What is \(p=\)% of \(y=\)?
\[x = \frac{p}{100}\cdot y\]
What percentage is \(x=\) of \(y=\)?
\[p = \frac{x}{y}\cdot 100\]
What is the percentage increase or decrease from \(s=\) to \(t=\)?
\[p = \left(\frac{t-s}{s}\right)\cdot 100\]
How much is \(x=\) \(p=\)%?
\[y = x\cdot\left(1 \pm \frac{p}{100}\right)\]
What percentage change does adding or subtracting \(d=\) to \(s=\) represent?
\[p = \frac{d}{s}\cdot 100\]
\(x=\) is \(p=\)% of what value?
\[y = \frac{x}{p/100} = \frac{100x}{p}\]
After how many years does inflation at \(r=\)% reduce purchasing power to \(p=\)% of its current value?
\[\left(1-\frac{r}{100}\right)^n = \frac{p}{100}\qquad\Rightarrow\qquad n = \frac{\log(p/100)}{\log(1-r/100)}\]
With inflation at \(r=\)% per year, what is the purchasing-power loss after \(n=\) years?
\[p = \left(1 - \left(1 - \frac{r}{100}\right)^n\right)\cdot 100\]
After how many years does a value grow to \(p=\)% of its starting value at an annual return of \(r=\)%?
\[\left(1+\frac{r}{100}\right)^n = \frac{p}{100}\qquad\Rightarrow\qquad n = \frac{\log(p/100)}{\log(1+r/100)}\]
What selling price leaves \(n=\)$ after a commission of \(p=\)%?
\[x - x\cdot\frac{p}{100} = n\qquad\Rightarrow\qquad x = \frac{n}{1-p/100}\]
With entry price \(e=\)$ per share, current price \(k=\)$, tax allowance \(F=\)$, and \(h=\) shares owned, how many can you sell within the allowance?
\[g = k - e,\qquad n = \frac{F}{k-e},\qquad V = n\cdot k = \frac{F\cdot k}{k-e}\]
\[n_{\text{whole}} = \left\lfloor\frac{F}{k-e}\right\rfloor\]
If an investment of \(x=\)$ was made when the share price was \(s=\)$, how many additional shares could be bought if the price falls to \(t=\)$?
\[n = \frac{x}{t} - \frac{x}{s}\]
How much must a stock rise to recover from a \(p=\)% drop?
\[x' = x\left(1-\frac{p}{100}\right),\qquad p' = \left(\frac{x-x'}{x'}\right)\cdot 100 = \frac{100p}{100-p}\]
What is a percentage?
A percentage means "per one hundred". So 1% literally means one part out of 100 equal parts, 25% means 25 parts out of 100, and 100% means the full quantity.
This is why percentages are written as fractions with denominator 100:
\[ p\% = \frac{p}{100} \]
From that single definition, all common percentage formulas follow.
Deriving the core formulas
Let \(y\) be a whole value and \(x\) be a part of that value.
By definition, "\(x\) is \(p\%\) of \(y\)" means:
\[ x = \frac{p}{100}\cdot y \]
Now solve this equation for the variable you need:
- Find the part \(x\): \(x = \frac{p}{100}\cdot y\)
- Find the percentage \(p\): \(p = \frac{x}{y}\cdot 100\)
- Find the whole \(y\): \(y = \frac{x}{p/100} = \frac{100x}{p}\)
Comparing fractions by normalization
Comparing fractions with different denominators can be cumbersome. For example, which is larger: \(\frac{7}{15}\) or \(\frac{5}{12}\)?
Convert each to "per 100":
\[ \frac{7}{15}\cdot 100 \approx 46.67\%,\qquad \frac{5}{12}\cdot 100 \approx 41.67\% \]
Since \(46.67\% > 41.67\%\), \(\frac{7}{15}\) is larger. This is the core reason percentages are practical: they put different ratios on the same scale.
How percentage change works
Suppose a value changes from \(s\) to \(t\). The absolute change is \(t-s\). To get the relative change, divide by the original \(s\):
\[ \text{relative change} = \frac{t-s}{s} \]
Convert to percent by multiplying by 100:
\[ p = \frac{t-s}{s}\cdot 100 \]
If \(p>0\), it is an increase. If \(p<0\), it is a decrease.
Example of percentage change
The change in the value of an amount can be expressed as a percentage increase or decrease relative to its original value. For example, if someone earning $500 per month receives a 5% raise, this means an additional $5 for every $100 of salary. So each $100 becomes $105.
\[ \text{New salary} = \frac{105}{100}\cdot 500 = 525 \]
On the other hand, if salary decreases by 5%, each $100 becomes $95.
\[ \text{New salary} = \frac{95}{100}\cdot 500 = 475 \]
In general:
- Increase by \(b\%\): \(\text{new value} = a\cdot\left(1+\frac{b}{100}\right)\)
- Decrease by \(b\%\): \(\text{new value} = a\cdot\left(1-\frac{b}{100}\right)\)
Why losses and gains are asymmetric
A frequent mistake is to think that a drop of \(p\%\) is reversed by a gain of the same \(p\%\). It is not.
After a drop of \(p\%\), a value \(x\) becomes:
\[ x' = x\left(1-\frac{p}{100}\right) \]
To recover to \(x\), the needed gain \(p'\%\) is measured on \(x'\):
\[ p' = \left(\frac{x-x'}{x'}\right)\cdot 100 = \frac{100p}{100-p} \]
For \(p=50\), this gives \(p'=100\). So a 50% loss needs a 100% gain to break even.
Frequently asked questions
How do I calculate a percentage of a number?
Divide the percentage by 100 and multiply by the base value. For example, 15% of 80 is \(0.15\cdot 80 = 12\).
How do I calculate percentage increase or decrease?
Subtract the starting value from the final value, divide by the starting value, and multiply by 100. A positive result indicates an increase, while a negative result indicates a decrease.
Why does a 50% loss require a 100% gain to recover?
After a 50% drop, only half of the original value remains. To get from 50 back to 100, the remaining amount must double, which corresponds to a 100% increase.
Is there a quick trick for mental percentage math?
Yes. You can swap the numbers: 36% of 50 equals 50% of 36. Often one side is much easier to compute mentally.