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 ==
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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 ==
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{{AHSME 50p box|year=1950|num-b=18|num-a=20}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 23:53, 11 October 2020

Problem

If $m$ men can do a job in $d$ days, then $m+r$ men can do the job in:

$\textbf{(A)}\ d+r \text{ days}\qquad\textbf{(B)}\ d-r\text{ days}\qquad\textbf{(C)}\ \frac{md}{m+r}\text{ days}\qquad\\ \textbf{(D)}\ \frac{d}{m+r}\text{ days}\qquad\textbf{(E)}\ \text{None of these}$

Solution

The number of men is inversely proportional to the number of days the job takes. Thus, if $m$ men can do a job in $d$ days, we have that it will take $md$ days for $1$ man to do the job. Thus, $m + r$ men can do the job in $\boxed{\textbf{(C)}\ \frac{md}{m+r}\text{ days}}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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