Complete question:
A chemical company makes two brands of antifreeze. The first brand is 40% pure antifreeze, and the second brand is 90% pure antifreeze. In order to obtain 160 gallons of a mixture that contains 60% pure antifreeze, how many gallons of each brand of antifreeze must be used?
Answer:
Amount of 40% pure antifreeze = y = 96gallons
Amount of 90% pure antifreeze = ((160-y) = (160 - 96) = 64gallons
Step-by-step explanation:
Brand A = 40% pure
Brand B = 90% pure
Therefore to obtain 160 gallons that contains 60% pure antibfreeze:
Let y = Amount of 40% pure antifreeze required
Therefore (160 - y) = Amount of 90% pure antifreeze required.
Expressing mathematically,
0.4y + (160 - y) (0.9) = 160×0.6
0.4y + 144 - 0.9y = 96
-0.5y = 96 - 144
-0.5y = - 48
y = 48 / 0.5
y = 96
Amount of 40% pure antifreeze = y = 96gallons
Amount of 90% pure antifreeze = ((160-y) = (160 - 96) = 64gallons
