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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?
Answer & Solution
Correct Answer:
Option
B
Let the quantity of the wine in the cask originally be x litres.
Then, quantity of wine left in cask after 4 operations
$$\eqalign{ & = \left[ {x{{\left( {1 - \frac{8}{x}} \right)}^4}} \right]\,{\text{litres}} \cr & \therefore \left( {{{\frac{{x\left( {1 - \left( {\frac{8}{x}} \right)} \right)}}{x}}^4}} \right) = \frac{{16}}{{81}} \cr & \Rightarrow {\left( {1 - \frac{8}{x}} \right)^4} = {\left( {\frac{2}{3}} \right)^4} \cr & \Rightarrow {\frac{{x - 8}}{x}} = \frac{2}{3} \cr & \Rightarrow 3x - 24 = 2x \cr & \Rightarrow x = 24 \,\text{litres} \cr} $$
Then, quantity of wine left in cask after 4 operations
$$\eqalign{ & = \left[ {x{{\left( {1 - \frac{8}{x}} \right)}^4}} \right]\,{\text{litres}} \cr & \therefore \left( {{{\frac{{x\left( {1 - \left( {\frac{8}{x}} \right)} \right)}}{x}}^4}} \right) = \frac{{16}}{{81}} \cr & \Rightarrow {\left( {1 - \frac{8}{x}} \right)^4} = {\left( {\frac{2}{3}} \right)^4} \cr & \Rightarrow {\frac{{x - 8}}{x}} = \frac{2}{3} \cr & \Rightarrow 3x - 24 = 2x \cr & \Rightarrow x = 24 \,\text{litres} \cr} $$
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