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> \frac{md}{m+r} </math> days and our is <math> \textbf{(C)}. </math>

Revision as of 22:48, 17 November 2011

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 $\frac{md}{m+r}$ days and our is $\textbf{(C)}.$