Confused about percentages

Question

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?

Answer 1

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 $0.95a$.

Answer 2

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.

Answer 3

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 $.

Answer 4

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.

Answer 5

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.