If $$60\% $$  of a first order reaction was completed in $$60\,\min, $$  $$50\% $$  of the same reaction would be completed in approximately $$\left( {{\text{log}}\,4 = 0.60,{\text{log}}\,5 = 0.69} \right)$$

Solution :
$$\eqalign{ & {\text{From first order reaction,}} \cr & {\text{Rate constant,}} \cr & k = \frac{{2.303}}{t}{\log _{10}}\frac{a}{{\left( {a - x} \right)}} \cr & {k_1} = \frac{{2.303}}{{{t_1}}}{\text{log}}\frac{{{a_1}}}{{{a_1} - {x_1}}}\,\,\,...{\text{(i)}} \cr & {k_2} = \frac{{2.303}}{{{t_2}}}{\text{log}}\frac{{{a_2}}}{{{a_2} - {x_2}}}\,\,\,...{\text{(ii)}} \cr & {x_1} = \frac{{60}}{{100}}{a_1},{t_1} = 60 \cr & {x_2} = \frac{{50}}{{100}}{a_2},{t_2} = ? \cr & {\text{From Eqs}}{\text{. (i) and (ii)}} \cr} $$
$$\frac{{2.303}}{{{t_1}}}{\text{log}}\frac{{{a_1}}}{{{a_1} - {x_1}}}$$     $$ = \frac{{2.303}}{{{t_2}}}{\text{log}}\frac{{{a_2}}}{{{a_2} - {x_2}}}$$
$$\frac{{2.303}}{{60}}{\text{log}}\frac{{{a_1}}}{{\left( {{a_1} - \frac{{60}}{{100}}{a_1}} \right)}}$$      $$ = \frac{{2.303}}{{{t_2}}}{\text{log}}\frac{{{a_2}}}{{\left( {{a_2} - \frac{{50}}{{100}}{a_2}} \right)}}$$
$$\eqalign{ & \frac{{2.303}}{{60}}{\text{log}}\frac{{100{a_1}}}{{40{a_1}}} = \frac{{2.303}}{{{t_2}}}{\text{log}}\frac{{100{a_2}}}{{50{a_2}}} \cr & \frac{1}{{60}}{\text{log}}\frac{{100}}{{40}} = \frac{1}{{{t_2}}}{\text{log}}\frac{{100}}{{50}} \cr & {t_2} = \frac{{60\,{\text{log}}\frac{{100}}{{50}}}}{{{\text{log}}\frac{{100}}{{40}}}} = \frac{{60\left( {{\text{log}}\,10 - {\text{log}}\,5} \right)}}{{\left( {{\text{log}}\,10 - {\text{log}}\,4} \right)}} \cr & \,\,\,\,\,\, = \frac{{60\left( {1 - 0.69} \right)}}{{\left( {1 - 0.60} \right)}} = \frac{{60 \times 0.31}}{{0.40}} \cr & \,\,\,\,\,\, = 1.5 \times 31 = 46.5 \approx 45\min \cr} $$