Bunuel
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?
A. 24 litres
B. 44 litres
C. 45 litres
D. 49 litres
D. 50 litres
Set the total original volume of wine as x litres. Each time we take away 8 litres of the solution, we lose \(\frac{8}{x}\) of the wine (and water too!), and we are left with \(1 - \frac{8}{x}\) of the wine. This ratio does not change as the total volume is always filled back to x. If we perform this 4 times in total we multiply x by this ratio 4 times, so we have the equation:
\(\frac{x * (1 - \frac{8}{x})^4}{x} = \frac{16}{81} \)
\( (1 - \frac{8}{x})^4 = 16/81\)
\(1 - \frac{8}{x} = 2/3\)
\(\frac{8}{x} = \frac{1}{3}\)
\(x = 24\)
Ans: A
A couple of numbers can help us see we are multiplying the volume of wine by a fixed ratio each step too;
Assuming the total volume was 50L and we took away 10L of the solution each time:
Wine vs Water
50 0
40 10, taking away 10L would be taking away 8 wine 2 water.
32 18, taking away 10L would be taking away 6.4 wine and 3.6 water.
25.6 ...
As we can see every time we take away 10L the wine portion is multiplied by a fixed ratio of 4/5.
I don't understand how you came up with such ratio ? And how is that ratio constant amongst all 4 cycles ?