Two cars leave a house 10 minutes apart and travel along the same road at 20 km/h in the same direction. A woman is walking towards the house along the same road. If she meets the two cars at an interval of 8 minutes, with what speed (in km/h) is she walking?

Introduction / Context:
This is a relative speed problem involving two cars leaving the same house at different times and a woman walking towards the house from the opposite direction. The key idea is that the time gap between the woman meeting the two cars is compressed compared to the starting time gap of the cars, because all are moving. By comparing these intervals, we can deduce the woman's walking speed.

Given Data / Assumptions:

• Two cars leave the house 10 minutes (1 / 6 hour) apart.
• Both cars travel at 20 km/h in the same direction away from the house.
• A woman walks towards the house from the opposite direction at constant speed v km/h.
• The woman meets the two cars at an interval of 8 minutes (2 / 15 hour).

Concept / Approach:
Let t1 be the time when the woman meets the first car and t2 when she meets the second car, measured from the time the first car leaves the house. The cars are separated in time by 10 minutes when leaving, but they are seen by the woman at an 8 minute interval. Using relative motion equations for positions of car and woman at t1 and t2 and the given time difference, we can eliminate the unknown distance and solve directly for v, the woman's speed.

Step-by-Step Solution:
Let the woman's speed be v km/h. Let distance of the woman from the house at t = 0 be x km. Position of woman at time t: x - v * t (towards house). Position of first car at time t: 20 * t (away from house). They meet when x - v * t1 = 20 * t1. Second car starts at t = 1 / 6 hour, position at time t: 20 * (t - 1 / 6). They meet when x - v * t2 = 20 * (t2 - 1 / 6). Also, t2 - t1 = 2 / 15 hour (8 minutes). From the two position equations, eliminating x gives (2 / 15) * (v + 20) = 10 / 3. Solving gives v + 20 = 25, so v = 5 km/h.

Verification / Alternative check:
Assuming v = 5 km/h, the relative speed between each car and the woman when they move towards each other is 20 + 5 = 25 km/h. The change from 10 minutes departure gap to 8 minutes meeting gap is consistent with the derived equation. A more detailed algebraic verification reproduces the 5 km/h value exactly.

Why Other Options Are Wrong:

• 4, 6 or 7 km/h do not satisfy the relationship between the 10 minute departure interval and the 8 minute meeting interval when used in the equations of relative motion.

Common Pitfalls:
Because there are three moving objects, many students attempt to guess based on proportions rather than writing equations. Another common error is to treat the 8 minutes as the same as the 10 minutes, ignoring the effect of the woman's motion. Carefully forming and simplifying the equations avoids these traps.

Final Answer:
The woman must walk at a speed of 5 km/h.