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# A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal tak

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Re: A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal tak [#permalink]
initial volume of Alcohol = 700, total volume = 700+ 175 = 875 , replaced quantity = 87.5 & n =2
Using, final volume =initial volume ( 1-(replaced quantity)/(total volume))^n = 700(1-87.5/875)^2 = 700×81/100 = 567
So water in final mixture = 875 – 567 = 308
Required percentage = 308/875×100 = 35.2

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Re: A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal tak [#permalink]
method II :
As Randy takes out 10% of the mixture
10% of alcohol is removed from 700 ml alcohol, 90% of alcohol is retained
So alcohol remaining = 700 × 90% × 90%
⟹ 700 × 0.9 × 0.9 = 567
We totally have 875 ml overall mixture and of this 567 ml is alcohol.
Remaining 875 – 567 = 308 is the amount of water.
Approximately 308 is 30% of 1000 so by this we know that 308 is more than 30%
Hence 35.2% is the percentage of water in the given mixture.
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A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal tak [#permalink]
total ammount in mixture = 700+175 = 875

10% taken out, and replaced with water
175-175(0.1)+875(0.1) = 245

Process is repeated again
245-245(0.1)+875(0.1) = 308

New value of water is 308
Ratio of new water total is 308/875 = 35.2
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Re: A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal tak [#permalink]
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Re: A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal tak [#permalink]
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