How do you reverse a percentage change?

Question

Is there any simple way to know how to reverse a percentage?

For example if I have 100 and it goes down by 10% I end up with 90. If I then add to it by 10% I end up with 99, not the 100 that you would think of. Is there a simple trick to quickly work out the reverse of a percentage change (even if it only works for some subset of all).

Answer 1

Yes, add $90$ divided by $90$ and multiply the result by $10$.

As another example: Let us say you subtract $5\%$ from $120$. That is, $\frac{120}{100} \cdot 5 = 6$. This gives $120 - 6 = 114$. Then to get back to $120$ you add $\frac{114}{95} \cdot 5 = 6$.

The way to think about this is the following. What you are doing is you are dividing the number into 100th pieces of equal length. Then, you are subtracting let us say $5\%$ or in other words $5$ pieces of equal length. Now, you are left with only $95$ pieces $(95\%)$ of equal length. To get back you divide the pieces by $95$ since you want them to be of equal length. And then multiply again by how many pieces you want to get back, in this case $5$.

Working with percentages is just working with ratios.

Answer 2

You may too consider all the percentage operations as multiplications by a factor :

• taking $10$% of $x\;$ is computing $\,\displaystyle \frac {10}{100}\cdot x=0.10\;x\quad$ ( replace '%' with $\displaystyle \frac 1{100}\;$)
• subtracting $10$% to $x\;$ is computing $\;\displaystyle x-\frac {10}{100}x=\left(1-\frac {10}{100}\right)x=(1-0.10)\;x=0.90\;x$
• of course adding $12$% to $x\;$ is simply computing $\;(1+0.12)\;x=1.12\;x$.

(note that subtracting $10$% and adding $12$% becomes multiplying by $\;0.90\cdot 1.12=1.008\;$ and allows us to observe that the order of the % operations doesn't matter!)

Reverting these operations will be done by dividing by the multiplicative factor :

• divide $0.10\;x\,$ by $\,0.10\,$ to get $x$ back
• to subtract $10$% to $\,x\;$ you computed $\;0.90\;x\;$ so divide this by $\,0.90\,$ to get the initial $\,x$.

Answer 3

First let's do the exact math, then we can see how to turn that into a trick that you can easily do in your head.

As other answers have explained, percentages can be a bit confusing because, even though we talk about adding or subtracting a percentage, mathematically one is actually multiplying by something. So if you are taking away $p$ percent you are multiplying by

$$ \times \left(1-\dfrac{p}{100}\right),$$

and if you are adding on $q$ percent you are multiplying by

$$ \times \left(1+\dfrac{q}{100}\right),$$

The exact, proper way to undo multiplying by something is to divide by that factor, so you are asking to find $q$ such that

$$ \left(1+\dfrac{q}{100}\right) = \dfrac{1}{\left(1- \dfrac{p}{100}\right)}.$$

The exact solution to this is

$$ q = 100 \times \left( \dfrac{1}{1-\dfrac{p}{100}} -1 \right).$$

But because you've asked for a quick simple trick we can use an approximation to get a "good enough" answer. There's a cool fact that when $|x| < 1$

$$ \dfrac{1}{1-x} = 1 + x + x^2 + x^3 + \dots$$

and when $x$ is small, say $|x| \le 0.1$, you don't need many terms from that infinite series to make a decent approximation. Applying this to our formula for $q$ we get

\begin{align} q & = 100 \times \left(1 + \frac{p}{100} + \frac{p^2}{100^2} + \frac{p^3}{100^3} + \dots - 1 \right)\\ & = p + \frac{p^2}{100} + \frac{p^3}{100^2} + \dots \\ \end{align}

which we will write as our final formula

$$ q \approx p + \frac{p^2}{100} + \left( \frac{p^2}{100}\right) \cdot p /100 $$

Let's see how this works with your "reversing a $10\%$ reduction" example. Our first approximation for the required percentage increase says "$p$", so we start with 10%. The next term says $\frac{p^2}{100}$, so we add on another $\frac{10\cdot 10}{100} = 1\%$. And to get the third term you take the second term ($\frac{p^2}{100}$) and multiply it by $p$ then divide by $100$, which gives an additional $1 * 10 /100 = 0.1$, for a final estimate of $11.1\%$. This is a very good approximation of the actual answer, which is $11.1111....\%$.

If $p$ is smaller the approximation is even better, and you can even skip the third term. Let's try it to see what it takes to reverse a 4% reduction:

• first term = $p$ = $4\%$.

• second term = $\frac{p^2}{100} = 0.16\%$

• third term = $0.16 * 4 /100 = 0.0064\%$

The exact answer is $4.16666666...\%$. All three terms gives us $4.1664\%$, but with just two terms we get $4.16\%$, which is probably close enough.

We should also see what happens when we have a bigger $p$ -- let's try reversing a $25\%$ reduction:

• first term = $25\%$

• second term = $\frac{25 \times 25}{100} = 6.25\%$

• third term = $6.25 \times 25 / 100 \approx 6\times25/100 = 1.5\%$

The exact answer is $33.3333...\%$. With two terms we get $31.25\%$, which is certainly better than just our first term of $25\%$, and adding our third term gives us about $32.75\%$. (You might have noticed that I started dropping a few digits in computing the third term, because we're just getting an approximation anyways so who cares.) Knowing we dropped a small amount in computing our third term I'd probably say "It's about $33\%$" and let it go at that.

To summarize: For small percentage decreases, say less than $10\%$, we can take just two terms and get a very good approximation. For less than $25\%$ we're in the right neighborhood with two terms, and if we bother to compute the third term we're close, and for values larger than $25\%$ you should probably just explicitly calculate what $\frac{1}{1 - p/100}$ is. For that last case I'd just fish out my phone and use its calculator.