A ratio compares multiple quanitites measured in the same units. We write ratios with a colon (:) in between the quanties. For example, if you had \(10\) marbles and \(6\) were red, the ratio of \(\text{red : blue}\) is ( \(6 : 4\) ). We read this as "the ratio of red to blue marbles is \(6\) to \(4\)".
You can write a ratio comparing more than one quantity. Consider a classroom of \(30\) students picking their favourite subject at school. We could write a ratio like:
\(\text{math : english : art}\)
\( 8 : 10 : 7\)
This means that there are \(8\) students whose favourite subject is math, \(10\) whose favourite is english and \(7\) whose favourite is art. Since we know there are \(30\) students, we can assign the remaining \(5\) to an other category:
\(\text{math : english : art : other}\)
\( 8 : 10 : 7 : 5\)
We can also include the total in the ratio:
\(\text{math : english : art : other : total}\)
\( 8 : 10 : 7 : 5 : 30\)
There are a lot of numbers in this ratio! You can pick the numbers you are interested in, such as the ratio of art to total:
\(\text{art : total}\)
\(7 : 30\)
We can also write ratios as fractions. The ratio of \(\text{math : english}\) is \(8 : 10\). You can write this as a fraction like:
\(\cfrac{\text{math}}{\text{english}} = \cfrac{8}{10}\)
Often we will reduce the ratio/fraction to simpliest terms. Recall that reduce fractions means to find an equilvant fraction that is in lowest terms. We should divide both the numerator and denominator by the greatest common factor.
Using the example above, we can divide top and bottom by \(2\):
\(= \cfrac{8 \div 2}{10 \div 2} \)
\(= \cfrac{4}{5} \)
We can interpret the ratio as for every \(4\) students that like math, \(5\) like english. The ratio is perserved as the number of students increases. For example, if there were \(40\) students that like math, then \(50\) would like english. Basically, the ratio is constant so each fraction has to be equivalent:
\( \cfrac{\text{math}}{\text{english}} = \cfrac{4}{5}\)
\( \cfrac{\text{math}}{\text{english}} = \cfrac{4}{5} = \cfrac{40}{x}\)
\( \cfrac{4 * ?}{5 * ?} = \cfrac{40}{x}\)
\( \cfrac{4 * 10}{5 * 10} = \cfrac{40}{50}\)
Let's start on the left side. The ratio has three terms with one missing. It is also equivalent to the right hand side that is also missing one term. Let's start by solving for \(x\). On the right hand side, let's take the \(1\) and \(x\). We choose the \(1\) because we also have the \(3\) on the other side of the equation. We also take the \(9\) since it goes with the \(x\).
Make equivalent fractions with each ratio:
\( \textcolor{red}{1} : 6 : \textcolor{blue}{x} = \textcolor{red}{3} : y : \textcolor{blue}{9} \)
\( \cfrac{\textcolor{red}{1}}{\textcolor{blue}{x}} = \cfrac{\textcolor{red}{3}}{\textcolor{blue}{9}} \)
From this fraction, we can see that we have to multiply by \(3\) to go from left to right to set the numerators equal to each other:
\(\cfrac{\textcolor{red}{1} *3}{\textcolor{blue}{x} *3} = \cfrac{\textcolor{red}{3}}{\textcolor{blue}{9}} \)
Then, we can solve for \(x\) by dividing both sides by its coefficient:
\(\cfrac{3x}{3} = \cfrac{9}{3}\)
\( x = 3 \)
You can see the two fractions are equivalent:
\(\cfrac{\textcolor{red}{1}}{\textcolor{blue}{3}} = \cfrac{\textcolor{red}{3}}{\textcolor{blue}{9}} \)
We will use the same process to solve for \(y\).
\( \textcolor{red}{1} : \textcolor{blue}{6} : x = \textcolor{red}{3} : \textcolor{blue}{y} : 9 \)
\( \cfrac{\textcolor{red}{1}}{\textcolor{blue}{6}} = \cfrac{\textcolor{red}{3}}{\textcolor{blue}{y}} \)
We still need to multiply by \(3\) to go from left to right. To solve for the denominator, we have the equation:
\(6*3 = y\)
\( y = 18\)
We can see that the two fractions are equivalent as such:
\(\cfrac{\textcolor{red}{1}}{\textcolor{blue}{6}} = \cfrac{\textcolor{red}{3}}{\textcolor{blue}{18}} \)
A Rate is a ratio that compares two quantities with different units. Rates are used to help us with conversions. For example, \(1\) hour is \(60\) minutes. A Unit Rate is a special rate where the second quantity is \(1\). For example, \(20 \; [\text{km/hr}]\) means \(20 \; [\text{km}]\) are travelled in \( 1 \; [\text{hr}]\).
Unit rates can be helpful for comparing different options. For example, you are at the grocery store and see two options for juice. One brand sells \(1 \; [\text{L}]\) for \($2.99\). Another brand sells \(700 \; [\text{mL}]\) for \($1.99\). Which price is better?
The price per litre ($/L) can be represented as a rate! The first option is per \(1 \; [\text{L}]\) already. Let's convert the second option:
\( \cfrac{$1.99}{700 \; [\text{mL}]} \)
\(= \cfrac{$1.99}{0.700 \; [\text{L}]} \)
\(= \cfrac{$2.84}{1 \; [\text{L}]} \)
Next, we can find the difference between the \(2\) rates to determine which price is better:
\(d = \text{rate}_1 - \text{rate}_2\)
\(d = $2.99 - $2.84\)
\(d = $0.15\)
Therefore, we can determine that the second option is cheaper by \(\boldsymbol{15}\) cents!
First, we can write equivalent rates to solve for the missing quantity:
\(\cfrac{20 \; [\text{c}]}{2 \; [\text{kWh}] } = \cfrac{x \; [\text{c}]}{10 \; [\text{kWh}]}\)
Next, we can multiply the top and bottom by \(5\) to convert the rate:
\(\cfrac{20 \; [\text{c}] * 5}{2 \; [\text{kWh}] * 5} = \cfrac{100 \; [\text{c}]}{10 \; [\text{kWh}] }\)
Therefore, we can determine that the cost for \(10 \; [\text{kWh}]\) is \(\boldsymbol{100 \; [\textbf{c}]}\).