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Pumps A, B, and C operate at their respective constant rates

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Pumps A, B, and C operate at their respective constant rates [#permalink] New post 10 Nov 2009, 14:14
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Mar 2012, 13:32, edited 1 time in total.
Edited the question and added the OA
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Re: Practice Test Question (Rates) [#permalink] New post 10 Nov 2009, 14:35
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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: Practice Test Question (Rates) [#permalink] New post 10 Nov 2009, 14:53
Good technique..had another way but a few more unnecessary steps.
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Re: Practice Test Question (Rates) [#permalink] New post 11 Nov 2009, 04:52
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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=Job


Also 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: GMAT Prep1: Pumps A,B,C [#permalink] New post 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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What am I doing wrong ?? to Bunuel... [#permalink] New post 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: What am I doing wrong ?? to Bunuel... [#permalink] New post 06 Mar 2010, 08:03
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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: What am I doing wrong ?? to Bunuel... [#permalink] New post 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: What am I doing wrong ?? to Bunuel... [#permalink] New post 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: Practice Test Question (Rates) [#permalink] New post 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=Job


Also 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: Practice Test Question (Rates) [#permalink] New post 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=Job


Also 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: What am I doing wrong ?? to Bunuel... [#permalink] New post 06 Jan 2012, 05:21
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 hour
Thanks Bunuel for amazing explanations.
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Re: What am I doing wrong ?? to Bunuel... [#permalink] New post 06 Jan 2012, 20:01
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 [#permalink] New post 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 [#permalink] New post 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 [#permalink] New post 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 [#permalink] New post 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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""HELP""Pumps A, B, and C operate at their respective rate. [#permalink] New post 19 Apr 2014, 05:16
I can't understand the reasoning behind this problem. Can you guys provide me an explanation?

Thanks you

Pumps A, B, and C operate at their respective rate. Pump A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill a certain tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill a certain tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?
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Re: ""HELP""Pumps A, B, and C operate at their respective rate. [#permalink] New post 19 Apr 2014, 08:28
Expert's post
plaintiff5 wrote:
I can't understand the reasoning behind this problem. Can you guys provide me an explanation?

Thanks you

Pumps A, B, and C operate at their respective rate. Pump A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill a certain tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill a certain tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?


Merging similar topics. Please refer to the discussion above.

Hope it helps.

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Re: ""HELP""Pumps A, B, and C operate at their respective rate.   [#permalink] 19 Apr 2014, 08:28
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