1954 AHSME Problems/Problem 30
Problem
and 
together can do a job in 
days; and 
can do it in four days; and and in 
days.
The number of days required for A to do the job alone is:
Solution
Let do 
of the job per day, do 
of the job per day, and do 
of the job per day. These three quantities have unit 
. Therefore our three conditions give us the three equations:
We divide the three equations by the required constant so that the coefficients of the variables become 1:
If we add these three new equations together and divide the result by two, we obtain an equation with left-hand side 
, so if we subtract 
(the value of which we know) from both equations, we obtain the value of , which is what we wish to determine anyways. So we add these three equations and divide by two:
![\[r_A+r_B+r_C=\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\right)\cdot\frac{\text{job}}{\text{day}}=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}.\]](http://latex.artofproblemsolving.com/c/8/4/c84e8dbd5827df45b177ffa8e4325dc8d51a33ab.png)
Hence:
This shows that does one third of the job per day. Therefore, if were to do the entire job himself, he would require 
days.
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Problem 31 | |
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