The radiator in a car is filled with a solution of 65 per cent antifreeze and 35 per cent water..

Question

The radiator in a car is filled with a solution of 65 per cent antifreeze and 35 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 5 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

The problem is actually in the wording of the prompt. No where is the word "coolant" defined.  In order to do this problem correctly, I am guessing you will have to define coolant as the solution of the antifreeze and water combined.  Defined that way, you will be draining out a 65/35 solution.  

So ... now you can rewrite your equation this way, where

 

.65(5), the amount of antifreeze currently in the radiator,

c = the amount of coolant that needs to be drained, and

.50(5), the amount of antifreeze you want in the radiator.

                         .65(5) - .65c = .50(5)

From here, you just solve for c, the total amount of coolant drained from the radiator.

 

Answer 2

5 liters of solution contains 65% of an antifreeze (A) and 35% of water (W), that is

5 * 0.65 = 3.25 L of (A)

We want 5L of 50% of (A), which will be 2.5L of (A) in the radiator.

3.25 - 2.5 = 0.75L of (A) must be removed.

To drain 0.75L of a solution that is 65% of (A) we must drain

0.75 / 0.65 ≈ 1.15 L

and add 1.15 L of (W)

Answer 3

radiator capacity is 5 liters.

currently 65% of 5 liters is antifreeze: .65 * 5 = 3.25 liters, so water is 5 - 3.25 = 1.75

since we want the percent concentration of water to coolant is 50/50: we need 2.5 liters of coolant and 2.5 liters of water.

So we remove this amount of coolant:

3.25 liters - 2.5 liters = .75 liters.