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  # Two water pumps, working simultaneously at their respective
  
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  Senior Manager
  Joined: 22 Sep 2005
  Posts: 262
  Two water pumps, working simultaneously at their respective  [#permalink]
  
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  Updated on: 12 Jan 2014, 04:44
  12
  00:00
  
  Difficulty:
  
  15% (low)
  
  Question Stats:
  
  84% (01:23) correct 16% (01:25) wrong based on 237 sessions
  
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  Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
  
  A. 5
  B. 16/3
  C. 11/2
  D. 6
  E. 20/3
  
  OPEN DISCUSSION OF THIS QUESTION IS HERE: two-water-pumps-working-simultaneously-at-their-respective-155865.html
  
  Originally posted by netcaesar on 14 Nov 2006, 12:56.
  Last edited by Bunuel on 12 Jan 2014, 04:44, edited 1 time in total.
  Renamed the topic, edited the question and added the OA.
  Manager
  Joined: 29 Aug 2006
  Posts: 156
  
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  14 Nov 2006, 13:06
  1
  4
  E from me...
  
  R(A) and T(A)= rate and time of 1st pump.
  R(B) and T(B)= rate and time of 2nd pump.
  
  R(A)= 1.5 R(B)
  
  Hence T(B)=1.5 T(A)
  
  1/T(A) + 1/T(B)= 1/4
  Substituting
  
  T(A)= 20/3
  Manager
  Joined: 12 Oct 2006
  Posts: 81
  Location: NJ, US
  
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  14 Nov 2006, 13:51
  E for me as well
  
  d=r*t
  
  1 = 7/2r*4
  
  1/10 = r
  
  Faster rate = 1/10*3/2 = 3/20
  
  1 = 3/20t
  
  20/3 = t
  Senior Manager
  Joined: 22 Sep 2005
  Posts: 262
  
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  17 Nov 2006, 11:31
  You are right.
  
  OA is E
  Senior Manager
  Joined: 20 Feb 2006
  Posts: 367
  
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  17 Nov 2006, 13:02
  3
  1
  Work = r * t
  
  Work A = 1 * 4
  
  Work B = 3/2 * 4
  
  Total work = (1*4) + (3/2 * 4) = 10
  
  10/rate B = 10/(3/2) or 10*2/3 = 20/3
  GMAT Club Legend
  Joined: 07 Jul 2004
  Posts: 4836
  Location: Singapore
  
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  19 Nov 2006, 18:49
  2
  Let the rate be:
  Pump A - 1 pool in A hours
  Pump B - 1 pool in B hours
  
  Then Pump A - 1/A pool in 1 hour
  And Pump B - 1/B pool in 1 hour
  
  Together: (A+B)/AB pool in 1 hour -> 1 pool in AB/(A+B) hours = 4 hours
  4A + 4B = AB -- (1)
  
  Assuming pump B is faster and A = 1.5B, then 4A+4B = AB becomes
  1.5B^2 = 10B
  B(1.5B - 10) = 0
  B = 0 or B = 20/3 hours
  
  So 1 pool in 20/3 hours
  
  Ans E
  Intern
  Joined: 06 Jan 2014
  Posts: 38
  Re: Two water pumps, working simultaneously at their respective  [#permalink]
  
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  11 Jan 2014, 15:09
  I felt like I understood "work" problems until I saw this one on my GMATPrep Exam.
  
  In my head the problem looks like this
  
  4hrs = (3/2)X + X
  
  or
  
  1/4 = (3/2)*(1/X) + (1/X)
  
  I simply dont know how to solve this problem....I'm confused!
  
  I've seen how others have solved this problem, but it just isn't clicking.
  
  Is someone willing to help?
  Math Expert
  Joined: 02 Sep 2009
  Posts: 51121
  Re: Two water pumps, working simultaneously at their respective  [#permalink]
  
  ### Show Tags
  
  12 Jan 2014, 04:47
  4
  7
  TroyfontaineMacon wrote:
  I felt like I understood "work" problems until I saw this one on my GMATPrep Exam.
  
  In my head the problem looks like this
  
  4hrs = (3/2)X + X
  
  or
  
  1/4 = (3/2)*(1/X) + (1/X)
  
  I simply dont know how to solve this problem....I'm confused!
  
  I've seen how others have solved this problem, but it just isn't clicking.
  
  Is someone willing to help?
  
  Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!
  
  Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
  
  A. 5
  B. 16/3
  C. 11/2
  D. 6
  E. 20/3
  
  Say the rate of the faster pump is x pool/hour, then the rate of the slower pump would be x/1.5=2x/3 pool/hour.
  
  Since, the combined rate is 1/4 pool/hour, then we have that x+2x/3=1/4 --> x=3/20 pool hour.
  
  The time is reciprocal of the rate, therefore it would take 20/3 hours the faster pump to fill the pool working alone.
  
  All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
  All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66
  
  OPEN DISCUSSION OF THIS QUESTION IS HERE: two-water-pumps-working-simultaneously-at-their-respective-155865.html
  _________________
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  Re: Two water pumps, working simultaneously at their respective  [#permalink]
  
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  14 Aug 2018, 23:32
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