From a cask filled with wine, 108.75 litres are first drawn and replaced with water. From this mixture 87 litres are drawn and replaced with water. The ratio of wine to water in the cask is now 3 : 2. How many litres of wine did the cask initially hold?

Correct option is D

Given:
The initial cask contains only wine.
In the first operation, 108.75 liters of wine are drawn and replaced with water.
In the second operation, 87 liters of the mixture (wine + water) are drawn and replaced with water.
After the second operation, the ratio of wine to water in the cask is 3:2.

Formula Used:

Final = Initial $$\left(1 - \frac{x}{c}\right)^n$$​​
x = quantity withdrawn
c = quantity withdrawn from
n = No. of times

Solution:

108.75 liters = $$\frac{435}{4}$$liters​​

Let the initial quantity of wine x.

​$$x\times \left(1- \frac{\frac{435}{4}}{x}\right) \times \left(1- \frac{87}{x}\right) = \frac{3x}{5x}$$

​​$$x\times \left(1- \frac{435}{4x}\right) \times \left(1- \frac{87}{x}\right) = \frac{3x}{5x}$$

$$\left(1- \frac{435}{4x}\right) \times \left(1- \frac{87}{x}\right) = \frac{3}{5}$$

Now, Put option (d) value = 435

​$$\left(1- \frac{435}{4\times 435}\right) \times \left(1- \frac{87}{435}\right) = \frac{3}{5}$$

​​$$\left(1- \frac{1}{4}\right) \times \left(1- \frac{1}{5}\right) = \frac{3}{5}$$

$$\left(\frac{4-1}{4}\right) \times \left(\frac{5-1}{5}\right) = \frac{3}{5}$$

$$\left(\frac{3}{4}\right) \times \left(\frac{4}{5}\right) = \frac{3}{5}$$

$$\frac{3}{5} = \frac{3}{5}$$

So, the initial quantity of wine 435.

Thus, the correct answer is (d).