A police patrol party travelling at 60 kmph crosses an escaping thief ...
Problem Statement
A police patrol party travelling at 60 kmph crosses an escaping thief travelling in the opposite direction at 48 kmph. The police party has to travel for a further 5 minutes before it can find a gap in the median where it can take a U-turn and start chasing the thief. After how much time after the police party crosses the thief does it catch him?
Solution
Let's assume that the distance between the police car and the thief at the time of their crossing is 'd'.
The relative speed of the police car with respect to the thief is the sum of their speeds, which is:
60 + 48 = 108 kmph
Since the police car has to travel for 5 more minutes before taking a U-turn, it covers a distance of:
Distance = Speed × Time
Distance = (60/60) × 5
Distance = 5 km
Therefore, the distance between the police car and the thief at the time of the U-turn is:
d - 5 km
Now, let's assume that the police car catches the thief after 't' hours of the U-turn.
So, the distance covered by the police car in 't' hours after the U-turn is:
Distance covered = Speed × Time
Distance covered = (60/60) × t
Distance covered = t km
At the same time, the distance covered by the thief in 't' hours after the U-turn is:
Distance covered = Speed × Time
Distance covered = (48/60) × t
Distance covered = 0.8t km
Since the police car catches the thief, the distance covered by both of them must be equal.
Therefore, we can equate the two distances:
d - 5 km + t = 0.8t
Solving for 't', we get:
t = 32
Therefore, the police car catches the thief after 32 minutes of the U-turn.
A police patrol party travelling at 60 kmph crosses an escaping thief ...
Correct answer is 50 not 32.