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At their respective rates, pump A, B, and C can fulfill an

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At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post Updated on: 29 Sep 2013, 09:37
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At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full tank, while C is used to fill water into the tank, in how many hours, the tank will be empty?

A. 2/3
B. 1
C. 3/4
D. 3/2
E. 2

Originally posted by Economist on 02 Apr 2009, 20:12.
Last edited by Bunuel on 29 Sep 2013, 09:37, edited 1 time in total.
Renamed the topic and edited the question.
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Re: PS: Pool fills  [#permalink]

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New post 03 Apr 2009, 07:51
10
13
Economist wrote:
At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full tank, while C is used to fill water into the tank, in how many hours, the tank will be empty?
A. 2/3
B. 1
C. 3/4
D. 3/2
E. 2


You can use a rates formula here, or convert each worker to the same amount of time (6 hours) :

A empties 3 tanks every 6 hours
B empties 2 tanks every 6 hours
C fills 1 tank every 6 hours
Together, they empty 4 tanks every 6 hours
So they empty 1 tank every 6/4 hours, and empty half of a tank in 3/4 hours. C.
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Re: PS: Pool fills  [#permalink]

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New post 16 Aug 2009, 12:09
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12.At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full tank, while C is used to fill water into the tank, in how many hours, the tank will be empty?
A. 2/3
B. 1
C. 3/4
D. 3/2
E. 2

let X = # of hours for the tank to empty

X*Rate A + X*Rate B - X*Rate C = 1/2 tank

X*(1/2) + X*(1/3) - X*(1/6) = 1/2
X = 3/4

It takes 3/4 hrs to empty the tank

ANSWER: C. 3/4
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Re: PS: Pool fills  [#permalink]

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New post 17 Aug 2009, 22:46
IanStewart wrote:
You can use a rates formula here, or convert each worker to the same amount of time (6 hours) :

A empties 3 tanks every 6 hours
B empties 2 tanks every 6 hours
C fills 1 tank every 6 hours
Together, they empty 4 tanks every 6 hours
So they empty 1 tank every 6/4 hours, and empty half of a tank in 3/4 hours. C.


that looks so simple and easy.. thanks Ian.
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Re: PS: Pool fills  [#permalink]

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New post 17 Aug 2009, 23:17
Quote:
You can use a rates formula here, or convert each worker to the same amount of time (6 hours) :

A empties 3 tanks every 6 hours
B empties 2 tanks every 6 hours
C fills 1 tank every 6 hours
Together, they empty 4 tanks every 6 hours
So they empty 1 tank every 6/4 hours, and empty half of a tank in 3/4 hours. C.


Good job Ian. You made it all simple. :)
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Re: PS: Pool fills  [#permalink]

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New post 24 Aug 2009, 23:30
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Effective rate(empty) : 1/2+1/3-1/6 = 4/6 = 2/3

So whole tank will be emptied in 3/2 hrs..and half will be in 3/4 hrs. C...
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Re: PS: Pool fills  [#permalink]

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New post 28 Sep 2013, 10:53
5
Rate(A)=1/2
Rate(B)=1/3
Rate(C)=1/6

A + B - C

1/2 + 1/3 - 1/6 = 4/6 (Combined rate per the question)

Rate * Time = Work

4/6 * Time = 1/2

Time = 3/4
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Re: PS: Pool fills  [#permalink]

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New post 12 Oct 2013, 00:14
TGC wrote:
Rate(A)=1/2
Rate(B)=1/3
Rate(C)=1/6

A + B - C

1/2 + 1/3 - 1/6 = 4/6 (Combined rate per the question)

Rate * Time = Work

4/6 * Time = 1/2

Time = 3/4


Hi,

I could understand this method.
But my doubt is
Work = Rate * time
Time = Work / Rate

So as per this problem
Rate(A)=1/2
Rate(B)=1/3
Rate(C)=1/6

So total time is:
(1/2)/(1/2) + (1/2) / (1/3) - (1/2) / (1/6) = -1/2

Where iam missing something?? Please help me

thanks,
Rrsnathan.
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Re: PS: Pool fills  [#permalink]

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New post 17 Oct 2013, 03:53
3
1
rrsnathan wrote:
TGC wrote:
Rate(A)=1/2
Rate(B)=1/3
Rate(C)=1/6

A + B - C

1/2 + 1/3 - 1/6 = 4/6 (Combined rate per the question)

Rate * Time = Work

4/6 * Time = 1/2

Time = 3/4


Hi,

I could understand this method.
But my doubt is
Work = Rate * time
Time = Work / Rate

So as per this problem
Rate(A)=1/2
Rate(B)=1/3
Rate(C)=1/6

So total time is:
(1/2)/(1/2) + (1/2) / (1/3) - (1/2) / (1/6) = -1/2


Where iam missing something?? Please help me

thanks,
Rrsnathan.


When adding rates you get combined rate not time. Also, why are you multiplying combined rate by 1/2?

At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full tank, while C is used to fill water into the tank, in how many hours, the tank will be empty?
A. 2/3
B. 1
C. 3/4
D. 3/2
E. 2

Rate of A = 1/2 tank/hour;
Rate of B = 1/3 tank/hour;
Rate of C = 1/6 tank/hour.

Combined rate when A and B are used to pump-out water, while C is used to fill water into the tank is 1/2+1/3-1/6=2/3 tank hour.

So, to empty the full tank 3/2 hours (reciprocal of rate) are needed. To empty the half-full tank half of that time would be needed: 1/2*3/2=3/4 hours.

Answer: C.

Hope it's clear.
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Re: At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post 26 Oct 2013, 10:08
t*[C's rate - (A's rate+B's rate)] + current tank size = 0
t*[1/6 - (1/2+1/3)] + .5 = 0
t = 3/4
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Re: At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post 04 Apr 2014, 02:16
2
Effective rate of A, B & C respectively

\(= \frac{1}{2} + \frac{1}{3} - \frac{1}{6}\)
(-ve as tank is getting empty)

\(= \frac{3+2-1}{6}\)

\(= \frac{4}{6}\)

\(= \frac{2}{3}\)

Time required for full tank \(= \frac{3}{2}\)

\(Time required for half tank = \frac{3}{2} * \frac{1}{2} = \frac{3}{4}\)

Answer = C
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Re: At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post 21 Sep 2015, 03:16
Time required to half the tank by each = (time for full tank)/2
Therefore
A=1hour
B=3/2 hour
C=6/2 hour
Now A and B empties and C adds
=> 1/A + 1/B - 1/C = 1+2/3-2/6 =8/6=4/3
hence 3/4 ans
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Re: At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post 13 Oct 2015, 06:19
Bunuel wrote:
rrsnathan wrote:
TGC wrote:
Rate(A)=1/2
Rate(B)=1/3
Rate(C)=1/6

A + B - C

1/2 + 1/3 - 1/6 = 4/6 (Combined rate per the question)

Rate * Time = Work

4/6 * Time = 1/2

Time = 3/4


Hi,

I could understand this method.
But my doubt is
Work = Rate * time
Time = Work / Rate

So as per this problem
Rate(A)=1/2
Rate(B)=1/3
Rate(C)=1/6

So total time is:
(1/2)/(1/2) + (1/2) / (1/3) - (1/2) / (1/6) = -1/2


Where iam missing something?? Please help me

thanks,
Rrsnathan.


When adding rates you get combined rate not time. Also, why are you multiplying combined rate by 1/2?

At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full tank, while C is used to fill water into the tank, in how many hours, the tank will be empty?
A. 2/3
B. 1
C. 3/4
D. 3/2
E. 2

Rate of A = 1/2 tank/hour;
Rate of B = 1/3 tank/hour;
Rate of C = 1/6 tank/hour.

Combined rate when A and B are used to pump-out water, while C is used to fill water into the tank is 1/2+1/3-1/6=2/3 tank hour.

So, to empty the full tank 3/2 hours (reciprocal of rate) are needed. To empty the half-full tank half of that time would be needed: 1/2*3/2=3/4 hours.

Answer: C.

Hope it's clear.


how did the reciprocal come. I mean 2/3 tank per hour into 3/2 I did not get that logic
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At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post 25 Nov 2015, 11:17
Let’s say work = 12 then:
Rate A = 6
Rate B = 4
Rate C = 2
There combined rate \(= 6+4-2=8\) and 50% of the tank \(=\frac{12}{2}=6\), \(hence 8*t=6, \frac{t=3}{4}\)
Answer C
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Re: At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post 25 Nov 2015, 12:00
Economist wrote:
At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full tank, while C is used to fill water into the tank, in how many hours, the tank will be empty?

A. 2/3
B. 1
C. 3/4
D. 3/2
E. 2


Let the volume of the tank be 12 units

Efficiency of tank A = -6 { -ve since they are pumpung out water}
Efficiency of tank B = -4 { -ve since they are pumpung out water}
Efficiency of tank C = 2

We are given the information that the tank is half filled, so 6 units is filled

Net/Effective work done by the 3 pipes is (-6)+ (-4) +(2) =>-8 units

So, the three pipes will drain out water and the time required to do the same from the half filled tank is 6/8 => 3/4

Answer is (C)

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At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post 25 Nov 2015, 18:16
t=time to empty tank
t(1/6-5/6)=-1/2➡t(-2/3)=-1/2
t=3/4 hour
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Re: At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post 06 Dec 2015, 12:40
lets set the equation as:
water in the tank + water pump in = water pump out

This will give you the following:
1/2 + 1/6(T) = (1/2)T + (1/3)T where T is the required time so that the tank is empty
solve for T,
you will get T = 3/4 hr

so the answer is C
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Re: At their respective rates, pump A, B, and C can fulfill an  [#permalink]

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New post 28 Mar 2018, 10:35
Economist wrote:
At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full tank, while C is used to fill water into the tank, in how many hours, the tank will be empty?

A. 2/3
B. 1
C. 3/4
D. 3/2
E. 2


We can calculate the net outflow rate (i.e., the sum of the rates of A and B minus the rate of C) as:

1/2 + 1/3 - 1/6 = 3/6 + 2/6 - 1/6 = 4/6 = 2/3

Since only ½ of a tank needs to be emptied, the time it takes to do that is (1/2)/(2/3) = 1/2 x 3/2 = 3/4 hour.

Answer: C
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Re: At their respective rates, pump A, B, and C can fulfill an &nbs [#permalink] 28 Mar 2018, 10:35
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