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 the wine left to that of water is 16:65. How much wine did the cask hold initially?

a. 24 litres
b. 18 litres
c. 32 litres OR
d. 42 litres

Please provide the clues to this question and how to do it and how to do it fast? Which category of maths does this query belong to?

Think in terms of how many liters of wine there are in the casket after each step. At step 0 (the beginning) there are x liters of wine. At step 1 you remove 8/x of the liters of wine, leaving (x-8)/x of the original amount. At step 2 you remove again (x-8)/x of the amount you had at step 1, so the amount of wine you have is now ((x-8)/x)^2 of the original. In general after step 'n' the amounts of wine and water you have are:


Wine\ = \ \frac {(x-8)^n}{x^{(n-1)}} \\
Water = \ x- \frac {(x-8)^n}{x^{(n-1)}} \\


The ratio of wine to water is then:

wine:water\ = \ \frac {(x-8)^n}{x^n-(x-8)^n}


You can try each of the 4 possible answers and see which one works.