WAEC 2025 · Paper 1 · Q15✱✱

Variation

If $y \propto \dfrac{1}{x^2}$ and $x = 3$ when $y = 4$ , find $y$ when $x = 2$ .

Worked solution (try it first)

$y = \dfrac{k}{x^2}$ , so $k = x^2y$ , which is $9 \times 4 = 36$ .
When $x = 2$ : $y = \dfrac{36}{4} = 9$ , option B.

Report a problem with this question