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JAMB Mathematics 2012 Paper

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Question : 13

Total: 48

If y varies directly as

\sqrt{n}

and y = 4 when n = 4, find y when n = 1

\frac{7}{9}

$\sqrt{17}$
$\frac{4}{3}$
$\frac{8}{3}$
$\frac{2}{3}$

Solution:

Concept:

Direct variation means

y = k\sqrt{n}

, where

k

is the constant of proportionality.

Explanation:

Given

y \propto \sqrt{n}

, write

y = k\sqrt{n}

.

Substitute

y = 4

and

n = 4

:

4 = k\sqrt{4}

4 = 2k

k = 2

So the relation is

y = 2\sqrt{n}

.

Now

n = 1\frac{7}{9} = \frac{16}{9}

.

Then

y = 2\sqrt{\frac{16}{9}}

y = 2\left(\frac{4}{3}\right)

y = \frac{8}{3}

Answer:

y = \frac{8}{3}

, which is option C.

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