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VeritasKarishma
ichha148
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 K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?
A. \(\frac{5}{M + K}\text{ hours}\)
B. \(6(M + K)\text{ hours}\)
C. \(\frac{300}{M + K}\text{ hours}\)
D. \(\frac{300}{M - K}\text{ hours}\)
E. \(\frac{60}{M - K}\text{ hours}\)
m22 q17
A. \(\frac{5}{M + K}\text{ hours}\)
B. \(6(M + K)\text{ hours}\)
C. \(\frac{300}{M + K}\text{ hours}\)
D. \(\frac{300}{M - K}\text{ hours}\)
E. \(\frac{60}{M - K}\text{ hours}\)
m22 q17
This is a relative speed question.
Distance to be covered together = 300 gallons (= 500 gallons - 200 gallons)
Relative speed (rate of work) = (K+M) gallons per minute OR 60*(K+M) gallons per hour (The rates get added because they are working in opposite directions)
Time taken = 300/60(K+M) hours = 5/(K+M) hours
Hi karishma
In relative speed questions -- why do we add the rates [M+K] that are working in opposite directions ? Could you perhaps explain why we are adding in this case and not subtracting
Also, you mentioned above that relative speeds are similar to relative rates...
But i thought relative rates are such that
-- if two people are working together on the project [rates in the same direction] - we can add the rates for a combined rates [ R1 + R2]
-- if one person is constructing and the other person is de-constructing a project [rates in the opposite direction] - the combined rate is obviously R1 - R2
Thank you !
On the same lines, think about it - one tank is at 500 gallons and another is at 200 gallons. They both need to reach the same level, one by reducing water and the other by gaining water. So they have to together bridge this gap of 300 gallons. Imagine the two tanks sitting side by side at different levels. They are both working in opposite directions to reach the same level such that both are contributing TOWARD removing the 300 gallons difference. Hence their rates will get added.
Always look at the logic of the question.
When one worker is making a wall while the other is destroying, their rates get subtracted because one is contributing towards work that has to be done while the other is taking away from it.
Here, both tanks are contributing toward work that has to be done.
Updated on: 22 Feb 2025, 00:02
Originally posted by GMATinsight on 18 Jul 2020, 21:03.
Last edited by Bunuel on 22 Feb 2025, 00:02, edited 2 times in total.
Last edited by Bunuel on 22 Feb 2025, 00:02, edited 2 times in total.
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ichha148
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 K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?
A. \(\frac{5}{M + K}\text{ hours}\)
B. \(6(M + K)\text{ hours}\)
C. \(\frac{300}{M + K}\text{ hours}\)
D. \(\frac{300}{M - K}\text{ hours}\)
E. \(\frac{60}{M - K}\text{ hours}\)
m22 q17
A. \(\frac{5}{M + K}\text{ hours}\)
B. \(6(M + K)\text{ hours}\)
C. \(\frac{300}{M + K}\text{ hours}\)
D. \(\frac{300}{M - K}\text{ hours}\)
E. \(\frac{60}{M - K}\text{ hours}\)
m22 q17
Answer: Option A
Please check the attached solution with a CORRECTION that last step should be T = 300/(m+k)*60 = 5/(m+K)
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Screenshot 2020-07-19 at 9.33.02 AM.png [ 533.96 KiB | Viewed 1910 times ]
Oppenheimer1945
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05 Aug 2024, 08:21
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500+kT=200+GT
=>T in min=300/(M+K)
T in hours= 5/(M+K)
=>T in min=300/(M+K)
T in hours= 5/(M+K)
