How many liters of a 30% alcohol solution must be mixed with 90 liters of a 60% solution to get a 40% solution

Question

How many liters of a 30% alcohol solution must be mixed with 90 liters of a 60% solution to get a 40% solution

Answer 1

Here is a very useful method when dealing with these types of problems of dilution.  It is called alligation, if you care to research it further.

You set up a simple grid, like a tic tac toe, with "what you have" on the left; "what you want in the middle; "what you need" on the right.  Then you subtract the middle value criss cross wise from the left side to get "what you need".  In this question, it would look like this:

 

HAVE...........WANT...........NEED......

30%................................ 20........

....................40%........................

60%.................................10.......

 

So, you need 20 parts (liters, mls, ounces, etc.) of the 30% solution and 10 parts of the 60%.  Since the question tells you that you must use 90 liters of the 60% (that is 9 x the calculated "parts") you will thus need 9 x the 20 "parts" or 180 liters of the 30% solution.

 

Once you do this a few times, it is a very simple method to use to apply to many types of dilution problems.

Answer 2

90 liters at 60% solution means

90*.60 = 54 liters alcohol and

90*.40 = 36 liters water.

 

Now you are adding an unknown amount of solution, call it x

 

.3x of it is alcohol and .7x of it is water

 

40% solution means the ratio of alcohol to total liquid is 40/100

So our new amount of alcohol will be 54 + .3x and our new total amount of liquid is 90 +x.

 

So (54+.3x)/(90 + x) = 40/100   cross multiply to get

5400+30x = 3600+40x

1800        =      10x

180          =        x

 

let's check.

We had 90 liquid which contained54 liters of alcohol

We are adding 180 liters of liquid which contains (.30)(180)=54 liters of alcohol

 

So we have 108 liters of alcohol in 270 liters of liquid...108/270 = .4 or 40%......so our answer is correct.

 

Hope this helped.