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Runners \(X\) and \(Y\) started an \(18\)-mile race at the same time. Runner \(X\) completed the course in \(6\) hours,

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by VJesus12 » Thu Jul 30, 2020 9:17 am

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Runners \(X\) and \(Y\) started an \(18\)-mile race at the same time. Runner \(X\) completed the course in \(6\) hours, and Runner \(Y\) finished \(2\) hours earlier. Runner \(Y\) ran an average of how many miles per hour faster than Runner \(X?\)

A. \(1\)

B. \(1\frac12\)

C. \(2\frac14\)

D. \(3\)

E. \(4\frac12\)

[spoiler]OA=B[/spoiler]

Source: Princeton Review
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Source: — Problem Solving |

VJesus12 wrote:
Thu Jul 30, 2020 9:17 am
Runners \(X\) and \(Y\) started an \(18\)-mile race at the same time. Runner \(X\) completed the course in \(6\) hours, and Runner \(Y\) finished \(2\) hours earlier. Runner \(Y\) ran an average of how many miles per hour faster than Runner \(X?\)

A. \(1\)

B. \(1\frac12\)

C. \(2\frac14\)

D. \(3\)

E. \(4\frac12\)

[spoiler]OA=B[/spoiler]

Solution:

The average speed of runner X is 18/6 = 3 mph, and the average speed of runner Y is 18/4 = 4.5 mph. Therefore, the average speed of runner Y is 4.5 - 3 = 1.5 mph faster than the average speed of runner X.

Answer: B

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