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): You can put this solution on YOUR website!
let x equal the amount of the alloy that is 20% copper.

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.

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