1957 AHSME Problems/Problem 20

Revision as of 08:55, 25 July 2024 by Thepowerful456 (talk | contribs) (added problem & see also box)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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