application problem

Question

one solution contains 40% alcohol. A second contains 60%. You need 20 L of 45% solution. How many L of the 40% should be used. I have to write an equation and solve

Answer 1

Hi Elizabeth, mixture problems are best set up in a table. The table will have three rows and three columns. Let "x" be the unknown number of liters needed of the 40% solution.

 

We do not know how many liters of the 60% solution are needed, but we do know the total number of liters of the 45% solution that is needed, or 20 liters.

 

If we need "x" number of 40% solution, then we can subtract x from the 20 liters we need of 45% solution to get the total number of 60% solution.

 

It would look something like this:

 

Percent      Liters         Totals      

------------------------------------------------

  40%             x           .40(x)

 

  60%       20 - x          .60(20 - x)

 

  45%          20            .45(20)

 

The equation becomes:

 

.40(x) + .60(20 - x) = .45(20)

 

Multiply all terms by 100 to clear the decimals:

 

40(x) + 60(20 - x) = 45(20)

 

40x + 1200 - 60x = 900

 

Rearrange terms giving:

 

60x - 40x = 1200 - 900

 

This gives:

 

20x = 300

 

x = 15, or we need 15 liters of 40% solution.

 

Quiestions?