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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A
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02 Feb 2017, 03:53
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
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08 Feb 2017, 10:20
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
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14 Aug 2017, 14:16
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
1/2c+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
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25 Aug 2017, 08:44
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
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21 Nov 2017, 07:50
\(\frac{1}{a} + \frac{1}{b} = \frac{5}{6}\)\(\frac{1}{a} + \frac{1}{c} = \frac{2}{3}\)\(\frac{1}{b} + \frac{1}{c} = \frac{1}{2}\)When A, B and C are working together  \(\frac{1}{a} + \frac{1}{b} + \frac{1}{a} + \frac{1}{c} + \frac{1}{b} + \frac{1}{c}\) \(=\) \(\frac{5}{6} + \frac{2}{3} + \frac{1}{2}\) \(= \frac{36}{18}\) = 22 (\(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\)) = 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
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23 May 2018, 07:25
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
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07 Oct 2018, 22:35
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A
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04 Jan 2019, 06:38
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




Re: Pumps A, B, and C operate at their respective constant rates. Pumps A &nbs
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