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

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: $$\begin{array}{rl}\frac{1}{a}& =m\ast \frac{1}{b+c}\\ \frac{1}{b}& =n\ast \frac{1}{a+c}\\ \frac{1}{c}& =x\ast \frac{1}{a+b}\end{array}$$ These equations can be rearranged into the following: $$\begin{array}{rl}\text{(i) }ma& =b+c\\ \text{(ii) }nb& =a+c\\ \text{(iii) }xc& =a+b\end{array}$$ Solving for in equation (i) gives us . Substituting this expression for into equation (ii) yields: $$\begin{array}{rl}nb& =\frac{b+c}{m}+c\\ mnb& =b+c+mc\\ (mn-1)b& =(m+1)c\\ b& =\frac{(m+1)c}{mn-1}\end{array}$$ Finally, substituting our expressions for and into equation (iii) yields our final answer: $$\begin{array}{rl}xc& =\frac{b+c}{m}+\frac{(m+1)c}{mn-1}\\ & =\frac{\frac{(m+1)c}{mn-1}+c}{m}+\frac{(m+1)c}{mn-1}\\ & =\frac{c}{m}(\frac{m+1+mn-1}{mn-1}+\frac{m^2+m}{mn-1})\\ & =\frac{c}{m}(\frac{m^2+mn+2m}{mn-1})\\ & =c(\frac{m+n+2}{mn-1})\end{array}$$ Thus, .

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.