Math word problem

Question

A radiator holds 5 gallons. The coolant in the radiator is a mixture of antifreeze and water. Presently, 40% of the mixture is pure antifreeze. The proper mixture to prevent freezing and rust formation is when antifreeze is 50% of the mixture. The radiator is now full to the top. In order to add antifreeze, some of the present mixture must be drained out to make room. How much mixture should be drained out to make just enough room to add the correct amount of pure antifreeze to give the resulting mixture in the radiator of 50%?

Answer 1

.40(x) + 1.00(5-x) = .50(5)

.4x +5 -x = 2.5

-,6x= 2.5- 5 = -2.5

x = 25/6 = 4 1/6 gal of 40% + 5/6 gal 100% yields 5 gallons of 50%

50-40=10

100-50=50

50/10 = 5 times as much 40% as 100% is needed

25/6 = 5 times 5/6

drain off 5/6 gallon