Question

$\frac{1}{2{\displaystyle \frac{1}{3}}}+\frac{1}{1{\displaystyle \frac{3}{4}}}$ is equal to

(a) $\frac{7}{14}$
(b) $\frac{12}{49}$
(c) $4\frac{1}{12}$
(d) None of these

Answer 1

(d) None of these

$$\begin{gathered} \frac{1}{2{\displaystyle \frac{1}{3}}}+\frac{1}{1{\displaystyle \frac{3}{4}}}=\frac{1}{{\displaystyle \frac{(3\times 2)+1}{3}}}+\frac{1}{{\displaystyle \frac{(4\times 1)+3}{4}}} \\ =\frac{1}{{\displaystyle \frac{7}{3}}}+\frac{1}{{\displaystyle \frac{7}{4}}}=\frac{3}{7}+\frac{4}{7}=\frac{3+4}{7}=\frac{7}{7}=1 \end{gathered}$$