An alloy of metal is 25% copper. Another alloy is 50% copper. How much of each alloy should be used to make 1000 grams of an alloy that is 45% copper?
use the amount of copper to balance the equation: use x grams of the 25% copper this will have .25x grams of pure copper if there are 1000 grams in the final alloy, that means 1000 - x grams will be the 50% alloy this will have .50(1000 - x) grams of copper the final alloy will have .45(1000) = 450 grams of copper the copper added must equal the copper in the final alloy, or .25x + .50(1000 - x) = 450 .25x + 500 - .50x = 450 50 = .25x x = 200 so 200 grams of 25% copper (50 grams copper) 800 grams of 50% copper (400 grams copper) total: 1000 grams of 45% copper (450 grams copper)
This is a system of equations problem: Let the weight of your first alloy be L, and the wieght of your second alloy be H. You know L + H = 1000g. You want your 1000 g to be 45% copper, so 1000 x 45% = 450 g of copper. To get the 450 g, it will need to be part L, part H. .25L + .5H = 450 L + H = 1000 L = 1000 - H Plug this into the first equation: .25 (1000 - H) + .5H = 450 250 - .25H + .5H = 450 .25H = 200 H = 800 g L = 1000 - H = 1000 - 800 = 200g
25% of alloy x is copper 50% of alloy y is copper You want to use x+y to make 45% of 1000 grams copper You may be able to continue from my re-stating the problem. I used to teach, so I don't just give answers!
