SOLUTION: James needs to mix a 10% saline solution with a 50% saline solution to create 200 mL of a 18% solution. How many mL of each solution must James use?
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Question 885491: James needs to mix a 10% saline solution with a 50% saline solution to create 200 mL of a 18% solution. How many mL of each solution must James use?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
a+b=200,
0.1*a+0.5*b=0.18*200
a=200-b
0.1*(200-b)+0.5*b=36
20-0.1b+0.5*b=36
0.4*b=16
b=40
a=200-b
a=160 b=40
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