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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. 2/3
D. 5/6
E. 1

    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

The question asks to find the value of 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: E.

Originally posted by Bunuel on 10 Nov 2009, 15:35.
Last edited by Bunuel on 18 Jan 2024, 13:26, edited 2 times 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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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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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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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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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: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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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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 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: E
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
found this great article on Magoosh for work/rate problems -
https://magoosh.com/gmat/gmat-work-rate-problems/
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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 .


We can also use LCM method to solve this time and work related question.

Let's assume the total Volume = LCM ( 6,3,2) = 6 L

*Note: if the time taken to complete the work are in fraction form, assume total work/volume as LCM of the numerators of the fractions.

Per work of pipe A and B = A+ B= 6/(6/5) = 5 L / hour
Per work of pipe A and C = A+ C= 6/(3/2) = 4 L / hour.

Per work of pipe B and C = B + C = 6/2 = 3 L/hour

On adding all three equations,
2( A + B + C ) = 5 + 4 + 3 = 12 L
A + B + C = 12/2 = 6L / hour.

So the time taken by pumps A, B, and C, operating simultaneously, to fill the tank = Total Volume/ Per hour work of (A,B and C) = 6/6 = 1 hour

Option E is the answer.

Thanks,
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
I din't saw anyone solving this way but it was my first thought

Let:
Rate A = A
Rate B = B
Rate C = C

The reciprocal of the Time is Rate, so:

A+B=5/6 --> B= 5/6 -A
A+C=2/3 --> C= 2/3 -A

So:
B+C=1/2 --> 5/6 -A + 2/3 -A = 1/2 --> A=1/2

Finally:
A+B+C=1/2 + 1/2 --> RATE of the 3 toghether is = 1

The reciprocal of the Rate is Time, so t=1

Option E

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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goncalogomes2000 wrote:
I din't saw anyone solving this way but it was my first thought

Let:
Rate A = A
Rate B = B
Rate C = C

The reciprocal of the Time is Rate, so:

A+B=5/6 --> B= 5/6 -A
A+C=2/3 --> C= 2/3 -A

So:
B+C=1/2 --> 5/6 -A + 2/3 -A = 1/2 --> A=1/2

Finally:
A+B+C=1/2 + 1/2 --> RATE of the 3 toghether is = 1

The reciprocal of the Rate is Time, so t=1

Option E

Hope it helps :)


When we have A+B, B+C and C+A and we are looking for A+B+C, an easier way is to add them all up and divide by 2 as done in many solutions before. Check them out too.
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Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]
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