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# 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. 1can No. 2
In the Beginningx gallons of watery gallons of milk
After doubling contents of Can 2x-y
water = x-y,
milk = 0
2y
water = y,
milk = y
After doubling contents of Can 12(x-y)
water = 2(3/4)(x-y),
milk = 2(1/4)(x-y)
2y-(x-y) ie (3y-x)
water = (1/2)(3y-x),
milk = (1/2)(3y-x)
After doubling contents of Can 22(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)

Now, we know that the number of gallons of Milk in Can 1 = number of gallons of Milk in Can 2

Hence: (1/4)(3x-5y) = (3/4)(3y-x)
Multiply by 4: 3x-5y = 3(3y-x)
Move x to one side and y to other: 6x = 14y
And so: x = (14/6)y

We ALSO know that the number of gallons of Water in Can 2 = number of gallons of Milk in Can 2 PLUS 1

Hence (5/4)(3y-x) = (3/4)(3y-x) +1
Multiply by 4: 5(3y-x) = 3(3y-x) + 4
Simplify: 2(3y-x) = 4
Replace "x" with "(14/6)y": 2(3y-(14/6)y) = 4
Simplify: (4/3)y = 4
Hence: y = 3

Now we know y=3, we also know that x = (14/6)y = 7

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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