Question 534007: A metal smith wishes to create an alloy of 20%,silverand12% gold. She has three raw materials to work with. One is 10%,silverand10% gold the second is 25%,silverand10% gold, and the final metal is 30%,silverand15% gold. In what proportions should she combine the three to produce the desired alloy?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A metal smith wishes to create an alloy of 20% silver and 12% gold.
One is 10% silver and 10% gold the second is 25% silver and 10% gold, and the final metal is 30% silver and 15% gold.
In what proportions should he combine the three to produce the desired alloy?
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Let a = amt of 1st alloy
let b = amt of 2nd
let c = amt of 3rd
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Let's assign a value to the resulting amt. 120 units is good value
Obviously besides gold and silver, a percentage of filler is in each alloy
The remaining amt when you subtract the silver and gold.
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Write an equation for each substance
Silver:
.10a + .25b + .30c = .20(120)
.10a + .25b + .30c = 24
Gold:
.10a + .10b + .15c = .12(120)
.10a + .10b + .15c = 14.4
Filler:
.80a + .65b + .55c = .68(120)
.80a + .65b + .55c = 81.6
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Subtract the 2nd equation from the 1st
.10a + .25b + .30c = 24
.10a + .10b + .15c = 14.4
----------------------------subtraction eliminates a
.15b + .15c = 9.6
:
Multiply the 2nd equation by 8, subtract the 3rd equation
.80a + .80b + 1.2c = 115.2
.80a + .65b + .55c = 81.6
-----------------------------subtraction eliminates a
.15b + .65c = 33.6
.15b + .15c = 9.6
------------------Subtraction eliminates b, find c
.50c = 24
c = 48 units of the 3rd alloy
Find b
.15b + .15(48) = 9.6
.15b = 9.6 - 7.2
.15b = 2.4
b = 2.4/.15.10a + .25b + .30c = 24
b = 16 units of the 2nd alloy
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Find a
.10a + .25b + .30c = 24
.10a + .25(16) + .30(48) = 24
.10a + 4 + 14.4 = 24
.10a = 24 - 18.4
.10a = 5.6
a = 56 units of the 1st alloy
:
a:b:c
56:16:48
reduce using 8 as a divisor
7:2:6 is the proportion of the 3 alloys
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I'll let you confirm that the math is correct here
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