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Question 256527: cara is preparing an acid solution. She needs 200 milliliters of 48% concentration solution. Cara has 60% and 40% concentration solutions in her lab. How many milliliters of 40% acid solution should be mixed with 60% acid solution to make the required amount of 48% acid solution
can you please solve it using matrices, when I try it change the problem to a matrix equation, and then try to solve it, but I never get the right answer.
Thank You!!
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! This is a mixture problem. Here is the table based on the given information:
ACID . . . . . . . .% . . . . . . . . . .ML . . . . . . . .%ML
60% . . . . . . . . .60 . . . . . . . . . m . . . . . . . .60m
40% . . . . . . . . .40 . . . . . . . .200-m . . . .. 8000-40m
MIX . . . . . . . .. . 48 . . . . . . .. 200 . . . . . . .9600
In the 3rd column add the first 2 to get the last number. We get
60m + 8000 - 40m = 9600
20m + 8000 = 9600
20m = 1600
m = 80
200-m = 120.
She needs 120ML of 40% acid.
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