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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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Suppose the chemist adds x grams of 70% iron alloy. Then the resulting alloy will have an iron content of 0.7*(x/(16+x)) + 0.2*(16/(16+x)) You want the iron content to be 0.3, so set the above equation equal to 0.3 and solve for the number of grams x required. 0.7x/(16+x) + 3.2/(16+x) = 0.3 0.7x + 3.2 = 4.8 + 0.3x I'll leave it to you from here
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