CAT 2020 Exam Shift 1 Solved Paper

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×t
Distance travelled in the first 5min=s×t/3.D
Distance to be travelled in the last 6‌min=2st/3
Required speed to cover that distance on time =‌

2st/3

2t/5

i.e., ‌

5s

3

Hence the percentage increase in its speed =(2/3)×100 i.e., 66‌

2

3

% or 67%