JAMB 2002 · UME · Q42

If $x$ varies directly as $\sqrt n$ and $x = 9$ when $n = 9$ , find $x$ when $n = \frac{17}{9}$ .

Worked solution (try it first)

$x = k\sqrt n$ .
Put in $x = 9$ , $n = 9$ : $9 = 3k$ , so $k = 3$ .
When $n = \frac{17}{9}$ : $\sqrt n = \dfrac{\sqrt{17}}{3}$ , because $\sqrt9 = 3$ .
So $x = 3 \times \dfrac{\sqrt{17}}{3}$
$= \sqrt{17}$ , option B.