Question

A solution of 78% alcohol is to be mixed with a solution of 23% alcohol to form 165 L of a 58% solution. How many liters of the 78% solution must be used?

Answer 1

Use algebra:
Let x = amount of 78% alcohol solution needed in Liters
Let y = amount of 23% alcohol solution needed in liters
What we are told. We will have 165 liters of the resulting 58% solution. With this given info we can now build ourselves a system of equations. We will need two equations as we have two unknowns (x, and y)
x + y = 165 Liters
For our 2nd equation, I choose to find out how much actual alcohol (100%) we have in our resulting mixture. Here is how that is done. 165 at 58% would result in (.58)(165) or 95.7 liters of alcohol. This fact gives us our 2nd equation which is:
.78x + .23y = 95.7 think about this a little (why is it true?)

Answer 2

Our system now looks like this:
1. x + y = 165 Liters
2. . .78x + .23y = 95.7 I don't like the decimals in No. 2 Multiply thru by 100
2. 78x + 23y = 9570 voila no decimals.
I choose to use the substitution method to solve this.
x + y = 165 , the x = 165-y Now in No. 2 I substitute this for the x getting:
78(165-y) + 23y =9570 the rest is algebra
12870 - 78y + 23y = 9570
12870 -55y = 9570
-55y = -3300
y = 60 Liters
x = 165-60 = 105 Liters
x is the desired answer requested by the problem. 105 Liters of the 78% solution must be used.