Question 466638
How many liters (L) of a 20% alcohol solution must be mixed with 60 L of a 90% alcohol solution to get a 50% alcohol solution?
...
x=liters of 20% alcohol solution to be mixed with 60 liters of 90% alcohol solution.
x+60 =liters of 50% alcohol solution
20%x+90%(60)=50%(x+60)
.2x+54=.5x+30
.3x=24
x=80
ans:
80 liters of a 20% alcohol solution must be mixed with 60 L of a 90% alcohol solution to get a 50% alcohol solution.