How many liters of 20% acid solution must be added to 10 liters of 50% acid solution to get a 30% solution?

Problem: How many liters of a $20\%$ acid solution must be added to $10$ liters of a $50\%$ acid solution to get a $30\%$ solution?

Steps

1. Define the variable
   $$\text{Let } x = \text{liters of the } 20\% \text{ acid solution to be added}$$
2. Set up the mixture equation based on the amount of pure acid
   $$0.20x + 0.50(10) = 0.30(x + 10)$$
3. Multiply by 100 to clear the decimals
   $$20x + 50(10) = 30(x + 10)$$
4. Distribute and simplify
   $$20x + 500 = 30x + 300$$
5. Subtract $20x$ from both sides
   $$500 = 10x + 300$$
6. Subtract 300 from both sides
   $$200 = 10x$$
7. Divide by 10
   $$x = 20$$