During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

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

(Created page with "==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 ...")
 
(Solution 2)
 
(10 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
==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?
+
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
Line 9: Line 8:
 
\textbf{(D)}\  52  \qquad
 
\textbf{(D)}\  52  \qquad
 
\textbf{(E)}\  54</math>
 
\textbf{(E)}\  54</math>
 +
 +
==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.
 +
 +
==Solution 2==
 +
We calculate the harmonic mean of Austin and Temple.
 +
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==
 +
2015 Problem 17
 +
{{AMC8 box|year=2009|num-b=13|num-a=15}}
 +
{{MAA Notice}}

Latest revision as of 19:25, 10 January 2022

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

We calculate the harmonic mean of Austin and Temple. 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

2015 Problem 17

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS