WAEC 2008 · Paper 2 · Q6

(a)
If $2^{x + y} = 16$ and $4^{x - y} = \frac{1}{32}$ , find the values of $x$ and $y$ .
Separate values with commas, e.g. 3, −2
(b)
$P$ , $Q$ and $R$ are related in such a way that $P \propto \dfrac{Q^2}{R}$ . When $P = 36$ , $Q = 3$ and $R = 4$ . Calculate $Q$ when $P = 200$ and $R = 2$ .

Worked solution (try it first)

(a)

Write both sides as powers of 2: $2^{x + y} = 2^4$ , so $x + y = 4$ .
$4^{x - y} = 2^{2(x - y)}$ and $\frac{1}{32} = 2^{-5}$ , so $2(x - y) = -5$ and $x - y = -\frac52$ .
Add the two equations: $2x = \frac32$ , so $x = \frac34$ .
Then $y = 4 - \frac34 = 3\frac14$ .

(b)

Write the variation with a constant: $P = \frac{kQ^2}{R}$ .
Find $k$ from $P = 36$ , $Q = 3$ , $R = 4$ : $36 = \frac{9k}{4}$ , so $k = 16$ .
So $P = \frac{16Q^2}{R}$ .
Put in $P = 200$ , $R = 2$ : $200 = \frac{16Q^2}{2} = 8Q^2$ .
So $Q^2 = 25$ and $Q = 5$ .

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