NECO 2023 · Paper 2 · Q1

Variation

$P$ varies directly as the square of $Q$ and inversely as the cube of $Z$ . When $P = 5$ , $Q = 3$ and $Z = 1$ . Find:

(i)
the relationship between $P$ , $Q$ and $Z$ ( $P$ in terms of $Q$ and $Z$ );
(ii)
$Z$ when $P = 3$ and $Q = 5$ .

Worked solution (try it first)

(i)

$P$ varies directly as $Q^2$ (on top) and inversely as $Z^3$ (underneath): $P = \dfrac{kQ^2}{Z^3}$ .
Put in $P = 5$ , $Q = 3$ , $Z = 1$ : $5 = \dfrac{9k}{1}$ , so $k = \frac59$ .
The relationship is $P = \dfrac{5Q^2}{9Z^3}$ .

(ii)

Put in $P = 3$ , $Q = 5$ : $3 = \dfrac{5 \times 25}{9Z^3}$
$= \dfrac{125}{9Z^3}$ .
So $27Z^3 = 125$ , $Z^3 = \dfrac{125}{27}$ and $Z = \sqrt[3]{\dfrac{125}{27}} = \dfrac53$ .

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