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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 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!
Re: Pumps A, B, and C operate at their respective constant rates. Pumps A &nbs [#permalink] 23 May 2018, 08:25

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