Fractional Ratio

On a school trip, \(\dfrac{1}{3}\) of the students go to a museum and the rest of the students go to an art gallery. What is the ratio of students who go to the museum to the art gallery?

Solution:

Since \(\dfrac{1}{3}\) of the students go to a museum, we can deduce that \(\dfrac{2}{3}\) of the students go to the art gallery since fractions add up to one.

We next write the fractions as a ratio, like this: \(\dfrac{1}{3}:\dfrac{2}{3}\).

Now, to simplify this ratio, we can multiply both sides by three. Thus, we end up with \(1:2\). Therefore, we can say that for every single student who goes to the museum, we have two students who go to the art gallery.

In a cinema, \(\dfrac{5}{6}\) of people are adults and the rest are children. What is the ratio of adults to children?

Solution:

Since \(\dfrac{5}{6}\) of the people are adults, \(\dfrac{1}{6}\) must be children since fractions add to one. Writing this as a ratio is \(\dfrac{5}{6}:\dfrac{1}{6}\).

Now, by multiplying both sides of the ratio by 6, we obtain \(5:1\). Thus, the ratio of adults to children is \(5:1\).

In a bag of sweets, \(\dfrac{1}{4}\) are red, \(\dfrac{1}{3}\) are green and the rest are orange. What is the ratio of red to green to orange sweets?

Solution:

The sum of red sweets and green sweets is \(\dfrac{1}{4}+\dfrac{1}{3}=\dfrac{3}{12}+\dfrac{4}{12}=\dfrac{7}{12}\).

Since the rest are orange, we can say that \(1-\dfrac{7}{12}=\dfrac{12}{12}-\dfrac{7}{12}=\dfrac{5}{12}\) are orange.

Putting our fractions into ratios, we have the ratio of red to green to orange sweets is \(\dfrac{1}{4}:\dfrac{1}{3}:\dfrac{5}{12}\). Now, to get integer values for our ratio, we need to multiply each element of the ratio by 12. We obtain \(3:4:5\). Thus, the ratio of red sweets to green sweets to orange sweets is \(3:4:5\).

In a reception class, the ratio of students to teachers is \(5:1\). What fraction of the class are students and what fraction are teachers?

Solution:

First, add together the different parts of the ratio. In this case, we add five and one to get six. This is the denominator of the fractions. Then, we can say that \(\dfrac{5}{6}\) are students and \(\dfrac{1}{6}\) of the class are teachers.

In a company, the ratio of female employees to male employees is \(2:3\). What fraction of the employees are male?

Solution:

First, add two and three to get five. Thus, \(\dfrac{2}{5}\) are female and \(\dfrac{3}{5}\) are male.

A piece of string is cut into three pieces in the ratio \(1:2:3\). Work out the fraction of the original piece that each of the three pieces makes up.

Solution:

\(1+2+3=5\). Thus, the smallest piece is \(\dfrac{1}{5}\), the middle piece is \(\dfrac{2}{5}\) and the largest piece is \(\dfrac{3}{5}\).

The ratio of boys to girls taking A-Level English in a school is \(3:7\). Work out the percentage of boys who take A-Level English.

Solution:

First, we can say that the fraction of boys taking A-level English is\(\dfrac{3}{10}\). Converting this to a percentage, we get 30%. Thus, 30% of students taking A-Level English are boys.

There are 300 people at the fair. The ratio of adults to children is \(1:2\). 20% of the children are under the age of 6 and get free entry. Work out the fraction of children who get free entry.

Solution:

First, we could say that the fraction of children is \(\(\dfrac{2}{3}\). Since there are 300 people at the fair, we can say that \(\dfrac{2}{3}\) of 300 are children. Since \(\dfrac{1}{3}\) of 300 is 100 since \(300\div 3=100\), we know that \(\dfrac{2}{3}\) of 300 is 200. Thus, 200 of the people at the fair are children. 20% of the children are under the age of 6, so we need to work out 20% of 200. Since 10% of 200 is 20, 20% of 200 is 40.

Thus, 40 attendees of the fair got free entry.

In the below triangle DEF, vector \(\overrightarrow{DE}=a\) and \(\overrightarrow{EF}=b\). Point \(A\) cuts the line \(DF\) such that \(\overrightarrow{DA}:\overrightarrow{AF}=2:3\). Work out \(\overrightarrow{DA}\).

Solution:

First, we can find the vector \(\overrightarrow{DF}\).

\[\overrightarrow{DF}=\overrightarrow{DE}+\overrightarrow{EF}=a+b\]

Now, since \(\overrightarrow{DA}:\overrightarrow{AF}=2:3\), we can say that,

\[\overrightarrow{DA}=\dfrac{2}{5}\overrightarrow{DF}=\dfrac{2}{5}(a+b)\]

Thus,

\[\overrightarrow{DA}=\dfrac{2}{5}(a+b)\]

In the below quadrilateral GHIJ, the point K cuts the vector \(\overrightarrow{IJ}\) in the ratio \(1:2\).

Given: \(\overrightarrow{GH}=a\), \(\overrightarrow{HI}=b\) and \(\overrightarrow{IJ}=c\), find an expression for the vector \(\overrightarrow{GK}\).

Solution:

First, note that the vector \(\overrightarrow{GI}=\overrightarrow{GH}+\overrightarrow{HI}=a+b\).

Now, K splits the vector \(\overrightarrow{IJ}\) in the ratio \(1:2\) and so we can say that,

\[\overrightarrow{IK}=\dfrac{1}{3}\overrightarrow{IJ}=\dfrac{1}{3}c\]

Thus,

\[\overrightarrow{GK}=a+b+\dfrac{1}{3}c\]