0
$\begingroup$

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 $p$?
  • 0
    This sounds like a homework question. Write an equation that describes how $d$ relates to $N$ and $G$2012-02-22

2 Answers 2

1

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

1

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:

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