If 60 \% , of a first order reaction was completed in 60 minutes,...

For a first order reaction $k=\frac{2.303}{t} \log \frac{a}{a-x}$ When t=60 and $x=60 \%$$\begin{aligned} k & =\frac{2.303}{60} \log \frac{100}{100-60} \\ & =\frac{2.303}{60} \log \frac{100}{40} \\ & =0.0153 \end{aligned}$$$Now,$$\begin{aligned} t^{1 / 2} & =\frac{2.303}{0.0153} \log \frac{100}{100-50} \\ & =\frac{2.303}{0.0153} \times \log 2 \\ & =\frac{2.303}{0.0153} \times 0.3010 \\ & =45.31 \mathrm{~min} \end{aligned}$$