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Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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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
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Last edited by Bunuel on 17 Feb 2016, 04:57, edited 2 times in total.
Edited the question and added the OA



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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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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Last edited by Bunuel on 10 Nov 2009, 14:49, edited 1 time in total.



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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10 Nov 2009, 14:53
Good technique..had another way but a few more unnecessary steps.



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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chicagocubsrule wrote: can you explain a little bit about adding the equations? Thanks Generally, if we are told that: A hours is needed for worker A (pump A etc.) to complete the job > the rate of A=\(\frac{1}{A}\); B hours is needed for worker B (pump B etc.) to complete the job > the rate of B=\(\frac{1}{B}\); C hours is needed for worker C (pump C etc.) to complete the job > the rate of C=\(\frac{1}{C}\); You can see that TIME to complete one job=Reciprocal of rate. eg 6 hours needed to complete one job (time) > 1/6 of the job done in 1 hour (rate). Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance. Time*Rate=Distance Time*Rate=JobAlso note that we can easily sum the rates:If we are told that A is completing one job in 2 hours and B in 3 hours, thus A's rate is 1/2 job/hour and B's rate is 1/3 job/hour. The rate of A and B working simultaneously would be 1/2+1/3=5/6 job/hours, which means that the will complete 5/6 job in hour working together. Time needed for A and B working simultaneously to complete the job=\(\frac{A*B}{A+B}\) hours, which is reciprocal of the sum of their respective rates. (General formula for calculating the time needed for two workers working simultaneously to complete one job). Time needed for A and C working simultaneously to complete the job=\(\frac{A*C}{A+C}\) hours. Time needed for B and C working simultaneously to complete the job=\(\frac{B*C}{B+C}\) hours. General formula for calculating the time needed for THREE workers working simultaneously to complete one job is: \(\frac{A*B*C}{AB+AC+BC}\) hours. Which is reciprocal of the sum of their respective rates: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\). We have three equations and three unknowns: 1. \(\frac{1}{A}+\frac{1}{B}=\frac{5}{6}\) 2. \(\frac{1}{A}+\frac{1}{C}=\frac{2}{3}\) 3. \(\frac{1}{B}+\frac{1}{C}=\frac{1}{2}\) Now the long way is just to calculate individually three unknowns A, B and C from three equations we have. But as we just need the reciprocal of the sum of relative rates of A, B and C, knowing the sum of \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{AB+AC+BC}{ABC}\) would be fine, we just take the reciprocal of it and bingo, it would be just the value we wanted. If we sum the three equations we'll get: \(2*\frac{1}{A}+2*\frac{1}{B}+2*\frac{1}{C}=\frac{5}{6}+\frac{2}{3}+\frac{1}{2}=2\) \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=1\), now we just need to take reciprocal of 1, which is 1. So the time needed for A, B, and C working simultaneously to complete 1 job is 1 hour. Hope it helps.
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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Ans e.
Let the rate of Pump A be a, Pump B b and Pump C c.
1/a + 1/b = 5/6
1/a + 1/c = 2/3
1/b + 1/c = 1/2
adding 2(1/a + 1/b + 1/c) = 5/6 + 2/3 + 1/2 = 5+4+3/6 = 2
=> 1/a + 1/b + 1/c = 1
or Pumps A,B and C take 1 hour to fill the tank.



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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06 Mar 2010, 07:42
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 We knw that a+b are taking 6/5 hrs i.e. 1.2 hrs to fill a tank... similarly B+C take 1.5 hrs and A+C take 2 hrs.... Adding all 3 equations we get A+b + B+c + A+C = 1.2+1.5+2 2A+2B+2C = 4.7hrs A+B+C = 2.35 hrs can u please guide where am I going wrong???
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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You are adding the times, when you should be adding the rates. Rates of: A+B = 5/6 (tank per hour) A+C = 2/3 (tank per hour) B+C = 1/2 (tank per hour) Combining these rates is like seeing "how much work would all of these pairs of machines do in 1 hour?" (A+B)+(A+C)+(B+C) = 5/6 + 2/3 + 1/2 = 5/6 + 4/6 + 3/6 = 12/6 (tank per hour) Therefore, 2A's, 2B's and 2C's working together would fill 2 tanks in an hour. A single A, B, and C working together would fill 1 tank in 1 hour.
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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Work done by A & B together in one hour = 5/6
Work done by B & C together in one hour = 2/3
Work done by C & A together in one hour = 1/2
=> 2(Work done by A , B & C in one hour) = 5/6 + 2/3 + 1/2
=> Work done by A , B & C in one hour = 1/2 [ 5/6 + 2/3 + 1/2] = 1
So together all three were able to finish the work in one hour.
Answer is E.



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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20 Apr 2011, 18:35
1/A + 1/B = 5/6 portion of tank in 1 hr 1/A + 1/C = 2/3 portion of tank in 1 hr 1/B + 1/C = 1/2 portion of tank in 1 hr 2(1/A + 1/B + 1/C) = 5/6 + 2/3 + 1/2 = (5 + 4 + 3)/6 = 2 So (1/A + 1/B + 1/C) = 1 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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Bunuel wrote: chicagocubsrule wrote: can you explain a little bit about adding the equations? Thanks Generally, if we are told that: A hours is needed for worker A (pump A etc.) to complete the job > the rate of A=\(\frac{1}{A}\); B hours is needed for worker B (pump B etc.) to complete the job > the rate of B=\(\frac{1}{B}\); C hours is needed for worker C (pump C etc.) to complete the job > the rate of C=\(\frac{1}{C}\); You can see that TIME to complete one job=Reciprocal of rate. eg 6 hours needed to complete one job (time) > 1/6 of the job done in 1 hour (rate). Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance. Time*Rate=Distance Time*Rate=JobAlso note that we can easily sum the rates:If we are told that A is completing one job in 2 hours and B in 3 hours, thus A's rate is 1/2 job/hour and B's rate is 1/3 job/hour. The rate of A and B working simultaneously would be 1/2+1/3=5/6 job/hours, which means that the will complete 5/6 job in hour working together. Time needed for A and B working simultaneously to complete the job=\(\frac{A*B}{A+B}\) hours, which is reciprocal of the sum of their respective rates. (General formula for calculating the time needed for two workers working simultaneously to complete one job). Time needed for A and C working simultaneously to complete the job=\(\frac{A*C}{A+C}\) hours. Time needed for B and C working simultaneously to complete the job=\(\frac{B*C}{B+C}\) hours. General formula for calculating the time needed for THREE workers working simultaneously to complete one job is: \(\frac{A*B*C}{AB+AC+BC}\) hours. Which is reciprocal of the sum of their respective rates: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\). We have three equations and three unknowns: 1. \(\frac{1}{A}+\frac{1}{B}=\frac{5}{6}\) 2. \(\frac{1}{A}+\frac{1}{C}=\frac{2}{3}\) 3. \(\frac{1}{B}+\frac{1}{C}=\frac{1}{2}\) Now the long way is just to calculate individually three unknowns A, B and C from three equations we have. But as we just need the reciprocal of the sum of relative rates of A, B and C, knowing the sum of \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{AB+AC+BC}{ABC}\) would be fine, we just take the reciprocal of it and bingo, it would be just the value we wanted. If we sum the three equations we'll get: \(2*\frac{1}{A}+2*\frac{1}{B}+2*\frac{1}{C}=\frac{5}{6}+\frac{2}{3}+\frac{1}{2}=2\) \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=1\), now we just need to take reciprocal of 1, which is 1. So the time needed for A, B, and C working simultaneously to complete 1 job is 1 hour. Hope it helps. Nice Explanation. Thanks for clearing the concepts.
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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21 Apr 2011, 02:00
Bunuel wrote: chicagocubsrule wrote: can you explain a little bit about adding the equations? Thanks Generally, if we are told that: A hours is needed for worker A (pump A etc.) to complete the job > the rate of A=\(\frac{1}{A}\); B hours is needed for worker B (pump B etc.) to complete the job > the rate of B=\(\frac{1}{B}\); C hours is needed for worker C (pump C etc.) to complete the job > the rate of C=\(\frac{1}{C}\); You can see that TIME to complete one job=Reciprocal of rate. eg 6 hours needed to complete one job (time) > 1/6 of the job done in 1 hour (rate). Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance. Time*Rate=Distance Time*Rate=JobAlso note that we can easily sum the rates:If we are told that A is completing one job in 2 hours and B in 3 hours, thus A's rate is 1/2 job/hour and B's rate is 1/3 job/hour. The rate of A and B working simultaneously would be 1/2+1/3=5/6 job/hours, which means that the will complete 5/6 job in hour working together. Time needed for A and B working simultaneously to complete the job=\(\frac{A*B}{A+B}\) hours, which is reciprocal of the sum of their respective rates. (General formula for calculating the time needed for two workers working simultaneously to complete one job). Time needed for A and C working simultaneously to complete the job=\(\frac{A*C}{A+C}\) hours. Time needed for B and C working simultaneously to complete the job=\(\frac{B*C}{B+C}\) hours. General formula for calculating the time needed for THREE workers working simultaneously to complete one job is: \(\frac{A*B*C}{AB+AC+BC}\) hours. Which is reciprocal of the sum of their respective rates: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\). We have three equations and three unknowns: 1. \(\frac{1}{A}+\frac{1}{B}=\frac{5}{6}\) 2. \(\frac{1}{A}+\frac{1}{C}=\frac{2}{3}\) 3. \(\frac{1}{B}+\frac{1}{C}=\frac{1}{2}\) Now the long way is just to calculate individually three unknowns A, B and C from three equations we have. But as we just need the reciprocal of the sum of relative rates of A, B and C, knowing the sum of \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{AB+AC+BC}{ABC}\) would be fine, we just take the reciprocal of it and bingo, it would be just the value we wanted. If we sum the three equations we'll get: \(2*\frac{1}{A}+2*\frac{1}{B}+2*\frac{1}{C}=\frac{5}{6}+\frac{2}{3}+\frac{1}{2}=2\) \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=1\), now we just need to take reciprocal of 1, which is 1. So the time needed for A, B, and C working simultaneously to complete 1 job is 1 hour. Hope it helps. Serioulsy.. Nice explanation.. It will clear the basics for solving Time and Rate questions..
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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Nice problem and nice solution. Sum all the given rates and divide by 2. Take the reciprocal of the result to get our required answer. Answer: 1 hourThanks Bunuel for amazing explanations.
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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This is an addition of rates problem A+B = 5/6 (NOTICE the inversion here) A+C = 2/3 B+C = 1/2 Add them, 2(A+B+C) = 12/6 A+B+C = 1
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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30 Jun 2012, 21:50
Given: Pumps A, B, and C operate at their respective constant rates. 1/A +1/ B = 5/6 1/A + 1/C = 2/3 1/B + 1/C = 1/2 Adding them up we get 2(1/A + 1/B + 1/C) = 5/6 + 2/3 + 1/2 = (5+4+3)/6 = 12/6 = 2 1/A + 1/B + 1/C = 1 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



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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13 Nov 2012, 22:22
\(\frac{1}{A}+\frac{1}{B}=\frac{5}{6}\) \(\frac{1}{A}+\frac{1}{C}=\frac{2}{3}\) \(\frac{1}{B}+\frac{1}{C}=\frac{1}{2}\) Note: time= reciprocal of rate and rate= reciprocal of time \(\frac{2}{A}+\frac{2}{B}+\frac{2}{C}=2\) \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{1}\) Answer: 1 hour
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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03 Mar 2013, 23:04
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+ B = 6/5 = 1.2 A+C = 3/2 = 1.5 B+C = 2 Sum all 2 (A+B+C) = formula for A+B+C = 1.2 * 1.5 * 2 / (1.5 * 2 + 1.2 * 2 + 1.5 * 1.2) = 1/2. For 2 (A+B+C) = 2 *1/2 = 1
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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27 Oct 2013, 03:13
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 Rate of A + B = 5/6 Rate of A + C = 2/3 Rate of B + C = 1/2 therefore 2(A + B + C) = 5/6 + 2/3 + 1/2 2(A + B + C) = 12/6 rate of A + B + C = 1 time is 1/rate = 1 hr



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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fastest approach... A+B+A+C+B+C = 2(A+B+C) this is equal to 5/6 +2/3 + 1/2 or 12/6 = 2. divide by 2 A+B+C will fill in 1 hour.



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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03 Nov 2016, 08:32
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) Bunuel  By this same logic, Why am I not able to solve by adding the Times of each of the combined machines and solving for combined rate (then taking the reciprocal to solve for time)? (A+B+A+C+B+C) = 6/5 + 3/2 + 2/1......2(A+B+C)=47/10 (A+B+C) = 47/20 With 3 full tanks done: Rate = 47/20 / 3 Rate = 47/60 Time = 60/47 I see this is not an answer choice, but why is my logic wrong here?



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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04 Nov 2016, 07:37
KFBR392 wrote: 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) Bunuel  By this same logic, Why am I not able to solve by adding the Times of each of the combined machines and solving for combined rate (then taking the reciprocal to solve for time)? (A+B+A+C+B+C) = 6/5 + 3/2 + 2/1......2(A+B+C)=47/10 (A+B+C) = 47/20 With 3 full tanks done: Rate = 47/20 / 3 Rate = 47/60 Time = 60/47 I see this is not an answer choice, but why is my logic wrong here? You CAN sum the rates but you CANNOT sum the times.
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A
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04 Nov 2016, 07:37



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