SOLUTION: How many gallons of a liquid that is 40% alcohol must be mixed with 8 gallons of one that is 30% alcohol to obtain a mixture that is 35% alcohol?

Question 508049: How many gallons of a liquid that is 40% alcohol must be mixed with 8 gallons of one that is 30% alcohol to obtain a mixture that is 35% alcohol?
Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!
Solution A
Amount = x gallons?
Concentration =40% = 0.4
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Solution B
Amount =8 gallons
Concentration =30% = 0.3
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Resultant Solution
Amount =( 8+x) gallons
Concentration = 35%=0.35
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[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant
=============================================================
Put the value in above equation
(x*0.4)+(8*0.3)=(8+x)*0.35
0.4x+2.4=2.8+0.35x
0.4x-0.35x =2.8-2.4
0.05x=0.4
Divide by 0.05 both sides of the equation
0.05x/0.05=0.4/0.05
x=8 gallons
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Solution A
Amount = 8 gallons?
Concentration =40%
============================================================
Solution B
Amount =8 gallons
Concentration =30%
=============================================================
Resultant Solution
Amount =( 8+8) gallons = 16 gallons
Concentration = 35%
=================================================================
Finally:-
8 gallons of a liquid that is 40% alcohol must be mixed with 8 gallons of one that is 30% alcohol to obtain a mixture of 16 gallons that is 35% alcoho

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