Question 346583: The amounts (by weight) of gold, silver, and lead in three alloys of these metals are in the ratio 4:3:2 in the first alloy, 3:5:1 in the second, and 2:2:5 in the third. It is desired to make an alloy containing equal amounts (by weight) of gold, silver, and lead. How many ounces of each alloy should be used for every 10 ounces of the new mixture?
thanks........
Answer by jrfrunner(365)   (Show Source):
You can put this solution on YOUR website! given
three alloys; A1,A2,A3
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weight equation 1: A1+A2+A3=10
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make up of each alloy: Gold, silve and lead ratio 4:3:2 in the first alloy, 3:5:1 in the second, and 2:2:5 in the third
want: It is desired to make an alloy containing equal amounts (by weight) of gold, silver, and lead
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Gold: 4/9*A1+3/9*A2+2/9*A3=10*1/3
silver: 3/9*A1+5/9*A2+2/9*A3=10*1/3
lead: 2/9*A1+1/9*A2+5/9*A3=10*1/3
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multiply the gold, silver and lead equations by 9 to eliminate fractions
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Gold equation 2: 4A1+3A2+2A3=30
Silver equation 3: 3A1+5A2+2A3=30
lead equation 4: 2A1+A2+5A3=30
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subtract equation 3 from equ 2: A1-2A2=0 or A1=2A2
mult eq 3 by 5 15A1+25A2+10A3=150
mult eq 4 by -2 -4A1-2A2-10A3=-60
add these equtions 11A1+23A2=90
substitute A1=2A2 into above equation
22A2+23A2=90 ---> 45A2=90 or A2=2
since A1=2A2=4
since A1+A2+A3=10 then A3=4
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Validate answers by taking the original equaitons substituting these values
ie lead: 2/9*A1+1/9*A2+5/9*A3=10*1/3
lead: 2/9*4+1/9*2+5/9*4=10*1/3
8/9+2/9+20/9=10/3
30/9=10/3 checks
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