SOLUTION: A mixture containing 8% salt is to be mixed with 40 mL of a mixture that is 20% salt, in order to obtain a solution that is 12% salt. How much of the first solution must be used?
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Question 215279: A mixture containing 8% salt is to be mixed with 40 mL of a mixture that is 20% salt, in order to obtain a solution that is 12% salt. How much of the first solution must be used? Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! .08x+.20*40=.12(40+x)
.08x+8=4.8-.12x
.08x-.12x=4.8-8
-.04x=-3.2
x=-3.2/.04
x=80 ml of 8% salt is used.
Proof:
.08*80+.20*40=.12(40+80)
6.4+8=.12*120
14.4=14.4
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