Question 327704
how many ounces of an alloy containing 30% gold on the world market must be mixed
 with an alloy containing 5% gold to obtain 25 ounces of an alloy containing 20% gold?
:
Let x = amt of 30% alloy
The resulting total is to be 25 oz, therefore:
(25-x) = amt of 5% alloy
:
A typical mixture equation:
.30x + .05(25-x) = .20(25)
.30x + 1.25 - .05x = 5
.30x - .05x = 5 - 1.25
.25x = 3.75
x = {{{3.75/.25}}}
x = 15 oz of 30% alloy required
then
25-15 = 10 oz of 5% alloy
:
:
Check solution
.30(15) + .05(10) = .20(25)
4.5 + .5 = 5