A runs <span class="math-tex">\(1 \frac{2}{3}\)</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?
Step-by-step Solution:
To solve this problem, we need to find the total length of the racecourse. The key condition is that both runners, A and B, finish the race in the exact same amount of time.
When the time taken is equal for two runners, the ratio of the distances they cover is equal to the ratio of their speeds.
DistanceA / DistanceB = SpeedA / SpeedB
We are given that A runs $1\frac{2}{3}$ times as fast as B.
$1\frac{2}{3} = \frac{5}{3}$
This means SpeedA = $(\frac{5}{3}) \times$ SpeedB. The ratio of their speeds is:
SpeedA / SpeedB = 5 / 3
Let 'D' be the total distance to the winning post (in meters).
Using the principle that the ratio of distances equals the ratio of speeds, we can set up the following equation:
D / (D - 80) = 5 / 3
Now, we cross-multiply to solve for D:
3 × D = 5 × (D - 80)
3D = 5D - 400
400 = 5D - 3D
400 = 2D
D = 200
The winning post must be 200 m away for A and B to reach it at the same time.