How much of an alloy that is 30% copper should be mixed with 600 ounces of an alloy that is 80% copper in order to get an alloy that is 40% copper?

Question

Owen Ambrose · Accepted Answer

The amount is #=2400# ounces

Let #x# ouncs be amount of the alloy And the total #=x+600# ounces Consequently, the equilibrium with regard to copper is #x*0.3 +600*0.8=(x+600)*0.4# #0.3x+480=0.4x+240# #0.1x=240# #x=2400# ounces

Kai Bowman · Answer

undefined To find out how much of the 30% copper alloy needs to be mixed with the 600 ounces of the 80% copper alloy to obtain an alloy that is 40% copper, we can set up the following equation:

Let $x$ be the amount of the 30% copper alloy to be mixed.

The total amount of copper in the resulting mixture will be the sum of the copper from the two alloys:

$$0.30x + 0.80(600) = 0.40(x + 600)$$

Solving for $x$:

$$0.30x + 480 = 0.40x + 240$$

$$480 - 240 = 0.40x - 0.30x$$

$$240 = 0.10x$$

$$x = \frac{240}{0.10} = 2400$$

Therefore, 2400 ounces of the 30% copper alloy should be mixed with the 600 ounces of the 80% copper alloy to obtain an alloy that is 40% copper.