1. If an alloy containing 30% silver is mixed with a 55% silver alloy to get 800 pounds of 40% alloy, how much of each mixture must he use?

:

Let x = amt of 55% alloy required:

Then

(800-x) = amt of 30% alloy required

:

Write a silver amt equation:

.55x + .30(800-x) = .40(800)

:

.55x + 240 - .3x = 320

:

.55x - .30x = 320 - 240

:

.25x = 80

:

x = 80/.25

:

x = 320 lb of 55% alloy

then

800 - 320 = 480 lb of 30% alloy

:

You can check solutions in the original equation