SOLUTION: How many liters of an antifreeze that is 18% alcohol must be mixed with an antifreeze that is 10% alcohol to produce an antifreeze that is 15% alcohol

Let "x" be a volume of the 18% antifreeze and "y" be a volume of the 18% antifreeze to mix.

Then the total volume is x+y;

The volume of the pure antifreeze in x liters of the 18% antifreeze is 0.18*x;

The volume of the pure antifreeze in y liters of the 10% antifreeze is 0.10*y;

The volume of the pure antifreeze in the mixture is 0.18x + 0.10y.

The condition on the concentration of the mixture says and requires 

 = 0.15,   or

0.18x + 0.1y = 0.15*(x+y),  or

0.18x - 0.15x = 0.15y - 0.1y,   or

0.3x = 0.5y,

Then y =  = ,  or   = .


In other words, the ratio of the volume of the 10% antifreeze to the volume of the 18% antifreeze must be ,  or  0.6.


For example, you can take 3 liters of the 10% antifreeze and 5 liters of the  18% antifreeze.

Let's check it:   =  =  = 0.15.   Correct.

Answer.  The ratio of the volume of the 10% antifreeze to the volume of the 18% antifreeze must be ,  or  0.6.