How many liters of a 30% alcohol solution must be mixed with 20 liters of a 80% solution to get a 70% solution?
1 Answer
Answer:
Explanation:
I'll assume the given percentages are those by volume (i.e.
We're asked to find the volume, in
To do this, we can set up sort of an algebraic equation that represents this situation.
Here, we have

#x# liters of#0.3# fractional volume solution 
#20# liters of#0.8# fractional volume solution
These two added together must be a solution with fractional volume
#ul(overbrace(x)^"volume of 30%"(0.3) + overbrace((20))^"volume of 80%"(0.8) = overbrace((x+20))^"total volume"(0.7)#
Now, we simply solve for
#0.3x + 16 = 0.7x + 14#
#0.4x = 2#
#color(red)(ul(x = 5#
So the required volume of the
#color(red)(ulbar(stackrel(" ")(" "5color(white)(l)"L"" "))#