Question

When one litre of water is added to a mixture of acid and water, the new mixture contains 20% acid. When one litre of acid is added to the new mixture, then the resulting mixture 1 contains $33\frac{1}{3}$ % acid. The percentage of acid in the original mixture was

• A. 20 %
  
• B. 22 %
  
• C. 24 %
  
• D. 25 %
  
• E. None of these

Answer 1

The correct option is D 25 %


Let the original mixture of acid/water be x litres.
Then, in (x + 1) litres of mixture, acid is 20%.
Quantity of acid
$=(x+1)\times \frac{20}{100}=\left(\frac{x+1}{5}\right)$
Again,$\frac{\frac{x+1}{5}+1}{x+2}\times 100=\frac{100}{3}$
or $\frac{x+1+5}{5(x+2)}\times 100=\frac{100}{3}$
or $x+6=\frac{5(x+2)}{3}$
or 3x+18=5x+10
18-10=5x-3x
or 8 =2x
or x =4L
And quantity of acid =$\frac{4+1}{5}=1L$
$\therefore$ Percentage of acid in original mixture
$=\frac{1}{4}\times 100$
=25%
Alternate Method
Here, we assume that total quatity of the mixture be 4L as after adding 1L of water, the quantity of acid in the mixture
becomes 20%, i.e. $\frac{1}{5}$, And adding 1L of acid in the mixture, the quantity of acid becomes $\frac{1+1}{6}=\frac{1}{3}$, i.e., 33.33%
$\therefore$ In original mixture, quantity of acid
$=\frac{1}{4}\times 100$
=25%