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1 vote

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 $: \text{A}$

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 $: \text{A}$

0 votes

$\textrm{We have $175$ ml water + $700$ ml alcohol = $875$ ml. }$

$\textrm{ Taking out 10%, (2 time repeat) alcohol content drops to $700*0.9*.9=567$ ml.}$

$\textrm{ Now, after water is refilled again, solution volume returns to 875ml.}$

$\textrm{We totally have $875$ ml overall mixture and of this $567$ ml is alcohol.}$

$\textrm{Remaining $875 – 567 = 308$ is the amount of water.}$

$\textrm{We have to find the % of the water in the mixture that is:}$

$\Rightarrow$ $\frac{308}{875}$

$\textrm{Hence 35.2% is the percentage of water in the given mixture. Option 1.}$

$\textrm{ Taking out 10%, (2 time repeat) alcohol content drops to $700*0.9*.9=567$ ml.}$

$\textrm{ Now, after water is refilled again, solution volume returns to 875ml.}$

$\textrm{We totally have $875$ ml overall mixture and of this $567$ ml is alcohol.}$

$\textrm{Remaining $875 – 567 = 308$ is the amount of water.}$

$\textrm{We have to find the % of the water in the mixture that is:}$

$\Rightarrow$ $\frac{308}{875}$

$\textrm{Hence 35.2% is the percentage of water in the given mixture. Option 1.}$