A runs 1(2/3) 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 1: Understand the problem A runs 1(2/3) times as fast as B. This means that for every 1 meter that B runs, A runs 1(2/3) meters. If A gives B a start of 80 meters, this means that B starts 80 meters ahead of A.

Step 2: Set up the equation We want to find the distance to the winning post (let's call this distance x) such that A and B reach it at the same time. Since speed = distance/time, we can set up the following equation:

B's distance / B's speed = A's distance / A's speed

Step 3: Substitute the given values into the equation B's distance is x, B's speed is 1 (we'll use 1 as B's speed for simplicity), A's distance is x + 80 (since A starts 80 meters behind B), and A's speed is 1(2/3). Substituting these values into the equation gives us:

x / 1 = (x + 80) / 1(2/3)

Step 4: Solve the equation To solve for x, we can first multiply both sides of the equation by 1(2/3) to get rid of the denominator on the right side:

1(2/3)x = x + 80

Then, subtract x from both sides to isolate the variable:

(2/3)x = 80

Finally, divide both sides by (2/3) to solve for x:

x = 80 / (2/3) = 120

So, the winning post must be 120 meters away for A and B to reach it at the same time.