Question 571243: By weight one alloy has 40% Aluminum, 38% Copper and the rest Cadmium. The second alloy has 52%, 20% and 28% Aluminum, Copper, and Cadmium respectively. The Aluminum, Copper, and Cadmium content of the third alloy is 61%, 12% and 27% respectively. How much of each alloy must be mixed to produce 1000 Lbs of alloy with the following content of 50.2% Aluminum, 23.8% Copper and 26% Cadmium? I am completely stumped on how to figure this one out. Can someone help Please?
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! let x = amount of 1st alloy , y = amount of 2nd alloy , and z = amount of 3rd alloy
you are given the amounts of the components in the final alloy
___ Al = (50.2%)(1000) = 502 lb
___ Cu = (23.8%)(1000) = 238 lb
___ Cd = (26%)(1000) = 260 lb
.4x + .52y + .61z = 502
.38x + .2y + .12z = 238
.22x + .28y + .27z = 260
solve the system of equations to find the amounts of the three alloys
you might want to think about Cramer's Rule
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