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Difference between revisions of "1950 AHSME Problems/Problem 27"

(Created page with "== Problem== A car travels <math>120</math> miles from <math>A</math> to <math>B</math> at <math>30</math> miles per hour but returns the same distance at <math>40</math> miles ...")
 
 
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==Solution==
 
==Solution==
  
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The car takes <math>120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{30 \text{ miles }}=4 \text{ hr}</math> to get from <math>A</math> to <math>B</math>. Also, it takes <math>120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{40 \text{ miles }}=3 \text{ hr}</math> to get from <math>B</math> to <math>A</math>. Therefore, the average speed is <math>\dfrac{240\text{ miles }}{7 \text{ hr}}=34\dfrac{2}{7}\text{ mph}</math>, which is closest to <math>\textbf{(B)}\ 34\text{ mph}</math>.
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The car takes <math>120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{30 \text{ miles }}=4 \text{ hr}</math> to get from <math>A</math> to <math>B</math>. Also, it takes <math>120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{40 \text{ miles }}=3 \text{ hr}</math> to get from <math>B</math> to <math>A</math>. Therefore, the average speed is <math>\dfrac{240\text{ miles }}{7 \text{ hr}}=34\dfrac{2}{7}\text{ mph}</math>, which is closest to <math>\boxed{\textbf{(B)}\ 34\text{ mph}}</math>.
  
 
==See Also==
 
==See Also==
  
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{{AHSME box|year=1950|num-b=26|num-a=28}}
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{{AHSME 50p box|year=1950|num-b=26|num-a=28}}
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[[Category:Introductory Algebra Problems]]
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[[Category:Rate Problems]]
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{{MAA Notice}}

Latest revision as of 13:49, 14 February 2016

Problem

A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:

$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37\text{ mph}$

Solution

The car takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{30 \text{ miles }}=4 \text{ hr}$ to get from $A$ to $B$. Also, it takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{40 \text{ miles }}=3 \text{ hr}$ to get from $B$ to $A$. Therefore, the average speed is $\dfrac{240\text{ miles }}{7 \text{ hr}}=34\dfrac{2}{7}\text{ mph}$, which is closest to $\boxed{\textbf{(B)}\ 34\text{ mph}}$.

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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All AHSME Problems and Solutions

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