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16 Sep 2014, 01:16
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Tanks X and Y contain 500 and 200 gallons of water, respectively. If water is being pumped out of tank X at a rate of \(m\) gallons per minute, and water is being added to tank Y at a rate of \(n\) gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?
A. \(\frac{5}{m + n}\text{ hours}\)
B. \(6(m + n)\text{ hours}\)
C. \(\frac{300}{m + n}\text{ hours}\)
D. \(\frac{300}{n - m}\text{ hours}\)
E. \(\frac{60}{n - m}\text{ hours}\)
A. \(\frac{5}{m + n}\text{ hours}\)
B. \(6(m + n)\text{ hours}\)
C. \(\frac{300}{m + n}\text{ hours}\)
D. \(\frac{300}{n - m}\text{ hours}\)
E. \(\frac{60}{n - m}\text{ hours}\)
16 Sep 2014, 01:16
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Official Solution:
Tanks X and Y contain 500 and 200 gallons of water, respectively. If water is being pumped out of tank X at a rate of \(m\) gallons per minute, and water is being added to tank Y at a rate of \(n\) gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?
A. \(\frac{5}{m + n}\text{ hours}\)
B. \(6(m + n)\text{ hours}\)
C. \(\frac{300}{m + n}\text{ hours}\)
D. \(\frac{300}{n - m}\text{ hours}\)
E. \(\frac{60}{n - m}\text{ hours}\)
Let's say \(t\) minutes are needed for the two tanks to contain equal amounts of water. Then we would have the equation \(500-mt=200+nt\). To find \(t\), we can solve for \(t\) as \(t=\frac{300}{m+n}\) minutes, or by converting to hours, \(\frac{1}{60} *\frac{300}{m+n} = \frac{5}{m+n}\) hours.
Answer: A
Tanks X and Y contain 500 and 200 gallons of water, respectively. If water is being pumped out of tank X at a rate of \(m\) gallons per minute, and water is being added to tank Y at a rate of \(n\) gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?
A. \(\frac{5}{m + n}\text{ hours}\)
B. \(6(m + n)\text{ hours}\)
C. \(\frac{300}{m + n}\text{ hours}\)
D. \(\frac{300}{n - m}\text{ hours}\)
E. \(\frac{60}{n - m}\text{ hours}\)
Let's say \(t\) minutes are needed for the two tanks to contain equal amounts of water. Then we would have the equation \(500-mt=200+nt\). To find \(t\), we can solve for \(t\) as \(t=\frac{300}{m+n}\) minutes, or by converting to hours, \(\frac{1}{60} *\frac{300}{m+n} = \frac{5}{m+n}\) hours.
Answer: A
General Discussion
14 Aug 2023, 08:05
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
