anne and nancy use a metal alloy that is 15% copper to make jewelry. how many ounces of a 10% alloy must be mixed with a 21% alloy to form 121 ounces of desired alloy? *my answer sheet says its 66 ounces but i don‘t know how to get it
Two things to track - amount of copper in the final mixture and total weight of the final mixture. Let x amount of 10% copper alloy Let y amount of 21% copper alloy Amount of copper in the final, 15%, mixture: (0.15)(121) 0.10x + 0.21y Weight of the final mixture: x + y 121 y 121-x (0.15)(121) 0.10x + 0.21(121 - x) (0.15)(121) 0.10x - 0.21x + 0.21(121) (0.15 - 0.21)(121) -0.11x x (0.06)(121)/0.11 66 y 121 - 66 55 66 oz of 10% alloy and 55 oz of 21% alloy will combine to form 121 oz of 15% alloy.
let x ounces of 10% to be used. You know you want to have 121 ounces in all so the amount of 21% will be what's left. so you get 121 - x ounces of 21% alloy used. The word equation for this is amount of copper in 10% + amount of copper in 21% amount of copper in 15% mixture Since you have x ounces of 10%, the amount of copper in that is 10% of x which can be written 0.10x. Same for the 15% and 21%. So the mathematical equation from the above word equation is 0.10x + 0.21(121 - x) 0.15(121) simplify and get 0.10x + 25.41 - 0.21x 18.15 -0.11x -7.26 divide both sides by -0.11 x 66 Hope the kinda wordy explanation helped you. I am assuming you wanted to know how, since you already had the answer.
Ounces of 10% alloy—x: 0.1x + 0.21(121 - x) 0.15(121) 0.1x + 25.41 - 0.21x 18.15 0.11x 7.26 x 66 Answer: 66 ounces of 10% alloy
