WAEC 2013 · Paper 2 · Q13

(a)
Given that $Y = \{x : -2 \le x \le 5\}$ and $W = \{x : 1 < x < 6\}$ , illustrate $Y \cap W$ on the number line.
Model answer
Open circle at 1, filled dot at 5, joined by a line.
(b)(i)
$x$ varies jointly as the square of $m$ and the cube of $n$ . When $x = 9$ , $m = \frac34$ and $n = \frac12$ . Determine the relationship between $x$ , $m$ and $n$ .
(b)(ii)
Calculate, correct to 3 significant figures, the value of: ( $\alpha$ ) $x$ when $m = \frac23$ and $n = \frac15$ ; ( $\beta$ ) $m$ when $x = 5$ and $n = \frac18$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$Y \cap W$ holds the numbers in both sets: greater than 1 (from $W$ ) and at most 5 (from $Y$ ).
So $Y \cap W = \{x : 1 < x \le 5\}$ .
On the number line, draw an open circle at 1 (1 is left out), a filled dot at 5 (5 is included), and join them.

Joint variation: $x = km^2n^3$ for some constant $k$ .
Put in $x = 9$ , $m = \frac34$ and $n = \frac12$ : $9 = k \times \frac{9}{16} \times \frac18$ , which is $9 = \frac{9k}{128}$ .
So $k = 128$ , and the relationship is $x = 128m^2n^3$ .

( $\alpha$ )** Put in $m = \frac23$ and $n = \frac15$ : $x = 128 \times \frac49 \times \frac{1}{125}$
$= \frac{512}{1125}$ .
$\frac{512}{1125} = 0.4551$ , so $x = 0.455$ to 3 significant figures.
( $\beta$ ) Make $m^2$ the subject: $m^2 = \frac{x}{128n^3}$ .
Put in $x = 5$ and $n = \frac18$ : $128 \times \frac{1}{512} = \frac14$ , so $m^2 = 5 \div \frac14 = 20$ .
$m = \sqrt{20} = 4.47$ to 3 significant figures.