Question 184965
Let x = the number of gallons of pure (100%) antifreeze needed to mix with the 10% antifreeze mixture to obtain 55 gallons of 20% antifreeze mixture.
x + (55-x)(0.1) = 55(0.2) 
x + 5.5 - 0.1x = 11
0.9x + 5.5 = 11
0.9x = 5.5
x = 6.11 (approx) gallons of pure antifreeze.