SOLUTION: A chemist needs a saline solution that is 20% sodium chloride but only has solutions that are 15% and 40% sodium chloride. If the chemist measures 150 mL of the 15% solution, how m
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-> SOLUTION: A chemist needs a saline solution that is 20% sodium chloride but only has solutions that are 15% and 40% sodium chloride. If the chemist measures 150 mL of the 15% solution, how m
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Question 313237: A chemist needs a saline solution that is 20% sodium chloride but only has solutions that are 15% and 40% sodium chloride. If the chemist measures 150 mL of the 15% solution, how many mililiters of the 40% solution should she add to make 20% solution? Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! .40X+.15*150=.20(150+X)
.40X+22.5=30+.20X
.40X-.20X=30-22.5
.20X=7.5
X=7.5/.20
X=37.5 ml OF THE 40% SOLUTION IS NEEDED.
PROOF:
.40*37.5+.15*150=.20(150+37.5)
15+22.5=.20*187.5
37.5=37.5
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