An old wine maker has a cask half-full of 3-year old wine and another cask, twice its size, which is 1/3 full of 3-year old wine. If both casks are filled with 1-year old wine and their contents are mixed, what part of the mixture is 3-year old wine? Thanks for any help.
You have pretty good sense of humour! However, it is very interesting question and I will take a crack at it!!. Here we go:
A cask half-full of 3-year old wine is 1/4 of a cask twice its size. Therefore: 1/4 + 1/3=7/12 of the large cask contains 3-year old wine. After filling both casks with 1-year old wine, the small cask is half full of 1-year old wine and the large cask is 2/3 full of 1-year old wine. Therefore: 1/4 + 2/3 =11/12 of the large cask contains 1-year old wine. Since 7/12 of the large cask contains 3-year old wine and 11/12 of it contains 1-year old wine, it follows that the total amount of wine, both new and old, 18/12. Thus (7/12) / (18/12) =7/18. Therefore, 7/18 of the mixture is 3-year old wine! That is my take on it!.
+37084 The 1/3 full one which is twice as large equals 2/3 of a smaller one
so you have 2/3 + 1/2 = 7/6 of smaller one of 3 yr old wine
you basically are filling THREE casks (the smaller one PLUS the 2x bigger one)
the part that will be 3 yr old will be 7/6 out of 18/6 (three casks)
(7/6) / (18/6) = 7/18ths of the new mixture will be 3 yr old wine.
+129840 Call the amount that the first cask can hold = A .... and call the amount that the second cask can hold = 2A
So the amount of 3 year old wine in the first cask = A/2.... and the amount of 3 year old wine in the second cask = (1/3)2A = (2/3)A
So....if he contents are mixed....the part that is 3 year old wine =
[ A/2 + (2/3) A ] / [ A + 2A] =
[ (3/6)A + (4/6)A ] / 3A =
[ (7/6)A] / 3A =
[ (7/6)] / 3 =
7 / 18
