# Formula for calculating transaction fees

Let's say a credit card processor wants to charge $(3\% + \text{US}\$ 2)$for all transaction. What formula do I use to make sure that after the charges are deducted, I get the figure I want. Example: If my product is$\text{US}\$100$ How much do I need to increase the price so that I will get $\text{US} \$100$even after the transaction is charges by the credit card processor. If it is purely$\% $-based, then I believe it is$\dfrac{\text{US}\$100}{1-0.03}$. But with the $\text{US}\$2$in play, I am not sure how to do it. Please help :) - ## 3 Answers Let$x$be the price we charge. Then there is a transaction fee of$3$percent, that is$0.03x$. There is an additional charge of$2$dollars. So we end up with a net amount of$x-0.03x-2$. We want this to be (say)$100$. Thus $$x-0.03x-2=100.$$ Simplify this first to$x-0.03x=102$, then to$0.97x=102$. We find that $$x=\dfrac{102}{0.97}.$$ - That's it. Thank you for your prompt respond :) – Lasker Jan 29 '13 at 7:43 You are welcome. I am sure it is clear to you what to do with a target amount different from$100$, and what to do when the fees go up. – André Nicolas Jan 29 '13 at 7:48 Yeah, that is clear. Thanks :) – Lasker Jan 29 '13 at 7:50 Okay just for my understanding, for \$100, they will charge you 3% (=\$3) + \$2 so you are finally left with \$95? If that is the case I understood your question correct. Let us assume you are at an ATM and click the "\$X" option. What you get is not \$X, but$\$X \cdot(100\%-3\%)-\$2$so you are left with$\$X\cdot0.97-\$2 $. If you actually want to have \$100 after they charged you, you want $\$X\cdot0.97-\$2$ to equal \$100. So you have to solve$\$X\cdot0.97-\$2 = \$100$ for $\$X$. The results in $$\X = \102/0.97\approx \105.15$$ - Thanks for your answer :) – Lasker Jan 29 '13 at 7:51$100=(1-\frac{p}{100})x-2$so $$x=\frac{102}{1-\frac{p}{100}}$$ To account for the$2$dollar surcharge, you can think of the price of the item as being$102\$ with surcharge, then apply what you have already figured out.

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Thanks for your answer :) – Lasker Jan 29 '13 at 7:50