WAEC 2014 · Paper 2 · Q1

(a)
Without using tables or a calculator, simplify $\dfrac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}$ , leaving the answer in standard form (scientific notation).
(b)
In the diagram (not drawn to scale), $\overline{EF}$ is parallel to $\overline{GH}$ . $AEB$ and $BHC$ are straight lines. If $\angle AEF = 3x^\circ$ , $\angle ABC = 120^\circ$ and $\angle CHG = 7x^\circ$ , find the value of $\angle GHB$ .
Not drawn to scale (as in the paper); here the angles are drawn at their true sizes, with x = 15.

Worked solution (try it first)

Remove the decimals: top $= 6 \times 32 \times 4 \times 10^{-4}$
$= 768 \times 10^{-4}$ .
Bottom $= 12 \times 8 \times 16 \times 10^{-6}$
$= 1536 \times 10^{-6}$ .
Divide: $\frac{768}{1536} \times 10^{-4 - (-6)} = 0.5 \times 10^2$
$= 50$
$= 5 \times 10^1$ .

Draw a line through $B$ parallel to $EF$ and $GH$ .
It splits $\angle ABC$ into two parts: $3x$ (alternate angles with $\angle AEF$ ) and $180^\circ - 7x$ (co-interior with $\angle CHG$ ).
So $3x + 180 - 7x = 120$ , $4x = 60$ and $x = 15$ .
$\angle GHB$ and $\angle CHG$ are on the straight line $BHC$ : $\angle GHB = 180^\circ - 7 \times 15^\circ$
$= 75^\circ$ .