SOLUTION: WORD PROBLEM: How much water should be added to 8 ML of 6% Saline solution to reduce the concentration to 4%? Please Help?

Question 54843: WORD PROBLEM:
How much water should be added to 8 ML of 6% Saline solution to reduce the concentration to 4%?
Please Help?
Found 2 solutions by stanbon, checkley71:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How much water should be added to 8 ML of 6% Saline solution to reduce the concentration to 4%?
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Amount of 6% solution is 8ML.
Amount of active ingredient is 0.06(8)= 0.48ML
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Amount of water to add is "x"ML
Amount of active ingredient is 0ML
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Amount of mixed solution is "8+x"ML
Amount of active ingredient is 0.04(8+x)= 0.32+0.4x
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EQUATION:
active ingre. + active ingre. = active ingre
0.48 + 0 = 0.32 + 0.04x
0.16 = 0.04x
x=4ML (amount of water that must be added)
Cheers,
Stan H.

Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
8*.06=(8+X).04
.48=.32+.04X
.04X=.48-.32
.04X=.16
X=.16/.04
X=4 ADDITIONAL L OF WATER NEEDED TO CHANGE THE MIX FROM 6% TO 4%.
PROOF
8*.06=(8+4).04
.48=12*.04
.48=.48
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