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Difference between revisions of "1950 AHSME Problems/Problem 29"

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==Problem==
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== Problem ==
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A manufacturer built a machine which will address <math>500</math> envelopes in <math>8</math> minutes. He wishes to build another machine so that when both are operating together they will address <math>500</math> envelopes in <math>2</math> minutes. The equation used to find how many minutes <math>x</math> it would require the second machine to address <math>500</math> envelopes alone is:
 
A manufacturer built a machine which will address <math>500</math> envelopes in <math>8</math> minutes. He wishes to build another machine so that when both are operating together they will address <math>500</math> envelopes in <math>2</math> minutes. The equation used to find how many minutes <math>x</math> it would require the second machine to address <math>500</math> envelopes alone is:
  
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\textbf{(D)}\ \dfrac{x}{2}+\dfrac{x}{8}=1 \qquad\\
 
\textbf{(D)}\ \dfrac{x}{2}+\dfrac{x}{8}=1 \qquad\\
 
\textbf{(E)}\ \text{None of these answers}</math>
 
\textbf{(E)}\ \text{None of these answers}</math>
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== Solution ==
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=== Solution 1 ===
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We first represent the first machine's speed in per <math>2</math> minutes: <math>125 \text{ envelopes in }2\text{ minutes}</math>. Now, we know that the speed per <math>2</math> minutes of the second machine is <cmath>500-125=375 \text{ envelopes in }2\text{ minutes}</cmath>
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Now we can set up a proportion to find out how many minutes it takes for the second machine to address <math>500</math> papers:
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<math>\frac{375}{2}=\frac{500}{x}</math>. Solving for <math>x</math>, we get <math>x=\frac{8}{3}</math> minutes.
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Now that we know the speed of the second machine, we can just plug it in each option to see if it equates. We see that <math>\boxed{\textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2}} </math> works.
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=== Solution 2 ===
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First, notice that the number of envelopes addressed does not matter, because it stays constant throughout the problem.
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Next, notice that we are talking about combining two speeds, so we use the formula <math>\frac{1}{\frac{1}{a}+\frac{1}{b}}=c</math>, where <math>a,b</math> are the respective times independently, and <math>c</math> is the combined time. Plugging in for <math>a</math> and <math>c</math>, we get <math>\frac{1}{\frac{1}{8}+\frac{1}{b}}=2</math>.
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Taking a reciprocal, we finally get <math>\boxed{\textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2}}</math>.
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== See Also ==
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{{AHSME 50p box|year=1950|num-b=28|num-a=30}}
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[[Category:Introductory Algebra Problems]]
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[[Category:Rate Problems]]
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{{MAA Notice}}

Latest revision as of 23:59, 11 October 2020

Problem

A manufacturer built a machine which will address $500$ envelopes in $8$ minutes. He wishes to build another machine so that when both are operating together they will address $500$ envelopes in $2$ minutes. The equation used to find how many minutes $x$ it would require the second machine to address $500$ envelopes alone is:

$\textbf{(A)}\ 8-x=2 \qquad \textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2} \qquad \textbf{(C)}\ \dfrac{500}{8}+\dfrac{500}{x}=500 \qquad \textbf{(D)}\ \dfrac{x}{2}+\dfrac{x}{8}=1 \qquad\\ \textbf{(E)}\ \text{None of these answers}$

Solution

Solution 1

We first represent the first machine's speed in per $2$ minutes: $125 \text{ envelopes in }2\text{ minutes}$. Now, we know that the speed per $2$ minutes of the second machine is \[500-125=375 \text{ envelopes in }2\text{ minutes}\] Now we can set up a proportion to find out how many minutes it takes for the second machine to address $500$ papers:

$\frac{375}{2}=\frac{500}{x}$. Solving for $x$, we get $x=\frac{8}{3}$ minutes. Now that we know the speed of the second machine, we can just plug it in each option to see if it equates. We see that $\boxed{\textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2}}$ works.

Solution 2

First, notice that the number of envelopes addressed does not matter, because it stays constant throughout the problem. Next, notice that we are talking about combining two speeds, so we use the formula $\frac{1}{\frac{1}{a}+\frac{1}{b}}=c$, where $a,b$ are the respective times independently, and $c$ is the combined time. Plugging in for $a$ and $c$, we get $\frac{1}{\frac{1}{8}+\frac{1}{b}}=2$.

Taking a reciprocal, we finally get $\boxed{\textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2}}$.

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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All AHSME Problems and Solutions

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