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The Mathematical Milkman


BMAD
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The school children were returning to their homes when they met the mathematical milkman, who propounds the following problem:
In one of the two cans there is milk which is so rich with cream that it becomes absolutely necessary to dilute it with a little water to make it wholesome.
Therefore, in the other can there is some pure spring water, now I proceed to pour spring water from can No. 1 into can No. 2 sufficient to double its contents, and then repour from No. 2 into No.1 enough of the mixture to double the contents.
Then to equalize matters, I again pour from No. 1 into No. 2 to double the contents of No. 2 and find the same number of gallons of milk in each can, although there is one more gallon of water in can No. 2 than there is milk, so I want you to tell me how much more water than milk is there in can No. 1?
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dm92 has the correct number. (just the wrong units.) :)

The answer is 3 gallons.

Working backward you start with 7 gallons water and 3 gallons milk.

It proceeds forward like this.

W1 M1 W2 M2

7 0 0 3

3 0 ==>

4 0 3 3

<== 2 2

6 2 1 1

3/2 1/2 ==>

9/2 3/2 5/2 3/2

W1 M1 W2 M2

Final:

M1=M2

W2-M2 = 1

W1-M1 = 3

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