WAEC 2020 · Paper 2 · Q2

(a)
Given that $P = \left(\dfrac{rk}{Q} - ms\right)^{\frac23}$ , (i) make $Q$ the subject of the relation; (ii) find, correct to two decimal places, the value of $Q$ when $P = 3$ , $m = 15$ , $s = 0.2$ , $k = 4$ and $r = 10$ .
(b)
Given that $\dfrac{x + 2y}{5} = x - 2y$ , find $x : y$ .
Show the answer
$3 : 1$

Worked solution (try it first)

(a)(i)

$Q$ is inside a bracket raised to the power $\frac23$ .
Undo it by raising both sides to the power $\frac32$ : $P^{\frac32} = \frac{rk}{Q} - ms$ .
Add $ms$ to both sides: $P^{\frac32} + ms = \frac{rk}{Q}$ .
Multiply both sides by $Q$ and divide by $\left(P^{\frac32} + ms\right)$ : $Q = \frac{rk}{P^{\frac32} + ms}$ .

(ii)

$P^{\frac32} = 3^{1.5}$
$= \sqrt{27}$
$\approx 5.196$ and $ms = 15 \times 0.2 = 3$ .
So $Q = \frac{10 \times 4}{5.196 + 3}$
$= \frac{40}{8.196}$
$\approx 4.880$ .
Correct to two decimal places: $Q = 4.88$ .

(b)

Multiply both sides by 5: $x + 2y = 5(x - 2y) = 5x - 10y$ .
Collect terms: $12y = 4x$ , so $x = 3y$ .
Divide both sides by $y$ : $\frac xy = 3$ .
So $x : y = 3 : 1$ .

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