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},</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{(A)}</math>. | + | == 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 <math>45</math> mph. | ||
| + | His average speed for the entire trip is: | ||
| + | |||
| + | <math>\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} </math> | ||
| + | |||
| + | == Solution == | ||
| + | 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>. | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 50p box|year=1957|num-b=19|num-a=21}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:AHSME]][[Category:AHSME Problems]] | ||
Latest revision as of 09:55, 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 
mph.
His average speed for the entire trip is:
Solution
Suppose the first half of the trip's distance is called 
. Then the time for the first half is 
and the second half's time is 
, so the total time is 
, so the speed is 
, so the answer is 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| 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. 
