Difference between revisions of "1950 AHSME Problems/Problem 19"
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==Solution== | ==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> \frac{md}{m+r} </math> days and our is <math> \textbf{(C)}. </math> | 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> | ||
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+ | {{AHSME box|year=1950|num-b=18|num-a=20}} |
Revision as of 10:57, 29 December 2011
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 days and our is
1950 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AHSME Problems and Solutions |