Author 
Message 
TAGS:

Hide Tags

Current Student
Joined: 26 May 2015
Posts: 33
Location: United States (CT)
Concentration: Technology, Entrepreneurship
WE: Information Technology (Consumer Products)

Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
03 Apr 2016, 16:13
Question Stats:
69% (01:37) correct 31% (01:38) wrong based on 828 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Aug 2009
Posts: 6785

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
03 Apr 2016, 19:28
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
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor




SC Moderator
Joined: 13 Apr 2015
Posts: 1706
Location: India
Concentration: Strategy, General Management
GPA: 4
WE: Analyst (Retail)

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
03 Apr 2016, 19:17
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




Manager
Joined: 08 Feb 2016
Posts: 73
Location: India
Concentration: Technology
GPA: 4

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
14 Feb 2017, 22:28
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.



SC Moderator
Joined: 13 Apr 2015
Posts: 1706
Location: India
Concentration: Strategy, General Management
GPA: 4
WE: Analyst (Retail)

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
15 Feb 2017, 03:23
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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8281
Location: Pune, India

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
15 Feb 2017, 05:32
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)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3497
Location: United States (CA)

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
16 Feb 2017, 12:13
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
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



VP
Joined: 07 Dec 2014
Posts: 1087

Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
16 Feb 2017, 18:52
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



Manager
Joined: 13 Dec 2013
Posts: 161
Location: United States (NY)
Concentration: Nonprofit, International Business
GMAT 1: 710 Q46 V41 GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
07 Apr 2017, 19:47
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!



Manager
Status: Aiming MBA!!
Joined: 19 Aug 2017
Posts: 126
Location: India
GPA: 3.75
WE: Web Development (Consulting)

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
05 Dec 2017, 05:10
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.



Manager
Joined: 11 Sep 2013
Posts: 162
Concentration: Finance, Finance

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
03 Jan 2018, 18:00
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)



Intern
Joined: 16 Oct 2017
Posts: 38

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
01 Mar 2018, 19:13
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!



DS Forum Moderator
Joined: 22 Aug 2013
Posts: 1343
Location: India

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
01 Mar 2018, 22:35
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



Intern
Joined: 16 Oct 2017
Posts: 38

Re: Working alone at its constant rate, pump X pumped out ¼ of the water
[#permalink]
Show Tags
02 Mar 2018, 14:53
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!




Re: Working alone at its constant rate, pump X pumped out ¼ of the water &nbs
[#permalink]
02 Mar 2018, 14:53






