Question

From a container of wine , 8 litres of wine is drawn and replace the same quantity with water. This is performed three more times, now the ratio of the quantity of wine to that of water in the container becomes 16 : 65. What is the initial quantity of wine in the container?

Answer 1

Let the initial quantity of wine be $x$ Litres

Let some $y$% of wine is removed every time

Then $\frac{yx}{100}=8$ $i.e.y=\frac{800}{x}$

So After performing this opertaion four times quantity of wine = $x\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$

Also quantity of water = $x-$ $x\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$

So According to question $x\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$ $:$ $x-$ $x\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$ $\times \frac{x-8}{x}$

Thus Solving we get $x=24$L

So, initial quantity of wine = 24L