A vessel contains pure wine. $10\% $ of wine is drawn and replaced with water. This operation is repeated $3$ times. Calculate the percent of wine drawn from the vessel.
Answer
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Hint:Initially the vessel contains $100\% $ wine. But after each stage, the percent of wine is decreasing. The decreased quantity is equal to the quantity of water added to it.If from something $x\% $ is drawn and replaced with anything else the resultant contains percent of the initial substance equal to $(100 - x)$.So using this concept we try to get the answer.
Complete step-by-step answer:
In the initial stage we have pure wine.
That is, the vessel contains $100\% $ wine.
Let percent of wine drawn from each round is ${v_1},{v_2},{v_3}$
For the first time $10\% $ is drawn.
This drawn part contains wine only.
$ \Rightarrow {v_1} = 10\% $
Since this is replaced with water, we now have $90\% $ of wine and $10\% $ of water.
Again (that is for the second time), $10\% $ is drawn.
This part contains $90\% $ wine and $10\% $ water.
That is $\dfrac{9}{{10}}$ part wine and $\dfrac{1}{{10}}$ part water.
This means in the second time $9\% $ of wine is taken from the vessel (since the remaining $1\% $ is water).
$ \Rightarrow {v_2} = 9\% $
Therefore, after the second round, total wine taken from the vessel is ${v_1} + {v_2} = 10\% + 9\% = 19\% $.
So the remaining solution contains $100 - 19 = 81\% $ wine and $19\% $ water.
Then for the third time, $10\% $ is drawn from the solution.
That $10\% $ contains $81\% $ wine and $19\% $ water.
That is, actual percent of wine taken in the third round is $81\% \,{\text{of }}10\% = \dfrac{{81}}{{100}} \times 10\% = \dfrac{{81}}{{10}}\% = 8.1\% $
So after three rounds, total amount of wine taken from the vessel is ${v_1} + {v_2} + {v_3} = 10 + 9 + 8.1 = 27.1\% $
Therefore the percent of wine drawn from the vessel after three rounds is $27.1\% $.
Note:It is important to check after each round, the percent of wine then. It is wrong if we think each time $10\% $ is drawn, then in total $30\% $ is drawn. Each time the percent of wine present in the vessel is varying.
Complete step-by-step answer:
In the initial stage we have pure wine.
That is, the vessel contains $100\% $ wine.
Let percent of wine drawn from each round is ${v_1},{v_2},{v_3}$
For the first time $10\% $ is drawn.
This drawn part contains wine only.
$ \Rightarrow {v_1} = 10\% $
Since this is replaced with water, we now have $90\% $ of wine and $10\% $ of water.
Again (that is for the second time), $10\% $ is drawn.
This part contains $90\% $ wine and $10\% $ water.
That is $\dfrac{9}{{10}}$ part wine and $\dfrac{1}{{10}}$ part water.
This means in the second time $9\% $ of wine is taken from the vessel (since the remaining $1\% $ is water).
$ \Rightarrow {v_2} = 9\% $
Therefore, after the second round, total wine taken from the vessel is ${v_1} + {v_2} = 10\% + 9\% = 19\% $.
So the remaining solution contains $100 - 19 = 81\% $ wine and $19\% $ water.
Then for the third time, $10\% $ is drawn from the solution.
That $10\% $ contains $81\% $ wine and $19\% $ water.
That is, actual percent of wine taken in the third round is $81\% \,{\text{of }}10\% = \dfrac{{81}}{{100}} \times 10\% = \dfrac{{81}}{{10}}\% = 8.1\% $
So after three rounds, total amount of wine taken from the vessel is ${v_1} + {v_2} + {v_3} = 10 + 9 + 8.1 = 27.1\% $
Therefore the percent of wine drawn from the vessel after three rounds is $27.1\% $.
Note:It is important to check after each round, the percent of wine then. It is wrong if we think each time $10\% $ is drawn, then in total $30\% $ is drawn. Each time the percent of wine present in the vessel is varying.
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