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Pumps A, B, and C operate at their respective constant rates. Pumps A

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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 02 Feb 2017, 04:53
1) A&B can do the job in \(\frac{6}{5}\) of an hour; and their rate when working together is \(\frac{1}{6/5}=\frac{5}{6}\)
A&C can do the job in \(\frac{3}{2}\) of an hour; and their rate when working together is \(\frac{1}{3/2}=\frac{2}{3}\)
B&C can do the job in 2 hours; and their rate when working together is \(\frac{1}{2}\)
2) If we combine all three rates \(\frac{5}{6}+\frac{2}{3}+\frac{1}{2}=\frac{5}{6}+\frac{4}{6}+\frac{3}{6}=\frac{12}{6}=2\) (2 tanks in an hour) then we have each machine represented twice. If we divide it by 2 (\(\frac{2}{2}=1\)), then we see that pumps A, B, and C when working together can fill one tank in one hour
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 08 Feb 2017, 11:20
chicagocubsrule wrote:
Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank.

A. 1/3
B. 1/2
C. 1/4
D. 1
E. 5/6


We are given that pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours. Recall that rate = work/time. If we consider the work (filling the tank) as 1, then the combined rate of pumps A and B is 1/(6/5). We can express this in the following equation in which a = the time it takes pump A to fill the tank when working alone (thus 1/a is pump A’s rate) and b = the time it takes pump B to fill the tank when working alone (thus 1/b is pump B’s rate):

1/a + 1/b = 1/(6/5)

1/a + 1/b = 5/6

We are next given that pumps A and C, operating simultaneously, can fill the tank in 3/2 hours. We can create another equation in which c = the time it takes pump C to fill the tank alone.

1/a + 1/c = 1/(3/2)

1/a + 1/c = 2/3

Finally, we are given that pumps B and C, operating simultaneously, can fill the tank in 2 hours. We can create the following equation:

1/b + 1/c = ½

Next we can add all three equations together:

(1/a + 1/b = 5/6) + (1/a + 1/c = 2/3) + (1/b + 1/c = ½)

2/a + 2/b + 2/c = 5/6 + 2/3 + 1/2

2/a + 2/b + 2/c = 5/6 + 4/6 + 3/6

2/a + 2/b + 2/c = 12/6

2/a + 2/b + 2/c = 2

Since we need to determine the combined rate of all three machines when filling 1 tank, we can multiply the above equation by ½:

(2/a + 2/b + 2/c = 2) x ½

1/a + 1/b + 1/c = 1

Since the combined rate of all 3 pumps is 1 and time = work/rate, the time needed to fill the tank when all 3 pumps are operating simultaneously is 1/1 = 1 hour.

Answer: D
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 14 Aug 2017, 15:16
corrected myself

a+b=5/6
a+c = 2/3
b+c = 1/2

so resolving by b
b= 1/2 -c
a = 2/3 - c
then first equation a+b=5/6 is

1/2-c+2/3 - c = 5/6
2c + 7/6 = 5/6
2c = 2/6
c= 1/6

then find others by substituting c into the rest equations
a = 1/2 b=1/3 and c=1/6 by summing them we get 1
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 25 Aug 2017, 09:44
When you look at the world do you just see one's and zero's 10101011 like the matrix? Solutions are ninja. Took me a while to figure this out.

Bunuel wrote:
chicagocubsrule wrote:
Pumps A, B, and C operate at their respective constant rates. Pumps A & B, operating simultaneously, can fill a certain tank in 6/5 hours; Pumps A & C, operating simultaneously, can fill the tank in 3/2 hours, and pumps B & C, operating simultaneously can fill the tank in 2 hours. How many hours does it take pumps A, B, & C, operating simultaneously, to fill the tank?

a) 1/3
b) 1/2
c) 1/4
d) 1
e) 5/6


A and B = 5/6 --> 1/A+1/B=5/6
A and C = 2/3 --> 1/A+1/C=2/3
B and C = 1/2 --> 1/B+1/C=1/2

Q 1/A+1/B+1/C=?

Add the equations: 1/A+1/B+1/A+1/C+1/B+1/C=5/6+2/3+1/2=2 --> 2*(1/A+1/B+1/A+1/C)=2 --> 1/A+1/B+1/A+1/C=1

Answer: D. (1)
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 21 Nov 2017, 08:50
\(\frac{1}{a} + \frac{1}{b} = \frac{5}{6}\)

\(\frac{1}{a} + \frac{1}{c} = \frac{2}{3}\)

\(\frac{1}{b} + \frac{1}{c} = \frac{1}{2}\)

When A, B and C are working together -

\(\frac{1}{a} + \frac{1}{b} + \frac{1}{a} + \frac{1}{c} + \frac{1}{b} + \frac{1}{c}\) \(=\) \(\frac{5}{6} + \frac{2}{3} + \frac{1}{2}\)

\(= \frac{36}{18}\)

= 2

2 (\(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\)) = 2

\(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\) = 1

Hence - Answer D
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 23 May 2018, 08:25
Using the percentages approach, which makes the calculation very easy.

Let RA be rate of pump A

Let RB be rate of pump B

Lett RC be rate of pump C

RA and RB takes 6/5hours which means that there combined rate is 83% (reciprocal of 6/5)
Similarly RB and RC = 50% (Reciprocal of 2)
and RA + RC = 66%(Reciprocal of 3/2)

The reciprocal tell you the amount of work the pumps will do in one hour if working together.

So this gives us three very simple equations

RA+RB=83%
RB+RC=50%
RA+RC=66%

Solving them we get
RA=50%
RB=33%
and RC=16%

Since we need to find the total time required when all three are working, we add the above and we get approximately 100%

This means that it will take 1 hour to do the work if all three pumps are working together.

What is great about this method is we can skip a lot of fractional equations and hence save time.

Cheers!
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 04 Jan 2019, 07:38
chicagocubsrule wrote:
Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank.

A. 1/3
B. 1/2
C. 1/4
D. 1
E. 5/6



Adding the respective rates given:

2 (1/A + 1/B + 1/C) = 5/6 + 2/3 + 1/2 = 2
(1/A + 1/B + 1/C) = 2/2 = 1 Hence, OA = D
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 03 Mar 2019, 17:36
If you write this as a r*t=W table it's very easy to notice that adding the combined rates gives 2 of each worker:
A+B --> r1 * 6/5 = 1
A+C --> r2 * 3/2 = 1
B+C --> r3 * 2 = 1

So you can add the 3 together (A+B+A+C+B+C) and get 2(A+B+C) = r1+r2+r3
We are looking for A+B+C = (r1+r2+r3)/2
A+B+C = (1/(6/5) + 1/(3/2) + 1/2)/2 --> (5/6 + 2/3 + 1/2)/2 --> (5/6 + 4/6 + 3/6)/2 --> 12/6/2 --> 12/12 = 1
So a combined rate of 1 means it takes 1 hour to do 1 Work, thus it's D.
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 31 May 2019, 08:15
I took total work as 12 units.
This means AB did 10 units in 1 hour
AC did 8 units and BC did 6 units per hour.
2(abc)=24
Abc=12 units.
Total work 12 units so it will take them 1 hour.
Am I doing it right?

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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 13 Jun 2019, 16:40
chicagocubsrule wrote:
Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank.

A. 1/3
B. 1/2
C. 1/4
D. 1
E. 5/6


Solution



Name, R, T, W
a+b, 5/6
a+c, 2/3
b+c, 1/2
Adding all of them: 2 (a+b+c) = 5/6 + 2/3 + 1/2 = (5+4+3)/6 = 2
a+b+c = 2/2 = 1
Time = 1/(a+b+c) = 1

ANSWER: D
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Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 24 Jun 2019, 09:15
I approached the problem by trying to solve for T = 1/Ra + Rb + Rc. (R= Rate)

Given Time of A and B = 6/5
W=RT
1=(Ra + Rb) * (6/5)
5/6=Ra + Rb
5/6 - Ra = Rb

Given Time of A and C = 3/2
W=RT
1=(Ra + Rc) * (3/2)
2/3=Ra + Rc
2/3 - Ra = Rc

Given Time of B and C = 2
W=RT
1=(Rb + Rc) * (2)
1/2=Rb + Rc
1/2 - Rb = Rc
Represent Rc in term of Ra so substitute Rb
1/2 - 5/6 - Ra = Rc

Now solve for T = 1/Ra + Rb + Rc. (R= Rate) ---This is where I ran into problems.

T=1/ (Ra +5/6 - Ra + 2/3 - Ra + 1/2 - 5/6 - Ra)
T=1/ (Ra +5/6 - Ra + 2/3 - Ra + 1/2 - 5/6 + Ra) ----Ra's cancel
T=1/ (5/6+ 2/3 + 1/2 - 5/6 )-----5/6 cancels
T=1/ (2/3+ 1/2 )
T = 1/ (7/6)
T= 6/7 which equals, the wrong answer.

The explanations above scrambled my brain. Can anyone help me unpack my error?

Bunuel can you opine?
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A  [#permalink]

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New post 22 Sep 2019, 08:34
just got this question in the official GMAT practice test. got it wrong and need to improve my work rates problems :(
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Re: Pumps A, B and C operate at their respective constant rates.  [#permalink]

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Re: Pumps A, B and C operate at their respective constant rates.   [#permalink] 09 Oct 2019, 05:55

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