Percentages

Percentages, fractions, and decimals are three different ways to talk about the same value. For example, \(\textcolor{colorprop}{50}\textcolor{colordef}{\pourcent}\), the fraction \(\dfrac{\textcolor{colorprop}{1}}{\textcolor{colordef}{2}}\), and the decimal \(0.5\) all mean "half." Being able to convert between these forms is a very useful skill.
Remark: A very important idea is that \(100\pourcent = \dfrac{100}{100} = 1\).
Since \(100\pourcent\) is just another way of writing the number \(1\), you can multiply any number by \(100\pourcent\) without changing its value. This is a very useful trick for converting decimals and fractions into percentages.

Percentages are one of the best ways to compare a part to a whole.
For example, if a class of \(\textcolor{colordef}{20}\) students has \(\textcolor{colorprop}{12}\) girls, what percentage of the class is female? To solve this, we can set up a proportion to find an equivalent fraction with a denominator of 100:$$\begin{aligned}\frac{\text{part}}{\text{whole}} = \frac{\textcolor{colorprop}{12}}{\textcolor{colordef}{20}} &= \frac{\textcolor{colorprop}{x}}{\textcolor{colordef}{100}}\\ \textcolor{colorprop}{x} &= \frac{\textcolor{colorprop}{12}}{\textcolor{colordef}{20}} \times \textcolor{colordef}{100} && \text{(multiplying both sides by }\textcolor{colordef}{100})\\ \textcolor{colorprop}{x} &= \textcolor{colorprop}{60}\\ \end{aligned}$$The percentage of girls in the class is \(\textcolor{colorprop}{60}\textcolor{colordef}{\pourcent}\). This means for every \(\textcolor{colordef}{100}\) students, \(\textcolor{colorprop}{60}\) would be girls.

• The \(\textcolor{colorprop}{\text{part}}\) is the number of correct answers: \(\textcolor{colorprop}{21}\).
• The \(\textcolor{colordef}{\text{whole}}\) is the total number of questions: \(\textcolor{colordef}{24}\).
• \(\begin{aligned}[t]\text{Percentage Score} &= \frac{\textcolor{colorprop}{21}}{\textcolor{colordef}{24}} \times \textcolor{colordef}{100}\pourcent\\ &= 0.875 \times 100\pourcent \\ &= \textcolor{colorprop}{87.5}\textcolor{colordef}{\pourcent}\end{aligned}\)

In Parliament A, there are \(\textcolor{colorprop}{26}\) women out of \(\textcolor{colordef}{50}\) members. In Parliament B, there are \(\textcolor{colorprop}{30}\) women out of \(\textcolor{colordef}{80}\) members. Hugo says, "Since there are more women in Parliament B, women have better representation there."
Is this statement a fair comparison?

Method 1: Using the formula$$\begin{aligned}\textcolor{colorprop}{\text{Number of girls}} &= \textcolor{colorprop}{60}\textcolor{colordef}{\pourcent} \times \textcolor{colordef}{200} \\ &= 0.60 \times \textcolor{colordef}{200} \\ &= \textcolor{colorprop}{120}\end{aligned}$$There are \(\textcolor{colorprop}{120}\) girls in the school.Method 2: Cross-Multiplication
Set up a proportion where \(x\) is the number of girls.$$\begin{aligned}\frac{\textcolor{colorprop}{60}}{\textcolor{colordef}{100}} &= \frac{\textcolor{colorprop}{x}}{\textcolor{colordef}{200}} \\ \textcolor{colordef}{100} \times \textcolor{colorprop}{x} &= \textcolor{colorprop}{60} \times \textcolor{colordef}{200} \\ \textcolor{colorprop}{x} &= \frac{12000}{100} = \textcolor{colorprop}{120}\end{aligned}$$

Method 1: Using the formula$$\begin{aligned}\textcolor{colordef}{\text{Total students}} &= \frac{\textcolor{colorprop}{14}}{\textcolor{colorprop}{40}\textcolor{colordef}{\pourcent}} \\ &= \frac{\textcolor{colorprop}{14}}{0.40} \\ &= \textcolor{colordef}{35}\end{aligned}$$There are \(\textcolor{colordef}{35}\) students in the class.Method 2: Cross-Multiplication
Set up a proportion where \(x\) is the total number of students.$$\begin{aligned}\frac{\textcolor{colorprop}{40}}{\textcolor{colordef}{100}} &= \frac{\textcolor{colorprop}{14}}{\textcolor{colordef}{x}} \\ \textcolor{colorprop}{40} \times \textcolor{colordef}{x} &= \textcolor{colorprop}{14} \times \textcolor{colordef}{100} \\ \textcolor{colordef}{x} &= \frac{1400}{40} = \textcolor{colordef}{35}\end{aligned}$$

Quantities often change by a certain percentage. For example:

• A store offers a 20\(\pourcent\) discount (a decrease).
• A salary increases by 7\(\pourcent\).
• A city's population grows by 10\(\pourcent\).

1. Calculate the decrease amount:$$\begin{aligned} \text{Decrease} &= 20\pourcent \text{ of }\dollar 50\\ &= 20\pourcent \times \dollar 50\\ & = 0.20 \times \dollar 50 \\ &= \dollar 10\\ \end{aligned}$$
2. Calculate the new price:$$\text{New Price} = \dollar 50 - \dollar 10 = \dollar 40$$

Percentage Change is a way to express a change in a quantity as a percentage of the original amount. This allows us to understand the scale of the change, whether it's an increase or a decrease.

The percentage change is \(+10\pourcent = +0.10\).$$\begin{aligned}\text{New amount} &= \dollar200 \times (1 + 0.10) \\ &= \dollar 200 \times 1.10 \\ &= \dollar 220\end{aligned}$$

The percentage change is \(-25\pourcent = -0.25\).$$\begin{aligned}\text{New amount} &= \dollar 200 \times (1 - 0.25) \\ &= \dollar 200 \times 0.75 \\ &= \dollar 150\end{aligned}$$

We start from the multiplier formula:$$\begin{aligned}\text{New Value} &= \text{Original Value} \times (1 + \text{Percentage Change}) \\ \frac{\text{New Value}}{\text{Original Value}} &= 1 + \text{Percentage Change} && \text{(Divide by Original Value)} \\ \frac{\text{New Value}}{\text{Original Value}} - 1 &= \text{Percentage Change} && \text{(Subtract 1)} \\ \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} &= \text{Percentage Change} && \text{(Combine into a single fraction)}\end{aligned}$$In this final step, the percentage change is in decimal form. We multiply by \(100\pourcent\) to express it as a percentage.

The weight increases, so we expect a positive result.$$\begin{aligned}\text{Percentage Change} &= \frac{28 - 25}{25} \times 100\pourcent \\ &= \frac{3}{25} \times 100\pourcent \\ &= +12\pourcent\end{aligned}$$This is a \(12\pourcent\) increase.

The price decreases, so we expect a negative result.$$\begin{aligned}\text{Percentage Change} &= \frac{420 - 500}{500} \times 100\pourcent \\ &= \frac{-80}{500} \times 100\pourcent \\ &= -16\pourcent\end{aligned}$$This is a \(16\pourcent\) decrease.

Let \(V_0, V_1, V_2\) be the successive values.

• \(V_1 = V_0 \times CM_1\) and \(V_2 = V_1 \times CM_2\).
• By substitution: \(V_2 = (V_0 \times CM_1) \times CM_2 = V_0 \times (CM_1 \times CM_2)\).
• By definition, \(V_2 = V_0 \times CM_{global}\). Therefore, \(CM_{global} = CM_1 \times CM_2\).

The multipliers are \(CM_1 = 1.08\) and \(CM_2 = 1.10\).$$\begin{aligned} CM_{global} &= 1.08 \times 1.10 = 1.188 \\ 1 + \frac{t}{100} &= 1.188 \\ \frac{t}{100} &= 0.188 \\ t &= 18.8\pourcent\end{aligned}$$The global increase is \(18.8\pourcent\). (Note: \(8 + 10 = 18\), but the real change is higher due to compounding).

We have \(V_1 = V_0 \times CM\). To find \(V_0\) from \(V_1\): \(V_0 = \frac{V_1}{CM} = V_1 \times \frac{1}{CM}\).
The multiplier required to go from \(V_1\) back to \(V_0\) is therefore \(\frac{1}{CM}\).

The initial multiplier is \(CM = 1 + 0.05 = 1.05\).
The reciprocal multiplier is:$$CM' = \frac{1}{1.05} \approx 0.9524$$Converting this to a rate \(t'\):$$\begin{aligned}1 + \frac{t'}{100} &= CM'\\ t'& = (CM' - 1) \times 100 \\ t'&\approx -4.76\pourcent\end{aligned} $$The population must decrease by approximately \(4.76\pourcent\) to return to its original level.