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:
[tex]0.1=\frac{x}{8}\\\\x=(0.1\times 8)=0.8[/tex]
Let the amount of pure antifreeze added in the solution be 'y'
Percent of solution to be made = 50 %
So,
[tex]0.5=\frac{0.8+y}{8+y}\\\\0.5(8+y)=(0.8+y)\\\\y=6.4[/tex]
Hence, the amount of pure antifreeze that must be added is 6.4 gallons
