Given,
The percentage of pure antifreeze in pure antifreeze is 85%.
The total amount of the mixture is 170 gallons.
The percentage of pure antifreeze required is 20%.
Consider,
x is the volume of the premium 85% antifreeze (in gallons) and y be the volume of water.
According to the question,
[tex]x+y=170\text{ gallons}[/tex]Similarly,
[tex]\begin{gathered} \frac{0.85}{170}=\frac{0.20}{x} \\ 0.85x=170\times0.2 \\ x=\frac{34}{0.85} \\ x=40 \end{gathered}[/tex]The amount of pure antifreeze is 40 gallons.
The amount of water is 170- 40 = 130 gallons.
Hence, the amount of the premium antifreeze is 40 gallon and water is 130 gallons is mixed.
