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Question : 14 of 86

Marks: +1, -0

The quantity

\frac{(0.6)^{0} - (0.1)^{-1}}{\left(\;\frac{3}{2^{3}}\right)^{-1}\left(\;\frac{3}{4}\right)^{3} + \left(-\;\frac{1}{3}\right)^{-1}}

is equal to

$\;\frac{24}{5}$
$-\;\frac{1}{2}$
$\;\frac{2}{3}$
$\;\frac{3}{2}$

Solution:

Concept:

Simplifying an expression involving powers and reciprocals.

Explanation:

Numerator:

(0.6)^0 = 1

,

(0.1)^{-1} = 10

, so

1 - 10 = -9

.

Denominator: First term:

\left(\frac{3}{2^{3}}\right)^{-1} = \left(\frac{3}{8}\right)^{-1} = \frac{8}{3}

.

Multiply by

\left(\frac{3}{4}\right)^{3} = \frac{27}{64}

gives

\frac{8}{3} \times \frac{27}{64} = \frac{216}{192} = \frac{9}{8}

.

Second term:

\left(-\frac{1}{3}\right)^{-1} = -3

.

Denominator sum:

\frac{9}{8} + (-3) = \frac{9}{8} - \frac{24}{8} = -\frac{15}{8}

.

Whole expression:

\frac{-9}{-15/8} = -9 \times \left(-\frac{8}{15}\right) = \frac{72}{15} = \frac{24}{5}

.

Answer:

\frac{24}{5}

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