Difference between revisions of "1950 AHSME Problems/Problem 19"
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| + | == Problem == | ||
If <math> m </math> men can do a job in <math> d </math> days, then <math> m+r </math> men can do the job in: | If <math> m </math> men can do a job in <math> d </math> days, then <math> m+r </math> men can do the job in: | ||
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<math> \textbf{(A)}\ d+r \text{ days}\qquad\textbf{(B)}\ d-r\text{ days}\qquad\textbf{(C)}\ \frac{md}{m+r}\text{ days}\qquad\\ \textbf{(D)}\ \frac{d}{m+r}\text{ days}\qquad\textbf{(E)}\ \text{None of these} </math> | <math> \textbf{(A)}\ d+r \text{ days}\qquad\textbf{(B)}\ d-r\text{ days}\qquad\textbf{(C)}\ \frac{md}{m+r}\text{ days}\qquad\\ \textbf{(D)}\ \frac{d}{m+r}\text{ days}\qquad\textbf{(E)}\ \text{None of these} </math> | ||
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| + | == Solution == | ||
| + | The number of men is inversely proportional to the number of days the job takes. Thus, if <math> m </math> men can do a job in <math> d </math> days, we have that it will take <math> md </math> days for <math> 1 </math> man to do the job. Thus, <math> m + r </math> men can do the job in <math>\boxed{\textbf{(C)}\ \frac{md}{m+r}\text{ days}}</math>. | ||
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| + | == See Also == | ||
| + | {{AHSME 50p box|year=1950|num-b=18|num-a=20}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:53, 11 October 2020
Problem
If 
men can do a job in 
days, then 
men can do the job in:
Solution
The number of men is inversely proportional to the number of days the job takes. Thus, if men can do a job in days, we have that it will take 
days for 
man to do the job. Thus, 
men can do the job in 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
