Question

How much pure antifreeze must be added to 8 gallons of 10% antifreeze to make a 50% antifreeze solution?

Answer 1

Answer: The amount of pure antifreeze that must be added is 6.4 gallons

Explanation:

We are given:

Percent of antifreeze given = 10 %

Amount of solution = 8 gallons

Let the amount of pure antifreeze be 'x'

Calculating the amount of pure antifreeze:

$0.1=\frac{x}{8}\\\\x=(0.1\times 8)=0.8$

Let the amount of pure antifreeze added in the solution be 'y'

Percent of solution to be made = 50 %

So,

$0.5=\frac{0.8+y}{8+y}\\\\0.5(8+y)=(0.8+y)\\\\y=6.4$

Hence, the amount of pure antifreeze that must be added is 6.4 gallons