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{(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}}
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==See Also==
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{{AHSME box|year=1950|num-b=28|num-a=30}}
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[[Category:Introductory Algebra Problems]]

Revision as of 13:20, 17 April 2012

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

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See Also

1950 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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All AHSME Problems and Solutions