Word Problems

Question

I am stumped...

 

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!

Answer 1

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.

Answer 2

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

Answer 3

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