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Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]

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10 Nov 2009, 15:14

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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.

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

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.

Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]

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06 Mar 2010, 08:42

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???
_________________

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.
_________________

Emily Sledge | Manhattan GMAT Instructor | St. Louis

Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]

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21 Apr 2011, 00:01

1

This post was BOOKMARKED

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.
_________________

Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]

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21 Apr 2011, 03:00

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..
_________________

Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]

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06 Jan 2012, 06:21

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.
_________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

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?

Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]

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04 Mar 2013, 00:04

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?

Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]

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27 Oct 2013, 04:13

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

Re: Pumps A, B, and C operate at their respective constant rates. Pumps A [#permalink]

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03 Nov 2016, 09:32

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)?

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)?