# How much wine did the cask hold originally?

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8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65.
How much wine did the cask hold originally?

posted Dec 21, 2015
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## 2 Answers

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40l.
The litres of wine drawn = 8+3*8 = 32
x being the quantity of wine the cask held originally,
16/65 = x-32/32
65x-2080 = 512
x = 39.88l
appr., x = 40 litres

answer Dec 21, 2015
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24 Litres!
Let the quantity of wine the cask held originally= X litres
Now, after first operation quantity of wine held by cask= X-8 litres
After second operation, the quantity of wine held by cask= (X-8) - (X-8)*8/X ...second term is the proportion of wine removed from the mixture of wine and water in second operation.
==> Held quantity = (X-8) - (8X-64)/X = (X^2-8X-8X+64)/X = (X^2-16X+64)/X
taking X^2 out of bracket term.... X^2(1-16/X+64/X^2)/X= X(1-2*8/X+(8/X)^2) = X(1-8/X)^2
Similarly, after third operation, quantity of wine held in mixture= X(1-8/X)^3
and after fourth operation, quanity held = X(1-8/X)^4
So, ratio of wine to total mixture...
(X(1-8/X)^4)/X = 16/81 =>> (1-8/X)^4 = (2/3)^4 =>> 1-8/X = 2/3 =>> (X-8)*3 = 2*X
Hence, X = 24 litres

answer Dec 22, 2015 by anonymous

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