Difference between revisions of "1950 AHSME Problems/Problem 29"
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==Problem== | ==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== | ||
+ | {{solution}} | ||
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+ | ==See Also== | ||
+ | {{AHSME box|year=1950|num-b=28|num-a=30}} | ||
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+ | [[Category:Introductory Algebra Problems]] |
Revision as of 14:20, 17 April 2012
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
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See Also
1950 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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 | ||
All AHSME Problems and Solutions |