CAT 2024 Slot 2 QA Question & Solution
Question
A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km per hour and reached its destination 3.5 hours later than its scheduled arrival time. Next day, it traveled two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km per hour to reach just on time. The scheduled arrival time of the bus is
Options
Solution
Let the scheduled time taken by the bus be $t$.
From the first statement (bus travelling at 60 km/h), the total distance travelled by the bus is:
$$
60 \cdot (t + 3.5)
$$
In the second scenario, the bus covered two-thirds of the distance in one-third of the time. This means the remaining one-third of the distance was covered in two-thirds of the time, giving the speed for the remaining distance as:
$$
\text{Speed} = \frac{\frac{1}{3}s t}{\frac{2}{3} t} = \frac{s}{2}
$$
It is given that this speed is 40 km/h, thereby giving the usual speed of the bus as:
$$
s = 80 , \text{km/h}
$$
Using the first relation:
$$
60 \cdot (t + 3.5) = 80 \cdot t
$$
Solving for $t$:
$$
60t + 210 = 80t \implies 20t = 210 \implies t = 10.5 , \text{hours}
$$
Thus, the bus usually takes 10.5 hours on its journey.
Starting at 9:00 AM, it will complete the journey at 7:30 PM.
Therefore, Option A is the correct answer.