CAT 2020 Quant Question [Slot 1] with Solution 16

Let the usual speed of the train be s and time taken at that speed be 't'.

Given by travelling at s/3, it reached 30 min late. Hence the usual time:

Distance travelled \( = s \times t\)

Distance travelled in the first \(5min = s \times t/3.\) D

Distance to be travelled in the last \(6min = 2st/3\)

Required speed to cover that distance on time \( = \frac{{2st/3}}{{2t/5}}\) i.e., \(\frac{{5s}}{3}\)

Hence the percentage increase in its speed \( = (2/3) \times \) 100 i.e., \(66\frac{2}{3}\% \) or \(67\% \)