Difference between revisions of "1965 AHSME Problems/Problem 38"
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== Solution == | == Solution == | ||
| − | <math>\ | + | |
| + | Let <math>a</math>, <math>b</math>, and <math>c</math> be the speeds at which <math>A</math>, <math>B</math> and <math>C</math> 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 <math>a</math> in equation (i) gives us <math>a=\frac{b+c}{m}</math>. Substituting this expression for <math>a</math> 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 <math>a</math> and <math>b</math> 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, <math>x=\boxed{\textbf{(E) }\frac{m+n+2}{mn-1}}</math>. | ||
== See Also == | == See Also == | ||
{{AHSME 40p box|year=1965|num-b=37|num-a=39}} | {{AHSME 40p box|year=1965|num-b=37|num-a=39}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:30, 20 July 2024
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{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, 
.
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 37 |
Followed by Problem 39 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
