# Fractions

We understand how to count the integers and how the order of operations work for them. But what happens if the items that we're working with are not in whole parts? If someone ate one slice of your pizza, how much do you have left? We certainly do not have 1 pizza, but neither do we have 0 pizzas.

**Fractions** are numbers defined by division, used to represent any number of equal parts of a whole. They are real numbers of form \(\dfrac { p }{ q }\) where \(p\) are \(q\) are integers. The number of parts is given by the number in the top of the fraction, called the **numerator**, and number of pieces that make up the whole, which tells us their size, is described by the **denominator**, the number on the bottom. Thus in the fraction \( \dfrac{2}{3} \), read as "two thirds", 2 is the numerator and 3 is the denominator. This tells us that there are two parts, and they each are one third of a whole.

Let us explore the concept of fractions, and understand how it is used to solve various problems.

#### Contents

## Visual Representation

The part shaded in blue below is a visual representation of two thirds:

Proceeding step-by-step we can see why \( \dfrac{2}{3} \) looks like the image above:

Begin with an entire object or unit. (\(1\))

Divide it according to the denominator (here we are dividing by \(3\)). \( \left(\dfrac{1}{3} \right) \)

Multiply that result according to the numerator (here we multiply by \(2\)). \( \left(\dfrac{2}{3} \right)\)

## Classification

A **proper fraction** is one with the numerator less than the denominator. Example: \( \dfrac{4}{5}. \)

An **improper fraction** is one with the numerator greater than the denominator. Example \( \dfrac{7}{4}. \)

A **mixed number** is written as with a whole number part and a fractional part. The fractional part of a mixed number is always a proper fraction. Example: \( 4\dfrac{1}{3}. \)

## Multiplying Fractions

To multiply fractions, we follow these steps:

- Multiply the numerators.
- Multiply the denominators.
- Simplify the fraction by dividing throughout by their GCD.

## Evaluate \( \dfrac{ 2}{3} \times \dfrac{ 9}{ 4 } \).

Multiplying the numerators and denominators, we get

\[ \dfrac{2}{3} \times \dfrac{9}{4} = \dfrac{ 2 \times 9 } { 3 \times 4} = \dfrac{ 18 } { 12 } . \]

Since \( \gcd(12, 18) = 6 \), hence the simplified fraction is

\[ \dfrac{18}{12} = \dfrac{ 18 \div 6 } { 12 \div 6 } = \dfrac{3}{2}. \]

Thus \( \dfrac{2}{3} \times \dfrac{9}{4} = \dfrac{3}{2} \). \( _\square \)

If the fraction is a mixed fraction, then we must first convert it into an improper fraction, before multiplying.

## Evaluate \( 1 \dfrac{2}{3} \times 2 \dfrac{ 3}{4} \).

Converting the mixed fractions into improper fractions, we have

\[ 1 \dfrac{2}{3} = \dfrac{ 1 \times 3 + 2 } { 3} = \dfrac{5}{3}, 2 \dfrac{3}{4} = \dfrac{ 2 \times 4 + 3 } { 4} = \dfrac{11}{4}. \]

Hence,

\[ 1 \dfrac{2}{3} \times 2 \dfrac{3}{4} = \dfrac{5}{3} \times \dfrac{ 11}{4} = \dfrac{5 \times 11 } { 3 \times 4 } = \dfrac{55}{12} . \]

Since the greatest common divisor of \(55\) and \(12\) is \(1,\) the final answer is \(\dfrac{55}{12} \). \( _\square\)

Note that \( 1 \dfrac{2}{3} \times 2 \dfrac{3} {4} \neq ( 1 \times 2) \dfrac{ 2 \times 3 } { 3 \times 4} = 2 \dfrac{1}{2} \). \( _\square \)

## Simplify: \[\dfrac{4}{5} \times 3 \dfrac{3}{2} \]

Converting the mixed fraction into an improper fraction, we have \( 3 \dfrac{3}{2} = \dfrac{3 \times 2 + 3}{2} = \dfrac{9}{2}.\) Hence,

\[ \dfrac{4}{5} \times 3 \dfrac{3}{2} = \dfrac{4}{5} \times \dfrac{9}{2} =\dfrac{4 \times 9}{5 \times 2}= \dfrac{36}{10}= \dfrac{18}{5},\]

where we simplified the fraction by dividing both the numerator and denominator by \( \gcd(36, 10) = 2 .\) \( _\square \)

## Evaluate \( \dfrac{2}{3} \times 1 \dfrac{3}{4} \times \dfrac{6}{5} .\)

Converting the mixed fraction into an improper fraction, we have \( 1 \dfrac{3}{4} = \dfrac{7}{4}. \) Hence,

\[ \dfrac{2}{3} \times \dfrac{7}{4} \times \dfrac{6}{5} = \dfrac{2 \times 7 \times 6}{3 \times 4 \times 5} = \dfrac{84}{60} = \dfrac{7}{5} ,\]

where we simplified the fraction by dividing both the numerator and denominator by \( \gcd(84, 60) =12 .\) \( _\square \)

## Adding Fractions

To add fractions, we follow these steps:

- Convert to a common denominator using the Lowest Common Multiple.
- Add the numerators.
- Simplify the fraction by dividing throughout by the Greatest Common Divisor.

## Evaluate \( \frac{2}{3} + \frac{4}{5} \).

Since the denominators are different, we have to make common denominator. The Lowest Common Multiple of 3 and 5 is 15, so we have \( \frac{2}{3} = \frac{ 2 \times 5 } { 3 \times 5 } = \frac{10}{15} \) and \( \frac{4}{5} = \frac{4 \times 3 } { 5 \times 3 } = \frac{ 12}{15} \).

Then, we add the numerators to get

\[ \frac{10}{15} + \frac{12}{15} = \frac{22}{15} . \]

Since \( \gcd(22, 15) = 1 \), hence this fraction is already simplified.

Thus, \( \frac{2}{3} + \frac{4}{5} = \frac{22}{15} = 1 \frac{7}{15} \).

The procedure is similar for mixed fractions, but we must convert to an improper fraction first.

Alternatively, it can be simpler to add the integers and the fractional parts separately.

## Evaluate \( 1 \frac{2}{3} + 2 \frac{5}{6} \).

Converting to improper fractions, we have \( 1 \frac{2}{3} = \frac{1 \times 3 + 2 } { 3} = \frac{5}{3} \) and \( 2 \frac{5}{6} = \frac{2 \times 6 + 5 } { 6} = \frac{17}{6} \).

Since the denominators are different, we make a common denominator of \(6\). This gives us

\[ \frac{5}{3} + \frac{17}{6} = \frac{ 5 \times 2 } { 3 \times 2 } + \frac{17}{6} = \frac{27} { 6} \]

Finally, we simplify the fraction. Since \( \gcd(27, 6) = 3 \), hence we divide throughout by 3 to get \( \frac{27 \div 3 } { 6 \div 3} = \frac{9}{2} \).

Thus, \( 1 \frac{2}{3} + 2 \frac{5}{6} = \frac{9}{2} = 4 \frac{1}{2} \). \(_\square\)

## Evaluate \(\displaystyle{ 2 \frac{3}{4} + 1\frac{1}{2} }.\)

Converting each term to improper fractions, we have \( 2 \frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{11}{3} \) and \( 1\frac{1}{2} = \frac{1\times2+1}{2}=\frac{3}{2}.\)

Since the denominators are different, we make a common denominator of \(6.\) This gives us

\[ 2 \frac{3}{4} + 1\frac{1}{2} = \frac{11}{3} + \frac{3}{2} = \frac{11 \times 2}{3 \times 2} + \frac{3 \times 3 }{2 \times 3} = \frac{22 + 9}{6} = \frac{31}{6} . \ _\square \]

## Evaluate \( \displaystyle{3\frac{1}{2} + 2\frac{2}{3} + 1\frac{3}{4} }.\)

Converting each term to improper fractions, we have \( 3\frac{1}{2} = \frac{3\times2+1}{2} = \frac{7}{2}, 2\frac{2}{3} = \frac{2\times3 + 2}{3} = \frac{8}{3}\) and \( 1\frac{3}{4} = \frac{1\times4+3}{4}=\frac{7}{4}.\)

Since the denominators are different, we make a common denominator of \(12.\) This gives us

\[ \begin{align} 3\frac{1}{2} + 2\frac{2}{3} + 1\frac{3}{4} &= \frac{7}{2} + \frac{8}{3} + \frac{7}{4} \\ &= \frac{7\times6}{2\times6} + \frac{8\times4}{3\times4}+\frac{7\times3}{4\times3} \\ &= \frac{42+32+21}{12} \\ &= \frac{95}{12}. \end{align} \]

Since \( \gcd(95,12) = 1 ,\) hence this fraction is already simplified.

Thus, \( 3\frac{1}{2} + 2\frac{2}{3} + 1\frac{3}{4} = \frac{95}{12} = 7\frac{11}{12}.\) \( _ \square\)

## Fraction Arithmetic

Given a sequence of operations on proper fractions, possibly including multiplication, division, addition, and subtraction, we first figure out the order to carry out the sequence of operations by following the usual rules for order of operations.

## What is \(\frac{1}{4} + \frac{2}{3} - \frac{3}{5}\)?

Solution:** Gathering the fractions using the least common multiple of the denominators, we have

\[\begin{align} \frac{1}{4} + \frac{2}{3} - \frac{3}{5} &= \frac{15}{60} + \frac{40}{60} - \frac{36}{60}\\ &= \frac{15 + 40 - 36}{60}\\ & = \frac{19}{60}. \end{align}\]

Since the greatest common divisor of \(19\) and \(60\) is 1, the final answer is \(\frac{19}{60} \). \( _\square\)

## What is \( \frac{1}{3} + \frac{2}{5} \div \frac{6}{5} \)?

Following order of operations, we must divide first, so we have \( \frac{1}{3} + \left( \frac{2}{5} \div \frac{6}{5} \right) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \). Thus the answer is \( \frac{2}{3} \). \( _\square \)

## What is \( \frac{20}{3} -\frac{1}{9} \div \left[ \frac{3}{7} \times \left( - \frac{1}{4} \right) \times \left( -\frac{2}{3} \right)^2 \right] \)?

Following order of operations, we first evaluate the quantity in the parenthesis, which gives \[\begin{align} \frac{20}{3} -\frac{1}{9} \div \left[ \frac{3}{7} \times \left( - \frac{1}{4} \right) \times \left( -\frac{2}{3} \right)^2 \right] &= \frac{20}{3} -\frac{1}{9} \div \left[ \frac{3}{7} \times \left( - \frac{1}{4} \right) \times \left( \frac{4}{9} \right) \right]\\ &= \frac{20}{3} -\frac{1}{9} \div \left[ \frac{3}{7} \times \left( - \frac{1}{9} \right) \right]\\ &= \frac{20}{3} + \frac{1}{9} \div \left[ \frac{3}{63} \right]\\ &= \frac{20}{3} + \left( \frac{1}{9} \times \frac{63}{3} \right)\\ &= \frac{20}{3} + \left( \frac{7}{3} \right)\\ &= \frac{27}{3}\\ & = 9. _\square \end{align}\]

## Complex Fractions

Main Article: Complex Fractions

A **complex fraction** is a fraction with another fraction in the numerator or denominator (or both). Below is an example of a complex fraction:

\[ \frac{1-\frac{1}{x}}{1+\frac{1}{x}} \]

Complex fractions should not be confused with fractions involving complex numbers.

## Simplify \[\dfrac{\;\dfrac{2}{3}\; }{\dfrac{2}{7}} .\]

We can multiply both the numerator and the denominator by 21, which is LCM of \(3\) and \(7,\) to the denominator,and numerator. And it follows that \[ \frac{\frac{2}{3}\times 21}{\frac{2}{7}\times 21} = \frac{2 \times 7}{2 \times 3} = \frac{14}{6} = \frac{7}{3} = 2 \frac{1}{3}. \ _ \square \]

## Fractions - Word Problems

To be able to solve word problems with fractions you must firstly be able to solve regular word problems by translating common language to math. The SAT Translating Word Problems gives a great insight of how this can be done. Please, give it a look if you're not familiar with word problems.

Fractions can be of great use in more realistic scenarios in which part of a total must be calculated. Chemical solutions, for example, use a great deal of fraction to calculate parts of a total, and just like problems that deal with solutions, any other problem which must differentiate or do any other process with parts of a total can usually be solved and simplified by the correct use of fractions.

Fractions can be given directly in a word problem (e.g.: \( \frac{7}{10} \) of the human body is composed of water) or by common language, with no numbers representing it. In that case you must be able to **search and identify keywords** which refer to fractions, which are usually the same ones that refer to division since they're technically the same operation.

The Final Math Exam will be 1 hour long. The teacher said that the whole test can be read in 5 minutes, each question answered in 2 minutes, and the work reviewed with the time that's left. If the test has 20 questions, then what fraction of the time to do the test can be used to review it?

Considering the problem has been read, we must identify what's been asked by it. The problem is requiring us to get the fraction of time that can be used to review from the total time the test has to be done. So here we need two things: The time to review the test (numerator) from the total test time (denominator).

\( \frac{Time \space to \space review \space the \space test}{Test \space time} \)

The time to review the test is not given, so we must figure that out. The total test time is 1 hour. Each hour has 60 minutes. Because everything else done in the test is given in minutes, we must have both units in the same measure to be able to correctly compare them. So the hours must be converted to minutes.

\( Test \space time = 1 \space hour \)

\( 1 \space hour = 60 \space minutes \)

\( \therefore Test \space time = 60 \space minutes \)

\( \frac{Time \space to \space review \space the \space test}{Test \space time} = \frac{Time \space to \space review \space the \space test}{60 \space minutes} \)

Now we must find the time to review the test. The time to review is the time that's left after everything else is done since, according to the problem, since "the work [is] reviewed with the time that's left." To find the time that's left, we must find the time that's used. 5 minutes of the test are going to be used to read it and 2 minutes to read each question. There are 20 questions, so \( 2 \space minutes \times 20 \) minutes are going to be used to do the questions. Out of the total of 1 hour, that's the time that'll be left:

\( Time \space to \space review \space the \space test = 60 \space minutes - 5 \space minutes - 2 \space minutes \times 20 \)

\( Time \space to \space review \space the \space test = 55 \space minutes - 40 \space minutes \)

\( Time \space to \space review \space the \space test = 15 \space minutes \)

Now we have the numerator and denominator.

\( \frac{Time \space to \space review \space the \space test}{Total \space test \space time} = \frac{15 \space minutes}{60 \space minutes} = \frac{1}{4} \)

So the answer is \( \frac{1}{4} \).

Problems with basic addition and subtraction tasks are relatively simple to analyze and identify the underlying procedures to solve for it. Other problems, however, may require a deeper contextual understanding so further correlations between given data may be identified.

There is a solution **A** of 150ml with 30% acidity. What should be the acidity concentration of solution **B** to be added to solution **A** so the resulting solution **C** will be 1L and 50% acidity? Round the answer to the nearest hundreth.

Tip: Use the following table for your help

Solution A | Solution B | Solution C | |

Volume | 150ml | ? | 1L |

Acidity % | 30% | ? | 50% |

## Fractions - Problem Solving

## How many \( \frac{1}{7}\text{'s}\) are there in \( 10 \frac{2}{5} \)?

Measuring one number by another is just division, so the question is equivalent to asking for \( 10\frac{2}{5} \div \frac{1}{7} \).

Converting the mixed number to an improper fraction and performing the operation, we get

\[ \begin{align} 10\frac{2}{5} \div \frac{1}{7} &= \frac{52}{5} \div \frac{1}{7} \\ &= \frac{52}{5} \times 7 \\ &= \frac{364}{5}. \end{align} \]