Question 709907
Try to start with the expression for what is in the radiator at start.


{{{(10*20)/10=20}}}
The righthand member is 20% antifreeze, and the left hand member is the quantity of pure antifreeze over 10 gallons of the liquid in the radiator.


In the numerator, you want to SUBTRACT g gallons of the 20% liquid and ADD g gallons of the 100% material.
{{{10*20-g*20+g*100}}}.
The denominator will stay unchanged, 10, for 10 gallons.  Also, you want this used as a ratio equalling 50, for 50% antifreeze.


{{{(10*20-g*20+g*100)/10=50}}}
Solve for g.


{{{(200+(100-20)g)=50*10}}}
{{{200+80g=500}}}
{{{80g=300}}}
g=30/8=15/4
{{{highlight(g=3&3/4)}}} gallons to remove and replace with 100% antifreeze.