Ametalworker has a metal alloy that is 30% copper and another alloy that is 70% copper. how many kilograms of each alloy should the metalworker combine to create 120 kg of a 54% copper alloy?

We will have to assume that both the 30% and 70% alloys "weigh" the same. (Have the same density.)

We'll call the 30 % copper "t" and the 70 % copper "s"

A) t + s = 120

B) .30t + .70s = .54 * 120

Multiplying equation A) by -.30 we get

A) -.30t -.30s = -36 then adding this to B)

B) .30t + .70s = 64.8 we get

.40s = 28.8

s = 72 kg and t = 48 kg

Double-Check

48 * .30 + 72 * .70 = 14.40 + 50.40 = 64.8 and

64.8 / 120 = .54 = 54 per cent

Correct!!!

Source: https://www.1728.org/mixture.htm

See Example A

Step-by-step explanation:

hi,

6x times 4 is 24x and 5 times 4 is 20, so the answer is 24x+20.

hope this and for a little humor i am groot.

well what did you do it matters what you did : )

step-by-step explanation: