To increase a number by a percentage I was taught the formula no x 1.percentage, so for example:

100 increased by 15%:
100 x 1.15 = 115

But today, I learned the hard way that doing the inverse of the above (i.e. no / 1.percentage), isn't the correct way of decreasing a number by a percentage (aka "applying a discount"). E.g:

100 decreased by 15%:
100 / 1.15 = 86.95652 <- WRONG
100 - (100 x 0.15) = 85 <- RIGHT

My Question: Why doesn't the formula no / 1.percentage work, when decreasing a number by a percentage?

Side Question: Is there a simpler formula to decrease a number by a percentage than no - (no x 0.percent)?

Note: My math knowledge is super basic, so maybe pretend you're explaining this to a twelve year old when answering. But if you want to give sophisticated answers for future readers, that's fine too, I might just not understand it though.

  • 3
    $\begingroup$ $86.95652$ is the answer to the question : If a product is worth $100 after a 15% increase, how much was it worth before the increase? $\endgroup$ Apr 1, 2019 at 18:35
  • $\begingroup$ Please don't use codeblocks for anything that is not actual code. There are tutorials available on using MathJax typesetting, such as that in the answers below. $\endgroup$
    – Nij
    Apr 2, 2019 at 7:35

7 Answers 7


The method of increasing a value $V$ by $p\%$ is actually adding $V\times\frac p{100}$: $$V_\text{new} = V + V\times\frac p{100} = V\times\left(1 + \frac p{100}\right)$$ For $p=15$ you have a nice multiplier $$\left(1 + \frac p{100}\right) = 1 + 0.15 = 1.15$$

Remember, however, that '1.p' is a shortcut or mnemonic, not a method. It wouldn't even work for $p$ greater than 99, say for $+120\%$. And it certainly won't work for $p <0$.

For decreasing you need to apply the method, which is: $$V_\text{new}=V - V\times\frac p{100} = V\times\left(1 - \frac p{100}\right)$$ so for 15-percent decrement you get: $$V_\text{new}=V - V\times\frac {15}{100} = V\times(1 - 0.15) = V\times 0.85$$


To answer your main question directly, forget percentages. Suppose you need to rise a value from 100 to 125. The new value is $$125 = 100\times\frac 54=100\times(1+\tfrac 14) $$ hence an increase by 25%. Now, if you want to reduce it back to 100, you get $$100=125\times \frac 45=125\times(1-\tfrac 15)$$ which is 20% decrease. Why was it 1/4 before, and 1/5 now? Because the same difference $\pm 25$ was taken relative to 100 in the former case and to 125 in the latter one. When we added a fourth part, we got five fourths of the initial value, so we needed to take away one of those five, i.e. a fifth part to get back.

And here you have the difference in percentages: 1/4=25%, while 1/5=20%.

So, a division like $value/1.p$ reduces the value by p% of the resulting value, not of the original one.


The reason that your suggested formula doesn't work is that "increase by $k$%" and "decrease by $k$%" do not undo one another like $+5$ and $-5$ or $\times 2$ and $\div 2$ do (we would say that they are not inverses).

For example $100$ increased by $50\%$ is $150$. However, $150$ decreased by $50\%$ is 75.

Therefore, if we want to decrease a number by a percentage, we can't just attempt to do the opposite operation of increasing by that percentage.

  • 5
    $\begingroup$ It's also important to understand this applies to any ratio and can be used to play psychological tricks. For example, if A is 50% faster than B, then B is only about 30% slower than A. You could use either depending on whether you want to emphasize or downplay this gap. $\endgroup$
    – MooseBoys
    Apr 1, 2019 at 20:04
  • $\begingroup$ it's exactly $\frac{1}{3}$ slower. $\endgroup$
    – user645636
    May 8, 2019 at 23:15

What you did is the right answer to a different problem. The general method, "multiply by $1+$ change" is a good one when used properly.

To increase a number by $15\%$ you multiply by $$ 1 + 0.15 = 1.15. $$ To decrease a number by $15\%$ you multiply by $$ 1 - 0.15 = 0.85. $$ But to undo a $15\%$ increase you have to "unmultiply" by $1.15$. So you divide by $1.15$. Since $$ \frac{1}{1.15} = 0.86956521739 \approx 0.87 = 1 - 0.13, $$ to undo a $15\%$ increase you make a $13\%$ decrease.


To find $15 \%$ of some number $N$, we solve an equation (proportion) $\frac{N}{100}=\frac{x}{15}$ or $x=0.15N$. That's why a $15%$ increase can be found as $N+0.15N=1.15N$ and decrease is found as $N-0.15N=0.85N$


It works the same way...

Increase : to increase of $15 \%$ means to multiply by $(1+0,15)=1,15$.

Decrease : to decrease of $15 \%$ means to multiply by $(1-0,15)=0,85$.

Unfortunately, the $15 \%$ factors get similar results; thus it can be misleading.

Consider instead an increase of $20 \%$. Why the "inverse" operation (i.e. to divide by $1,2$ to decrease does not produce the correct result ?

With a $20 \%$ increase the updated price will be $120$.

Now, what is the amount of a decrease of the updated price by $20 \%$ ? It is not $20$ but $24$.

And we have that $120 \times (1-0,2)=96$ while $\dfrac {120}{1,2}=100$.


If we want to increase $n$ by $p$%, we work out what $p$% of $n$ is, and then add it to $n$ itself: $$0.p\times n+n=(0.p+1)\times n=1.p\times n$$This is how you arrived to your formula.

If we want to decrease $n$ by $p$%, we work out what $p$% of $n$ is, and then subtract it from $n$ itself:$$n-0.p\times n=(1-0.p)\times n$$This is how to derive the formula for a decrease.


Consider the literal meaning of "per cent"

Increase $X$ by 15%: $$ X + X \times \frac{15}{100} = \frac{100X}{100} + \frac{15X}{100} = \frac{(100+15)X}{100} = \frac{115X}{100} = 1.15X $$

Decrease X by 15%: $$ X - X \times \frac{15}{100} = \frac{100X}{100} - \frac{15X}{100} = \frac{(100-15)X}{100} = \frac{85X}{100} = 0.85X $$


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