WAEC 2012 · Paper 2 · Q11

(a)
In the diagram, $ABCD$ is a rectangular garden $(3n - 1)\text{ m}$ long and $(2n + 1)\text{ m}$ wide. A wire mesh $135\text{ m}$ long is used to mark its boundary and to divide it into 8 equal plots (3 lines along the length and 5 across). Find the value of $n$ .
(b)
A cylinder with base radius $14\text{ cm}$ has the same volume as a cube of side $22\text{ cm}$ . Calculate the ratio of the total surface area of the cylinder to that of the cube. $\left[\text{Take }\pi = \frac{22}{7}\right]$
Show the answer
$73 : 77$ (about $0.95 : 1$ )

Worked solution (try it first)

Read the diagram carefully: the mesh runs along the length 3 times (the two long sides and one line between them) and across the width 5 times (the two short sides and three lines between them).
So the total length of mesh is $3(3n - 1) + 5(2n + 1) = 135$ .
Expand: $9n - 3 + 10n + 5 = 135$ .
So $19n + 2 = 135$ , $19n = 133$ and $n = 7$ .

Volume of the cube $= 22^3 = 10\,648\text{ cm}^3$ .
The cylinder has the same volume: $\frac{22}{7} \times 14^2 \times h = 10\,648$ , so $616h = 10\,648$ and $h = \frac{121}{7}\text{ cm}$ .
Total surface area of the cylinder $= 2\pi r(r + h)$
$= 2 \times \frac{22}{7} \times 14 \times \left(14 + \frac{121}{7}\right)$
$= 88 \times \frac{219}{7}$
$= \frac{19\,272}{7}\text{ cm}^2$ .
Total surface area of the cube $= 6 \times 22^2 = 2904\text{ cm}^2$ .
Ratio $= \frac{19\,272}{7} : 2904$
$= 19\,272 : 20\,328$
$= 73 : 77$ .