SOLUTION: 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 freezin

Question 103148: 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 a resulting mixture in the radiator of 50%?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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 a resulting mixture in the radiator of 50%?
:
Let x = amt to be removed; & amt of pure antifreeze to be added
:
.40(5-x) + 1.0x = .50(5)
:
2 - .4x + 1x = 2.5
:
.6x = 2.5 - 2
:
.6x = .5
:
x = .5/.6
:
x = gal removed, and gal of pure antifreeze added:
:
:
Check solution using decimals = .833:
:
5 - .833 = 4.167 gal of 40% solution
:
.4(4.167) + 1.0(.833) = .5(5)
1.67 + .833 = 2.50, confirms our solution

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