# 1965 AHSME Problems/Problem 38

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## Problem

$A$ takes $m$ times as long to do a piece of work as $B$ and $C$ together; $B$ takes $n$ times as long as $C$ and $A$ together; and $C$ takes $x$ times as long as $A$ and $B$ together. Then $x$, in terms of $m$ and $n$, is:

$\textbf{(A)}\ \frac {2mn}{m + n} \qquad \textbf{(B) }\ \frac {1}{2(m + n)} \qquad \textbf{(C) }\ \frac{1}{m+n-mn}\qquad \textbf{(D) }\ \frac{1-mn}{m+n+2mn}\qquad \textbf{(E) }\ \frac{m+n+2}{mn-1}$

## Solution 1

Let $a$, $b$, and $c$ be the speeds at which $A$, $B$ and $C$ work, respectively. Also, let the piece of work be worth one unit of work. Then, using the information from the problem along with basic rate formulas, we obtain the following equations: \begin{align*} \frac{1}{a}&=m*\frac{1}{b+c} \\ \frac{1}{b}&=n*\frac{1}{a+c} \\ \frac{1}{c}&=x*\frac{1}{a+b} \end{align*} These equations can be rearranged into the following: \begin{align*} \text{(i) } ma&=b+c \\ \text{(ii) } nb&=a+c \\ \text{(iii) } xc&=a+b \\ \end{align*} Solving for $a$ in equation (i) gives us $a=\frac{b+c}{m}$. Substituting this expression for $a$ into equation (ii) yields: \begin{align*} nb&=\frac{b+c}{m}+c \\ mnb&=b+c+mc \\ (mn-1)b&=(m+1)c \\ b&=\frac{(m+1)c}{mn-1} \end{align*} Finally, substituting our expressions for $a$ and $b$ into equation (iii) yields our final answer: \begin{align*} xc&=\frac{b+c}{m}+\frac{(m+1)c}{mn-1} \\ &=\frac{\frac{(m+1)c}{mn-1}+c}{m}+\frac{(m+1)c}{mn-1} \\ &=\frac{c}{m}(\frac{m+1+mn-1}{mn-1}+\frac{m^2+m}{mn-1}) \\ &=\frac{c}{m}(\frac{m^2+mn+2m}{mn-1}) \\ &=c(\frac{m+n+2}{mn-1}) \end{align*} Thus, $x=\boxed{\textbf{(E) }\frac{m+n+2}{mn-1}}$.

If we let $A$, $B$, and $C$ work at the same speed, then it is clear that $m=n=x=2$. After plugging in $m=n=2$ into all of the answer choices, we see that the only two choices which give a value $x=2$ are choices (A) and (E). Now suppose that $A$ and $B$ work at the same speed, but $C$ does no work at all. Then, $m=n=1$, but $x$ is undefined. After plugging in $m=n=1$ into choices (A) and (E), we see that the only choice which is undefined is choice $\boxed{\textbf{(E) }\frac{m+n+2}{mn-1}}$.