1965 AHSME Problems/Problem 38

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: 1a=m1b+c1b=n1a+c1c=x1a+b These equations can be rearranged into the following: (i) ma=b+c(ii) nb=a+c(iii) xc=a+b Solving for $a$ in equation (i) gives us $a=\frac{b+c}{m}$. Substituting this expression for $a$ into equation (ii) yields: nb=b+cm+cmnb=b+c+mc(mn1)b=(m+1)cb=(m+1)cmn1 Finally, substituting our expressions for $a$ and $b$ into equation (iii) yields our final answer: xc=b+cm+(m+1)cmn1=(m+1)cmn1+cm+(m+1)cmn1=cm(m+1+mn1mn1+m2+mmn1)=cm(m2+mn+2mmn1)=c(m+n+2mn1) Thus, $x=\boxed{\textbf{(E) }\frac{m+n+2}{mn-1}}$.


Solution 2 (Answer choices)

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 $\boxed{\textbf{(E) }\frac{m+n+2}{mn-1}}$.


See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 37
Followed by
Problem 39
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