8 litres of wine is drawn from a cask full of wine and is then filled

After the replacement process has been performed 4 TIMES, left in the casket is the following fraction:
\(\frac{wine}{total} = \frac{16}{16+65} = \frac{16}{81}\)

Since \((\frac{2}{3})^4 = \frac{16}{81}\), each of the 4 replacement processes must leave behind \(\frac{2}{3}\) of the wine.
Put another way, each replacement process must reduce the amount of wine in the casket by \(\frac{1}{3}\).

Thus, the 8 liters drawn off in the first replacement process must constitute \(\frac{1}{3}\) of the original amount of wine:
\(8 = \frac{1}{3}w\)
\(24 = w\)


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Let the initial volume of wine is V

Volume of wine left after first replacement= \(V(1-\frac{8}{V})\)
Volume of wine left after fourth replacement= \(V(1-\frac{8}{V})^4\)

\(V(1-\frac{8}{V})^4\)/V=\(\frac{16}{81}\)

\((1-\frac{8}{V})^4\)=\(\frac{16}{81}\)

\((1-\frac{8}{V})\)=\(\frac{2}{3}\)
V= 24 L

Although this question can be solved in the conventional method using expressions, a faster approach in this question would be to observe the numbers and pick options accordingly.

The withdrawal and replacement operations are being carried out four times. Each time, 8 litres are withdrawn and replaced. At the end of four operations, the ratio of wine to water is 16:65.

This means that the total volume left in the cask is a multiple of 81 (16+65). We know 81 = \(3^4\).
So, let’s pick up those options which are multiples of 3.

Let’s start with option A. Out of 18 litres, if 8 litres is removed, the concentration of wine left after 4 operations = \(\frac{5}{9} * \frac{5}{9} * \frac{5}{9} * \frac{5}{9}\). This will definitely not give us \(\frac{16}{65}\). Option A can be eliminated.

Let’s try option B. Out of 24 litres, if 8 litres is removed, the concentration of wine left after 4 operations = \(\frac{2}{3} * \frac{2}{3} * \frac{2}{3} * \frac{2}{3}\) = \(\frac{16}{81}\). Clearly, this is,

\(\frac{Wine}{Total}\) = \(\frac{16}{81}\) i.e. \(\frac{Wine}{Wine + Water}\) = \(\frac{16}{16 + 65}\). Therefore, the ratio of wine to water is 16:65, which is what is mentioned in the question.

So, the correct answer option is B.

As you can see, this approach where we try out options based on a certain logic worked really well and did not take too much time as well. So, in such questions, instead of always relying on an expression to help you solve the question, try using logic and the answer options instead.

Hope this helps!
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Dear Aravind,

Can you please explain how did you get the ratio of 5/9 and 2/3 while working on options A & B?

When we have a situation where, from a solution containing 'xi' and 'yi' liters of two components X and Y, 'z' liters is drawn out and replaced by 'z' liters of Y and this process is repeated 'n' times, the remaining quantity of X(xr) is given by the following formula which is explained in the attachment:

xr = xi{(xi+yi-z)/(xi+yi)}^n

This is a generalized formula which can be applied to solve several variations of this problem (e.g. when both components are present in the original solution which I have come across in other similar problems)

In this particular case, 'yi' (water) is zero since the cask is filled entirely with wine. To simplify matters, let's say the volume of the cask is T so xi=xi+yi=T. So the formula becomes:

xr=T{(T-z)/T}^n=T{(T-8)/T}^4

The fraction of wine that is present in the final solution after the process is done 4 times is T{(T-8)/T}^4/T={(T-8)/T}^4 which is equal to 16/(65+16)=16/81=(2/3)^4....> (T-8)/T=2/3...> T=24

\(Cfinal=Cinitial(Vinitial/Vfinal)^{(repeated)}\)
Cinitial of wine = 100% = 1 ("a cask full of wine")
Cfinal of wine = 16/(65+16) = 16/81 ("ratio wine to water")
Vinitial in the cask = t-8 (liquid removed)
Vfinal in the cask = t-8+8 (liquid removed and water added)

find initial volume: \(Cfinal=Cinitial(Vinitial/Vfinal)^4=… 16/81=1(t-8/t)^4… 2/3=(t-8/t)… t=24\)

Answer (B)

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