Question

Mark (✓) against the correct answer

${\left\{6^{-1}+{\left(\frac{3}{2}\right)}^{-1}\right\}}^{-1}=?$

(a) $\frac{2}{3}$
(b) $\frac{5}{6}$
(c) $\frac{6}{5}$
(d) none of these

Answer 1

(c) $\frac{6}{5}$

We have:
${\left\{6^{-1}+{\left(\frac{3}{2}\right)}^{-1}\right\}}^{-1}={\left(\frac{1}{6}+\frac{2}{3}\right)}^{-1}$
= ${\left(\frac{1+4}{6}\right)}^{-1}$ [since L.C.M. of 3 and 6 is 6]
= ${\left(\frac{5}{6}\right)}^{-1}$
= ${\left(\frac{6}{5}\right)}^1=\left(\frac{6}{5}\right)$ $\left[s\mathrm{ince}{\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$