Question

The radiator in a car is filled with a solution of 70 per cent antifreeze and 30 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 4 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?

Answer 1

let's just look at the antifreeze,

right now the volume of antifreeze is .7(4) L or
2.8 L
Let the amount of the current fluid we are draining be x L
Now 70% of that is antifreeze, so the amount of antifreeze we lose is .7x
So the actual amount of antifreeze left in the rad
is 2.8 - .7x L
But that be enough to produce our required 50% solution, if we pour in x L of straight water

2.8-.7x + 0(x) = .5(4) --> my equation only shows antifreeze
-.7x = 2-2.8
.7x = .8
x = .8/.7 = 8/7 = appr 1.14 L

Answer 2

reduce by v, then the volume left if 4-v

antifreeze conc*volume+solutionconc*volume=.5*4

1*v+.7(4-v)=2
solve for v

Answer 3

The radiator in a car is filled with a solution of 70 per cent antifreeze and 30 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the radiator is 3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?