WAEC 2016 · Paper 2 · Q9✱

(a)
Solve: $3\log_{10}2 - 2\log_{10}3 = 1 + \log_{10}\left(\frac1x\right)$ .
(b)
The height of a cylindrical water container is $8\text{ m}$ . It took an athlete, running with a speed of $3\text{ km/h}$ , 3 minutes to run round the container once, keeping a constant distance of one metre from the container. Calculate, correct to the nearest whole number, the: (i) radius of the container; (ii) volume of the container. $\left[\text{Take }\pi = \frac{22}{7}\right]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$3\log_{10}2 - 2\log_{10}3 = \log_{10}\frac{8}{9}$ , and $1 + \log_{10}\frac1x = \log_{10}\frac{10}{x}$ .
So $\frac89 = \frac{10}{x}$ and $x = \frac{90}{8} = 11\frac14$ .

In 3 minutes at 3 km/h the athlete runs $3000 \times \frac{3}{60} = 150$ m.
He runs 1 m outside the container, on a circle of radius $r + 1$ : $2 \times \frac{22}{7} \times (r + 1) = 150$ , so $r + 1 \approx 23.86$ and $r \approx 22.86$ , which is 23 m to the nearest whole number.

Volume $= \pi r^2 h$
$= \frac{22}{7} \times 23^2 \times 8$
$\approx 13\,300.6$ , which is 13,301 m³ to the nearest whole number.