Fountain of Knowledge: get 2 oz w/ 7-oz, 11-oz tumblers

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kazafz

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Joined
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Hey there people, I recently found this maths question on the web:

You have a soda fountain but only two unmarked containers (one 7 ounces and one 11 ounces) that can be filled or emptied or poured back and forth as needed. Your goal is to get precisely the target amount (2 ounces) in one of the containers.

I tried to solve this question but all the methods i tried doesn't seem to work...so if anyone can help me it'll be awesome!

MY METHOD:
Let "a" be the 7 ounces container and "b" be the 11 ounces container:

a b
7 0
0 7
7 7
3 11

The numbers represent the ounces of soda water so for the first step i pour the 7 ounces full then i pour it into the 11 ounces container and etc. but the closest i would get is only 3 ounces but the answer wants 2 ounces...

EDIT: I finally figured it out lol i should've continued with my calculation.

Heres the real method:

a b
7 0
0 7
7 7
3 11
3 0
0 3
7 3
0 10
7 10
6 11
6 0
0 6
7 6
2 5
 
Re: Fountain of Knowledge

Hey there people, I recently found this maths question on the web:

You have a soda fountain but only two unmarked containers (one 7 ounces and one 11 ounces) that can be filled or emptied or poured back and forth as needed. Your goal is to get precisely the target amount (2 ounces) in one of the containers.

I tried to solve this question but all the methods i tried doesn't seem to work...so if anyone can help me it'll be awesome!

MY METHOD:
Let "a" be the 7 ounces container and "b" be the 11 ounces container:

a b
7 0
0 7
7 7
3 11

The numbers represent the ounces of soda water so for the first step i pour the 7 ounces full then i pour it into the 11 ounces container and etc. but the closest i would get is only 3 ounces but the answer wants 2 ounces...

Assume that the containers are cylindrical.

Step 1--Fill each container.

Step 2--Tilt each container over and carefully proceed to empty it until the water level simultaneously touches the lower lip of the container's open end, and the point where the bottom of the container meets the top inner surface wall of the container (the diagonal across the rectangular profile of the container). At this time, the container is exactly half full, there being 3.5 ounces in the smaller container and 5.5 ounces in the larger container.

Step 3--Pour the 3.5 ounces in the smaller container into the 11 ounce container leaving you with 9 ounces in the larger container and an empty 7 ounce container.

Step 4--Pour 7 ounces from the larger container into the 7 ounce container leaving you with 2 ounces in the large container.

Thy Fountain of Knowledge has been expanded.
 
Re: Fountain of Knowledge

kazafz said:
Hey there people, I recently found this maths question on the web:

You have a soda fountain but only two unmarked containers (one 7 ounces and one 11 ounces) that can be filled or emptied or poured back and forth as needed. Your goal is to get precisely the target amount (2 ounces) in one of the containers.

MY METHOD:
Code: 
a                               b

7                               0
0                               7
7                               7
3                               11
3                               0
0                               3
7                               3
0                               10
7                               10
6                               11
6                               0
0                               6
7                               6
2                               11.............Small typo corrected[/color]
Click to expand...
 
Hello, kazafz

Has anyone seen the "billard ball" solution to these problems?

You have a soda fountain but only two unmarked containers, one 7 oz, and one 11 oz.
that can be filled or emptied or poured back and forth as needed.
Your goal is to get precisely the target amount, 2 oz, in one of the containers.
Click to expand...

I'll demonstrate the method with a classic problem . . .

We have a 3-ounce container and a 5-ounce container.
. . Measure out exactly 4 ounces.


Lay out a triangular grid.
Code: 
                1   2   3   4   5
          3 * - * - * - * - * - * 3
           / \ / \ / \ / \ / \ /
        2 * - * - * - * - * - * 2
         / \ / \ / \ / \ / \ /
      1 * - * - * - * - * - * 1
       / \ / \ / \ / \ / \ /
      * - * - * - * - * - *
      0   1   2   3   4   5

Measure off 5 units on the horizontal axis, 3 units on the "vertical" axis.
. . (But the order doesn't matter.)


Start at "0" and move to the furthest vertex. .**
Code: 
            *   .   .   .   .   *
           .   .   .   .   .   .
          .   .   .   .   .   .
         .   .   .   .   .   .
        .   .   .   .   .   .
       .   .   .   .   .   .
      o - - - - - - - - - o
                        (5,0)

Note the "coordinates" of that point: (5, 0)


The ball bounces off and strikes another wall.
Code: 
                  (2,3)
            *   .   *   .   .   *
           .   .   . \ .   .   .
          .   .   .   \   .   .
         .   .   .   . \ .   .
        .   .   .   .   \   .
       .   .   .   .   . \ .
      *   .   .   .   .   *

The coordinates of that point are: (2, 3).


The ball bounces off and strikes another wall, and another.
Code: 
            *   .   *   .   .   *
           .   .   /   .   .   .
     (0,2)*   .   /   .   .   .
         . \ .   /   .   .   .
        .   \   /   .   .   .
       .   . \ /   .   .   .
      *   .   *   .   .   *
            (2,0)

The coordinates are: (2,0) and (0,2).


The ball bounces off and strikes another wall, and another.
Code: 
                          (4,3)
            *   .   .   .   o   *
           .   .   .   .   . \ .
     (0,2)* - - - - - - - - - * (5,2)
         .   .   .   .   .   .
        .   .   .   .   .   .
       .   .   .   .   .   .
      *   .   .   .   .   *

The coordinates are: (5,2) and (4,3).

We found a point with a coordinate of 4 . . . stop!


List the coordinates.
They indicate the steps for the solution.

\(\displaystyle \begin{array}{cc}(5,0) & \text{Fill the "5"} \\ (2,3) & \text{Pour "5" into "3"} \\ (2,0) & \text{Empty "3"} \\ (0,2) & \text{Pour "5" into "3"} \\ (5,2) & \text{Fill "5"} \\ (4,3) & \text{Pour "5" into "3'} \end{array}\)

And there are 4 ounces in the 5-ounce container (in six steps).


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

We could have moved to the nearest vertex.

\(\displaystyle \text{The coordinates would be: }\;\begin{array}{cc}(0,3)& \text{Fill "3"} \\ (3,0) & \text{Pour "3" into "5"} \\ (3,3) & \text{Fill "3"} \\ (5,1) & \text{Pour "3" into "5"} \\ (0,1) & \text{Empty "5"} \\ (1,0) & \text{Pour "3" into "5"} \\ (1,3)& \text{Fill "3'} \\ (4,0) & \text{Pour "3" into "5"} \end{array}\)

But this takes eight steps.


I'm still unsure if there is a Rule for the shortest solution.
. . For now, I just solve it both ways.

 
Hey soroban thats a very interesting way of solving it! But one thing I don't quite get from the solution is that how do you know which way the (5,0) heads off to at first? Because can't it go the other way too?
 
Hello again, kazafz

My diagrams are not to scale.
Those triangles are equilateral ... all angles are 60*.

Code: 
                  *   .   .   .   .   *
                o   .   .   .   .   .
              .   \   .   .   .   .
            .   .   \   .   .   .
          .   .   .   \   .   .
        .   .   .   60* \60*.
      o - - - - - - - - - o
                    60* .
                      .
                    .

Recall: angle of incidence = angle of reflection.

 
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