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Given the value $N$ and the percentage $p$, where $N = G + G \times (p/100)$, $d$ is the percentage you need to subtract of $N$ to get the base value $G$.

  • Is there a general formula for determining $d$?
  • Is there a formula independent of $G$ and $N$, where $d$ can be calculated only from $p$?
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This sounds like a homework question. Write an equation that describes how $d$ relates to $N$ and $G$… –  Seamus Feb 22 '12 at 13:32

2 Answers 2

$$N=G(1+\frac{p}{100})$$ $$N(1-\frac{d}{100})=G$$ Putting $G$ of the second equation in the first equation you have: $$(1+\frac{p}{100})(1+\frac{d}{100})=1$$ So:$$d=\frac{100p}{100+p}$$

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So if I understood you correctly, for any numbers $N$, $G$ and $p$ for which $N=G+G\frac{p}{100}$ you want to compute $d$ such that $G=N-N\frac{d}{100}$. This should be achievable quite easily by substituting the formula for $G$ into that of $N$ (or vice versa) and eliminating terms:

$$N = N - N\frac{d}{100} + N\frac{p}{100} - N\frac{dp}{10000}$$ $$1 = 1 - \frac{d}{100} + \frac{p}{100} - \frac{dp}{10000}$$ $$d + \frac{dp}{100} = p$$ $$d = \frac{p}{1+\frac{p}{100}}$$

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