SOLUTION: How much of an alloy that is 20% copper should be mixed with 100 ounces of an alloy that is 70% copper in order to get an alloy that is 40% copper.
Question 1112012: How much of an alloy that is 20% copper should be mixed with 100 ounces of an alloy that is 70% copper in order to get an alloy that is 40% copper. Answer by Theo(13342) (Show Source):
the amount of copper in that alloy is equal to .2 * x.
the amount of copper that is in the alloy that is 70% copper would be .7 * 100 = 70.
when you add x ounces to 100 ounces, you get a total of x + 100 ounces in the final mixture.
that final mixture will contain 40% copper.
the amount of copper that is in that final mixture will be .4 * (x + 100).
your equation becomes:
.2 * x + .7 * 100 = .4 * (100 + x).
simplify to get:
.2 * x + 70 = 40 + .4 * x
subtract .4 * x from both sides of the equation and subtract 70 from both sides of the equation to get:
.2 * x - .4 * x = 40 - 70
simplify to get:
-.2 * x = -30
divide both sides of this equation by -.2 to get:
x = -30 / -.2
this results in x = 150.
you would need to mix 150 ounces of 20% copper with 100 ounces of 70% copper to get a 250 ounce mixture that is 40% copper.
.2 * 150 = 30 ounces of copper in the 20% mixture.
.7 * 100 = 70 ounces of copper in the 70% mixture.
final mixture is 150 + 100 = 250 ounces
total copper in the final mixture is 30 + 70 = 100 ounces.
100 / 250 = .4 which is equal to 40% copper in the final mixture.
