Early vs late arrival; find distance Ram travels a certain distance at 3 km/h and arrives 15 min late. At 4 km/h he arrives 15 min early. What is the distance?

Introduction / Context:
When two different speeds produce equal and opposite schedule deviations, set up equations with the same scheduled time and eliminate it to solve for distance directly.

Given Data / Assumptions:

• At 3 km/h: arrival is 15 min late.
• At 4 km/h: arrival is 15 min early.
• Let scheduled time be T hours and distance be D km.

Concept / Approach:
Form two equations: D/3 = T + 0.25 and D/4 = T − 0.25. Subtract to remove T and solve for D.

Step-by-Step Solution:

Verification / Alternative check:
Compute T: From D/4 = T − 0.25 ⇒ 6/4 = T − 0.25 ⇒ T = 1.75 h. Check D/3 = 2 h = T + 0.25.

Why Other Options Are Wrong:
7.2, 4.5, 12 contradict the equal 15-min deviations with the given speeds.

Common Pitfalls:
Using minutes in one equation and hours in the other; always convert consistently.

Final Answer:
6 km