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8 litres are drawn from a cask filled with 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 the total solution is 16:81.

How much wine did the cask hold originally?


24 litres


45 litres


49 litres


44 litres

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Option(A) is correct

Let the quantity of the wine in the cask originally be $x$ litres.
Using formula: 

$\text{Final Amount of solute that is not replaced} =$ 
\[\text{Initial Amount} \times \left(
\dfrac{\text{Vol. after removal}}{\text{Vol. after replacing}}\right)^n \text{ ---(1)}\]
$\text{Final ratio of solute not replaced to total} =$ 
\[\text{Initial ratio} \times \left(\dfrac{\text{Vol. after removal}}{\text{Vol. after replacing}}\right)^n \text{ ---(2)}\]
Considering $2^{nd}$ formula here,

Then ratio of wine to total solution in cask after 4 operations:
\Rightarrow 1 \times \left(\dfrac{x-8}{x}\right)^4&=\dfrac{16}{81}\\
\Rightarrow \dfrac{x-8}{x}&=\dfrac{2}{3}\\
\Rightarrow 3x-24&=2x\\
\Rightarrow x=24 \text{ litres.}

(3) Comment(s)


Why initial ratio is 1:1??


Initially, the cask only has wine. Which means 100% of the solution is wine. So, 1:1.


I need a thorough explanation of this, mind if you could dig deeper on its relevance and implications., thanks!