madmax1000 wrote:
I am thinking about that for over one hour now...
I don't understand why 80% can be the answer to the question what percentage of the original alcohol was replaced?
0.5(100−x)+0.25x=0.3*100 → x=80
First we take out 80 liters (x) of 100 liters solution. 20 Liters remain, of which 10 are alcohol. Then we add to that 20 liters 80 liters, which consist only of 25% alcohol (20 liters pure alcohol). Therefore the new solution logically consists of 70 (10+60) liters water and 30 (10+20) liters alcohol.
Now 80 liters or 80% of the of 100, is what we take out and add of the WHOLE solution. First we take 80 liters out than we add 80 liters.
In my opinion 30 liters of alcohol we being replaced by water so 3/5.
Can somebody explain that to me comprehensibly?
Thanks a lot.
Say, you have a 100 ml glass of 50% water + 50% alcohol solution.
So water is 50 ml in it and alcohol is 50 ml in it homogeneously mixed.
Now, of the 100 ml, say you take out, 20 ml solution (i.e. you take out 20% of the solution). How much water and how much alcohol did you take out? Since the solution has 50-50 alcohol and water, the 20 ml will have 10 ml water and 10 ml alcohol, right?
So of the 50 ml alcohol originally, you took out 10 ml alcohol. This means you took out 20% of the original alcohol.
Of the 50 ml water originally, you took out 10 ml water. This means you took out 20% of the original water.
So by taking out 20% of the original solution (which was 100 ml), you took out 20% of original water (which was 10 ml of 50 ml) and 20% of original alcohol (which was 10 ml of 50 ml).
Does this make sense?
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Karishma
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