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: \begin{align} (2\text{ days})(r_A+r_B)&=1\text{ job},\nonumber\\ (4\text{ days})(r_B+r_C)&=1\text{ job},\nonumber\\ (2.4\text{ days})(r_C+r_A)&=1\text{ job}.\nonumber \end{align} We divide the three equations by the required constant so that the coefficients of the variables become 1: \begin{align} r_A+r_B&=\frac{1}{2}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ r_B+r_C&=\frac{1}{4}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ r_C+r_A&=\frac{5}{12}\cdot\frac{\text{job}}{\text{day}}.\nonumber \end{align} 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: Hence: \begin{align} r_A &= (r_A+r_B+r_C)-(r_B+r_C)\nonumber\\ &=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}-\frac{1}{4}\cdot\frac{\text{job}}{\text{day}}\nonumber\\ &=\frac{1}{3}\cdot\frac{\text{job}}{\text{day}}.\nonumber \end{align} 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

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