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Working alone at its constant rate, pump X pumped out ¼ of the water
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03 Apr 2016, 15:13
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Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank? A) 6 B) 12 C) 15 D) 18 E) 24 I had trouble solving these kind of questions in less than 3 minutes. If you guys have any inputs on how to solve it quickly, please share them.
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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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03 Apr 2016, 18:28
SVTTCGMAT wrote: Working alone at it's constant rate, pump X pumped out 1/4 of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at it's constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at it's constant rate, to pump out all of the water that was pumped out of the tank? A) 6 B) 12 C) 15 D) 18 E) 24
I had trouble solving these kind of questions in less than 3 minutes. If you guys have any inputs on how to solve it quickly, please share them. Hi, when you look at the Q, you realize speed of all three combined and speed of two pump individually is given.. convert all into time for entire work1) pump Xit fills up 1/4 in 2 hr, so it will fill up entire into 2*4= 8 hr 1 hr work = 1/82) combinedcombined three do 11/4 = 3/4 in 3 hr, so all three will fill entire tank in 3*4/3=4hr 1 hr work = 1/43) pump ypump y can do 3/4 in 18 hr so it will do complete in 18*4/3=24h... 1 hr work = 1/24combined 1 hr work should be sum of 1 hr work of x,y and z so\(\frac{1}{4}= \frac{1}{8} +\frac{1}{24} +\frac{1}{z}\).. \(\frac{1}{z}=\frac{1}{4} (\frac{1}{8} + \frac{1}{24}) =\frac{(631)}{24}= \frac{2}{24} =\frac{1}{12}\)... so z can do it in 12 hrB
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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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03 Apr 2016, 18:17
Rate of pump X = 1/8
3 hours are required to pump out the remaining (3/4)ths of tank > 1 hr to pump out 1/4
Rate of X + Rate of Y + Rate of Z = 1/4
Rate of Y + Rate of Z = 1/4  1/8 = 1/8
Y takes 18 hours to pump out the remaining (3/4)ths of tank > 6 hrs per (1/4)ths > 24 hrs to pump out fully.
Rate of Y = 1/24
1/24 + Rate of Z = 1/8
Rate of Z = 1/8  1/24 = 1/12
Time required to pump out all the water by Z = 12 hrs
Answer: B




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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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14 Feb 2017, 21:28
Hi Vyshak, chetan2u & BunuelCan you please point out the flaw in my approach ? X does 1/4th of work in 2 hours. So X does unit work in 8 hrs.Now 3/4th work remains. X,Y & Z do 3/4th work in 3 hours. So, when X,Y & Z work together, they do unit work in 4 hours. (3*(4/3)) Y individually does unit work in 18 hrs.Combining the data colored into below equation, (XYZ) / (XY + YZ + XZ) = 4 ; where X,Y & Z are the individual times in which work is done (8*18*Z) / (8*18 + 18*Z + 8*Z) = 4. Z comes out to be 14.4 ~ 15. SVTTCGMAT wrote: Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?
A) 6
B) 12
C) 15
D) 18
E) 24
I had trouble solving these kind of questions in less than 3 minutes. If you guys have any inputs on how to solve it quickly, please share them.



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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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15 Feb 2017, 02:23
ajay2121988 wrote: Can you please point out the flaw in my approach ?
X does 1/4th of work in 2 hours. So X does unit work in 8 hrs.
Now 3/4th work remains. X,Y & Z do 3/4th work in 3 hours. So, when X,Y & Z work together, they do unit work in 4 hours. (3*(4/3))
Y individually does unit work in 18 hrs.
Combining the data colored into below equation, (XYZ) / (XY + YZ + XZ) = 4 ; where X,Y & Z are the individual times in which work is done
(8*18*Z) / (8*18 + 18*Z + 8*Z) = 4. Z comes out to be 14.4 ~ 15.
Y takes 18 hours to perform (3/4)th of work > Y takes 24 hours to do the full work alone.



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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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15 Feb 2017, 04:32
SVTTCGMAT wrote: Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?
A) 6
B) 12
C) 15
D) 18
E) 24
I had trouble solving these kind of questions in less than 3 minutes. If you guys have any inputs on how to solve it quickly, please share them. Take one line of the question at a time: "pump X pumped out ¼ of the water in a tank in 2 hours." 1/4th was pumped out in 2 hrs. So entire 4/4 tank would be pumped out in 4*2= 8 hrs. Hence rate of work of X is 1/8th tank every hr. "Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours." The 3 pumped 3/4th tank in 3 hrs. So they pumped 1/4th tank every hour. Note here that the rate of tank X is 1/8 and that of all 3 tanks is 1/4. So basically rate of tank Y + rate of tank Z is 1/8 tank per hour. "If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water," Rate of pump Y = (3/4)/18 = 1/24 So out of a total rate of 1/8 for pumps Y and Z, 1/24 (i.e. a third) belongs to pump Y. So two thirds i.e. 2/24 = 1/12 would belong to pump Z. So pump Z will take 12 hrs to pump out the water. Answer (B)
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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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16 Feb 2017, 11:13
SVTTCGMAT wrote: Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?
A) 6
B) 12
C) 15
D) 18
E) 24 We are given that Working alone at its constant rate, pump X pumped out 1/4 of the water in a tank in 2 hours. Since rate = work/time, the rate of pump X is (1/4)/2 = 1/8. Since 1/4 of the water is pumped out of the tank, 3/4 is left to be pumped out. We are also given that all 3 pumps pumped the remaining 3/4 of the water out in 3 hours; thus the combined rate of all three pumps is (3/4)/3 = 3/12 = 1/4. We are finally given that pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water. Thus, the rate of pump Y = (3/4)/18 = 3/72 = 1/24. If we let z = the time it takes pump Z to pump out all the water, then the rate of pump Z = 1/z and create the following equation: 1/8 + 1/24 + 1/z = 1/4 Multiplying the entire equation by 24z gives us: 3z + z + 24 = 6z 24 = 2z 12 = z Answer: B
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Working alone at its constant rate, pump X pumped out ¼ of the water
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16 Feb 2017, 17:52
SVTTCGMAT wrote: Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?
A) 6
B) 12
C) 15
D) 18
E) 24 let z=time of z to pump out entire tank alone rate of x=(1/4)/2=1/8 rate of y=(3/4)/18=1/24 3(1/8+1/24+1/z)=3/4 z=12 hours B



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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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07 Apr 2017, 18:47
Rate of pump X = (1/4)/2 = 1/8 of the water per hour Remaining amount = 1  1/4 = 3/4 (1/X)+(1/Y)+(1/Z)=(3/4)/3=3/12=1/4 of the water per hour 1/Y=(3/4)/18=3/72=1/24 1/Z= (1/4)  (1/24)  (1/8) = (6/24)  (3/24)  (1/24) = 2/24 = 1/12 Time = 12h
Kudos if you agree or comment if you have a better method!



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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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05 Dec 2017, 04:10
SVTTCGMAT wrote: Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?
A) 6
B) 12
C) 15
D) 18
E) 24 Solving this question quickly requires to go through each statement thoroughly and interpret an equation from all of those. I am following the same strategy to solve this question, Quote: Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. This means that pump X can pump out the entire water in a tank in 8 hours. Quote: Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. The rest of the water here means 3/4th of the water in the tank. Thus, the complete water will be pumped out by (x+y+z) in 4 hours. Quote: If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water. This means that 3/4th of the water in the tank is pumped out in 18 hours. Thus, the entire water will be pumped out by Y in 24 hours. Q: how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?1/X + 1/Y + 1/Z = 1/4 1/8 + 1/24 + 1/Z = 1/4 1/Z = 1/4  1/8  1/24 = (6  3 1)/24 = 2/24 = 1/12 Thus, 12 hours is the answer! Bingo.



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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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03 Jan 2018, 17:00
in 1 hr pump x will pump 1/4. In 3 hr it will pump 3/4. Therefore, in 3hr Y will pump 1/4 or 25%
So, To fill 1/4 or 25% pump Y takes 3 hrs. For filling 100% it will take 12 hrs. The actual rate will be 2*12 = 24hrs. ( We are multiplying by 2 because the calculations above is for half of the tank)



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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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01 Mar 2018, 18:13
ScottTargetTestPrep wrote: SVTTCGMAT wrote: Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?
A) 6
B) 12
C) 15
D) 18
E) 24 We are given that Working alone at its constant rate, pump X pumped out 1/4 of the water in a tank in 2 hours. Since rate = work/time, the rate of pump X is (1/4)/2 = 1/8. Since 1/4 of the water is pumped out of the tank, 3/4 is left to be pumped out. We are also given that all 3 pumps pumped the remaining 3/4 of the water out in 3 hours; thus the combined rate of all three pumps is (3/4)/3 = 3/12 = 1/4. We are finally given that pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water. Thus, the rate of pump Y = (3/4)/18 = 3/72 = 1/24. If we let z = the time it takes pump Z to pump out all the water, then the rate of pump Z = 1/z and create the following equation: 1/8 + 1/24 + 1/z = 1/4 Multiplying the entire equation by 24z gives us: 3z + z + 24 = 6z 24 = 2z 12 = z Answer: B Everything makes sense except for the 1/4 at the end. Where did that come from? The question is asking how many hours it would've taken pump z to pump ALL the water, so wouldn't that be a whole number? 1? 1/8 + 1/24 + 1/z = 1? Why would it be 1/4 instead? That is only a quarter of the water (which is how much X pumped out)... Please help!



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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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01 Mar 2018, 21:35
OCDianaOC wrote: ScottTargetTestPrep wrote: SVTTCGMAT wrote: Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?
A) 6
B) 12
C) 15
D) 18
E) 24 We are given that Working alone at its constant rate, pump X pumped out 1/4 of the water in a tank in 2 hours. Since rate = work/time, the rate of pump X is (1/4)/2 = 1/8. Since 1/4 of the water is pumped out of the tank, 3/4 is left to be pumped out. We are also given that all 3 pumps pumped the remaining 3/4 of the water out in 3 hours; thus the combined rate of all three pumps is (3/4)/3 = 3/12 = 1/4. We are finally given that pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water. Thus, the rate of pump Y = (3/4)/18 = 3/72 = 1/24. If we let z = the time it takes pump Z to pump out all the water, then the rate of pump Z = 1/z and create the following equation: 1/8 + 1/24 + 1/z = 1/4 Multiplying the entire equation by 24z gives us: 3z + z + 24 = 6z 24 = 2z 12 = z Answer: B Everything makes sense except for the 1/4 at the end. Where did that come from? The question is asking how many hours it would've taken pump z to pump ALL the water, so wouldn't that be a whole number? 1? 1/8 + 1/24 + 1/z = 1? Why would it be 1/4 instead? That is only a quarter of the water (which is how much X pumped out)... Please help! Hi Here's how the 1/4 on the Right hand side comes from: In 2 hours, X pumped out 1/4 of water. So how much water is left (in fraction) = 3/4 (threefourth). Now all three pumps (X, Y, Z) pumped out remaining water (3/4 th) in 3 hours. See, since they pumped 3/4 th water in 3 hours, it means they will pump out 1/4 th water in 1 hour, OR you could say together the three pumps will pump out complete water in 1*4 = 4 hours (1/4 water in = 1 hour full water in = 1*4 = 4 hours) This means, when X, Y, Z are working together  they take 4 hours to pump out full tank, Or you could say they pump out 1/4 of water in 1 hour. Thus, their respective waters pumped out in 1 hour would equal to 1/4 of tank drained out. So, per hour work of X + per hour work of Y + per hour work of Z = 1/4



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Re: Working alone at its constant rate, pump X pumped out ¼ of the water
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02 Mar 2018, 13:53
Oooh, I understand now. Thanks! We're taking the 1/4 answer that we previously got when we combined X, Y, and Z... Thanks amanvermagmat!




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