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Thread: Final amount after X seconds

  1. #1
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    Drammen
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    Final amount after X seconds

    Hi
    I have 3 trasparent Containers A, B and C wich can contain water. It is possible to have an negative amount of water.

    Container A start off with a random amount ( a ) of water. The amount of water can be negative or possitive.
    Container B starts of with a fixed amount ( b ) of water wich never changes. The amount of water can be negative or possitive.
    Container C starts off empty.

    Every second i take a bucket and fill it with the exact amount of water wich is in Container B, and then pour it intoo Container C.
    Every second i take a bucket and fill it with the exact amount of water wich is in Container C, and pour it intoo Container A.

    ...
    After x number of seconds ( t ), how much water is in container A?
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  2. #2
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    Lexington, MA (USA)
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    Re: Final amount after X seconds

    Hello, CakeSpear!

    Did you make any sketches?


    I have 3 trasparent Containers A, B and C which can contain water.
    It is possible to have an negative amount of water.

    Container A start off with a random amount of water $\displaystyle a$.
    Container B starts of with a fixed amount of water $\displaystyle b$which never changes.
    Container C starts off empty.

    Every second I take a bucket and fill it with the exact amount of water which is in Container B,
    . . and then pour it into Container C.
    Every second I take a bucket and fill it with the exact amount of water which is in Container C,
    . . and then pour it into Container A.
    . . and so on.

    After $\displaystyle t$seconds, how much water is in container A?

    Note that the water is not transferred from container to container.
    The additional water comes from an outside source.


    Here is what I found . . .


    The containers start like this:
    Code:
          |       |
          |       |
          |       |                   |       |
          |       |                   |       |
          |       |                   |       |
          |       |                   |       |
    t = 0 |       |                   |       |
          | - - - |     |       |     |       |
          |:::::::|     | - - - |     |       |
          |:::a:::|     |:::b:::|     |       |
          * - - - *     * - - - *     * - - - *
              A             B             C

    $\displaystyle b$ is added to container C.
    Code:
          |       |
          |       |
          |       |                   |       |
          |       |                   |       |
          |       |                   |       |
          |       |                   |       |
    t = 1 |       |                   |       |
          | - - - |     |       |     |       |
          |:::::::|     | - - - |     | - - - |
          |:::a:::|     |:::b:::|     |:::b:::|
          * - - - *     * - - - *     * - - - *
              A             B             C

    Then $\displaystyle b$ is added to container A.
    Code:
          |       |
          |       |
          |       |                   |       |
          |       |                   |       |
          |       |                   |       |
          |       |                   |       |
    t = 2 | - - - |                   |       |
          |:::::::|     |       |     |       |
          |::a+b::|     | - - - |     | - - - |
          |:::::::|     |:::b:::|     |:::b:::|
          * - - - *     * - - - *     * - - - *
              A             B             C

    $\displaystyle b$ is added to Container C.
    Code:
          |       |
          |       |
          |       |                   |       |
          |       |                   |       |
          |       |                   |       |
          |       |                   |       |
    t = 3 | - - - |                   | - - - |
          |:::::::|     |       |     |:::::::|
          |::a+b::|     | - - - |     |::2b:::|
          |:::::::|     |:::b:::|     |:::::::|
          * - - - *     * - - - *     * - - - *
              A             B             C

    Then $\displaystyle 2b$ is added to Container A.
    Code:
          |       |
          |       |
          |       |                   |       |
          |       |                   |       |
          | - - - |                   |       |
          |:::::::|                   |       |
    t = 4 |:::::::|                   | - - - |
          |::a+3b:|     |       |     |:::::::|
          |:::::::|     | - - - |     |::2b:::|
          |:::::::|     |:::b:::|     |:::::::|
          * - - - *     * - - - *     * - - - *
              A             B             C

    $\displaystyle b$ is added to Container C.
    Code:
          |       |
          |       |
          |       |                   |       |
          |       |                   |       |
          | - - - |                   |       |
          |:::::::|                   | - - - |
    t = 5 |:::::::|                   |:::::::|
          |::a+3b:|     |       |     |:::::::|
          |:::::::|     | - - - |     |:::3b::|
          |:::::::|     |:::b:::|     |:::::::|
          * - - - *     * - - - *     * - - - *
              A             B             C

    Then $\displaystyle 3b$ is added to Container A.
    Code:
          |       |
          |       |
          | - - - |                   |       |
          |:::::::|                   |       |
          |:::::::|                   |       |
          |:::::::|                   | - - - |
    t = 6 |::a+6b:|                   |:::::::|
          |:::::::|     |       |     |:::::::|
          |:::::::|     | - - - |     |:::3b::|
          |:::::::|     |:::b:::|     |:::::::|
          * - - - *     * - - - *     * - - - *
              A             B             C

    We see a pattern developing.

    . . $\displaystyle \begin{array}{cc}\text{Time} & \text{Container A} \\ \hline 0 & a \\ 2 & a+b \\ 4 & a+3b \\ 6 & a+6b \\ \vdots & \vdots \end{array}$


    The coefficients of $\displaystyle b$ are triangular numbers: .$\displaystyle 0,1,3,6,10, 15, \hdots$


    If time is an even number of seconds, $\displaystyle t\,=\,2k$

    . . then: .$\displaystyle A \;=\;a + \frac{k(k+1)}{2}b$
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  3. #3
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    Drammen
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    Re: Final amount after X seconds

    Thank you for taking the time to illustrate and explain this.
    You really make it seem more logical the way you explain it

    But i dont understand why time is increased by 2 for each step.
    In my case time should increase by only 1 each step.
    In that case would the new formula be
    a + ( ( ( t+1 ) / 2 ) * b )
    ???

    And might i ask how you create such nice Image Tables/ Lists?
    I would be very usefull for me when trying to solve thing on my own
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