1954 AHSME Problems/Problem 30

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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: (2 days)(rA+rB)=1 job,(4 days)(rB+rC)=1 job,(2.4 days)(rC+rA)=1 job. We divide the three equations by the required constant so that the coefficients of the variables become 1: rA+rB=12jobday,rB+rC=14jobday,rC+rA=512jobday. 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: rA=(rA+rB+rC)(rB+rC)=712jobday14jobday=13jobday. 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.

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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