Solving a percent mixture problem

Question

a chemical company makes two brands of antifreeze. The first brand is 55% pure antifreeze, and the second brand is 80% pure antifreeze. In order to obtain 50 gallons of a mixture that contains 70% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

Maddie,

 

Let x be the number of gallons of the 55% antifreeze solution. Then 50 - x is the number of gallons of 80% antifreeze. The amount of antifreeze is the concentration (percent) times the number of gallons. The amounts in both solutions added together is the amount in the final mixture. When you write this down mathematically, you obtain an equation in x:

 

.55x + .80(50 - x) = .70(50).

 

Solve this:

 

.55x + 40 - .80x = 35

 

-.25x = -5

 

x = -5/-.25 = 20

 

20 gallons of 55% antifreeze and 30 gallons of 80% antifreeze must be used to obtain 50 gallons of 70% antifreeze. You can check this by using these answers in the original problem statement.