1965 AHSME Problems/Problem 38
Problem
takes times as long to do a piece of work as and together; takes times as long as and together; and takes times as long as and together. Then , in terms of and , is:
Solution 1
Let , , and be the speeds at which , and 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 in equation (i) gives us . Substituting this expression for 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 and 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, .
Solution 2 (Answer choices)
If we let , , and work at the same speed, then it is clear that . After plugging in into all of the answer choices, we see that the only two choices which give a value are choices (A) and (E). Now suppose that and work at the same speed, but does no work at all. Then, , but is undefined. After plugging in into choices (A) and (E), we see that the only choice which is undefined is choice .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 37 |
Followed by Problem 39 | |
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