# Thread: Mixture Problem

1. ## Mixture Problem

No matter what I try, I just cannot figure this one out. Like I mentioned on the other thread, I'm not having any trouble solving them, just modeling them.

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 3.6 L, how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?

2. Let x be the liters of 60% antifreeze coolant and y be liters of pure water.

Then: x + y = 3.6
.6x + 0y = .5(3.6)

Solve the system for x and y, and of course your answer will be y.

3. let x = number of liters to drain and fill ...

3.6(60%) - x(60%) + x(0%) = 3.6(50%)

solve for x

4. Originally Posted by icemanfan
Let x be the liters of 60% antifreeze coolant and y be liters of pure water.

Then: x + y = 3.6
.6x + 0y = .5(3.6)

Solve the system for x and y, and of course your answer will be y.
Thank you so much!

I got the solution .6L. The only thing I don't quite understand is the set up. I mean, of course I know it's right. I was just wondering why you put everything where you did? I understand the x+y=3.6 just fine, it's the 0y I don't exactly understand. Because aren't you adding water? I don't know why I can't get this modeling down.. once I see it done it looks so easy.

5. Originally Posted by Chinnie15
it's the 0y I don't exactly understand. Because aren't you adding water?
No, because the water you are adding doesn't have any antifreeze in it. The second equation I wrote represents a relationship that equates the total amount of antifreeze (in liters) on both sides.