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Mathematical type problem
Finally, the water in the third bucket is still 6/7, 12÷6/7= 14.

Pour into the first barrel: 14× 1/7=2 liters.

The first barrel: before pouring, there is 12-2= 10 liter, which is 2/3 of the original.

So the first barrel originally had: 10÷2/3= 15 liters.

15× 1/3=5 liters.

After the second barrel is poured into the third barrel, there is 12 liter, which is 3/4 of that before being poured into the third barrel.

So there is 12÷3/4= 16 liters before pouring into the third bucket.

16-5= 1 1

The first barrel initially contained 15 liter, the second barrel initially contained 1 1 liter, and the third barrel initially contained 10 liter.

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