Runner A runs 1 2/3 times as fast as runner B. If A gives B a head start of 80 m, how far from the start should the winning post be placed so that A and B reach it at the same time?

Introduction / Context:
This race question checks understanding of relative speed and the effect of a head start. Runner A is faster than B, but B starts the race some distance ahead. We must find the race length that balances this advantage so both finish together. This type of question appears often in aptitude tests on races and games.

Given Data / Assumptions:
- A runs 1 2/3 times as fast as B, so vA = 5/3 of vB. - B starts 80 m ahead of A. - Both start running at the same time and move with constant speeds. - We must find the race distance D so that both reach the winning post together.

Concept / Approach:
When two runners start together but one has a fixed head start, both finish together when their times are equal. Time is distance divided by speed. A runs the full distance D, while B needs to cover only D - 80 m. Using the speed ratio, we set their times equal and solve for D. This is a direct application of proportional reasoning with speeds.

Step-by-Step Solution:
Step 1: Let vB = v units. Then vA = (5/3) * v. Step 2: Let D be the total distance from the common start point of A to the winning post. Step 3: Time taken by A to finish = D / vA = D / ((5/3) * v) = 3D / (5v). Step 4: Since B starts 80 m ahead, B covers only D - 80 m. Step 5: Time taken by B = (D - 80) / v. Step 6: They finish together, so their times are equal: 3D / (5v) = (D - 80) / v. Step 7: Cancel v to get 3D / 5 = D - 80. Step 8: Multiply both sides by 5 to obtain 3D = 5D - 400. Step 9: Rearrange: 5D - 3D = 400, so 2D = 400 and D = 200 m.

Verification / Alternative check:
If D = 200 m, A runs 200 m, B runs 200 - 80 = 120 m. A is 5/3 times as fast as B, so the time ratio TA : TB = 200 / vA : 120 / vB = 200 / (5v/3) : 120 / v = 200 * 3 / (5v) : 120 / v = 600 / (5v) : 120 / v = 120 / v : 120 / v. Times are equal, confirming that both reach the winning post together when the race length is 200 m.

Why Other Options Are Wrong:
- 300 m: With this distance A would need more time than required to match B, so A would finish earlier than B. - 270 m: This does not satisfy the equality of times when the ratio 5/3 is applied. - 160 m: This distance is too short and would not allow A to catch B despite being faster.

Common Pitfalls:
Students sometimes mix up which runner covers D and which covers D minus the head start. Another frequent mistake is to treat the 1 2/3 factor as 1.23 or as 123 percent without converting it correctly to an exact fraction of 5/3. Always convert mixed numbers to improper fractions and build the equation by equating the times, not the distances.

Final Answer:
The winning post should be 200 m from the starting point of A.