Try this: 1) note that weight of pure antifreeze before mixing and after mixing is the same. So, if 'x' is weight of pure antifreeze in 50% solution, it is possible to make up equation before mixing: 0.5x+0.2*90. 2) there are 0.2*90=18 gal. of pure antifreeze in the 20% solution. If 'x' gal. is the weight of pure antifreeze in 50% sol. and 18 gal. is the weight of pure antifreeze in 20% sol., it is possible to make up an equation after mixing: 0.4(x+18). 3) using the both parts: 0.5x+0.2*90=0.4(x+18) ⇒ x=54 gal. of pure weight. 4) to find 50% solution of 54 gal. pure weight just 54:0.5=108 gal. Answer: 108 gal.
