One day Mihir started 30 minutes late from home and reached his office 55…

Concept: For a fixed distance, speed and time are inversely proportional (Speed × Time = Distance = constant). If the speed changes to a fraction of the usual speed, the time taken changes to the reciprocal of that fraction. When someone starts late and also travels at a different speed, the extra lateness at arrival (over and above the start delay) is caused only by the change in time taken for the journey itself.

Application:

1. Let the usual speed be S and the usual time taken be t minutes, so distance = S × t (constant).

2. The new speed is 25% less than usual, i.e. (3/4)S, so the new time taken = distance ÷ new speed = (S × t) ÷ ((3/4)S) = (4/3)t minutes.

3. Extra time consumed on the road due to the slower speed = (4/3)t − t = (1/3)t minutes.

4. He started 30 minutes late but reached 55 minutes late, so the extra time actually consumed on the road (beyond the start delay) = 55 − 30 = 25 minutes.

5. Equating the two: (1/3)t = 25, so t = 75.

Cross-check:

With t = 75 minutes, the new time taken = (4/3) × 75 = 100 minutes, i.e. 25 minutes more than usual. Starting 30 minutes late and then taking 25 extra minutes on the road means he reaches 30 + 25 = 55 minutes late — exactly matching the given data.

Hence, Mihir usually takes 75 minutes to reach his office.