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

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\textbf{(E) }\ \frac{m+n+2}{mn-1} </math>
 
\textbf{(E) }\ \frac{m+n+2}{mn-1} </math>
  
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== Solution ==
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== Solution 1 ==
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<math>\fbox{E}</math>
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 +
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} \
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\frac{1}{b}&=n*\frac{1}{a+c} \
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\frac{1}{c}&=x*\frac{1}{a+b}
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\end{align*}
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These equations can be rearranged into the following:
 +
\begin{align*}
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\text{(i) } ma&=b+c \
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\text{(ii) } nb&=a+c \
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\text{(iii) } xc&=a+b \
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\end{align*}
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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*}
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nb&=\frac{b+c}{m}+c \
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mnb&=b+c+mc \
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(mn-1)b&=(m+1)c \
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b&=\frac{(m+1)c}{mn-1}
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\end{align*}
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Finally, substituting our expressions for <math>a</math> and <math>b</math> into equation (iii) yields our final answer:
 +
\begin{align*}
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xc&=\frac{b+c}{m}+\frac{(m+1)c}{mn-1} \
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&=\frac{\frac{(m+1)c}{mn-1}+c}{m}+\frac{(m+1)c}{mn-1} \
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&=\frac{c}{m}(\frac{m+1+mn-1}{mn-1}+\frac{m^2+m}{mn-1}) \
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&=\frac{c}{m}(\frac{m^2+mn+2m}{mn-1}) \
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&=c(\frac{m+n+2}{mn-1})
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\end{align*}
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Thus, <math>x=\boxed{\textbf{(E) }\frac{m+n+2}{mn-1}}</math>.
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 +
 
 +
== Solution 2 (Answer choices) ==
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If we let <math>A</math>, <math>B</math>, and <math>C</math> work at the same speed, then it is clear that <math>m=n=x=2</math>. After plugging in <math>m=n=2</math> into all of the answer choices, we see that the only two choices which give a value <math>x=2</math> are choices (A) and (E). Now suppose that <math>A</math> and <math>B</math> work at the same speed, but <math>C</math> does no work at all. Then, <math>m=n=1</math>, but <math>x</math> is undefined. After plugging in <math>m=n=1</math> into choices (A) and (E), we see that the only choice which is undefined is choice <math>\boxed{\textbf{(E) }\frac{m+n+2}{mn-1}}</math>. 
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== 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}}
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 10:39, 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 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=m∗1b+c1b=n∗1a+c1c=x∗1a+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(mn−1)b=(m+1)cb=(m+1)cmn−1 Finally, substituting our expressions for $a$ and $b$ into equation (iii) yields our final answer: xc=b+cm+(m+1)cmn−1=(m+1)cmn−1+cm+(m+1)cmn−1=cm(m+1+mn−1mn−1+m2+mmn−1)=cm(m2+mn+2mmn−1)=c(m+n+2mn−1) 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 (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

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