CAT 2024 Slot 2 · Quant · Question 17

Let the route be D km long and the scheduled travel time be T hours.

On the first day the bus took \frac{D}{60} hours, which was 3.5 hours more than scheduled, so \frac{D}{60}=T+3.5.

On the second day, two-thirds of the route took one-third of the scheduled time, so the remaining one-third of the route had to be covered in the remaining T-\frac{T}{3}=\frac{2T}{3} hours.

That remaining distance is \frac{D}{3} km, covered at 40 km per hour, taking \frac{D}{120} hours.

Since the bus arrived exactly on time, \frac{D}{120}=\frac{2T}{3}, which gives D=80T.

Substituting into the first-day equation, \frac{80T}{60}=T+3.5, that is \frac{4T}{3}=T+3.5.

So \frac{T}{3}=3.5 and T=10.5 hours.

The bus starts at 9 am, so the scheduled arrival is 10 hours 30 minutes later, at 7:30 pm — the third option.

Shortcut. The second day's information is purely about proportions: covering \frac{2}{3} of the route in \frac{1}{3} of the time leaves \frac{1}{3} of the route for \frac{2}{3} of the time, so the closing speed of 40 km/h fixes D=80T at once. Only then is the 60 km/h run needed.