NECO 2023 · Paper 2 · Q10

Solid mensuration

A bucket full of water is $40\text{ cm}$ in diameter at the open end, $24\text{ cm}$ in diameter at the bottom and $32\text{ cm}$ deep. The bucket is emptied completely into a cylindrical drum of diameter $56\text{ cm}$ . Find the level of water in the drum, to the nearest whole number. $\left[\pi = \frac{22}{7}\right]$

(a)
Depth of water (cm)

Worked solution (try it first)

The bucket is a frustum: a cone with its tip cut off.
Its volume is the big cone minus the small cone that was removed.
Let the full cone have height $H$ .
The radii are 20 cm (top) and 12 cm (bottom), and the small cone's height is $H - 32$ .
By similar triangles, $\frac{H}{20} = \frac{H - 32}{12}$ , so $12H = 20H - 640$ and $H = 80$ cm.
The small cone is $80 - 32 = 48$ cm high.
Volume of water $= \frac13\pi(20^2 \times 80) - \frac13\pi(12^2 \times 48)$
$= \frac13 \times \frac{22}{7} \times (32\,000 - 6912)$
$\approx 26\,282.7\text{ cm}^3$ .
(The frustum formula $\frac{\pi h}{3}(R^2 + Rr + r^2)$ gives the same.)
The drum's base is $\frac{22}{7} \times 28^2 = 2464\text{ cm}^2$ .
The water keeps its volume, so its depth is $\frac{26\,282.7}{2464} \approx 10.67$ cm, which is 11 cm to the nearest whole number.

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