Question 730945
You remove some volume, q of the 20% antifreeze.  The amount of pure antifreeze in the tank is then 40*0.20-q*0.20 mass of pure antifreeze.  Next q volume of the 100% antifreeze is added, so the amount of pure antifreeze in the tank becomes 40*0.20-q*0.20+q*1.00 mass of pure antifreeze.  


All that was done so that the volume of antifreeze solution in the tank is again 40 liters of volume.  NOW, we expect the result to be 0.40 or 40% antifreeze for that 40 liters.
{{{(40*0.20-q*0.20+q*1.00)/40=0.40}}}, and you need to solve for q.