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WAEC Mathematics 2019 Paper

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Question : 12

Total: 50

H varies directly as p and inversely as the square of y. If H = 1, p = 8 and y = 2, find H in terms of p and y.

H = $\;\frac{p}{4y^2}$
H = $\;\frac{2p}{y^2}$
H = $\;\frac{p}{2y^2}$
H = $\;\frac{p}{y^2}$

Solution:

Concept:

H varies directly as

p

and inversely as

y^2

. This means

H

equals a constant

k

multiplied by

\frac{p}{y^2}

.

Explanation:

Write the variation equation as:

H = \frac{kp}{y^2}

Substitute the given values

H = 1

,

p = 8

, and

y = 2

:

1 = \frac{k \times 8}{2^2}

1 = \frac{8k}{4}

1 = 2k

Solve for

k

:

k = \frac{1}{2}

Put

k = \frac{1}{2}

back into the equation:

H = \frac{\frac{1}{2} \times p}{y^2}

Simplify:

H = \frac{p}{2y^2}

Answer:

H = \frac{p}{2y^2}

Correct option: C

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