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Pump A can empty a pool in A minutes, and pump B can empty
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Updated on: 24 Jan 2014, 06:19
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Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool? A. \(\frac{A+B1}{2}\) B. \(\frac{A(B+1)}{A+B}\) C. \(\frac{AB}{(A+B)}\) D. \(\frac{AB}{(A+B)} 1\) E. \(\frac{A(B1)}{(A+B)}\)
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Originally posted by gmatgambler on 24 Jan 2014, 06:13.
Last edited by Bunuel on 24 Jan 2014, 06:19, edited 2 times in total.
Edited the question.




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Re: Pump A can empty a pool in A minutes, and pump B can empty
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24 Jan 2014, 06:29
gmatgambler wrote: Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?
A. \(\frac{A+B1}{2}\)
B. \(\frac{A(B+1)}{A+B}\)
C. \(\frac{AB}{(A+B)}\)
D. \(\frac{AB}{(A+B)} 1\)
E. \(\frac{A(B1)}{(A+B)}\) The rate of pump A is \(\frac{1}{A}\) pool/minute; The rate of pump B is \(\frac{1}{B}\) pool/minute. Their combined rate is \(\frac{1}{A} + \frac{1}{B} = \frac{A+B}{AB}\) pool/minute. In one minute that A works alone it does (rate)*(time) = \(\frac{1}{A} * 1= \frac{1}{A}\) part of the job, thus \(1\frac{1}{A}=\frac{A1}{A}\) part of the job is remaining to be done by A and B together. To do \(\frac{A1}{A}\) of the job both pumps working together will need (time) = (job)/(combined rate) = \(\frac{A1}{A}*\frac{AB}{A+B}=\frac{(A1)B}{A+B}\). So, the total time is \(1+\frac{(A1)B}{A+B}=\frac{A(B+1)}{A+B}\). Answer: B.
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Re: Pump A can empty a pool in A minutes, and pump B can empty
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24 Jan 2014, 06:35



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Re: Pump A can empty a pool in A minutes, and pump B can empty
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30 Mar 2014, 14:32
I did this question using smart numbers
A = 6 B = 3
Therefore in 1 minute A did 1/6 of the job. 5/6 of the job remains
Together A and B work at a rate of 1/2.
Thus 1/2 (x) = 5/6
x = 5/3 minutes
Now we need to add the minute worked by A so total of 5/3 + 1 = 8/3 minutes
Replacing we can see that in (B)
(6)(4) / 9 = 8/3
Therefore it is the correct answer choice
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Re: Pump A can empty a pool in A minutes, and pump B can empty
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30 Mar 2014, 19:24
jlgdr wrote: I did this question using smart numbers
A = 6 B = 3
Therefore in 1 minute A did 1/6 of the job. 5/6 of the job remains
Together A and B work at a rate of 1/2.
Thus 1/2 (x) = 5/6
x = 5/3 minutes
Now we need to add the minute worked by A so total of 5/3 + 1 = 8/3 minutes
Replacing we can see that in (B)
(6)(4) / 9 = 8/3
Therefore it is the correct answer choice
Gimme some freaking Kudos! Cheers J Taking numbers is a great strategy but try to take easier numbers. Say, 1, 0 etc as long as you meet all conditions mentioned. I would take A = B = 1 (since there are no symmetric options in A and B such as A/(A+B) and B(A + B), I can take both A and B equal) Now I need to do no calculations. A starts and works for a minute. In that time, the pool is empty. B joins but has nothing to do. Total time taken to empty the pool is 1 min. Only option (B) gives 1.
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Re: Pump A can empty a pool in A minutes, and pump B can empty
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01 Apr 2014, 09:32
Method 1:The time taken by both pumps working together to do the work = AB/(A+B) minutes Now, since Pump A has already worked for 1 minute, 1*(1/A) of the work has already been done. Time taken to do the remaining work = [AB/(A+B)]*(11/A) = B(A1)/(A+B) minutes Total time taken from the start = [B(A1)/(A+B)] + 1 = A*(B+1)/(A+B) Option (B). Method 2:Let t be the time taken from the start. Then pump A works for t minutes to finish the work and pump B for (t1) minutes => t(1/A) + (t1)(1/B) = 1 => t = A*(B+1)/(A+B) Option (B). Method 3:We know that the time taken to do the work by both pumps working together right from the beginning is AB/(A+B) minutes. Lets flip the question around to say that the two pumps worked together to do the work, except pump B was off in the first minute. So we must add the time it would take the two pumps to do the work that pump B would have done in this one minute to the total time. In this one minute, pump B would have done 1*(1/B) of the work = 1/B of the work. Total time taken is therefore = AB/(A+B) + AB*(1/B)/(A+B) = A*(B+1)/(A+B) Option (B).
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Solve Work Problem using Numerical approach
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Updated on: 18 Apr 2014, 00:25
Q. Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool? Answer Choices: (a) A+B1/2 (b) A(B+1)/A+B (c) AB/A+B (d) AB/A+B 1/1 (e) A(B1)/A+B  METHOD ONE: Algebraic approach This approach to this question involves some tricky algebra. Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems  when two machines or people work together, we add the rates. In the first minute, pump A works alone and drains an amount of 1/A (that is, one "Ath" of a pool). This leaves an amount of The time it will take the two pumps, working at the combined rate, to drain this, is: That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!) Answer = B METHOD TWO: Numerical approach Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3. Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left. Then pump B kicks in  A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool? Add the first minute for total time. If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time? Only answer (B) works, so that the correct answer.
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Originally posted by royQV on 17 Apr 2014, 09:57.
Last edited by royQV on 18 Apr 2014, 00:25, edited 1 time in total.



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Re: Solve Work Problem using Numerical approach
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17 Apr 2014, 18:05
royQV wrote: Q. Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?
METHOD ONE: Algebraic approach
This approach to this question involves some tricky algebra.
Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems  when two machines or people work together, we add the rates.
In the first minute, pump A works alone and drains an amount of 1/A (that is, one "Ath" of a pool). This leaves an amount of
The time it will take the two pumps, working at the combined rate, to drain this, is:
That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!)
Answer = B
METHOD TWO: Numerical approach
Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3.
Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left.
Then pump B kicks in  A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool?
Add the first minute for total time.
If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time?
Only answer (B) works, so that the correct answer. Can you please post the OA? Rate of Pump A \(= \frac{1}{a}\) Rate of Pump B \(= \frac{1}{b}\) Work done by Pump A in 1 min \(= \frac{1}{a} * 1 = \frac{1}{a}\) Work remaining (To be done by both pumps) \(= 1  \frac{1}{a} = \frac{a1}{a}\) Combined rate of Pump A & B\(= \frac{1}{a} + \frac{1}{b}\) Time required to complete the remaining work = t \(\frac{a+b}{ab} * t = \frac{a1}{a}\) \(t = \frac{b(a1)}{a+b}\) Total time required = \(\frac{b(a1)}{a+b} + 1\) \(= \frac{a(b1)}{a+b}\)
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Re: Solve Work Problem using Numerical approach
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18 Apr 2014, 01:30
royQV wrote: Q. Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?
Answer Choices:
(a) A+B1/2
(b) A(B+1)/A+B
(c) AB/A+B
(d) AB/A+B 1/1
(e) A(B1)/A+B

METHOD ONE: Algebraic approach
This approach to this question involves some tricky algebra.
Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems  when two machines or people work together, we add the rates.
In the first minute, pump A works alone and drains an amount of 1/A (that is, one "Ath" of a pool). This leaves an amount of
The time it will take the two pumps, working at the combined rate, to drain this, is:
That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!)
Answer = B
METHOD TWO: Numerical approach
Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3.
Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left.
Then pump B kicks in  A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool?
Add the first minute for total time.
If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time?
Only answer (B) works, so that the correct answer. Merging similar topics. Please refer to the discussion above. Hope it helps. P.S. Please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Pay attention to rules 1, 3, and 8. Thank you.
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Re: Pump A can empty a pool in A minutes, and pump B can empty
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07 Aug 2017, 22:24
gmatgambler wrote: Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?
A. \(\frac{A+B1}{2}\)
B. \(\frac{A(B+1)}{A+B}\)
C. \(\frac{AB}{(A+B)}\)
D. \(\frac{AB}{(A+B)} 1\)
E. \(\frac{A(B1)}{(A+B)}\) Since A empties a pool in A minutes. In 1 minute A empties 1/A fraction of the pool. B empties a pool in B minutes. In 1 minute B empties 1/B fraction of the pool. So, 1/A + (1/A + 1/B)n = 1 where is n is the time in minutes for which A and B worked together. (A+b)/AB *n = 1 1/A n = AB/(A+B)  B /(A+B)= (AB B)/(A+B) SO, total time taken to empty a pool = 1+ (AB B)/(A+B) = (A+B+ ABB)/(A+B) = A(B+1)/(A+B) Answer B
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Re: Pump A can empty a pool in A minutes, and pump B can empty
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17 Aug 2017, 06:49
The fastest time in which tank can be emptied is by working together which is AB /(A+B)
But in this case it would be greater than AB/(A+B) since they have not worked together all time.
out of option B , C , D & E only option "B" is greater than AB /(A+B)



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Re: Pump A can empty a pool in A minutes, and pump B can empty
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