A man reaches his office late by 15 minutes if he drives his car at 30 km/hr,…

Concept: In a “reach on time” speed-time-distance problem with two trial speeds (one giving a late arrival, one giving an early arrival), treat the on-time travel duration T and the distance D as unknowns. Each trial speed gives a travel time equal to T adjusted by the stated lateness/earliness, and distance = speed × time must be the SAME D in both cases — equating the two expressions for D pins down T and D, after which the required speed is simply D ÷ T.

Application:

1. Let T (hours) be the time needed to reach exactly on time, and D (km) the distance to the office.

2. At 30 km/hr he is 15 min (0.25 h) late, so his travel time is T + 0.25 h, giving D = 30(T + 0.25).

3. At 40 km/hr he is 5 min (1/12 h) early, so his travel time is T − 1/12 h, giving D = 40(T − 1/12).

4. Equate the two expressions for D: 30(T + 0.25) = 40(T − 1/12) → 30T + 7.5 = 40T − 3.333 → 10T = 10.833 → T = 13/12 h (65 minutes).

5. Substitute back: D = 30(13/12 + 1/4) = 30 × 4/3 = 40 km.

6. Required speed = D ÷ T = 40 ÷ (13/12) = 480/13 ≈ 36.92 km/hr.

Cross-check: using the second equation, D = 40(13/12 − 1/12) = 40 × 1 = 40 km, which matches — the distance and time are consistent both ways.

Note: 480/13 is not a whole number, so the exact required speed (≈ 36.92 km/hr) does not fall exactly on any offered whole-number option. Since the question asks for the value closest to the required speed, 36 km/hr is the intended answer — it differs from the exact figure by under 1 km/hr, whereas every other offered speed is off by roughly 9 km/hr or more.