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Pumping alone at their

This topic has 4 expert replies and 1 member reply
boomgoesthegmat Senior | Next Rank: 100 Posts Default Avatar
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Pumping alone at their

Post Wed May 04, 2016 4:38 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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 Wed May 04, 2016 6:10 am
    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
    Let the tank = 18 liters.
    Since the first pipe takes 3 hours to fill 1/2 of the 18-liter tank, the rate for the first pipe = w/r = (1/2 * 18)/3 = 3 liters per hour.
    Since the second pipe takes 6 hours to fill 2/3 of the 18-liter tank, the rate for the second pipe = w/r = (2/3 * 18)/6 = 2 liters per hour.
    Since the combined rate for the two pipes = 3+2 = 5 liters per hour, the time for the two pipes together to fill the tank = w/r = 18/5 = 3.6 hours.

    The correct answer is B.

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    Danny@GMATAcademy Junior | Next Rank: 30 Posts Default Avatar
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    Post Sat Jan 07, 2017 1:14 pm
    Here are a couple of ways to solve this problem:

    Method 1 (add the rates): https://www.youtube.com/watch?v=zqeIIbSdEc4
    Method 2 (Time Formula): https://www.youtube.com/watch?v=1LeaTvlAA-U

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    Last edited by Danny@GMATAcademy on Sat Jan 07, 2017 1:23 pm; edited 1 time in total

    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

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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

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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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