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

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== Solution ==
 
== Solution ==
<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:
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\begin{align*}
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\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:
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\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:
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\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:
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\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>.
  
 
== 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 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: 1a=m1b+c1b=n1a+c1c=x1a+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(mn1)b=(m+1)cb=(m+1)cmn1 Finally, substituting our expressions for $a$ and $b$ into equation (iii) yields our final answer: xc=b+cm+(m+1)cmn1=(m+1)cmn1+cm+(m+1)cmn1=cm(m+1+mn1mn1+m2+mmn1)=cm(m2+mn+2mmn1)=c(m+n+2mn1) Thus, $x=\boxed{\textbf{(E) }\frac{m+n+2}{mn-1}}$.

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 37
Followed by
Problem 39
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