Runner A completes 51/3 laps around a track in the same amount of time it takes Runner B to complete 6 laps. B's pace is 7.5 minutes per mile. How many seconds longer will it take Runner A to run 1 mile compared to B.

Guest Dec 18, 2018

#3**+1 **

6 laps / 5 1/3 laps =12.5% is runner B faster than runner A.

If runner B run a mile in 7.5 minutes, then:

7.5 x 1.125 =8.4375 - minutes for runner A to run a mile.

8.4375 - 7.5 =0.9375 - minutes longer that takes runner A to run a mile.

0.9375 x 60 =56.25 seconds longer for runner A to run the same mile.

Guest Dec 18, 2018

#1**+1 **

Runner A's speed in laps, 17 laps per x.

Runner B's speed in laps, 6 lags per x.

The ratio is \(\dfrac{7.5}{z} = \dfrac{17}{6}\). Cross multiplying, we have \(17z = 45 \Rightarrow z = 2.647 \). 2.647 minutes is roughly 159 seconds. 7.5 minutes is 450 seconds.

450 - 159 = \(\boxed{291}\) seconds.

- PM

PartialMathematician Dec 18, 2018

#2**+1 **

Realize that the ratio is inverse.

The MORE laps in time, the better.

The LESS time for a mile, the better.

MORE does not equal LESS, so we need to get the reciprocal of LESS to have \(MORE = \dfrac{1}{LESS}\).

PartialMathematician
Dec 18, 2018

#3**+1 **

Best Answer

6 laps / 5 1/3 laps =12.5% is runner B faster than runner A.

If runner B run a mile in 7.5 minutes, then:

7.5 x 1.125 =8.4375 - minutes for runner A to run a mile.

8.4375 - 7.5 =0.9375 - minutes longer that takes runner A to run a mile.

0.9375 x 60 =56.25 seconds longer for runner A to run the same mile.

Guest Dec 18, 2018