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: Plug in the given values B's distance is x (the total distance to the winning post), and B's speed is 1 (we'll use this as the base speed). A's distance is x - 80 (the total distance minus the head start that B gets), and A's speed is 1(2/3). Plugging 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 fraction:

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

However, distance cannot be negative. This means that there's been a mistake in our calculation. Let's go back and check our work.

Step 5: Correct the mistake The mistake was made in step 3. The correct equation should be:

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

This is because A has to run the total distance x plus the 80 meter head start that B gets in order to reach the winning post.

Step 6: Solve the corrected equation Following the same steps as before, we get:

1(2/3)x = x + 80

(2/3)x = 80

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.