WAEC 2012 · Paper 2 · Q6✱✱

(a)
Simplify: $\left[\dfrac{3m + 1}{6} - \dfrac{m + 2}{5}\right] \div \dfrac{9m - 7}{15}$ .
(b)
The external diameter of a tyre is $52\text{ cm}$ . How many revolutions, correct to four significant figures, will the tyre make to cover a distance of $2.05\text{ km}$ ? $\left[\text{Take }\pi = \frac{22}{7}\right]$
(c)
The diagram is a square cardboard $PRSU$ of side $10\text{ cm}$ , with $Q$ on $PR$ , $T$ on $US$ , $V$ on $PU$ and $QT \parallel PU$ . $|PQ| = 4\text{ cm}$ , $|PV| = 8\text{ cm}$ and $|UT| = 4\text{ cm}$ . If a hole is made at random on the cardboard, what is the probability that it is made on $\triangle QVT$ ?

Worked solution (try it first)

Combine the bracket over the common denominator 30: $\frac{5(3m + 1) - 6(m + 2)}{30} = \frac{15m + 5 - 6m - 12}{30}$
$= \frac{9m - 7}{30}$ .
Dividing by a fraction means multiplying by it upside down: $\frac{9m - 7}{30} \times \frac{15}{9m - 7}$ .
The $(9m - 7)$ cancels, leaving $\frac{15}{30} = \frac12$ .

Radius $= 26\text{ cm}$ , so one revolution covers the circumference: $2 \times \frac{22}{7} \times 26 = \frac{1144}{7}\text{ cm}$ .
The distance is $2.05\text{ km} = 205\,000\text{ cm}$ .
Revolutions $= 205\,000 \div \frac{1144}{7}$
$= \frac{1\,435\,000}{1144}$
$\approx 1254.4$ .
Correct to 4 significant figures: 1254 revolutions.

The square has area $10 \times 10 = 100\text{ cm}^2$ .
In $\triangle QVT$ , take $QT$ as the base: it runs straight across the square, so $|QT| = 10\text{ cm}$ .
The height is the distance from $V$ to the line $QT$ , which is $|PQ| = 4\text{ cm}$ .
Area $= \frac12 \times 10 \times 4$
$= 20\text{ cm}^2$ .
Probability $= \frac{20}{100} = \frac15$ .