Approximately 0.89 quarts solution must be drained and replaced with pure antifreeze to create the desired solution.
Explanation
Suppose, [tex]x[/tex] quarts solution must be drained and replaced with pure antifreeze.
So, the amount of solution in the radiator after draining out [tex]=(8-x)[/tex] quarts, which is 10% antifreeze solution.
That drained out solution is replaced with [tex]x[/tex] quarts 100% antifreeze solution.
David needs to have a 20% solution of antifreeze in the radiator, which can contain 8 quarts solution. So, the equation will be.......
[tex]10\%*(8-x)+100\%*(x)=20\%*(8)\\ \\ 0.10(8-x)+1.00(x)=0.20(8)\\ \\ 0.80-0.10x+1.00x=1.60\\ \\ 0.90x=1.60-0.80\\ \\ 0.90x=0.80\\ \\ x=\frac{0.80}{0.90}=0.888... \approx 0.89[/tex]
So, the amount of solution that must be drained and replaced with pure antifreeze will be approximately 0.89 quarts.
