Question

How many gallons of a

50

​%

antifreeze solution must be mixed with

90

gallons of

20

​%

antifreeze to get a mixture that is

40

​%

​antifreeze? Use the​ six-step method.

Answer 1

Let x be the number of gallons of 50% antifreeze.

Amount of pure antifreeze added is 0.5x gallons.

Amount of pure antifreeze in 20% solution is 0.2×90=18 gallons.

Volume of final solution is x+90 gallons.

Amount of pure antifreeze in final solution is 0.4(x+90).

Mixture: 0.5x+18=0.4(x+90).

0.5x+18=0.4x+36, 0.1x=18, x=18/0.1=180 gallons.

180 gallons of 50% concentration have to be added.