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He originally had a total 54 gallons of milk in three churns and he wanted to make sure each churn contained 18 gallons of milk.
In order to do this, he did the following:
First he poured 1/4 of the first churn in the second churn.
He then poured 1/2 of the second churn into the third churn.
Finally he poured 1/3 of the third churn into the first churn.
How many gallons did each churn contain before Farmer Stone started pouring?
Hint: Work backwards with each churn containing 18 gallons.
Answer:
12, 33 and 9 gallons respectively for churns 1, 2 and 3.
The solution may require more than one read!
Before After Therefore
>> Pour 1 C1 = 12 C1 = 9 C2 = +3
>> Pour 1 C2 = 33 C2 = 36
>> Pour 2 C2 = 36 C2 = 18 C3 = +18
>> Pour 3 C3 = 27 C3 = 18 C1 = +9
>> Pour 3 C1 = 9 C1 = 18
To explain a little further. After pour 3, C3 must have contained 18 gallons, so it must have contained 27 before the pour. Similarly after pour 2, C2 must have contained 18 gallons, so must have contained 36 before the pour. After pour 3, C1 must have contained 18 gallons, and it received 9 gallons from pour 3, so must have had 9 before the pour. So pour 1 left 9 gallons in C1, which means C1 contained 12 before the pour and C2 received 3 gallons. We know that C2 had 36 gallons before pour 2, so must have started with 33 gallons. QED!
Puzzle 164
Three countrymen met at a cattle market. "Look here," said Hodge to Jakes, "I'll give you six of my pigs for one of your horses, and then you'll have twice as many animals here as I've got."
"If that's your way of doing business," said Durrant to Hodge, "I'll give you fourteen of my sheep for a horse, and then you'll have three times as many animals as I."
"Well, I'll go better than that," said Jakes to Durrant; "I'll give you four cows for a horse, and then you'll have six times as many animals as I've got here."
No doubt this was a very primitive way of bartering animals, but it is an interesting little puzzle to discover just how many animals Jakes, Hodge, and Durrant must have taken to the cattle market.
[Ref: ZUCK] At A Cattle Market. Amusements In Mathematics by Henry Ernest Dudeney (1917).
Hint: A little algebra might help.
Answer:
Jakes must have taken 7 animals to market, Hodge must have taken 11, and Durrant must have taken 21. There were thus 39 animals altogether.
This can be solved with a little algebra, where J  Jakes, H = Hodge and D = Durrant:
From the clues:
2 x (H  6 + 1) = J + 6  1 [1]
3 x (D  14 + 1) = H + 14  1 [2]
6 x (J 4 + 1) = D + 4  1 [3]
These can be rearranged to give:
2H  J = 15 [4] 3D  H = 52 [5] 6J  D = 21 [6]
We can now use 2 x [5] + [4] to give:
6D  J = 119 [7]
We can now use [7] + 6 x [6] to give:
35J = 245 J = 7
We can then use J = 7 in [4] and [6] to give H = 11, and D = 21. QED.
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