Question 1190141: A truck radiator holds 5 gal and is filled with water. A gallon of water is removed from the radiator and replaced with a gallon of antifreeze; then a gallon of the mixture is removed from the radiator and again replaced by a gallon of antifreeze. This process is repeated indefinitely. How much water remains in the tank after this process is repeated 3 times? 5 times? n times?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Each time we drain out 1 gallon of mix, we lose 1/5 = 0.20 = 20% of its contents. This means we keep the remaining 80% of it.
If we started with 5 gallons of water, then after one drain cycle we'd have 5*0.8 = 4 gallons of water left. We dumped 1 gallon of water, to have it be replaced with 1 gallon of antifreeze.
After another draining, we lose another 20% of the mix and keep the remaining 80%
Water: 80% of 4 = 0.80*4 = 3.2 gallons remain
Antifreeze: 80% of 1 = 0.80*1 = 0.8 gallons remain
After the second draining, we have 3.2 gallons of water left. We can write that as 5*0.8^2 since
5*0.8^2 = 0.80*(5*0.8)
After doing this n times, we'd have 5*(0.8)^n gallons of water left.
Effectively, we're just applying the 80% multiplier n times to find out how much water is left after each drain cycle.
Plug in n = 3 to find that the expression above results in 2.56
If n = 5, then that expression is equal to 1.6384 gallons.
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