SOLUTION: a car radiator contains 10 liters of a 30 percent anti freeze solution how many liters would have to be replaced with pure antifreeze if the resulting is to contain 50 percent?

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Question 158680: a car radiator contains 10 liters of a 30 percent anti freeze solution how many liters would have to be replaced with pure antifreeze if the resulting is to contain 50 percent?
Answer by ankor@dixie-net.com(12701) (Show Source):
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a car radiator contains 10 liters of a 30 percent anti freeze solution how many liters would have to be replaced with pure antifreeze if the resulting is to contain 50 percent?
:
Let x = amt to be replaced by pure antifreeze
;
.30(10-x) + 1.0x = .50(10)
:
3 - .3x + 1.0x = 5
:
+.7x = 5 - 3
x =
x = 2.857 liters to be replaced
;
:
Check:
.3(10-2.857) + 2.857 = .5(10)
.3(7.143) + 2.857 = 5
2.1429 + 2.857 = 4.9999 ~ 5