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05 Aug 2022, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
39% (03:04) correct
61%
(04:03)
wrong
based on 59
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A rectangular tank has the dimensions of its base as 30 metres by 20 metres and a height of 10 metres. There are two taps attached to the tank such that each tap working alone at a constant rate can fill the tank completely in 60 hours and 90 hours respectively. One of the walls of the tank has holes along the height of the tank at a regular distance of 2 metres and the first such hole is 2 metres above the base of the tank. The rate of water outflow from each hole is 10m3 per hour. If both the taps are opened simultaneously in the empty tank, approximately how many hours will it take to fill the tank completely?
A. 36
B. 42
C. 48
D. 54
E. 60
Are You Up For the Challenge: 700 Level Questions
A. 36
B. 42
C. 48
D. 54
E. 60
Are You Up For the Challenge: 700 Level Questions
05 Aug 2022, 11:39
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Bunuel
Volume of the tank is 30*20*10 = 6000m^3.
Tap 1 fills in 60 hours, so 100m^3 per hour.
Tap 2 fills in 90 hours, so (200/3)m^3 per hour.
Combined inflow rate is (500/3)m^3 per hour.
If there were no holes, it would take 36 hours. Eliminate A.
There are four holes, one at a height of 2m, one at a height of 4m, one at a height of 6m, an one at a height of 8m.
From here, there are two ways to approach the rest of the question. The first way is to understand how test-writers design questions, use some logic, test the other extreme (we already tested no holes, so we need to test what happens if all four holes impact the net inflow the entire time), eliminate four answer choices, get the right answer quickly, and move on. The second way is to get bogged down in the actual math to prove the one right answer. I strongly prefer the first way. You do whatever is best for you.
First.
We've already said that it will take longer than 36 hours, because that's how long it would take with no holes. Eliminate A.
What if all four holes were right at the bottom of the tank? The combined inflow would be (500/3) and the combined outflow would be 4*10, so the net inflow would be (500/3)-40 = (500/3)-(120/3) = 380/3. How long would it take to fill 6000? (6000*3)/380 = 18000/380 = 1800/38 = 900/19. That's right around 47. That's if all four holes are right at the bottom, but they aren't; we will be able to fill faster for most of the time, so it will take less than 47 hours. Eliminate C, D, and E.
Answer choice B.
Second.
Volume prior to the first hole is 30*20*2 = 1200m^3.
At a fill rate of (500/3)m^3 per hour, that will take 3600/500 = 36/5 = 7.2 hours.
Volume between the first and second holes is another 1200m^3.
Fill rate during this time is (500/3)-10 = (500/3)-(30/3) = 470/3.
At a fill rate of (470/3)m^3 per hour, that will take 3600/470 = 360/47. That's close enough to 360/48, which is easier to calculate. 360/48 = 7.5 hours.
Volume between the second and third holes is another 1200m^3.
Fill rate during this time is (500/3)-20 = (500/3)-(60/3) = 440/3.
At a fill rate of (440/3)m^3 per hour, that will take 3600/440 = 360/44 = 180/22 = 90/11, which is pretty close to 8.2 hours.
Volume between the third and fourth holes is another 1200m^3.
Fill rate during this time is (500/3)-30 = (500/3)-(90/3) = 410/3.
At a fill rate of (410/3)m^3 per hour, that will take 3600/410 = 360/41. That's close enough to 360/40, which is easier to calculate. 360/40 = 9 hours.
Volume between the fourth hole and the top is another 1200m^3.
Fill rate during this time is (500/3)-40 = (500/3)-(120/3) = 380/3.
At a fill rate of (380/3)m^3 per hour, that will take 3600/380 = 360/38. That's close to halfway between 340/40 and 360/36, so let's call it 9.5 hours.
7.2+7.5+8.2+9+9.5 = 41.4
Answer choice B.
General Discussion
10 Nov 2024, 12:02
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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