SOLUTION: how many mL of sterile water will need to be mixed with 60 mL of a 60% solution to form a 40 % solution? what will the total volume of the 40% solution ?

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Question 1158615: how many mL of sterile water will need to be mixed with 60 mL of a 60% solution to form a 40 % solution? what will the total volume of the 40% solution ?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
                      %           vol         solute

STERILE WATER          0           x             0

SOLUTION              60          60            60*60

MIXTURE RESULT        40          x+60          40(x+60)

%2860%2A60%29%2F%28x%2B60%29=40
-
3600=40%28x%2B60%29
40x%2B2400=3600
40x=800
highlight%28x=20%29-----ml. of the sterile water.

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Algebraically....

x mL of 0% solution, plus 60 mL of 60% solution, equals (x+60) mL of 40% solution:

0%28x%29%2B.60%2860%29+=+.40%28x%2B60%29

Solve using relatively simple algebra.

And here is an informal way to solve this kind of mixture problem quickly and easily.

You are starting with 60% solution and adding 0% solution, stopping when you get to 40%.
40 is one-third of the way from 60 to 0.
Therefore, 1/3 of the mixture is what you are adding.
That means the ratio of 60% solution to 0% solution is 2:1.
Since the amount of 60% solution is 60 mL, the amount of 0% solution required is half of that, which is 30mL.

Another variation of the informal approach is to reason that the target 40% is "twice as close" to 60% as it is to 0%; and therefore the mixture needs to have twice as much of the 60% solution as it has of the 0% solution. So the 60 mL of the 60% solution is twice as much as the amount you need of the 0% solution, leading again to the answer that you need 30 mL of the 0% solution.