Adding and Removing Non-Compounded Percentages does not produce the same result?

If I take the value 100 and I want to and 10% tax to it and then a 7% tax to it, I am doing the following:

\begin{align*} 100 \times \left(1 + \frac{10}{100}\right) &= 110\\ 100 \times \left(1 + \frac{7}{100}\right) &= 107\\ 100 + 10 + 7 &= 117. \end{align*}

If I remove 10% and then 7% from 117, I do not get 100, I get 98.7094.... My formula is setup like this:

GrandTotal = 117

AjustedTotal = GrandTotal

Value = GrandTotal - (GrandTotal/(1 + (percent/100))

If I am doing more than 1 percentage, I run into problems such as 10% and 7%.

Am I overdoing this and what am I doing wrong. Basically, I want to take a value and add taxes to it and then remove the same taxes to retrieve my original value.

I added a question about this on stackoverflow here, but I am stuck at programming the non-compounding part and the mixture of both.

• 17% of 117 is bigger than 17% of 100. – Willie Wong Apr 13 '11 at 21:41
• For small changes the rise and fall are almost the same, so this will work. The error comes from the multiplication of the two percentages, so if they are small you can (maybe) ignore it. $(1+1\%)(1-1\%)=1\cdot 1 + 1 \cdot 1\% 1 1\% \cdot 1 - 1\% \cdot 1\%=0.9999$. It is the last term that makes it not come out. – Ross Millikan Dec 12 '11 at 15:20

Let's take a simpler example first. If you increase 400 by 25% you get 500. If you reduce 500 by 25% you get 375. With equations this is $$400 \times (1+ 0.25) = 500$$ $$500 \times (1- 0.25) = 375$$ so it is clear that this multiplication does not work, and we should not expect it to since $(1+ 0.25) \times (1- 0.25) = 0.9375$ not $1$. So to undo the increase (from 400 to 500), instead we should divide as follows: $$500 \div (1+ 0.25) = 400$$ and since $1 \div (1+ 0.25) = (1 - 0.20)$, we would have $$500 \times (1 - 0.20) = 400$$ so a 25% increase must be undone by a 20% decrease. You seem to understand this in your equation for value.

The second issue is your lack of compounding the taxes. If you did compound then it is correct that $$100 \times (1 + 0.10) \times (1 + 0.07) = 117.70$$ and you would undo it with $$117.70 \div (1 + 0.10) \div (1 + 0.07) = 100$$ but that does not work here.

So instead you start with $$100 \times (1 +0.10 + 0.07) = 117$$ and to undo it you do $$117 \div (1 +0.10 + 0.07) = 100$$

• I'm confused when you say lack of compounding. Can you give me an example of compounding and non-compounding taxes given a total of 100 and a 10% tax and a 7% tax. – Xaisoft Apr 14 '11 at 4:45
• @Xaisoft: The part starting "The second issue..." shows what happens with compounding, while the part starting "So instead..." shows what happens without compounding. – Henry Apr 14 '11 at 8:58
• Ah, OK. My math stinks :). What if It was a mixture, for example I had 3 percentage values, 10, 7,and 3 and 10 and 7 were compounding , but 3 was not, What would the formula for that look like? – Xaisoft Apr 14 '11 at 11:24
• It gets messy: $100 \times(1 + ((1 + 0.10) \times (1 + 0.07) -1) + ((1+0.03)-1))$ which simplifies to $100 \times((1 + 0.10) \times (1 + 0.07) + 0.03)= 120.70$ – Henry Apr 14 '11 at 12:03
• I edited my post and put a programming question on stackoverflow implementing this. It is still confusing. I can't get the non-compounding to work and a mixture of both. – Xaisoft Apr 14 '11 at 17:35

Essentially, you've computed your after-tax value using the formula $$\array{T &=& P + (0.10)P + (0.07)P\\&=& (1.17)P}$$ where T is the after-tax price, and P is the before-tax price.

If you want to get back to P from T, just solve for P in the above equation to get $$P = \frac{T}{1.17}$$

Look at a simpler problem. Add 50% to 100. You get 150. Now subtract 50%. You get 75. Whenever you add or subtract a percentage, the value you add or subtract always depends on the value which you are adjusting. Once you've added your percentage, then your total value is larger than when you started, so if you try to subtract the same percentage, you will be subtracting more than you added.

If you first add a percentage, say 10%, and then you subtract 10% of that total, you don't get back to the original amount.

Say you start with 100; adding 10% gives you 100+10 =110. Subtracting 10% of that gives you 110-11 = 99, less than what you started with.

(That's why if you first get a 10% paycut, and then you get a 10% raise, you are not back where you started).

This is because adding 10% is the same as multiplying by $1 + \frac{10}{100}$. Subtracting 10% is the same as multiplying by $1 - \frac{10}{100}$. But $$\left( 1 + \frac{10}{100}\right)\times\left(1 - \frac{10}{100}\right) = 1 - \frac{100}{10000} = 1 - \frac{1}{100} \neq 1.$$

So you certainly cannot "cancel" the adding of a percentage by simply subtracting that percentage later. (Note, however, that the result of first adding r% and then subtracting s% is the same as first subtracting s% and then adding r%; it's just not the same as adding (r-s)%)

In your case, you are trying to add 17% and then subtract 17%. This means $$\left(100\times\left(1 +\frac{17}{100}\right)\right)\times\left(1 - \frac{17}{100}\right) = 100\times\left(1 - \frac{289}{10000}\right).$$ In general, if you first add r% and then subtract r% of the total, you will end up below your original amount by r% of r%, because you are doing $$X\times\left( 1 + \frac{r}{100}\right)\left(1 - \frac{r}{100}\right) = X \times\left( 1 - \left(\frac{r}{100}\right)^2\right).$$

In my first example, it was 10% of 10%, which is 1%, hence 99 instead of 100. Here, it's 17% of 17%, which is about 2.89%, hence you end up at 97.11, or 2.89% less than 100.