Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 575 mL of a solution that is 60% alcohol. How much of each solution should you use?( In mL)

9514 1404 393

Step-by-step explanation:

Let x represent the amount of 80% alcohol. Then (575 -x) is the amount of 30% alcohol. The total amount of alcohol in the mix is ...

  0.30(575 -x) +0.80(x) = 0.60(575)

  172.5 +0.50x = 345

  0.50x = 172.5 . . . . . . . . subtract 172.5

  x = 345 . . . . . . . . . . . . multiply by 2; amount of 80% needed

  575 -x = 230 . . . . amount of 30% needed

You should use 230 mL of 30% alcohol and 345 mL of 80% alcohol.

Additional comment

You can work a problem like this by writing two equations in two unknowns. The variables would be the amounts of each solution (w=weak, s=strong), and the equations would reflect the total amount and the amount of alcohol in the mix.

You will notice that if you solve this by substitution, substituting for the "weak" variable, you get ...

which is the same equation we used above.

When you simplify this and isolate the variable, the coefficient of the variable is positive. This makes the arithmetic less prone to error. If you substitute for the "strong" variable, then the coefficients come out negative. That still works, but you need to spend extra effort to get the signs right.

Essentially, our choice of a single variable for the "strong" solution results in a single 2-step equation that is easily solved. A lot of mixture problems can be solved with this approach.