SOLUTION: how many liters of pure alcohol must be added to 15 liters of 20% solution to obtain a mixture which is 30 % alcohol?

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Question 513002: how many liters of pure alcohol must be added to 15 liters of 20% solution to obtain a mixture which is 30 % alcohol?
Answer by Maths68(1474) About Me  (Show Source):
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Solution A
Amount = x
Concentration = 100%=1
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Solution B
Amount = 15
Concentration = 20%=0.2
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Solution Final
Amount=15+x
Concentraction = 30% =0.3
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(Amount A * Concentration A)+(Amount B * Concentration B)=(Amount final * Concentration final)
(1*x)+(15*0.2)=(15+x)*0.3
x+3=4.5+0.3x
x-0.3x=4.5
0.7x=4.5
0.7x/0.7=4.5/0.7
x=6.428
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6.428 liters of pure alcohol must be added to 15 liters of 20% solution to obtain a mixture which is 30 % alcohol.