Question

A chemical company makes two brands of antifreeze. The first brand is 60% pure antifreeze, and the second brand is 80% pure antifreeze. In order to obtain 40 gallons of a mixture that contains 75% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

Answer:

10 gallons of the 1st brand and 30 gallons of the second brand.

Step-by-step explanation:

The equation I came up with is 0.6x+0.8y=0.75(x+y), with x and y representing the total amount of mixture of each brand. Meaning, 0.6x is the amount of pure antifreeze in the first brand, 0.8y is pure antifreeze in the second brand, and 0.75(x+y) is the amount of pure antifreeze there's supposed to be when you mix the two brands together.

Now, the question also says that the mixture is 40 gallons. The total mixture of the 2 brands is basically x+y = 40, then you can minus x from both sides and get y = 40-x, then you substitute it into the original equation:  0.6x+0.8(40-x)=0.75(x+40-x), x = 10, the amount of mixture from the first brand, which also means that 30 gallons is the amount of mixture from the second brand.