SOLUTION: a chemist needs a 30% iron alloy. how many grams of a 70% iron alloy must be mixed with 16 grams of a 20% iron alloy to obtain the required 30% alloy

Question 303480: a chemist needs a 30% iron alloy. how many grams of a 70% iron alloy must be mixed with 16 grams of a 20% iron alloy to obtain the required 30% alloy
Found 3 solutions by mananth, stanbon, richwmiller:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
a chemist needs a 30% iron alloy. how many grams of a 70% iron alloy must be mixed with 16 grams of a 20% iron alloy to obtain the required 30% alloy
let 70% alloy required be x grams
20% alloy 16 grams
Total alloy 30% x+16 grams
0.7x+0.2*16 = 0.3(x+16)
0.7x+3.2 =0.3x+4.8
0.7x-0.3x=4.8-3.2
0.4x=1.6
x= 1.6 / 0.4
4 grams of 70% alloy
Ananth

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
a chemist needs a 30% iron alloy. how many grams of a 70% iron alloy must be mixed with 16 grams of a 20% iron alloy to obtain the required 30% alloy
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Equations:
alloy + alloy = alloy
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0.70x + 0.20*16 = 0.30(x+16)
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Multiply thru by 100 and solve for "x":
70x + 20*16 = 30x + 30*16
40x = 10*16
x = 4 grams (amt. of 70% alloy needed for the mixture)
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Cheers,
Stan H.
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Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
.7x+.2*16=.3*(x+16)
x=4
.7*4+.2*16=.3*20
.28+.32=.60
ok
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