Sorry 20 where ?

Could you be more precise sir ?

Since alcohol is not added and stays in the same amount, it's volume is 10*0.2=2 liters
\(\frac{2}{10+x}\)=\(\frac{1}{5}*\frac{1}{4}\), where 1/5 is 20% and 1/4 is 75% reduced amount
\(2*20=10+x\), \(x=30\). Answer is C

Hi , Can this be termed as a right method too?

20% of alcohol in 10 li = 2 l
reducing this by 75 % makes the alochol content to be 0.5l
Now water content in the mixture is 9.5l which is 95 %
8 + X / 10 +x = 95/100
the answer of x litres of water be 30 .

You are reducing the actual alcohol content and we are asked that the concentration be reduced by \(75\)%

Consider this:

\(20\)% in \(10\) lit, so alcohol = \(2\)lit
Concentration is \(20\)% now, but when it is reduced by \(75\)%, then the new concentration will be \(\frac{1}{4}*20\)% \(= 5\)%

So now after the water is added the \(2\)lit alcohol will represent \(5\)% of the total mixture.
if \(5\)% = \(2\)lit, then \(100\)% will be \(40\)lit

So water added \(= 40 - 10 = 30\)lit

Answer C

I got the point , thank you so much

After reducing alcohol you can apply short cut, every 0.5 l alcohol requires 9.5 l water. Hence, when alcohol is 2 l, water is 38 l. The difference between old volume of water and the required volume is 30 l.

IMO, you have good intuition


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