Weighted AveragesDate: 11/30/2001 at 22:32:26 From: Tyler Subject: Weighted averages I cannot figure out the correct formula to use. I think if I knew how to set it up I could do it. Problem: An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice? Can you help me? Date: 11/30/2001 at 23:50:22 From: Doctor Paul Subject: Re: Weighted averages You have 5 quarts of "drink" that is 10% orange juice. Thus the 5 quarts of drink contain 5*10% = 5*.1 = .5 quarts of orange juice, and hence must contain 4.5 quarts of other stuff. Now we're going to add 100% orange juice and we want to know how much to add so that the new mixture will be 40% orange juice. Well, what if you add 1 quart of 100% orange juice? What would that do to the concentration of orange juice in the mixture? It would mean that we now have 1.5 quarts of pure orange juice mixed in a carton that now contains 6 quarts of liquid. Thus the concentration of orange juice in this 6-quart jug would be 1.5/6 = .25 = 25% So we haven't added enough, but maybe you see what's going on. In a more general situation: If we add x quarts of pure orange juice to the original mixture that contains .5 quarts of OJ and 4.5 quarts of something else, then what we have is a mixture of 5+x quarts that contains .5+x quarts of OJ. So after we've added x quarts of pure OJ, the concentration of OJ will be (.5+x)/(5+x) We want to know when this will be equal to .4 So solve for x: (.5+x)/(5+x) = .4 .5+x = 2 + .4x .6x = 1.5 x = 1.5/.6 = 2.5 Thus you should add 2.5 quarts of pure OJ to make the concentration of the new mixture 40% OJ. I hope this helps. Please write back if you'd like to talk about this some more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/