Question : A 10-litre solution of milk and water contains 8 litres of milk. 2 litres of the solution is replaced by pure milk and mixed. The process is repeated two more times. How much milk (in litres) is present in the mixture so obtained?
Option 1: 8.976
Option 2: 8.597
Option 3: 8.796
Option 4: 8.679
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Correct Answer: 8.976
Solution : Given, A 10-litre solution of milk and water contains 8 litres of milk. 2 litres of the solution is replaced by pure milk and mixed. Final proportion = Initial proportion$\times(\frac{\text{Volume before replacement}}{\text{Volumne after replacement}})^n$ Where, n = number of times the process is repeated. Final proportion of water = $\frac{2}{10}\times(\frac{8}{10})^3=\frac{1}{5}\times(\frac{4}{5})^3=\frac{64}{625}$ Now, initial proportion of mixture = $\frac{2}{10}=\frac15$ ⇒ Volume before replacement = 10 – 2 = 8 litres ⇒ Volume after replacement = 8 + 2 = 10 litres ⇒ Proportion of milk = $\frac{625-64}{625}=\frac{561}{625}$ ⇒ Quantity of milk = $\frac{561}{625}\times 10=8.976$ litres Hence, the correct answer is 8.976 litres.
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