Wine problem - ratio and mixture

Question

Question

$8$ 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 the water is $256:625$. How much wine did the cask hold originally?

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My attempt

$4$ times $8$ litres are drawn and filled with water. So water contain is $32$ litres. Let originally there was $x$ litres of wine.

$$x:32 = 256:625$$

solving it gives $x = 13.1072$

But the answer given is $40$

Please help me solve the question and please explain.

Answer 1

Here is my answer assuming that $256:625$ is the ratio of the quantity of wine now left in cask to the total volume.

Let the initial amount of wine be $x$. The purity of wine after the first drawn is $\dfrac{x-8}{x}$. Then the purity of wine after the fourth drawn is $\left(\dfrac{x-8}{x}\right)^4$. Thus \begin{align} \left(\frac{x-8}{x}\right)^4&=\frac{256}{625}\\ \left(\frac{x-8}{x}\right)^4&=\left(\frac{4}{5}\right)^4\\ \frac{x-8}{x}&=\frac{4}{5}\\ x&=\large\color{blue}{40}. \end{align}

Answer 2

Here is an alternative way of explaining it. Let $V$ be the initial volume of the wine. Also let $V_n$ be the volume of the wine after the $n$-th drawn. One can use a recursive formula to determine $V_n$. $$ V_n=V_{n-1}-8\cdot\frac{V_{n-1}}{V},\tag1 $$ where $V_{n-1}$ denote the volume of wine just after the previous drawn. Rewrite $(1)$ as $$ V_n=V_{n-1}\left(1-\frac{8}{V}\right).\tag2 $$ Starting from $n=4$, we follow the recursive formula in $(2)$ until we arrive at the equation: $$ V_4=V_0\left(1-\frac{8}{V}\right)^4.\tag3 $$ But $V_0=V$, thus the volume for wine after fourth drawn is \begin{align} V_4&=V\left(1-\frac{8}{V}\right)^4\\ \frac{V_4}{V}&=\left(1-\frac{8}{V}\right)^4.\tag4 \end{align} Again, we assume that $256:625$ is the ratio of the quantity of wine after fourth drawn to the total volume, then using $(4)$ yields \begin{align} \frac{256}{625}&=\left(1-\frac{8}{V}\right)^4\\ V&=\color{blue}{40}. \end{align}

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Addendum :

If $256:625$ is the ratio of the quantity of wine after fourth drawn to the water only, then using $(4)$ we will obtain \begin{align} \frac{256}{256+625}&=\left(1-\frac{8}{V}\right)^4\\ V&\approx\color{blue}{30.089}. \end{align} Now, the choice is yours.