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

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==Solution==
 
==Solution==
 
The way to Temple took <math>\frac{50}{60}=\frac56</math> hours, and the way back took <math>\frac{50}{40}=\frac54</math> for a total of <math>\frac56 + \frac54 = \frac{25}{12}</math> hours. The trip is <math>50\cdot2=100</math> miles. The average speed is <math>\frac{100}{25/12} = \boxed{\textbf{(B)}\ 48}</math> miles per hour.
 
The way to Temple took <math>\frac{50}{60}=\frac56</math> hours, and the way back took <math>\frac{50}{40}=\frac54</math> for a total of <math>\frac56 + \frac54 = \frac{25}{12}</math> hours. The trip is <math>50\cdot2=100</math> miles. The average speed is <math>\frac{100}{25/12} = \boxed{\textbf{(B)}\ 48}</math> miles per hour.
 +
 +
==Solution 2==
 +
This question simply asks for the harmonic mean of <math>60</math> and <math>40</math>, regardless of how far Austin and Temple are.
 +
Plugging in, we have: <math>\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{\textbf{(B)}\ 48}</math> miles per hour.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=13|num-a=15}}
 
{{AMC8 box|year=2009|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:16, 8 August 2018

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.

Solution 2

This question simply asks for the harmonic mean of $60$ and $40$, regardless of how far Austin and Temple are. Plugging in, we have: $\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{\textbf{(B)}\ 48}$ miles per hour.

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
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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