ERAChas - A Simple Coolant Math Problem

[patrickt]

Chas, RodKnock's engine is now two years old and he knows he needs to change the coolant but doesn't want to hassle with burping the notorious FE air pocket that always develops, flushing the the heater core, etc. so he came up with a wonderful "simple" method for changing the coolant the "EZ way." Rod is going to simply open the petcock at the bottom of the radiator and drain one gallon of old coolant in to a catch basin while simultaneously filling his surge tank with one gallon of new coolant so that it is always full and no air gets in to the system. Just one gallon, mind you; no more. He will then take the car for a ride to thoroughly circulate the coolant, let the car cool, then drain another gallon of old coolant out while simultaneously replacing it with a new gallon. But Rod is smart, he realizes that his system holds four gallons and that while the first drain and fill replaced 25% of that old coolant with new coolant, the second drain actually had one quarter of a gallon of new coolant in it, so he was really only replacing three quarters of a gallon of old coolant. After two "drain and fills" he now has a system with two and one quarter gallons old coolant and one and three quarters new coolant. Rod says he will be happy if he can replace 90% of his old coolant with new coolant. How many times will he have to repeat his "drain one gallon while adding one gallon and then drive" task in order to achieve his goal of a 90% coolant change? Show your work for full credit.