SOLUTION: Deon has 12 milliliters of a 5% solution. He also has a solution that has a concentration of 30%. How many milliliters of the 30% solution does Deon need to add to the 5% solution

Question 568749: Deon has 12 milliliters of a 5% solution. He also has a solution that has a concentration of 30%. How many milliliters of the 30% solution does Deon need to add to the 5% solution to obtain a 20% solution?
Found 2 solutions by stanbon, mananth:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Deon has 12 milliliters of a 5% solution. He also has a solution that has a concentration of 30%. How many milliliters of the 30% solution does Deon need to add to the 5% solution to obtain a 20% solution?
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Equation:
active + active = active
0.05*12 + 0.30x = 0.20(12+x)
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Multiply thru by 100 to get:
5*12 + 30x = 20*12 + 20x
10x = 15*12
x = (3/2)12
x = 18 ml (amt of 30% solution needed)
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Cheers,
Stan H.
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Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
------------------------ percent ---------------- quantity
Solution Type I 12 ---------------- 5 ml
solution type II 30 ---------------- x ml
Mixture 20 ---------------- 5 + x ml
The sum of individual contents of the two types of components =content in mixture
12*5+30x=20(5+x)
60+30x =100+20x
30x-20x=100-60
10x=40
/10
x=4 ml of solution type II

m.ananth@hotmail.ca

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