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One water pump can fill half of a certain empty tank in 3 hours. Anoth
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04 Jan 2012, 09:17
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Question Stats:
75% (01:51) correct 25% (02:08) wrong based on 491 sessions
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One water pump can fill half of a certain empty tank in 3 hours. Another pump can fill half of the same tank in \(3\frac{1}{2}\) hours. Working together, how long will it take these two pumps to fill the entire tank?
Re: One water pump can fill half of a certain empty tank in 3 hours. Anoth
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04 Jan 2012, 09:59
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36mba wrote:
One water pump can fill half of a certain empty tank in 3 hours. Another pump can fill half of the same tank in \(3\frac{1}{2}\) hours. Working together, how long will it take these two pumps to fill the entire tank?
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04 Jan 2012, 10:02
1st pump takes (3+3)=6 hours to fill the full tank 2nd pump takes (3.5+3.5)=7 hours to fill the full tank Therefore, 1/6 + 1/7= 1/T [T= time it will take if they work together to fill the tank] solving for T= 3 3/13=D
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04 Jan 2012, 10:45
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1st pump, 1/2 tank -> 3 hours, full -> 6 hours. This implies rate of 1/6 tank/hr 2nd pump, 1/2 tank -> 3.5 hours, full -> 7 hours. This implies rate of 1/7 tank/hr
Combined rate 1/6 + 1/7 = 13/42 For full tank -> 1/(13/42) = 42/13 = 3 3/13 -> D
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Re: One water pump can fill half of a certain empty tank in 3 hours. Anoth
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16 Feb 2015, 11:20
36mba wrote:
One water pump can fill half of a certain empty tank in 3 hours. Another pump can fill half of the same tank in \(3\frac{1}{2}\) hours. Working together, how long will it take these two pumps to fill the entire tank?
(A) \(1\frac{7}{13}\)
(B) \(1\frac{5}{8}\)
(C) \(3\frac{1}{4}\)
(D) \(3\frac{3}{13}\)
(E) \(3\frac{1}{2}\)
1/A = .5/3
1/B = .5/ 3.5
1/A+1/B = .5/3+.5/3.5 = 1/6+1/7 = 13/42
so rate is 13/42 Work =1 Time = 1/ (13/42) = 42/13
Re: One water pump can fill half of a certain empty tank in 3 hours. Anoth
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16 Feb 2015, 18:01
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Hi All,
This question is an example of a "Work Formula" question; since it involves just two entities (in this case, water pumps), we can us the Work Formula to quickly get to the correct answer.
We're told that 2 different water pumps can fill HALF of a tank in 3 hours and 3.5 hours, respectively. That means that the two pumps could fill the ENTIRE tank in 6 hours and 7 hours, respectively.
Work = (A)(B)/(A+B)
A = 6 hours B = 7 hours
(6)(7)/(6+7) = 42/13 = 3 3/13 hours to fill the entire tank when working together.
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30 Jan 2017, 03:05
1) First we need to find the rates of each pump individually: \(r*t=w, r1*3=1/2, r1=1/2*1/3=1/6; r2*7/2=1/2, r2=1/2*2/7=1/7\). 2) Then we need to combine the two rates, since the pumps are working simultaneously: \(1/6+1/7=13/42\) 3) Now we can find the time it takes the two of them to fill the tank: \(13/42*T=1\); \(T=42/13=3 \frac{3}{13}\)
Re: One water pump can fill half of a certain empty tank in 3 hours. Anoth
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02 Feb 2017, 11:07
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36mba wrote:
One water pump can fill half of a certain empty tank in 3 hours. Another pump can fill half of the same tank in \(3\frac{1}{2}\) hours. Working together, how long will it take these two pumps to fill the entire tank?
(A) \(1\frac{7}{13}\)
(B) \(1\frac{5}{8}\)
(C) \(3\frac{1}{4}\)
(D) \(3\frac{3}{13}\)
(E) \(3\frac{1}{2}\)
We are given that a water pump can fill half of a certain tank in 3 hours; thus, the rate of the pump is (1/2)/3 = 1/6. We are given that another pump can fill 1/2 of the same tank in 3½, or 7/2, hours. Thus, the rate of the second pump is (1/2)/(7/2) = 2/14 = 1/7. If we let t = the time it takes the two pumps working together to fill the entire tank, and if we let 1 equal the work (i.e., filling the entire tank) needed to be completed, we can create the following equation and determine t:
(1/6)t + (1/7)t = 1
Multiplying the entire equation by 42, we have:
7t + 6t = 42
13t = 42
t = 42/13 = 3 3/13
Answer: D
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