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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.

 Dec 18, 2018

Best Answer 

 #3
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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.

 Dec 18, 2018
 #1
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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

 Dec 18, 2018
 #2
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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
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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

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