Question 139566
Dr. Bennington took an alloy containing 10% silver and mixed it with an alloy containing 40% silver to get a 60 ounce alloy containing 30% silver. How many ounces of the 10% alloy did he use?
:
Let x = amt of 10% alloy required
then
(60-x) = amt of 40% silver
:
.10x + .40(60 - x) = .30(60)
:
.1x + 24 - .4x = 18
:
.1x - .4x = 18 - 24
:
-.3x = -6
x = {{{(-6)/(-.3)}}}
x = +20 oz of 10% alloy required