Question 511886
(1)
How many liters of a 40% and 80% acids must be mixed in order to make 20 liters of a 55% acid solution? 

Solution A
Amount = x 
Concentration =40% = 0.4

Solution B
Amount = 20-x 
Concentration =80% = 0.8

Resultant Solution
Amount =20 
Concentration =55%=0.55

[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant

(0.4)x+(0.8)(20-x)=(20)(0.55)
0.4x+16-0.8x=11
0.4x-0.8x=11-16
-0.4x=-5
-0.4x/-0.4=-5/-0.4
x=12.5

Solution A
Amount = x =12.5 liters 
Concentration =40% = 0.4

Solution B
Amount = 20-x =20-12.5=7.5 liters
Concentration =80% = 0.8

Result
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12.5  liters of a 40% and 7.5 liters of 80% acids must be mixed in order to make 20 liters of a 55% acid solution
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(2)
A 12 liter mixture is 60% alcohol. if 8 liters of a 25% alcohol mixture are added, what percent mixture will be formed.

Solution A
Amount =12 liters
Concentration =60% = 0.6

Solution B
Amount = 8 liters 
Concentration =25% = 0.25

Resultant Solution
Amount =(8+12)=20 liters
Concentration =x

[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant
(12*0.6)+(8)(0.25)=20x
7.2+2=20x
9.2=20x
9.2/20=20x/20
0.46=x
x=0.46
x=46/100= 46%

Result
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If we add  12 liter 60% alcohol solution to  8 liters of a 25% alcohol solution, a  mixture of 20 liters of 46% alcohol will be formed.