Antifreeze question

Question

A chemical company makes two brands of antifreeze. The first brand is 45% pure antifreeze and the second brand is 90% pure antifreeze. In order to obtain 180 gallons of a mixture that contains 70% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

Hello Cheryl, first get the ratio in which the brands should be mixed. Simple calculation show that the ratio of first brand (45%) to the second brand(90%) is 4:5.

That done, the amount of gallons for the first brand should be (4/9×180)=80, while that for the second brand (90%) should be (5/9×180)=100. Thank you

Answer 2

The trick with these problems is to focus on the amount of the prime ingredient - in this case pure antifreeze.

The target is 180 gal of 70% antifreeze content, or 180 x .7 = 126 gal of pure A/F.

Solution A is 45%. For A gals of that solution, the A/F content would be 0.45 x A, or 0.45A.

Similarly, for solution B, B gals x 0.90 would contain 0.9B gals of pure A/F.

So in total 0.45A + 0.9B = 126

We also know that A + B = 180 total solution. B = 180 - A

Using substitution, we get 0.45A + 0.9(180 - A) = 126

0.45A + 162 - 0.9A = 126 -> 36 = 0.45A -> A = 80 gal.

B = 180 - 80 = 100 gal.

Check: 0.45(80) + 0.9(100) = 36 + 90 = 126