Question 244989
here is another problem all worked out for you. 
For your problem just change the numbers
This time, suppose you work in a lab. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?

Let x stand for the number of liters of 10% solution, and let y stand for the number of liters of 30% solution. (The labeling of variables is, in this case, very important, because "x" and "y" are not at all suggestive of what they stand for. If we don't label, we won't be able to interpret our answer in the end.) For mixture problems, it is often very helpful to do a grid:

        	liters sol'n 	percent acid 	total liters acid
      10% sol'n 	x 	0.10 	0.10x
      30% sol'n 	y 	0.30 	0.30y
      mixture 	x + y = 10 	0.15 	(0.15)(10) = 1.5

Since x + y = 10, then x = 10 – y. Using this, we can substitute for x in our grid, and eliminate one of the variables:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

        	liters sol'n 	percent acid 	liters acid
      10% sol'n 	10 – y 	0.10 	0.10(10 – y)
      30% sol'n 	y 	0.30 	0.30y
      mixture 	x + y = 10 	0.15 	(0.15)(10) = 1.5

When the problem is set up like this, you can usually use the last column to write your equation: The liters of acid from the 10% solution, plus the liters of acid in the 30% solution, add up to the liters of acid in the 15% solution. Then:

      0.10(10 – y) + 0.30y = 1.5
      1 – 0.10y + 0.30y = 1.5
      1 + 0.20y = 1.5
      0.20y = 0.5
      y = 0.5/0.20 = 2.5

Then we need 2.5 liters of the 30% solution, and x = 10 – y = 10 – 2.5 = 7.5 liters of the 10% solution. (If you think about it, this makes sense. Fifteen percent is closer to 10% than to 30%, so we ought to need more 10% solution in our mix.)