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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?

Help is greatly appreciated!

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.

Regards

Edmondo C.
x = oz of 15% Cu
y = oz of 24% Cu

x + y = 90  =>  y = 90 - x

(0.15 x + 0.24 y)/90 = 18.3%
Multiply both sides by 100:
(15 x + 24 y)/90 = 18.3
Multiply both sides by 90:
15 x + 24 y = 90(18.3)
Substitute 90 - x for y:
15 x + 24 (90 - x) = 90 (18.3)
15 x - 24 x = 90 (18.3 - 24)
- 9 x = 90 (-5.7)
x = 10 (5,7) = 57 oz
Let F = amount of 15% alloy
Let T = amount of 24% alloy
F     +     T  =90
.15F +  .24T=90×.183
Rewrite as
F      +     T    =  90
150F+240T    =90×183=16470
multiply first equation by 240 and have
240F  +  240T= 21600
150F  +  240T= 16470
subtract
90F               = 5130
F=57 and so T =33