JAMB 1986 · UME · Q16

Variation

If $x$ varies directly as $y^3$ , and $x = 2$ when $y = 1$ , find $x$ when $y = 5$ .

Worked solution (try it first)

$x$ varies directly as $y^3$ , so $x = ky^3$ .
Put in $x = 2$ , $y = 1$ : $k = 2$ .
When $y = 5$ : $5^3 = 125$ , so $x = 2 \times 125 = 250$ , option D.

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