The Mathematical Milkman of Puzzleland - Solution
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?
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?
Our 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,
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.
| 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 = 2(3/4)(x-y), milk = 2(1/4)(x-y) | 2y-(x-y) i.e. (3y-x) water = (1/2)(3y-x), milk = (1/2)(3y-x) |
| After doubling contents of Can 2 | 2(x-y)-(3y-x) i.e. (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.
Puzzle Author: Loyd, Sam
See this puzzle without solution
Discuss this puzzle at the Math is Fun Forum