A man travels an unknown distance at speed v km/h. If he had moved 3 km/h faster, his time would decrease by 40 minutes; if he had moved 2 km/h slower, his time would increase by 40 minutes. What is the distance (in km)?

Problem restatement
The same journey is done at three hypothetical speeds: v, v + 3, and v − 2 km/h. The time differences are each 40 minutes (2⁄3 hour). Find the distance.

Given data

• Let distance = D km; normal speed = v km/h.
• At (v + 3) km/h, time is 2⁄3 hour less.
• At (v − 2) km/h, time is 2⁄3 hour more.

Concept/Approach
Use the identity Time = Distance ÷ Speed for the same D. Set up two equations from the given time differences and solve for v, then compute D.

Step-by-step calculation
D ÷ v − D ÷ (v + 3) = 2⁄3D ÷ (v − 2) − D ÷ v = 2⁄3From the first: D × [1÷v − 1÷(v + 3)] = 2⁄3 ⇒ D × 3 ÷ [v(v + 3)] = 2⁄3 ⇒ D = [2v(v + 3)] ÷ 9From the second: D × [1÷(v − 2) − 1÷v] = 2⁄3 ⇒ D × 2 ÷ [v(v − 2)] = 2⁄3 ⇒ D = v(v − 2) ÷ 3Equate D: [2v(v + 3)] ÷ 9 = v(v − 2) ÷ 3 ⇒ 2(v + 3) = 3(v − 2) ⇒ v = 12 km/hDistance D = v(v − 2) ÷ 3 = 12 × 10 ÷ 3 = 40 km

Verification/Alternative
At v = 12: time = 40 ÷ 12 = 3ℓ⁄3 hAt v + 3 = 15: time = 40 ÷ 15 = 2ℓ⁄3 h ⇒ 40 min lessAt v − 2 = 10: time = 40 ÷ 10 = 4 h ⇒ 40 min more

Common pitfalls
Avoid confusing 40 minutes with 0.4 hour; use 2⁄3 hour.

Final Answer
40 km