One day, Ajay started 25 minutes late from home and reached his office 50…
2026
One day, Ajay started 25 minutes late from home and reached his office 50 minutes late, while driving 25% slower than his usual speed. How much time in minutes does Ajay usually take to reach his office from home?
- A.
72
- B.
65
- C.
69
- D.
75
Attempted by 5 students.
Show answer & explanation
Correct answer: D
Concept: For a fixed distance, travel time is inversely proportional to speed: if speed falls to a fraction k of the usual speed, the time taken becomes 1/k times the usual time, so the increase in time equals (1/k minus 1) times the usual time.
Application:
Let the usual time Ajay takes be t minutes, at his usual speed s.
Driving 25% slower means his speed that day was three-quarters of s, so by the inverse relationship above his time that day was t divided by three-quarters, i.e. four-thirds of t.
The extra time taken on the road that day equals four-thirds of t minus t, which is one-third of t.
Ajay started 25 minutes late but reached only 50 minutes late, so the journey itself (excluding the late start) accounted for 50 minus 25, i.e. 25 extra minutes.
Equating the two: one-third of t equals 25, which gives t equals 75.
Cross-check: At the usual speed, a distance of 75 times s (s being the usual speed) is covered in 75 minutes. At three-quarters of that speed, the same distance takes 75s divided by 0.75s, which is 100 minutes -- exactly 25 minutes more, matching the 50 minus 25 equals 25 minute figure from the question. So Ajay usually takes 75 minutes.
