8 liters 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 16 : 65. How much wine did the cask originally hold?Options30 ltrs26 ltr24 ltr32 ltr

The problem is a classic example of a successive replacement problem. Here are the steps to solve it:

1. We know that each time we perform the operation, we are replacing 8 liters of wine with water. This means that after each operation, the amount of wine left in the cask is reduced by a certain fraction.

2. The fraction of wine left after each operation is (original volume - volume replaced) / original volume. Since 8 liters are replaced each time, the fraction of wine left is (original volume - 8) / original volume.

3. After four operations, the fraction of wine left is ((original volume - 8) / original volume) ^ 4. According to the problem, this is equal to 16 / (16 + 65) = 16 / 81.

4. Therefore, ((original volume - 8) / original volume) ^ 4 = 16 / 81. Taking the fourth root of both sides, we get (original volume - 8) / original volume = cubed root of (16 / 81).

5. Solving the above equation for the original volume, we get original volume = 8 / (1 - cubed root of (16 / 81)).

6. Calculating the right side of the equation, we get the original volume approximately equal to 32 liters.

So, the cask originally held 32 liters of wine.