If 60% of first order reaction was completed in 60 min, 50% of the same reaction would be completed in approximately:

$\begin{array}{l}\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\\ \mathrm{k}=\frac{2.303}{60}\log \frac{100}{100-60}=\frac{2.303}{60}\log \frac{100}{40}\\ \mathrm{k}=\frac{2.303}{60}\log 2.5=\frac{2.303\times 0.397}{60}\\ \mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\\ \mathrm{t}=\frac{2.303\times 60}{2.303\times 0.39}\log \frac{100}{50}\\ 46.15\approx 45\min \end{array}$