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.

Step-by-step 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} (100)/(100)x+(20)/(100)(40-x)=(40)/(100)(40) \\ (100x+800-20x)/(100)=(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.