Question

How many gallons of a 50% antifreeze solution must be mixed with 60 gallons of 30% antifreeze to get a mixture that is 40% antifreeze? Use the six-step method

Answer 1

I don't know what the "six-step method" is supposed to be, so I'll just demonstrate the typical method for this problem.

Let x be the amount (in gal) of the 50% antifreeze solution that is required. The new solution will then have a total volume of (x + 60) gal.

Each gal of the 50% solution used contributes 0.5 gal of antifreeze. Similarly, each gal of the 30% solution contributes 0.3 gal of antifreeze. So the new solution will contain (0.5 x + 0.3 * 60) gal = (0.5 x + 18) gal of antifreeze.

We want the concentration of antifreeze to be 40% in the new solution, so we need to have

(0.5 x + 18) / (x + 60) = 0.4

Solve for x :

0.5 x + 18 = 0.4 (x + 60)

0.5 x + 18 = 0.4 x + 24

0.5 x - 0.4 x = 24 - 18

0.1 x = 6

x = 6/0.1 = 60 gal