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Pumps A, B, and C operate at their respective constant rates. Pumps A
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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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Originally posted by chicagocubsrule on 10 Nov 2009, 14:14.
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
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11 Nov 2009, 04:52
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
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06 Mar 2010, 08:03
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
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Updated on: 10 Nov 2009, 14:49
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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Originally posted by Bunuel on 10 Nov 2009, 14:35.
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
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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
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20 Feb 2010, 07:07
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
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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
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20 Apr 2011, 14:33
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
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20 Apr 2011, 23:01
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
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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
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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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03 Jun 2020, 11:01
An alternative approach I prefer to use on most work/rate problems since I’m not a fan of fractions.
Assign the work a number of units. So 1 job in this case would equal 6 units work (based on LCM of the rates 5/6, 2/3 and 1/2)
Therefore multiplying the rates by these units of work gives us: A+B = 5 units of work per hour (5/6*6) B+C = 3 units of work per hour (1/2*6) C+A = 4 units of work per hour (2/3*6)
Adding all of them would give us 12 which is equal to 2A + 2B + 2C = 12 therefore A+B+C=6 units of work per hour
Or one could substitute by finding one of the values (for example A = 5B and B = 3C therefore plugging in third equation —> C + (5B) = 4 —> C + 5  (3C) = 4 —> C + 5  3 + C = 4 —> 2C + 2 = 4 —> C = 1 Adding this to the rate of A+B gives us A+B+C = 6 units of work per hour.
Now since one task was 6 units of work and combined rate is 6 units/hour therefore it would take one hour to complete the task.
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A
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04 Jun 2020, 23:15
My strategy was:
1) 1/A + 1/B =5/6 so 1/A= 5/61/A 2) 1/B + 1/C = 1/2 so 1/C= 1/2  1/B 3)1/A + 1/C= 2/3 this last equation ill use to find B
1/B=1/3 and i will use this value to find 1/A and 1/C
then I will add 1/A + 1/B + 1/C and the result will be 1



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A
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08 Jun 2020, 23:27
Ra = Rate at which pump A works Rb = Rate at which pump B works Rc = Rate at which pump C works
Work done by pump A & B , W = (Ra + Rb)* 6/5 Work done by pump A & C, W = (Ra+Rc) * 3/2 Work done by pump B & C, W = (Rb + Rc) * 2
Ra +Rb = 5/6 Ra + Rc = 2/3 Rb + Rc = 1/2
2*Ra + 2*Rb + 2*Rc = 5/6 + 2/3 + 1/2 Ra + Rb + Rc = 1/2[ 5/6 + 2/3 + 1/2 ] = 1/2 [ 1/3(5/2+2)+1/2] = 1/2[ 3/2 + 1/2] = 1/2[ 2] = 1
Thus, Ra + Rb + Rc = 1 Time taken by all the pump to fill the tank is obtain by
W = R*T T = R / W = 1 hour
IMO(D)



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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A
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03 Jul 2020, 04:43
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 let assume A, B, C be the rate of pipe if they fill the tank indvidually A+B= \(\frac{5}{6}\)eqn 1 A+C= \(\frac{2}{3}\)eqn 2 B+C= \(\frac{1}{2}\)eqn 3 now, adding all the 3 equations A+B+A+C+B+C=\(\frac{5}{6}\)+\(\frac{2}{3}\)+\(\frac{1}{2}\) 2(A+B+C)= 2 A+B+C= 1 ANS D




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