JAMB 2011 · UTME · Q16

Variation

If $x$ varies directly as the square root of $y$ and $x = 81$ when $y = 9$ , find $x$ when $y = 1\frac79$ .

Worked solution (try it first)

$x = k\sqrt y$ .
Put in $x = 81$ , $y = 9$ : $81 = 3k$ , so $k = 27$ .
Change the mixed number: $1\frac79 = \frac{16}{9}$ , so $\sqrt y = \frac43$ .
So $x = 27 \times \frac43 = 36$ , option D.

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