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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?

So I've tried these equations but I'm pretty sure they're not correct:

N(0.45 + 6) = .78

6 + 0.45 = .78N

0.45N + 6 = .78N

Ambrose
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2 Answers2

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Suppose $x$ mg of 45%Ni alloy can be combined with 6 mg pure Ni to form a 78%Ni alloy. Focussing on the non-Ni portion, we know that $$ 0.55x = 0.22(x+6)$$

And so $5x = 2(x+6)$ and the solution is straightforward.

Joffan
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hint: $.45N + 6 = .78(N+6)$. Can you continue?

DeepSea
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  • Wow thanks alot! This problem had been wreaking havoc on my mind all day and now it's finally been solved, thanks again your help, it definitely brought my mind at ease :) – Ambrose Mar 22 '15 at 22:12
  • I am glad it helped you find peace again with math. – DeepSea Mar 22 '15 at 22:15