# SOLUTION: How many grams of pure silver must be mixed with 450grams of an alloy that is 10% copper in order to produce a sterling silver that is 7.5% copper?

Algebra ->  Algebra  -> Equations -> SOLUTION: How many grams of pure silver must be mixed with 450grams of an alloy that is 10% copper in order to produce a sterling silver that is 7.5% copper?      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Algebra: Equations Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Equations Question 113878: How many grams of pure silver must be mixed with 450grams of an alloy that is 10% copper in order to produce a sterling silver that is 7.5% copper?Found 2 solutions by Edwin McCravy, filboi:Answer by Edwin McCravy(8908)   (Show Source): You can put this solution on YOUR website!``` How many grams of pure silver must be mixed with 450 grams of an alloy that is 10% copper in order to produce a sterling silver that is 7.5% copper? We have three things to consider: 1. the first stuff, 2. the second stuff, and 3. the final stuff that we get after we mix the first stuff with the second stuff. Each of those "stuffs" contains some pure copper, and/or some pure silver, and/or some other stuff. So we mix: 1. x grams of the first stuff that is 100% silver and therefore 0% copper and 0% other metal. with 2. 450 grams of the second stuff that is 10% copper and therefore 90% other metal and 0% silver and end up with 3. x+450 grams of the final stuff that is 7.5% copper and therefore 92.5% silver and other metal. Note: it isn't necessary to deal with anything but the copper, but I think it makes it easier to deal with all the ingredients. Make this chart: grams of grams of grams of grams of stuff silver copper other metal first stuff second stuff -------------------------------------------------------------- final stuff Fill in the number of grams of the first, second and final stuff. That is, x grams of the first stuff, 450 grams of the second stuff, and therefore x+450 grams of the final stuff: grams of grams of grams of grams of stuff silver copper other metal first stuff x second stuff 450 -------------------------------------------------------------- final stuff x+450 Next fill in the number of grams of silver only, by multiplying the grams of the first and second stuff by the percentage of silver expressed as a decimal. The first stuff is 100% silver. The second stuff is 0& silver. Then since the second stuff contains no silver, the final stuff contains exactly the same amount of silver as the first stuff. grams of grams of grams of grams of stuff silver copper other metal first stuff x 1.00x second stuff 450 0.00(450) ---------------------------------------------------------- final stuff x+450 1.00x Next fill in the number of grams of copper only, by multiplying the grams of each stuff by their percentage of copper expressed as a decimal. grams of grams of grams of grams of stuff silver copper other metal first stuff x 1.00x 0.00x second stuff 450 0.00(450) 0.10(450) ---------------------------------------------------------- final stuff x+450 1.00x 0.075(x+450) Next fill in the number of grams of other metal only, by multiplying the grams of the first and final stuff by their percentage of silver expressed as a decimal. Then since the first stuff contains no "other metal", the final stuff contains exactly the same amount of other metal as the second stuff. grams of grams of grams of grams of stuff silver copper other metal first stuff x 1.00x 0.00x 0.00x second stuff 450 0.00(450) 0.10(450) 0.90(450) --------------------------------------------------------------- final stuff x+450 1.00x 0.075(x+450) 0.90(450) Now we can make the "stuff equation" out of the first column: x + 450 = x + 450 But that equation doesn't do us any good. We can make the "silver equation" out of the second column: 1.00x + 0.00(450) = 1.00x But that equation becomes x = x and does us no good either. We can make the "copper equation" out of the third column: .00x + .10(450) = 0.075(x+450) Now that equation shows promise!! We can make the "other metal equation" out of the fourth column: .00x + .90(450) = .90(450) That equation becomes 405 = 405 and does us no good. So we only have one equation that does us any good, the copper equation: .00x + .10(450) = 0.075(x+450) or .10(450) = 0.075(x+450) Solve that for x and we get x = 150 grams of pure silver Edwin``` Answer by filboi(35)   (Show Source): You can put this solution on YOUR website!I would assume that this problem only involves silver & copper. I think adding other metal in the equation is a little bit "over-analyzing" it since there is no mention of any other metal apart from the two. Now to lower the concentration of copper, we would have to add more silver in the alloy so we're solving this problem from the silver's perspective. Let s = pure silver to be added 100% - 10%(copper content) = 90% = 0.90 = silver content in the original alloy 100% - 7.5%(copper content) = 92.5% = 0.925 = silver content in the new alloy Our equation is: s + (0.90)(450 - s) = (0.925)(450) Solving for s will give us 112.5g