# Thread: Final amount after X seconds

1. ## 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?

2. ## 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 $b$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 $t$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

$b$ 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: . $A \;=\;a + \frac{k(k+1)}{2}b$

3. ## 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