SOLUTION: The radiator in Natalie�s car contains 6.3 L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should she drain and replace with pure antifreeze s

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Question 449124: The radiator in Natalie�s car contains 6.3 L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should she drain and replace with pure antifreeze so that there will be a mixture of 50% antifreeze?
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
------ percent ---------------- quantity
Antifreez1 I 80 ---------------- 40 gallons
Antifreeze II 10 ---------------- x gallons
Total 30 ---------------- 40 + x gallons

30*6.3+100*x=50(6.3+x)
189+100x =315+50x
100x-50x =315-189
50x=126
/50
x=2.52 liters 100% Antifreeze

Answer by ikleyn(53614) About Me  (Show Source):
You can put this solution on YOUR website!
.
The radiator in Natalie's car contains 6.3 L of antifreeze and water. This mixture is 30% antifreeze.
How much of this mixture should she drain and replace with pure antifreeze so that there will be a mixture of 50% antifreeze?
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        In the post by @mananth, the solution is incorrect, because his starting equation is written incorrectly.
        I came to bring a correct solution. 


Let x be the volume of the original mixture to drain and replace with pure antifreeze.


Then the balance equation for pure antifreeze content is

    0.3*(6.3-x) + x = 0.5*6.3.


Simplify and find x

    0.3*6.3 - 0.3x + x = 0.5*6.3,

    -0.3x + x = 0.5*6.3 - 0.3*6.3,

         0.7x =  (0.5*-0.3)*6.3

         0.7x =     0.2*6.3

            x =     0.2*9 = 1.8.


ANSWER.  1.8 liters of the original mixture should be drained and replaced by the pure antifreeze.

Solved correctly.