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Pump A can empty a pool in A minutes, and pump B can empty [#permalink]
24 Jan 2014, 06:13
4
This post was BOOKMARKED
00:00
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Difficulty:
65% (hard)
Question Stats:
57% (03:11) correct
43% (02:23) wrong based on 191 sessions
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?
Re: Pump A can empty a pool in A minutes, and pump B can empty [#permalink]
24 Jan 2014, 06:29
2
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
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. \(\frac{A+B-1}{2}\)
B. \(\frac{A(B+1)}{A+B}\)
C. \(\frac{AB}{(A+B)}\)
D. \(\frac{AB}{(A+B)} -1\)
E. \(\frac{A(B-1)}{(A+B)}\)
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 \(1-\frac{1}{A}=\frac{A-1}{A}\) part of the job is remaining to be done by A and B together.
To do \(\frac{A-1}{A}\) of the job both pumps working together will need (time) = (job)/(combined rate) = \(\frac{A-1}{A}*\frac{AB}{A+B}=\frac{(A-1)B}{A+B}\).
So, the total time is \(1+\frac{(A-1)B}{A+B}=\frac{A(B+1)}{A+B}\).
Re: Pump A can empty a pool in A minutes, and pump B can empty [#permalink]
24 Jan 2014, 06:35
1
This post received KUDOS
Expert's post
Bunuel wrote:
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. \(\frac{A+B-1}{2}\)
B. \(\frac{A(B+1)}{A+B}\)
C. \(\frac{AB}{(A+B)}\)
D. \(\frac{AB}{(A+B)} -1\)
E. \(\frac{A(B-1)}{(A+B)}\)
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 \(1-\frac{1}{A}=\frac{A-1}{A}\) part of the job is remaining to be done by A and B together.
To do \(\frac{A-1}{A}\) of the job both pumps working together will need (time) = (job)/(combined rate) = \(\frac{A-1}{A}*\frac{AB}{A+B}=\frac{(A-1)B}{A+B}\).
So, the total time is \(1+\frac{(A-1)B}{A+B}=\frac{A(B+1)}{A+B}\).
Re: Pump A can empty a pool in A minutes, and pump B can empty [#permalink]
30 Mar 2014, 19:24
Expert's post
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. _________________
Re: Pump A can empty a pool in A minutes, and pump B can empty [#permalink]
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)]*(1-1/A) = B(A-1)/(A+B) minutes Total time taken from the start = [B(A-1)/(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 (t-1) minutes => t(1/A) + (t-1)(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). _________________
Solve Work Problem using Numerical approach [#permalink]
17 Apr 2014, 09:57
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+B-1/2
(b) A(B+1)/A+B
(c) AB/A+B
(d) AB/A+B -1/1
(e) A(B-1)/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 "A-th" 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. _________________
--------------------------
"The will to win, the desire to succeed, the urge to reach your full potential..."
Re: Solve Work Problem using Numerical approach [#permalink]
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 "A-th" 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) \(= 1 - \frac{1}{a} = \frac{a-1}{a}\)
Combined rate of Pump A & B\(= \frac{1}{a} + \frac{1}{b}\)
Re: Solve Work Problem using Numerical approach [#permalink]
18 Apr 2014, 01:30
Expert's post
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+B-1/2
(b) A(B+1)/A+B
(c) AB/A+B
(d) AB/A+B -1/1
(e) A(B-1)/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 "A-th" 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.
Re: Pump A can empty a pool in A minutes, and pump B can empty [#permalink]
23 Aug 2015, 19:20
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