Question

A chemical company makes two brands of antifreeze. The first brand is 65% pure antifreeze, and the second brand is 90% pure antifreeze. In
order to obtain 120 gallons of a mixture that contains 80% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

With amounts measured in gallons, let

x = amount of 65% antifreeze

y = amount of 90% antifreeze

1 gal of the 65% brand contains 0.65 gal of pure antifreeze; x gal would contain 0.65x gal. Similarly, y gal of the 90% brand contains 0.90y gal of pure antifreeze.

To obtain 120 gal of 80% antifreeze solution (which contains 0.80•120 = 96 gal of pure antifreeze), we must have

x + y = 120 … … … … … [total volume of antifreeze solution]

0.65x + 0.90y = 96 … [total volume of pure antifreeze]

Solve the first equation for y :

y = 120 - x

Substitute this into the second equation and solve for x :

0.65x + 0.90 (120 - x) = 96

0.65x + 108 - 0.90x = 96

0.25x = 12

x = 48

Solve for y :

y = 120 - 48

y = 72