Anne and Nancy use a metal alloy that is 18.3% copper to make jewelry. How many ounces of a 15% alloy must be mixed with a 24% alloy to form 90 ounces of the desired alloy?
Although there are several valid calculation procedures for this question (see other answers),
chemists prefer simplify the calculations needed by using a general formula that applies to all mixing operations.
Two quantities (h and l ) can be combined or mixed to obtain a desired intermediate quantity i in accordance with:
L= h-i /(h-i) +(i-l) = h-i / h-l and H= i-l / h-l
Where l and h are the lower and higher quantity and H and L are the parts of the higher and lower quantities respectively to be mixed to obtain one part of the desired quantity.
In our case:
h=24; l= 15; and i =18.3 , hence L=5.7/9 and H-3.3/9
And to prepare 90 parts (oz.) of the product:
90L = 5.7 x 90/9 = 57oz. and 90H = 3.3x 90/9 =33 oz.