How many liters of a 50% antifreeze solution must be mixed with 6 liters of a 75% antifreeze solutions to make a mixture that is 60% antifreeze

Answer 1

Let x be the liters of antifreeze and do an antifreeze volume balance: (before mixing) = (after mixing)

.5x + .75(6) = .6(x+6) (we assume that the volumes add (not completely true as antifreeze dissolves in water.)

.5x + 4.5 = .6x + .3.6

-.1x = -.9

x = 9 liters