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?
Algebra.Com
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) (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
RELATED QUESTIONS
How many mg of a metal containing 45% nickel must be combined with 6 mg of pure nickel to (answered by josmiceli)
how many mg of a metal containing 45% nickel must be combined with 6 mg of pure nickel to (answered by htmentor)
How many mg of a metal containing 45% nickel must be combined with 6 mg of pure nickel to (answered by jorel1380)
How many mg of a metal containing 45% nickel must be combined with 6 mg of pure nickel to (answered by josmiceli)
How many mg of a metal containing 45%
nickel must be combined with 6 mg of pure
nickel... (answered by ikleyn)
How many mg of a metal containing 45% nickel must be combined with 6 mg of pure nickel to (answered by Fombitz)
How many mg of metal containing 45% nickel must be combined with 6 mg of pure nickel to... (answered by stanbon)
How many mg of a metal containing 45% nickel must be combined with 6mg of pure nickel to... (answered by josmiceli,nospmoht,HydroKid)
How many kg of a metal containing 20% nickel must be combined with 9 kg of pure nickel to (answered by mananth)