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I use the following method to calculate $b$, which is $a$ increased by $x$ percent:

$\begin{align} a = 200 \end{align}$

$\begin{align} x = 5\% \text{ (represented as } \frac{5}{100} = 0.05 \text{)} \end{align}$

$\begin{align} b = a \cdot (1 + x) \ = 200 \cdot (1 + 0.05) \ = 200 \cdot 1.05 \ = 210 \end{align}$

Now I want to calculate $c$, which is also $a$ but decreased by $x$ percent.

My instinct is to preserve the method, but to use division instead of multiplication (being the inverse operation):

$ \begin{align} c = \frac{a}{1 + x} \ = \frac{200}{1 + 0.05} \ = \frac{200}{1.05} \ = 190.476190476 \ \end{align} $

The result looks a bit off? But also interesting as I can multiply it by the percent and I get back the initial value ($190.476190476 \cdot 1.05 = 200$).

I think the correct result should be 190 (without any decimal), using:

$ \begin{align} c = a \cdot (1 - x) \ = 200 \cdot (1 - 0.05) \ = 200 \cdot 0.95 \ = 190 \end{align} $

What's the difference between them? What I'm actually calculating?

Dan
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5 Answers5

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The difference between them is with respect to what variable the percentage is being taken. In your first calculation, when you divide, you are saying "$a$ is $x\%$ more than $c$", while on the latter, you are saying "$c$ is $x\%$ less than $a$". These are not the same! In the first case, the percentage is with respect to $c$ (hence why you multiply $c$ and $1+x$). In the second, the percentage is with respect to $a$ (hence why you multiply $a$ and $1-x$). Hope this is clear.

Ezra
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It is true that: increase by $x$ percent then decrease the result by $x$ percent does not get you back where you started.

Let's do a case where it is clearer, say $x=100$. Start with $20$. Increase by $100$ percent. Well, $100$ percent of $20$ is exactly $20$, so this means increase by $20$. The result is $$ 20 + 20 = 40 . $$

Now, starting at that $40$, let's decrease that by $100$ percent. Well, $100$ percent of $40$ is $40$, so we have to decrease by $40$. The result is: $$ 40 - 40 = 0 . $$ So we certainly did not arrive back where we started.

Look at this. We increased by $100$ percent of $20$, then decreased by $100$ percent of $40$. Of course $100$ percent of $20$ is not the same as $100$ percent of $40$. So the amount increased is not the same as the amount decreased.

GEdgar
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It is easier to translate these expressions from English to Math than to think in terms of multiplication and division. The latter way leads to memorizing an arbitrary rule which won't stay with you.

We are saying we want to increase $a$ by $x\%$ where it is understood that "by $x\%$" means $x\%$ of $a$. This gives us a straight forward translation $$a+\frac{x}{100}a=a(1+\frac{x}{100})$$

When $x=5$ we get $1.05a$ as you calculated.

If we decrease $a$ by $x\%$, we get $$a-\frac{x}{100}a=a(1-\frac{x}{100})$$ and when $x=5$ we get $0.95a$.

John Douma
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  • Talking about translation from English, some people use "$n$ times more than $a$" to mean only $100(n-1)%$ of $a$ more, or use "$n$ times fewer/less than $a$" to mean only $100(1-1/n)%$ of $a$ fewer/less. This always bothers me. – peterwhy Jun 17 '22 at 17:45
  • @peterwhy The secret to percentages is to realize that the symbol "$%$" is simply an alias for $\frac{1}{100}$. That, the ability to manipulate fractions and the knowledge that "of" means "times" is all that is needed to master these problems. – John Douma Jun 17 '22 at 17:48
  • I forgot to mention that I usually use the notation 1.percent (1.05) straight away, because I find it easier to type that on a calculator. Basically, if I want to find the increase by 5%, I type n * 1.05 to get the result. This is the reason why I though the decrease should be n / 1.05 (instead of n * 0.95). – Dan Jun 17 '22 at 17:48
  • @Dan We do that when we know $n$ and we want to know the number that was increased by $5%$ to get $n$. One of the points others have made is that the number that was increased to get our number is not the same number we get if we decrease by the same percentage. This is why understanding the mathematical statement is more important than memorizing a rule. – John Douma Jun 17 '22 at 17:50
  • @JohnDouma Got it, it's more clear now. Thanks for your time! – Dan Jun 17 '22 at 17:52
  • @Dan - I address where division might be used in my answer. It might be a good supplement to this one. – JonathanZ Jun 17 '22 at 17:59
  • @JonathanZsupportsMonicaC I appreciate your answer as well. Good explanation! PS: I don't have 15 reputation to be able to upvote. – Dan Jun 17 '22 at 18:03
  • @Dan - No problem, and glad it helped. I don't think you can check more than one answer anyways, no matter how many points you have. Good luck! – JonathanZ Jun 17 '22 at 18:07
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I think there are two issues at play here.

Issue #1: Increase vs. decrease. That determines whether you use $1 + \frac{p}{100}$ or $1-\frac{p}{100}$, respectively.

Using $I$ for "Initial amount" and $F$ for "Final amount":

$$F = I \times (1 \pm \frac{p}{100}) $$

Issue #2: Initial vs. Final. Do you know the initial (i.e. before the change) value and want to find the final (after the change) value, or do you know what the value is after the change, and want to find the initial value? That determines whether you multiply or divide.

If you know the initial value, use the same equation above. If you know the final value, you will actually re-use the same equation again (there's really only one equation for percentages!), but it will be re-written as

$$ I = \frac{F}{1 \pm \frac{p}{100}}$$

I think that last case is where you have seen solutions that use division. Just realize that it's not the difference between increase and decrease that determines that, it's the difference of whether you want to find the value before or after the change.

JonathanZ
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Notice that what you have done is basically exploit the first-order approximation of the Taylor series of $\frac{1}{1-x}$: $$\displaystyle\frac{1}{1-x}=1+x+x^2+x^3+… for |x|\lt1$$ $≈1+x $; for $x<<1 $.

insipidintegrator
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