In a 100 m race, runner A can beat B by 25 m and B can beat C by 4 m. In the same 100 m race, by how many metres can A beat C?

Introduction / Context:
This race question links two separate pairwise results to compare A and C directly. It is a classic exam pattern where one runner beats another by a certain distance, and that second runner beats a third. The challenge is to convert these distance advantages into speed ratios and then combine them to find the final lead.

Given Data / Assumptions:
- In a 100 m race, when A finishes 100 m, B has covered only 75 m. - In the same type of 100 m race, when B finishes 100 m, C has covered only 96 m. - All runners move at constant speeds throughout the race. - We must find how many metres A is ahead of C when A has completed the 100 m race.

Concept / Approach:
Distances covered in the same time are proportional to speeds. From the first statement we find the ratio of speeds vB and vA. From the second we get vC and vB. Multiplying these ratios gives vC relative to vA. Once we know vC / vA, we find how far C runs in the time when A covers 100 m and compute the difference.

Step-by-Step Solution:
Step 1: From A beating B by 25 m in 100 m race, we have vB / vA = 75 / 100 = 3 / 4. Step 2: From B beating C by 4 m in 100 m race, we have vC / vB = 96 / 100 = 24 / 25. Step 3: Multiply the ratios to get vC / vA = (vC / vB) * (vB / vA) = (24 / 25) * (3 / 4) = 72 / 100 = 18 / 25. Step 4: When A runs 100 m, the time taken is T = 100 / vA. Step 5: In the same time, C covers distance = vC * T = vC * 100 / vA = 100 * (vC / vA) = 100 * 18 / 25 = 72 m. Step 6: Therefore A is ahead of C by 100 - 72 = 28 m.

Verification / Alternative check:
Take vA = 25 units. Then vB = 3/4 of 25 = 18.75 units, and vC is 24/25 of vB, which is 18 units. Time for A to run 100 m is 100 / 25 = 4 time units. In 4 units, C covers 18 * 4 = 72 m. The lead is again 28 m, so the numerical result is consistent and the method is correct.

Why Other Options Are Wrong:
- 21 m and 26 m: These come from incorrect combination of ratios or partial subtraction of the original leads. - 29 m: Slightly larger than the correct value, usually caused by rounding or arithmetic slips.

Common Pitfalls:
Many learners simply add the two gaps, 25 m and 4 m, and claim that A beats C by 29 m, which is wrong. Others subtract them and get 21 m. The correct approach is always to convert leads into speed ratios, multiply those ratios, and only then convert back to distances for the new comparison. Remember that race results are about speeds, not about direct addition of distances.

Final Answer:
In the 100 m race, A can beat C by 28 m.