A runs <span class="math-tex">\(1 \dfrac{2}{3}\)</span> times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?
1. Find the Ratio of Speeds - A runs 1 2/3 times as fast as B. First, convert the mixed number to a fraction: 1 2/3 = 5/3.
- This means the ratio of their speeds, Speed(A) : Speed(B), is 5 : 3.
2. Relate Speeds to Distances - Since both runners finish the race in the same amount of time, the ratio of the distances they cover must be the same as the ratio of their speeds.
- So, the ratio of distances covered, Distance(A) : Distance(B), is also 5 : 3.
3. Use the Head Start to Find the Difference - A gives B a head start of 80 m. This means that for them to finish together, A must cover 80 m more than B in the same amount of time.
- This 80 m difference corresponds to the difference in their distance ratio, which is 5 - 3 = 2 parts.
- If 2 parts = 80 m, then 1 part = 40 m.
4. Calculate the Length of the Race - The total length of the race is the full distance that runner A covers.
- Since A's distance corresponds to 5 parts, the total length is:
- Total Distance = 5 parts × 40 m/part = 200 m.
Final Answer: The winning post must be 200 m far.
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