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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, 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
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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
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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
1/2c+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, 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
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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  \(\frac{1}{a} + \frac{1}{b} + \frac{1}{a} + \frac{1}{c} + \frac{1}{b} + \frac{1}{c}\) \(=\) \(\frac{5}{6} + \frac{2}{3} + \frac{1}{2}\) \(= \frac{36}{18}\) = 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, 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
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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, 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
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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
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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
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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
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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
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