Question

Mark the correct alternative in the following question:

$$\begin{gathered} \left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}= \\ \left(\mathrm{a}\right)\frac{19}{64}\left(\mathrm{b}\right)\frac{64}{19}\left(\mathrm{c}\right)\frac{27}{16}\left(\mathrm{d}\right)\frac{-19}{64} \end{gathered}$$

Answer 1

Since,

$$\begin{gathered} \left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3} \\ =\left\{{\left(\frac{3}{1}\right)}^3-{\left(\frac{2}{1}\right)}^3\right\}÷{\left(\frac{4}{1}\right)}^3\left(\mathrm{As},x^{-1}=\frac{1}{x}\right) \\ =\left\{3^3-2^3\right\}÷4^3 \\ =\left\{27-8\right\}÷64 \\ =19÷64 \\ =\frac{19}{64} \end{gathered}$$

Hence, the correct alternative is option (a).