Question

Mark (✓) against the correct answer

${\left\{{\left(\frac{3}{4}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}=?$

(a) $\frac{3}{8}$
(b) $\frac{-3}{8}$
(c) $\frac{8}{3}$
(d) $\frac{-8}{3}$

Answer 1

(b) $\frac{-3}{8}$

${\left\{{\left(\frac{3}{4}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}={\left(\frac{4}{3}-\frac{4}{1}\right)}^{-1}$
= ${\left(\frac{4-12}{3}\right)}^{-1}$ [ since L.C.M. of 1 and 3 is 3]
= ${\left(\frac{-8}{3}\right)}^{-1}$
= ${\left(\frac{3}{-8}\right)}^1$ $\left[s\mathrm{ince}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-1}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^1\right]$
= $\left(\frac{3\times -1}{-8\times -1}\right)=\frac{-3}{8}$