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WAEC Mathematics 2020 Paper

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Question : 3

Total: 48

Simplify; [(

\frac{16}{9}

^{\frac{-3}{2}}

^{\frac{-3}{4}}

^{\frac{1}{3}}

$\frac{3}{4}$
$\frac{9}{16}$
$\frac{3}{8}$
$\frac{1}{4}$

Solution:

Concept:

Use the laws of exponents:

a^{-n}=\frac{1}{a^n}

and

\left(a^m\right)^n=a^{mn}

.

Explanation:

Write the negative exponents as reciprocals first:

\left[\left(\frac{16}{9}\right)^{-\frac32}\times 16^{-\frac34}\right]^{\frac13}=\left[\left(\frac{9}{16}\right)^{\frac32}\times\left(\frac{1}{16}\right)^{\frac34}\right]^{\frac13}

Simplify

\left(\frac{9}{16}\right)^{\frac32}

:

\left(\frac{9}{16}\right)^{\frac32}=\left(\left(\frac34\right)^2\right)^{\frac32}=\left(\frac34\right)^3=\frac{27}{64}

Simplify

16^{-\frac34}

:

16^{-\frac34}=\frac{1}{16^{\frac34}}=\frac{1}{\left(2^4\right)^{\frac34}}=\frac{1}{2^3}=\frac18

Multiply the results inside the bracket:

\frac{27}{64}\times\frac18=\frac{27}{512}

Now take the cube root:

\left(\frac{27}{512}\right)^{\frac13}=\frac{3}{8}

Answer:

\frac{3}{8}

, which is option C.

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