Concentration % 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 4.1 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?

So basically all I understand is that whatever the equation is will be set equal to .5(4.1)

Re: Concentration % Question??

You have the right idea: we want 0.5 * 4.1 = 2.05 liters of antifreeze in the radiator when finished.

The radiator currently contains 0.7 * 4.1 = 2.87 liters of antifreeze.

So we must remove 2.87 - 2.05 = 0.82 liters of antifreeze.

To accomplish this, we must remove 0.82 / 0.70 liters of the 70% mix.

Re: Concentration % Question??

Another way: Let L be the amount of radiator fluid remove, in liters. Each liter contains .7 liters of anti-freeze and .3 liters of water so you will have removed .7L liters of anti-freeze. Initially, the radiator contained 4.1(.7)= 2.87 liters of anti-freeze so after draining, it will contain contain 2.87- .7L liters of anti-freeze. Adding water to bring the total fluid back to 4.1 liters, you want $\displaystyle \frac{2.87- .7L}{4.1}= 0.5$.

Solve that for L.

Re: Concentration % Question??

Hello, samiileigh94!

Yet another approach . . .

Quote:

The radiator in a car is filled with a solution of 70% antifreeze and 30% water.

The manufacturer of the antifreeze suggests that for summer driving, it should br 50% antifreeze.

If the capacity of the raditor is 4.1 liters, how much coolant (in liters) must be drained

and replaced with pure water to reduce the antifreeze concentration to 50%?

We consider the amount of *water* at each stage.

We start with 4.1 liters which is 30% water.

. . It contains: $\displaystyle 0.3(4.1) \,=\,1.23$ liters of water. . + 1.23

We remove $\displaystyle x$ liters of the mixture which is 30% water.

. . So we remove: $\displaystyle 0.3x$ liters of water. .- 0.3x

Then we add $\displaystyle x$ liters of pure water.

. . This has, of course, $\displaystyle x$ liters of water. .+ x

The result is 4.1 liters which is 50% water.

. . The final mixture contains: $\displaystyle 0.5(4.1) = 2.05$ liters of water. .2.05

And there is our equation . . . .$\displaystyle 1.23 - 0.3x + x \:=\:2.05$