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An alloy weighing 30 lbs. is 11% tin. The alloy was made by mixing a 15% tin alloy and a 9% tin alloy. How many pounds of each alloy were used to make the 11% alloy?
Weegy: 3.3 let x = lbs of 15% alloy 30 - x = lbs of 9% alloy so 0.15x + 0.09(30 - x) = 0.11(30)=3.3 User: An alloy of copper is 10% copper and weighs 25 pounds. A second alloy is 18% copper. How much (to the nearest tenth lb.) of the second alloy must be added to the first alloy to get a 13% mixture. Weegy: ok, so the total weight of the first alloy is 25 pounds and since it is 10% copper, that means that 2.5 pounds is copper. We want to have 13% copper. [ remember that percentage is the following: (part/whole)*100 so to match our problem: (total copper/total weight)*100 Well right now we have 2.5 pounds of cppper and a total weight of 25. We are going to add some of the 2nd alloy, but we dont' know how much, so let's say x. that means we have a total of 25 + x pounds of total weight and : 2.5 + .18(x) total copper. So we have: (2.5+.18x)/(25+x) = 13/100 because it's supposed to equal 13% cross multiply: 250 + 18x = 325 + 13x 5x = 75 x = 15 therefore we must add 15 pounds of alloy 2 ] (More)
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Asked 5/16/2012 9:04:35 AM
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