Difference between revisions of "2009 AMC 8 Problems/Problem 14"

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==Problem==
 
==Problem==
  
Austin and Temple are <math> 50</math> miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging <math> 60</math> miles per hour. Leaving the car with her daughter, Bonnie rod a bus back to Austin along the same route and averaged <math> 40</math> miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
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Austin and Temple are <math> 50</math> miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging <math> 60</math> miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged <math> 40</math> miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
  
 
<math> \textbf{(A)}\  46  \qquad
 
<math> \textbf{(A)}\  46  \qquad

Revision as of 17:39, 31 January 2015

Problem

Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?

$\textbf{(A)}\  46  \qquad \textbf{(B)}\   48  \qquad \textbf{(C)}\   50  \qquad \textbf{(D)}\   52  \qquad \textbf{(E)}\   54$

Solution

The way to Temple took $\frac{50}{60}=\frac56$ hours, and the way back took $\frac{50}{40}=\frac54$ for a total of $\frac56 + \frac54 = \frac{25}{12}$ hours. The trip is $50\cdot2=100$ miles. The average speed is $\frac{100}{25/12} = \boxed{\textbf{(B)}\ 48}$ miles per hour.

See Also

2009 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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