Given that,
Total mixture $ = 175 \; \text{ml} + 700 \; \text{ml} $
$ = 875 \; \text{ml} $
Gopal takes out $10\%$ of the mixture and substitutes it by water of the same amount.
We know that,
A container contains $\text{x} \; \text{units}$ of the liquid from which $\text{y} \; \text{units}$ are taken out and replaced by water.
Again from this mixture $\text{y} \; \text{units}$ are taken out and replaced by water. If this process is repeated $\text{‘n’}$ times.
Then,
The liquid left in the container after
$\frac{ \text{n}^{th} \text{operation}}{ \text{Original quantity of the liquid in the container}} = \left( \frac{\text{x -y}}{\text{x}} \right)^{\text{n}}$
$\Rightarrow \text{Quantity of liquid left after n}^{th} \text{operation} = \text{x} \ast \left( 1 – \frac{\text{y}}{\text{x}} \right)^{\text{n}} $
Here, $\text{n} = 2$
Final quantity of alcohol in the mixture $ = 700 \times \left( 1 – \frac{70}{700} \right)^{2} $
$ = 700 \times \left( 1 – \frac{1}{10} \right)^{2} $
$ = 700 \times \frac{9}{10} \times \frac{9}{10} = 7 \times 81 = 567 \; \text{ml} $
Therefore, Final quantity of water in the mixture $ = 875 – 567 = 308 \; \text{ml} $
Hence, the percentage of water in the mixture $ = \frac{308}{875} \times 100 \% $
$ = 0 \cdot 352 \times 100 \% $
$ = 35 \cdot 2 \% $
Correct Answer $: \text{A}$