Difference between revisions of "1965 AHSME Problems/Problem 38"

Revision as of 10:30, 20 July 2024

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

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: $\begin{aligned} \frac{1}{a} & = m * \frac{1}{b + c} \\ \frac{1}{b} & = n * \frac{1}{a + c} \\ \frac{1}{c} & = x * \frac{1}{a + b} \end{aligned}$ These equations can be rearranged into the following: $\begin{aligned} (i) m a & = b + c \\ (ii) n b & = a + c \\ (iii) x c & = a + b \end{aligned}$ Solving for $a$ in equation (i) gives us $a=\frac{b+c}{m}$ . Substituting this expression for $a$ into equation (ii) yields: $\begin{aligned} n b & = \frac{b + c}{m} + c \\ m n b & = b + c + m c \\ (m n - 1) b & = (m + 1) c \\ b & = \frac{(m + 1) c}{m n - 1} \end{aligned}$ Finally, substituting our expressions for $a$ and $b$ into equation (iii) yields our final answer: $\begin{aligned} x c & = \frac{b + c}{m} + \frac{(m + 1) c}{m n - 1} \\ = \frac{\frac{(m + 1) c}{m n - 1} + c}{m} + \frac{(m + 1) c}{m n - 1} \\ = \frac{c}{m} (\frac{m + 1 + m n - 1}{m n - 1} + \frac{m^{2} + m}{m n - 1}) \\ = \frac{c}{m} (\frac{m^{2} + m n + 2 m}{m n - 1}) \\ = c (\frac{m + n + 2}{m n - 1}) \end{aligned}$ Thus, $x=\boxed{\textbf{(E) }\frac{m+n+2}{mn-1}}$ .

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.

@@ Line 10: / Line 10: @@
 == Solution ==
-<math>\fbox{E}</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 ==
 {{AHSME 40p box|year=1965|num-b=37|num-a=39}}
 {{MAA Notice}}

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Difference between revisions of "1965 AHSME Problems/Problem 38"

Revision as of 10:30, 20 July 2024

Problem

Solution

See Also