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?

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

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

To solve this problem, we need to find out how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 percent.

Let's start by calculating the initial amount of antifreeze in the radiator. The radiator holds 4 liters, and it is filled with a solution that is 70 percent antifreeze, which means that 70 percent of the 4 liters is antifreeze.

Initial amount of antifreeze = 70% of 4 liters
= 0.70 * 4 liters
= 2.8 liters

Now, let's calculate how much antifreeze needs to be drained in order to reach a 50 percent concentration. We want to know how much antifreeze should remain after draining, which is 50 percent of the total volume of the radiator.

Desired amount of antifreeze = 50% of 4 liters
= 0.50 * 4 liters
= 2 liters

The difference between the initial amount and the desired amount of antifreeze will tell us how much antifreeze needs to be drained.

Amount of antifreeze to be drained = Initial amount of antifreeze - Desired amount of antifreeze
= 2.8 liters - 2 liters
= 0.8 liters

Therefore, you need to drain 0.8 liters of the coolant from the radiator.

Finally, to reach the desired 50 percent concentration, you need to replace the drained coolant with pure water. Since the coolant concentration is now 30 percent (70 percent antifreeze initially - 50 percent desired), you would need to replace the drained coolant with the same amount of pure water.

Amount of pure water to be added = Amount of antifreeze drained
= 0.8 liters

Hence, you would need to drain 0.8 liters of the coolant and replace it with 0.8 liters of pure water to reduce the antifreeze concentration to 50 percent in the radiator.