A \( 20 \% \) acid solution is mixed with a \( 70 \% \) acid solution to get 50 liters of a \( 40 \% \) solution. How much of the \( 70 \% \) solution is used? 25 liters 20 liters 30 liters

To find out how much of the \( 70\% \) solution is used, let’s set up the problem using variables. Let \( x \) be the amount of the \( 70\% \) solution, and thus \( 50 - x \) will be the amount of the \( 20\% \) solution. The equation for the total amount of acid in the mixture can be set up as follows: \[ 0.7x + 0.2(50 - x) = 0.4(50) \] Simplifying this gives: \[ 0.7x + 10 - 0.2x = 20 \] Combining like terms: \[ 0.5x + 10 = 20 \] Subtracting \( 10 \) from both sides: \[ 0.5x = 10 \] Dividing by \( 0.5 \): \[ x = 20 \] So, \( 20 \) liters of the \( 70\% \) solution is used. Now, let's check our options: 25 liters, 20 liters, and 30 liters. The answer is **20 liters**!