Adding a percentage to a number and then subtracting the same percentage to get the same number

Question

I wonder if you can help me? I have found similar answers to this question but they don't seem to work.

I am after a formula or an excel formula so I can add a percentage to a number, then when I discount the same percentage off the larger figure I get back to exactly the same figure.

The reason for this is because I sell products at a price, some of my customers have special terms when they purchase so I add a percentage onto the selling price to cover these terms. Then when this percentage is discounted off the inflated cost, I need the selling total to go back to the original selling price.

For instance: If I sell at £1.00, Customer A has terms of 10%, so I need to add 10% onto 1.00, which is £1.10. Then when I take 10% off £1.10, which is £0.11p, this returns to £0.99p, which is lower than my original selling price.

If anyone could help with a formula to explain the answer it would be amazing! I would need the formula to work with different selling prices and different percentages.

Thanks in advance! Adam

Answer 1

You will need to modify the percentages. If you add $x\%$ to a price $P$ then your new price is $Q=P+\frac{Px}{100}$. You now want to subtract $\frac{Px}{100}$ from $Q$, but that is $x\%$ of $P$, not $x\%$ of $Q$. As a proportion of $Q$, it is $$\frac{\frac{x}{100}}{1+\frac{x}{100}}=\frac{x}{100+x},$$ so this means that instead of subtracting $x\%$ you need to subtract $y\%$ of $Q$, where $$y=\frac{100x}{100+x}.$$ For example, if you added $10\%$ you must subtract $\frac{1000}{110}\%=9.\overline{09}\%$.

Answer 2

If you take $99\%$ off $€100$, you get $€1$. Then if you add $99\%$ to $€1$, you get $€1.99$, much less than $€100$ because you didn't consider $99\%$ of the same thing.

In general adding $x\%$ is multiplying by $1+\frac x{100}$, so if you want the reverse of that you need to divide by $1+\frac x{100}$, which is the same as multiplying by $\frac{1}{1+\frac x{100}}$. Now

$$\frac{1}{1+\frac x{100}}=1-\frac{x}{100+x}$$ so the reverse of adding $x\%$ is subtracting $(100\cdot \frac{x}{100+x})\%$

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Example: $x=10$; the reverse operation of $+10\%$ is $-(100\cdot \frac{10}{100+10})\%\simeq -9.09\%$