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Pumping alone at their

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boomgoesthegmat Senior | Next Rank: 100 Posts Default Avatar
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Pumping alone at their

Post Wed May 04, 2016 4:38 am
Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A) 3.25
B) 3.6
C) 4.2
D) 4.4
E) 5.5

OA: B

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Post Tue Jan 10, 2017 11:09 pm
boomgoesthegmat wrote:
Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A) 3.25
B) 3.6
C) 4.2
D) 4.4
E) 5.5

OA: B
The rate of inlet pipe 1 = 3/(1/2) = 6 hours. It can alone fill the empty tank in 6 hours.

The rate of inlet pipe 2 = 6/(2/3) = 9 hours. It can alone fill the empty tank in 9 hours.

Once both are in operation, the part of the tank filled by them is given by:

1/6 + 1/9 = 5/18

=> The empty tank would be filled by both in 18/5 = 3.60 hours.

Answer: B

-Jay
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Post Mon Jan 16, 2017 4:56 pm
boomgoesthegmat wrote:
Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A) 3.25
B) 3.6
C) 4.2
D) 4.4
E) 5.5
We are given that one inlet pipe fills an empty tank to 1/2 capacity in 3 hours. Since rate = work/time, the rate of the one inlet pipe is (1/2)/3 = 1/6.

We are also given that a second inlet pipe fills the same empty tank to 2/3 capacity in 6 hours. Thus, the rate is (2/3)/6 = 1/9.

We need to determine how many hours it will take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity.

If we let t = the time in hours the two inlet pipes are working together, then the work of the first inlet pipe = (1/6)t and the work of the second inlet pipe = (1/9)t.

Since the tank is filled, we can set total work to 1 and create the following equation:

(1/6)t + (1/9)t = 1

Multiplying the entire equation by 18, we obtain:

3t + 2t = 18

5t = 18

t = 18/5 = 3.6

Answer:B

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Post Tue Jan 10, 2017 11:09 pm
boomgoesthegmat wrote:
Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A) 3.25
B) 3.6
C) 4.2
D) 4.4
E) 5.5

OA: B
The rate of inlet pipe 1 = 3/(1/2) = 6 hours. It can alone fill the empty tank in 6 hours.

The rate of inlet pipe 2 = 6/(2/3) = 9 hours. It can alone fill the empty tank in 9 hours.

Once both are in operation, the part of the tank filled by them is given by:

1/6 + 1/9 = 5/18

=> The empty tank would be filled by both in 18/5 = 3.60 hours.

Answer: B

-Jay
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Locations: New York | Singapore | Doha | Lausanne | and many more...

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Post Mon Jan 16, 2017 4:56 pm
boomgoesthegmat wrote:
Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A) 3.25
B) 3.6
C) 4.2
D) 4.4
E) 5.5
We are given that one inlet pipe fills an empty tank to 1/2 capacity in 3 hours. Since rate = work/time, the rate of the one inlet pipe is (1/2)/3 = 1/6.

We are also given that a second inlet pipe fills the same empty tank to 2/3 capacity in 6 hours. Thus, the rate is (2/3)/6 = 1/9.

We need to determine how many hours it will take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity.

If we let t = the time in hours the two inlet pipes are working together, then the work of the first inlet pipe = (1/6)t and the work of the second inlet pipe = (1/9)t.

Since the tank is filled, we can set total work to 1 and create the following equation:

(1/6)t + (1/9)t = 1

Multiplying the entire equation by 18, we obtain:

3t + 2t = 18

5t = 18

t = 18/5 = 3.6

Answer:B

_________________

Scott Woodbury Stewart Founder & CEO
GMAT Quant Self-Study Course - 500+ lessons 3000+ practice problems 800+ HD solutions
5-Day Free Trial 5-DAY FREE, FULL-ACCESS TRIAL TTP QUANT

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Post Sat Jan 07, 2017 1:20 pm
Hi All,

This question can be solved in a couple of different ways, but it's essentially just a Work Formula question.

Work = (A)(B)/(A+B) where A and B are the respective times it takes for two entities to individually complete a task.

To start, we have to figure out how long it takes each pipe to fill the FULL tank....

First pipe = fills 1/2 the tank in 3 hours... so it fills the FULL tank in 6 hours
Second pipe = fills 2/3 the tank in 6 hours... so it fills the FULL tank in 9 hours

Working together, the two pipes will fill the tank in (6)(9)/(6+9) = 54/15 = 3 9/15 hours = 3.6 hours

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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