How many mg if metal containing 45% nickel must be combined with 6 mg of pure nickel to form an alloy containing 78% nickel?
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Let x =mg of metal containing 45% nickel to be added
after mixing with 6 mg of pure nickel,total amount of alloy is x + 6
Therefore the mixture contains this amount of nickel:
0.45x + 6 =0.78(x + 6)
For x,
0.45x + 6 = 0.78x + 4.68
x = 1.32/0.33
x = 4 mg
after mixing with 6 mg of pure nickel,total amount of alloy is x + 6
Therefore the mixture contains this amount of nickel:
0.45x + 6 =0.78(x + 6)
For x,
0.45x + 6 = 0.78x + 4.68
x = 1.32/0.33
x = 4 mg
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