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Difference between revisions of "1957 AHSME Problems/Problem 20"

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Suppose the first half of the trip's distance is called <math>x</math>. Then the time for the first half is <math>\dfrac{x}{50}, and the second half's time is </math>\dfrac{x}{45}<math>, so the total time is </math>\dfrac{x}{50}+\dfrac{x}{45}<math>, so the speed is </math>\dfrac{2x}{\frac{x}{50}+\frac{x}{45}}=\dfrac{2x}{\frac{19x}{450}}=2x \cdot \dfrac{450}{19x}=\dfrac{900}{19}=47\dfrac{7}{19}<math>, so the answer is </math>\boxed{(A)}$.
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== Problem ==
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A man makes a trip by automobile at an average speed of 50 mph. He returns over the same route at an average speed of <math>45</math> mph.
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His average speed for the entire trip is:
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<math>\textbf{(A)}\ 47\frac{7}{19}\qquad
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\textbf{(B)}\ 47\frac{1}{4}\qquad
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\textbf{(C)}\ 47\frac{1}{2}\qquad
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\textbf{(D)}\ 47\frac{11}{19}\qquad
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\textbf{(E)}\ \text{none of these} </math>
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== Solution ==
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Suppose the first half of the trip's distance is called <math>x</math>. Then the time for the first half is <math>\dfrac{x}{50},</math> and the second half's time is <math>\dfrac{x}{45}</math>, so the total time is <math>\dfrac{x}{50}+\dfrac{x}{45}</math>, so the speed is <math>\dfrac{2x}{\frac{x}{50}+\frac{x}{45}}=\dfrac{2x}{\frac{19x}{450}}=2x \cdot \dfrac{450}{19x}=\dfrac{900}{19}=47\dfrac{7}{19}</math>, so the answer is <math>\boxed{\textbf{(A)}}</math>.
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== See Also ==
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{{AHSME 50p box|year=1957|num-b=19|num-a=21}}
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{{MAA Notice}}
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[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 08:55, 25 July 2024

Problem

A man makes a trip by automobile at an average speed of 50 mph. He returns over the same route at an average speed of $45$ mph. His average speed for the entire trip is:

$\textbf{(A)}\ 47\frac{7}{19}\qquad \textbf{(B)}\ 47\frac{1}{4}\qquad \textbf{(C)}\ 47\frac{1}{2}\qquad \textbf{(D)}\ 47\frac{11}{19}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Suppose the first half of the trip's distance is called $x$. Then the time for the first half is $\dfrac{x}{50},$ and the second half's time is $\dfrac{x}{45}$, so the total time is $\dfrac{x}{50}+\dfrac{x}{45}$, so the speed is $\dfrac{2x}{\frac{x}{50}+\frac{x}{45}}=\dfrac{2x}{\frac{19x}{450}}=2x \cdot \dfrac{450}{19x}=\dfrac{900}{19}=47\dfrac{7}{19}$, so the answer is $\boxed{\textbf{(A)}}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AHSME Problems and Solutions

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