 14^2 = 616\text{ cm}^2$ .
Depth of water $= \frac{\text{volume}}{\text{base area}}$
$= \frac{3285\frac13}{616}$
$= 5\frac13$ .
The water is $5\frac13\text{ cm}$ (about $5.33\text{ cm}$ ) deep.