<PRE><B>Question 309557</B>
How many liters of a 10% alcohol solution must be mixed with 40L of a 50% solution to get a 40% solution?

let 10% alcohol required be x L
Concentration of 10% alcohol in the mix = 0.1x
.
40 L of 50% alcohol
concentration of this alcohol in the mix = 40*0.5= 20
.
Mixture concentration = 40% . quantity = (x+40)
= 0.4(x+40)
.
0.1x +20 =0.4(x+40)
0.1x +20 = 0.4x +16

-0.3x = -4
x= 13.33 liters


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