Question

how many gallons of a 10% acid solution must be mixed with 5 gal of a 20% acid solution to make a 12% acid solution? gal

Explanation:

Step1: Set up the equation

Let $x$ be the number of gallons of the 10% acid solution. The amount of acid in the 10% solution is $0.1x$, the amount of acid in the 5 - gallon 20% solution is $0.2\times5$, and the amount of acid in the final 12% solution is $0.12(x + 5)$. So the equation is $0.1x+0.2\times5=0.12(x + 5)$.

Step2: Expand and simplify the equation

$0.1x + 1=0.12x+0.6$.

Step3: Solve for $x$

Subtract $0.1x$ from both sides: $1 = 0.02x+0.6$. Then subtract 0.6 from both sides: $0.4 = 0.02x$. Divide both sides by 0.02, we get $x=\frac{0.4}{0.02}=20$.

Answer:

20