Tip #2: From a solution of volume V containing x percent of a substance, if y liters are replaced by water, the new percentage of the substance in the solution = x (V-y)/ V
Question: 8 liters are drawn from a cask full of pure 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?
Let Vw be the volume of wine in the cask originally.
Let Vt be the total volume of liquid in the cask.
Let X be the percentage of wine in the cask originally.
X = original volume of wine/ total volume of liquid in the cask = Vw / Vt
When 8 liters are drawn out, the volume of wine is reduced by 8 X liters while the total volume of liquid remains unchanged as it is re-filled with water.
Let X’ be the new percentage of wine in the cask after this operation
X’ = (original volume of wine – 8 X) / total volume of liquid in the cask
= [Vw – 8 (Vw/ Vt)] / Vt
= X (Vt – 8) / Vt
After 4 such replacement operations, X’’’’ = X [(Vt – 8)/ Vt] ^ 4
From the problem, X’’’’ = 16/ (16 + 65) = 16/ 81
Also, since originally the cask was full of pure wine, X = 1
[(Vt – 8)/ Vt] ^ 4 = 16/ 81
=> Vt = 24 liters
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