SOLUTION: How many 6 mg of 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 251730: How many 6 mg of 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 checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
I'M ASSUMING THE 6 MG OF 45% NICKEL IS MISS-INFORMATION.
.45*X+6=.78(X+6)
.45X+6=.78X+4.68
.45X-.78X=4.68-6
.33X=1.32
X=1.32/33
X=4 MG OF 45% NICKEL IS USED.
PROOF:
.45*4+6=.78(4+6)
1.8+6=.78*10
7.8=7.8