WAEC 2019 · Paper 1 · Q12

$H$ varies directly as $p$ and inversely as the square of $y$ . If $H = 1$ , $p = 8$ and $y = 2$ , find $H$ in terms of $p$ and $y$ .

Worked solution (try it first)

Directly as $p$ , inversely as $y^2$ : $H = \dfrac{kp}{y^2}$ .
Put in $H = 1$ , $p = 8$ , $y = 2$ : $1 = \dfrac{8k}{4}$ , so $1 = 2k$ and $k = \frac12$ .
So $H = \dfrac{p}{2y^2}$ , option C.