Difference between revisions of "1950 AHSME Problems/Problem 27"
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==Solution== | ==Solution== | ||
− | 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>. | + | 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 50p box|year=1950|num-b=26|num-a=28}} | {{AHSME 50p box|year=1950|num-b=26|num-a=28}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | [[Category:Rate Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:49, 14 February 2016
Problem
A car travels miles from to at miles per hour but returns the same distance at miles per hour. The average speed for the round trip is closest to:
Solution
The car takes to get from to . Also, it takes to get from to . Therefore, the average speed is , which is closest to .
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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