Mixture Problem... An auto tech needs a radiator to

Question

have a 40% antifreeze solution. The radiator currently
is filled with 4 gallons of a 25% antifreeze solution.
How much of the antifreeze mixture should be
drained from the car if the mechanic replaces it with
pure antifreeze?

Answer 1

If he drains x gallons, then

.25(4-x) + 1.00x = .40(4)
1 - .25x + x = 1.6
.75x = .6
x = .8

So, after draining .8 gals, he has 3.2 gals of 25% antifreeze. That means he has .8 gals of antifreeze in the radiator.

Now he adds .8 gals of pure antifreeze. Then he has 1.6 gals of antifreeze in 4 gals of mixture. That's .16/4 = 40% antifreeze.