Question

The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 4.3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50%?

Answer 1

The radiator contains 0.6*4.3=2.58l of antifreeze and 0.4*4.3=1.72l of water. If x litres of solution are drained then the  amounts of antifreeze and water become 0.6(4.3-x) and 0.4(4.3-x). We now add x litres of water, so the water content becomes 0.4(4.3-x)+x=1.72-0.4x+x=1.72+0.6x. The radiator is full again, so the proportion of water is now (1.72+0.6x)/4.3=0.5 or 50%. 1.72+0.6x=0.5*4.3=2.15. x=(2.15-1.72)/0.6=0.43/0.6=0.717l approx. Therefore about 0.72 litres of the solution need to be drained and replaced by pure water.

Check: The water content is 1.72+0.43=2.15 litres and the antifreeze content is 0.6(4.3-0.43/0.6)=0.6*3.583=2.15 litres, which is a 50% solution of antifreeze.