Question

Pure acid is to be added to a 20% acid solution to obtain 40 L of a 40% acid solution.What amounts of each should be used?

Answer 1

Given:

Pure acid is to be added to a 20% acid solution to obtain 40 L of a 40% acid solution.

To find:

The amount of each solution.

Explanation:

Let x be the amount of liter of pure acid solution.

Let 40-x be the amount of liter of 20% acid solution.

So, the equation can be written as,

$100\text{ \% of }x+20\text{ \% of }(40-x)=40\text{ \% of }(40)$

Solving we get,

$\begin{gathered} \frac{100}{100}x+\frac{20}{100}(40-x)=\frac{40}{100}(40) \\ \frac{100x+800-20x}{100}=\frac{1600}{100} \\ 100x+800-20x=1600 \\ 80x=1600-800 \\ 80x=800 \\ x=10 \end{gathered}$

Since x = 10,

Therefore,

$\begin{gathered} 40-x=40-10 \\ =30 \end{gathered}$

Final answer:

The pure solution is 10 liter.

20% acid solution is 30 liter.