Question 1129355
how many pounds of a 16% aluminum alloy must be mixed with 300lb of a 37% aluminum alloy to make a 22% aluminum alloy?
:
let x = 16% alloy
:
The mixture equation
.16x + .37(300) = .22(x+300)
.16x + 111 = .22x + 66
.16x - .22x = 66 - 111
-.06x = -45
x = -45/-.06
x = +750 lb of 16% alloy required
:
:
Check this yourself
.16(750) + .37(300) = .22(750+300)