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 race question uses the idea of giving a start, which is another way of expressing relative speeds. If one runner can give fixed starts to two different runners over the same race distance, then their speeds are related. We must find how much start the intermediate runner, B, can give to the slowest runner, C, in the same 100 m race.

Given Data / Assumptions:
- Race distance is 100 m. - A can give B a start of 10 m, meaning when A runs 100 m, B runs only 90 m in the same time. - A can give C a start of 28 m, meaning when A runs 100 m, C runs only 72 m in the same time. - Speeds are constant for all runners throughout the race. - We must find how many metres start B can give C so that both finish 100 m together.

Concept / Approach:
Again we use relative speeds. From the two given starts we find vB and vC in terms of vA, then get vC / vB. Once we know the speed ratio of B and C, we imagine B running the full 100 m and find the distance C would cover in the same time. The difference between 100 m and that distance is the start B can give C.

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

Verification / Alternative check:
Assume vB = 5 units and vC = 4 units from the 5 : 4 ratio. Time for B to cover 100 m is 100 / 5 = 20 units. In 20 units, C covers 4 * 20 = 80 m. If C starts 20 m ahead, C needs to cover only 80 m to finish 100 m, and both will finish together. This confirms the logic and the computed start of 20 m.

Why Other Options Are Wrong:
- 18 m and 27 m: These values come from incorrect manipulation of ratios or from approximations that ignore exact fractional values. - 9 m: This is far too small and does not match the proportional relationship between the speeds.

Common Pitfalls:
One common error is to subtract 10 m and 28 m directly to obtain 18 m, assuming that B can give C the difference between the two starts. Another is to confuse which runner is getting the head start in the second stage of the problem. Always convert starts to speed ratios first and then apply those ratios to the full race distance to find the correct start between B and C.

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