Question

How much of an alloy that is 30% copper should be mixed with 50 ounces of an alloy that is 60% copper in order to get an alloy that is 40% ​copper?

Answer 1

Answer: 180 i think

Step-by-step explanation: .5*200+.2X=.3(200+X)

100+.2X=60+.3X

.2X-.3X=60-100

-.1X=-40

X=-40/-.1

X=400 OUNCES OF 20% COPPER.

PROOF

.5*200+.2*400=.3(200+400)

100+80=.3*600

100+80=180

180=180

Answer 2

Answer: 100 ounces of the alloy.

Step-by-step explanation: To solve, we create an equation:

First, though, let's define some variables and values -

x = amount of the alloy we are trying to find

0.3 = 30% copper

0.6 = 60% copper

50 = 50 ounces of the alloy

0.3(x) + 0.6(50) = 0.4(x+50) - original equation

0.3x + 0.6*50 = 0.4x + 0.4*50 - expanded

subtract 0.3x and 0.4*50 on both sides to get...

0.2*50 = 0.1x - simplified

calculate...

10 = 0.1x - simplified

divide both sides by 0.1 to get...

100 = x --> x = 100