Mixture Problem with Removing part

Question

How much antifreeze which is 30% alcohol must be removed
from a 48-ounce container and replaced with water to make 48 ounces of a
solution which is 20% alcohol?

I am not getting jist of above problem as to how to solve it?

Answer 1

Hint: Initially, $30\%$ of the $48$ ounce solution is alcohol. How much alcohol does it contain? After some of the alcohol has been replaced by water, $20\%$ of the $48$ ounce solution is alcohol. How much alcohol remains? Therefore, how much alcohol has been replaced by water? Keep in mind that the amount of alcohol that has been replaced is $30\%$ of the anti-freeze that has been replaced.

Answer: (using the hint) Initially, the amount of alcohol is $30\%$ of the $48~\text{oz}$ solution, which is $$0.3 \cdot 48~\text{oz} = 14.4~\text{oz}$$ The final amount of alcohol is $20\%$ of the $48~\text{oz}$ solution, which is $$0.2 \cdot 48~\text{oz} = 9.6~\text{oz}$$ The amount of alcohol that has been removed is $$14.4~\text{oz} - 9.6~\text{oz} = 4.8~\text{oz}$$ Since the amount of alcohol that is removed is $30\%$ of the amount of anti-freeze that has been removed and replaced by water, the amount of anti-freeze that has been replaced by water is $$\frac{10}{3} \cdot 4.8~\text{oz} = 16~\text{oz}$$

Alternate Solution: Let $x$ be the amount of anti-freeze that has been removed. Then the volume of anti-freeze that remains before the water is added is $48~\text{oz} - x$, of which $30\%$ is alcohol. After the water is added, the new volume is $48~\text{oz}$, of which $20\%$ is alcohol. Since adding water does not change the amount of alcohol in the solution, \begin{align*} 0.3(48~\text{oz} - x) & = 0.2 \cdot 48~\text{oz}\\ 14.4~\text{oz} - 0.3x & = 9.6~\text{oz}\\ 4.8~\text{oz} & = 0.3x\\ 16~\text{oz} & = x \end{align*} which agrees with the previous result.