# Calculating discount not working

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.

• $86.95652$ is the answer to the question : If a product is worth $100 after a 15% increase, how much was it worth before the increase? Apr 1, 2019 at 18:35 • 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. – Nij Apr 2, 2019 at 7:35 ## 7 Answers 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: $$V_\text{new}=V - V\times\frac {15}{100} = V\times(1 - 0.15) = V\times 0.85$$ EDIT 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. • 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. Apr 1, 2019 at 20:04 • it's exactly$\frac{1}{3}\$ slower.
– 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%: $$X - X \times \frac{15}{100} = \frac{100X}{100} - \frac{15X}{100} = \frac{(100-15)X}{100} = \frac{85X}{100} = 0.85X$$