# 1954 AHSME Problems/Problem 30

## Problem $A$ and $B$ together can do a job in $2$ days; $B$ and $C$ can do it in four days; and $A$ and $C$ in $2\frac{2}{5}$ days. The number of days required for A to do the job alone is: $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8$

## Solution

Let $A$ do $r_A$ of the job per day, $B$ do $r_B$ of the job per day, and $C$ do $r_C$ of the job per day. These three quantities have unit $\frac{\text{job}}{\text{day}}$. 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 $r_A+r_B+r_C$, so if we subtract $r_B+r_C$ (the value of which we know) from both equations, we obtain the value of $r_A$, which is what we wish to determine anyways. So we add these three equations and divide by two: $$r_A+r_B+r_C=\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\right)\cdot\frac{\text{job}}{\text{day}}=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}.$$ 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 $A$ does one third of the job per day. Therefore, if $A$ were to do the entire job himself, he would require $\boxed{\textbf{(B)}\ 3}$ days.

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