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Math Help - 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 a.
    Container B starts of with a fixed amount of water bwhich 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 tseconds, 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

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

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

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

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

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

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

    We see a pattern developing.

    . . \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 b are triangular numbers: . 0,1,3,6,10, 15, \hdots


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

    . . then: . 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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