Difference between revisions of "1950 AHSME Problems/Problem 29"
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| − | ==Problem== | + | == Problem == |
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{(E)}\ \text{None of these answers}</math> | \textbf{(E)}\ \text{None of these answers}</math> | ||
| − | ==Solution== | + | == Solution == |
| − | ===Solution 1=== | + | === Solution 1 === |
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> | 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> | ||
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: | 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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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. | 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. | ||
| − | ===Solution 2=== | + | === Solution 2 === |
First, notice that the number of envelopes addressed does not matter, because it stays constant throughout the problem. | 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 <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>. | 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>. | ||
| − | Taking a reciprocal, we finally get <math>\boxed{\textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2}} </math> | + | Taking a reciprocal, we finally get <math>\boxed{\textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2}}</math>. |
| − | ==See Also== | + | |
| + | == See Also == | ||
{{AHSME 50p box|year=1950|num-b=28|num-a=30}} | {{AHSME 50p box|year=1950|num-b=28|num-a=30}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | [[Category:Rate Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:59, 11 October 2020
Problem
A manufacturer built a machine which will address 
envelopes in 
minutes. He wishes to build another machine so that when both are operating together they will address envelopes in 
minutes. The equation used to find how many minutes 
it would require the second machine to address envelopes alone is:
Solution
Solution 1
We first represent the first machine's speed in per minutes: 
. Now, we know that the speed per minutes of the second machine is ![\[500-125=375 \text{ envelopes in }2\text{ minutes}\]](http://latex.artofproblemsolving.com/0/3/1/0310b05862e443272838a35017e922903ff57766.png)
Now we can set up a proportion to find out how many minutes it takes for the second machine to address papers:
. Solving for , we get 
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 
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 
, where 
are the respective times independently, and 
is the combined time. Plugging in for 
and , we get 
.
Taking a reciprocal, we finally get .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
