Question

In a 1-mile race, the winner crosses the finish line 10 feet ahead of the second-place runner and 23 feet ahead of the third-place runner. assuming that each runner maintains a constant speed throughout the race, by how many feet does the second-place runner beat the third-place runner? (5280 feet in 1 mile.)

Answer 1

5280 - 10 = 5270 ft, traveled by the 2nd place runner
5280 - 23 = 5257 ft, traveled by the 3rd place runner
Use a ratio equation5280/5270 = x/5257
Cross multiply:
5270x = 5280 * 5257x
= 27756960/5270x
= 5266.975 traveled by the 3rd place runner when the 2nd place runner crosses the finish line.
5280 - 5266.975
= 13.025 ft is the distance the 2nd place runner beats the 3rd place runner.
This makes sense because they were 13 ft apart when the winner crossed the line, the 3rd place runner is 13 ft behind the 2nd runner because he runs slower.