The Mathematical Milkman of Puzzleland - Solution

Puzzles -> Sam Loyd Puzzles

The Puzzle: 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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The Solution . . .

Suppose, in the beginning there was x gallons of spring water in can No 1 and y gallons of milk in can No. 2 then,

	can No. 1	can No. 2
In the Beginning	x gallons of water	y gallons of milk
After doubling contents of Can 2	x-y water=x-y, milk=0	2y water=y, milk=y
After doubling contents of Can 1	2(x-y) water=(3/4)(x-y), milk=(1/4)	2y-(x-y) ie (3y-x) water=(1/2)(3y-x), milk=(1/2)(3y-x)
After doubling contents of Can 2	2(x-y)-(3y-x) ie (3x-5y) water= (3/4)(3x-5y), milk= (1/4)(3x-5y)	2(3y-x) water= (5/4)(3y-x), milk= (3/4)(3y-x)

These two equations give x=7 gallons and y=3 gallons.

Hence, there was initially 7 gallons of water in can No. 1 and 3 gallons of milk in can No 2.

After all the mixings there would be 4½ gallons of water and 1½ gallons of milk in can No. 1 and 2½ gallons of water and 1½ gallons of milk in can No 2.

Hence, there is 3 more gallons of water than milk in can No 1.

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