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

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70% (02:36) correct
30% (01:43) wrong based on 822 sessions

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: Practice Test Question (Rates) [#permalink]
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

Re: Practice Test Question (Rates) [#permalink]
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.

What am I doing wrong ?? to Bunuel... [#permalink]
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??? _________________

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: Practice Test Question (Rates) [#permalink]
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. _________________

Re: Practice Test Question (Rates) [#permalink]
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.. _________________

Re: What am I doing wrong ?? to Bunuel... [#permalink]
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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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 [#permalink]
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?

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

Re: Pumps A, B, and C operate at their respective constant rates [#permalink]
07 May 2015, 17:05

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