Settle a simple algebra problem please.?

Question

How much of a 20% copper alloy should be mixed with 200 oz, 50% copper alloy in order to get an alloy of 30%?

Answer 1

let x oz of 20% copper alloy is to be mixed with 200 oz of 50% to get 30 %

x(20/100) + 200(50/100) = (200+x)(30/100)

20x + 10000 = 6000 + 30x

10x = 4000

x = 400 oz

Answer 2

This is a mixture problem.

Let
x = amount of 20% alloy to be added

.2x + .5(200) = .3(x + 200)
.2x + 100 = .3x + 60
40 = .1x
x = 400 oz.

Answer 3

Everybody is right, assuming the alloy is of the same metals.