In a 100 m race, runner A can give B a start of 10 m and can give C a start of 28 m. In the same race, how many metres can B give C?

Introduction / Context:
This question is a repeat of the earlier race setup with A, B, and C, emphasising again how head starts and relative speeds are related. Understanding this structure is useful because very similar problems appear in many exams with only slight number changes.

Given Data / Assumptions:
- Standard race length is 100 m. - A can give B 10 m start, so A runs 100 m while B runs 90 m in the same time. - A can give C 28 m start, so A runs 100 m while C runs 72 m in the same time. - Speeds remain constant for each runner. - We must find B's possible start to C in a 100 m race.

Concept / Approach:
A giving different starts to B and C lets us calculate the speed ratio of B and C. Once we know vB : vC, we can imagine B running the whole 100 m and calculate how far C travels in that time. The difference between 100 m and that distance is the head start that B can safely give to C without losing.

Step-by-Step Solution:
Step 1: From A and B, vB / vA = 90 / 100 = 9 / 10. Step 2: From A and C, vC / vA = 72 / 100 = 18 / 25. Step 3: Therefore vC / vB = (18 / 25) / (9 / 10) = (18 / 25) * (10 / 9) = 4 / 5. Step 4: So vB / vC = 5 / 4. Step 5: Let B run the full 100 m. Time taken is TB = 100 / vB. Step 6: In time TB, C runs distance = vC * TB = 100 * (vC / vB) = 100 * 4 / 5 = 80 m. Step 7: Hence B can give C a start of 100 - 80 = 20 m.

Verification / Alternative check:
Assume vB = 5 units and vC = 4 units. Then time for B to run 100 m is 100 / 5 = 20 time units. In 20 units, C covers 4 * 20 = 80 m. If C begins 20 m ahead, C only needs 80 m to reach 100 m, so both runners finish together. This confirms that 20 m is the correct start from B to C.

Why Other Options Are Wrong:
- 18 m and 27 m: These do not correspond to any correct manipulation of the speed ratios and would not make the finishing times equal. - 9 m: Much too small a start, and C would finish behind B, contradicting the equal finishing time requirement with that start.

Common Pitfalls:
Students often subtract the starts given by A to B and C, obtaining 18 m, without working through the speed ratio. Another confusion comes from mixing which distance corresponds to which runner in the ratio. Always structure the solution around vB / vA and vC / vA, then divide to get vB / vC and only then apply it to the actual race distance.

Final Answer:
In the 100 m race, B can give C a start of 20 m.