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WAEC Mathematics 2019 Paper

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Question : 11

Total: 50

If m : n = 2 : 1, evaluate

\;\frac{3m^2-2n^2}{m^2+mn}

$\;\frac{4}{3}$
$\;\frac{5}{3}$
$\;\frac{3}{4}$
$\;\frac{3}{5}$

Solution:

Concept:

Ratios imply a constant multiplier for each term, so let

m = 2k

and

n = k

.

Explanation:

Given

m : n = 2 : 1

, take

m = 2k

and

n = k

.

Substitute into

\dfrac{3m^2 - 2n^2}{m^2 + mn}

:

= \dfrac{3(2k)^2 - 2(k)^2}{(2k)^2 + (2k)(k)}

= \dfrac{3(4k^2) - 2k^2}{4k^2 + 2k^2}

= \dfrac{12k^2 - 2k^2}{6k^2} = \dfrac{10k^2}{6k^2}

= \dfrac{10}{6} = \dfrac{5}{3}

Answer:

The value is

\dfrac{5}{3}

, which is option B.

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