Two ladies start simultaneously from Cities A and B, 24 km apart, walking toward each other. The first lady walks 2 km/h faster than the second and reaches B exactly 1 hour before the second reaches A. What are their speeds (km per hour)?

Introduction / Context:Both travellers start together and traverse the entire 24 km, one going from A to B and the other from B to A. Let the slower speed be v km/h and the faster be v + 2. The faster arrives 1 hour earlier, giving a time-difference equation that can be solved for v directly.

Given Data / Assumptions:

• Total one-way distance for each = 24 km.
• Faster speed = v + 2; slower speed = v.
• Time difference between arrivals = 1 h.

Concept / Approach:Arrival times: T_fast = 24/(v+2), T_slow = 24/v. Since the faster arrives earlier by 1 h: T_slow − T_fast = 1. Solve for v by clearing denominators.

Step-by-Step Solution:

Verification / Alternative check:Times: 24/6 = 4 h; 24/8 = 3 h; difference = 1 h as required.

Why Other Options Are Wrong:Other pairs do not differ by exactly 2 km/h or do not yield a 1-hour arrival gap over 24 km.

Common Pitfalls:Using relative speed meeting-time equations instead of whole-journey arrival times; mixing roles of the two speeds.

Final Answer:8 km, 6 km