SOLUTION: How many mg of a metal containing 45% nickel must be combined with 6 mg of pure nickel to form an alloy containing 78% nickel?

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Question 198833: How many mg of a metal containing 45% nickel must be combined with 6 mg of pure nickel to form an alloy containing 78% nickel?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(mg of nickel) / (mg of alloy) = 78%
Let x= mgs of metal with 45% nickel
.45x+%2B+6%29+%2F+%28x+%2B+6%29+=+.78
.45x+%2B+6+=+.78%2A%28x+%2B+6%29
.78x+-+.45x+=+6+-+.78%2A6
.33x+=+1.32
x+=+4
4 mg of the metal must be used
check:
.45x+%2B+6%29+%2F+%28x+%2B+6%29+=+.78
.45%2A4+%2B+6%29+%2F+%284+%2B+6%29+=+.78
1.8+%2B+6%29+%2F+10+=+.78
7.8%2F10+=+.78
.78+=+.78
OK