A starts from P to Q (51.75 km) at 3.75 km/h. One hour later, B starts from Q to P at 4.25 km/h. When and where do they meet?

Introduction / Context:
This is a meet-with-head-start problem. One traveler leaves earlier from one end; the other starts later from the opposite end. We form a single distance equation with their speeds and the staggering of start times.

Given Data / Assumptions:

• Distance P–Q = 51.75 km.
• A: 3.75 km/h from P at t = 0.
• B: 4.25 km/h from Q at t = 1 hour.

Concept / Approach:
Let T be hours after A starts when they meet (T ≥ 1). Then A covers 3.75T and B covers 4.25(T − 1). Their sum equals 51.75 km.

Step-by-Step Solution:
3.75T + 4.25(T − 1) = 51.75.8T − 4.25 = 51.75 → 8T = 56 → T = 7 h.Distance from P = 3.75 * 7 = 26.25 km → from Q = 51.75 − 26.25 = 25.50 km.

Verification / Alternative check:
In 6 hours (T − 1), B covers 25.5 km; A covers 26.25 km in 7 hours; sums to 51.75 km.

Why Other Options Are Wrong:
26.25 km from Q confuses ends; 25.30 km is a rounding/inexact arithmetic.

Common Pitfalls:
Ignoring the 1-hour delay for B or summing distances to a value different from the total.

Final Answer:
25.50 km from Q