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?
~~~~~~~~~~~~~~~~~~~~~~~~



        In the post by @mananth, the solution is incorrect, because his starting equation is written incorrectly.

        I came to bring a correct solution.



<pre>
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.


<U>ANSWER</U>.  1.8 liters of the original mixture should be drained and replaced by the pure antifreeze.
</pre>

Solved correctly.