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Question 803690: Please help me solve this : A chemist can purchase a 10% saline solution in 500 cubic centimeter containers, a 20% saline solution in 500 cubic centimeter containers, and a 50% saline solution in 1,000 cubic centimeter containers. He needs 12,000 cubic centimeters of 30% saline solution. How many containers of each type of solution should he purchase in order to form this solution?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A chemist can purchase a 10% saline solution in 500 cubic centimeter containers, a 20% saline solution in 500 cubic centimeter containers, and a 50% saline solution in 1,000 cubic centimeter containers.
He needs 12,000 cubic centimeters of 30% saline solution.
How many containers of each type of solution should he purchase in order to form this solution?
:
It will greatly simplify things if we just use the 10% and the 50% solutions
:
Let x = amt (in cu/cm) of 50% solution
the resulting amt is 12000 cu/cm, therefore
(12000-x) = amt of 10% solution
:
A typical mixture equation
:
.10(12000-x) .50x = .30(12000)
1200 - .10x + .50x = 3600
-.10x + .50x = 3600 - 1200
.40x = 2400
x = 2400/.4
x = 6000 cu/cm of the 50% solution
then also
6000 cu/cm of the 10% solution
:
The requirement then
6000/1000 = 6 ea containers of the 50% solution
and
6000/500 = 12 ea containers of the 10% solution
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