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JAMB Mathematics 2011 Paper

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Question : 14

Total: 46

If x varies directly as square root of y and x = 81 when y = 9, Find x when y = 1

\frac{7}{9}

20 $\frac{1}{4}$
27
2 $\frac{1}{4}$
36

Solution:

Concept:

x

varies directly as

\sqrt{y}

, so

x = k\sqrt{y}

for a constant

k

.

Explanation:

Given

x = 81

when

y = 9

:

81 = k\sqrt{9}

\sqrt{9} = 3

, so

81 = 3k

k = \frac{81}{3} = 27

Thus,

x = 27\sqrt{y}

.

Now convert

y = 1\frac{7}{9}

to an improper fraction:

y = \frac{16}{9}

.

Substitute into

x = 27\sqrt{y}

:

x = 27\sqrt{\frac{16}{9}}

x = 27 \times \frac{4}{3}

x = 9 \times 4 = 36

.

Answer:

x = 36

(Option D).

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