Number & algebra - Numeracy fundamentals - UniSkills

This page explains some key ideas about number and algebra. As you go through the content, you’ll find activities to help you practise and strengthen your understanding.

If you would like more example questions on these topics, especially if you are preparing for the Literacy and Numeracy Test for Initial Teacher Education (LANTITE), you might like to visit the LANTITE page. If you’re interested in learning more advanced algebra skills, you can also check out the Algebra module.

As you work through the examples, remember that there is often more than one way to solve a problem. In some cases, we’ve shown different methods, but for many examples, only one way is explained. If you already know another correct method, that’s great! You don’t need to change how you do it. What matters most is that you use a method that makes sense to you and gives the right answer.

In brief, this page covers the following:

How to use BIMDAS to solve problems with more than one operation.
How to round decimal numbers.
How to perform calculations involving percentages.
How to carry out calculations involving fractions.
How to perform calculations involving ratios, proportions and rates.
How to calculate averages.
How to write and apply simple algebraic formulas

Order of operations (BIMDAS)

Understanding the correct order to complete calculations in a maths problem is very important. If you don’t follow the right order, you might get the wrong answer. It is also important to use the correct order when writing expressions, so that other people - and even your calculator - can understand and solve them the way you meant.

The rule for this is called the order of operations, and one way to remember it is with the acronym BIMDAS. This stands for:

Brackets - Always do the calculations inside brackets first. When writing expressions, use brackets to make sure certain parts are done first. If you’re not sure, it’s okay to add brackets anyway - it won’t cause any problems.
Indices - After brackets, solve any indices (also called exponents). These are not explained in this section, but if you want to learn more, check out the Algebra module.
Multiply, Divide - Next, do any multiplication or division. These two are treated equally, so just go from left to right and do whichever comes first.
Add, Subtract - Lastly, do any addition or subtraction. Like multiplication and division, these are also equal in priority, so work from left to right and solve whichever comes first.

Example: Michelle is a maths teacher. She wants to buy a licence for some maths software that her students will use during a $10$ week school term. The licence costs $\$5$ per student as a one-time registration fee. After that, it costs $\$2$ per student each week. Michelle has $27$ students in her class. How much will the total cost be?

Solution: To work this out, we first need to calculate the cost for one student. Each student pays $\$5$ for registration, plus $\$2$ per week for $10$ weeks. So the cost for one student is $5 + 2 \times 10$ (we don’t need brackets here because BIMDAS tells us that multiplication happens before addition, so the total weekly cost is calculated first).

Next we need to multiply the cost per student by $27$ as there are $27$ students in the class. To make sure the cost per student is calculated first, we use brackets, so the formula should be $27 \times (5 + 2 \times 10)$

You can enter this into a calculator, or work it out by hand. Either way, the total cost is $\$675$

Another way to write the formula is $27 \times 5 + 27 \times 2 \times 10$ which gives the same total of $$675$

Once you have worked through the example above, have a go at using BIMDAS to solve the problems in the following activity:

Decimal numbers

A decimal number has two parts. The part on the left side of the decimal point is the whole number part. The part on the right side is the fractional part.

If the whole number part is $0$ we usually write the $0$ before the decimal point to make the number clearer. For example, we write $0.35$ not just $.35$

If the fractional part is $0$ we usually leave out the decimal point and write the number as a whole number. For example, instead of writing $2.0$ we would usually just write $2$

Being able to round decimal numbers is important as often it is not practical or necessary to write every digit. In some cases, like with the number $\pi$ (Pi), it is actually impossible to write the full number because the decimal goes on forever without repeating. These are called irrational numbers.

The number of decimal places you round to can depend on the situation. For example:

when working with money, you usually round to two decimal places;
if a question gives values to three decimal places, you should round your answer to three decimal places;
sometimes, it makes more sense to round to the nearest whole number, such as when counting people, or;
you might also be told exactly how many decimal places to round to.

When solving a problem with more than one step, it is important to keep all the exact values while you are woirking. You should only round your final answer. This helps make sure your answer is accurate.

When rounding a decimal number to a certain number of decimal places, or to the nearest whole number, follow these steps:

First, find the ‘round-off place’. If you are rounding to $n$ decimal places then your round-off place is the number in the $n^{th}$ decimal place. If you are rounding to the nearest whole number, then your round-off place is the number to the left of the decimal point.
Next, look at the digit immediately to the right of your round-off place.
If this number is $5$ or more, round up. This means you increase the digit in the round-off place by $1$ and remove all digits after it. If this number is less than $5$, round down. This means you keep the digit in the round-off place the same and remove all digits after it.

Example: Suppose you want to round $4.576$ to two decimal places.

Solution: Following the steps above, we have:

Since we are rounding to two decimal places the round-off place is the number in the second decimal place, which in $4.576$ is $7$
The number immediately to the right of this is the number in the third decimal place, which is $6$
Since $6$ is greater than $5$ we increase the number in the round-off place from $7$ to $8$ and remove all digits after it. So we round up to $4.58$

Once you have worked through the example above, have a go at the rounding problems in the following activity:

Percentages

A percentage is a way of showing a number out of $100$. It is written using the % symbol. For example, if the tax rate is $30\%$, it means you pay $\$30$ in tax for every $\$100$ you earn.

Here are some important facts about percentages:

You can find a percentage of a quantity. This includes working out percentage increases and decreases. You can also show one quantity as a percentage of another.
Sometimes, the percentage can be more than $100\%$. This can happen when calculating a percentage increase or when comparing one quantity to another.

Percentages are often used to understand and compare data. They are also used to show things like discounts in sales, commissions, and interest rates.

Calculating percentages

A percentage shows an amount out of $100$. To calculate a percentage of a quantity you need to divide the percentage by $100$ and multiply by the quantity.

Example: If $15\%$ of the students in a particular school are likely to need extra help with their numeracy each year, and there are $360$ students in the school this year, how many students are likely to need extra help?

Solution: To calculate $15\%$ of $360$ you need to divide $15$ by $100$ and multiply by $360$:

$15 \div 100 \times 360 = 54$

So $54$ students are likely to need extra help with their numeracy. You can always do the operations in a different order if you prefer, as long as it gives you the correct answer and makes sense to you.

Building on this idea, you will often need to work out the result of a percentage increase or decrease. For example, you may want to find:

the new price of an item after a $20\%$ discount, or
the amount of money in your bank account after $3\%$ interest has been added.

There are two main ways to find the new value after a percentage is applied:

Find the percentage of the original value, using the method explained earlier.
- If it’s an increase, add this amount to the original value.
- If it’s a decrease, subtract this amount from the original value.
Add or subtract the percentage from 100%, depending on whether it’s an increase or decrease.
- The calculate this new percentage of the original value, using the same method as before.

Both methods one and two will give you the same final result. You can choose the one that feels easier for you or the situation.

Example 1: Suppose you want to find the total cost of a $\$350$ computer, once $10\%$ GST has been added.

Solution: One way of calculating this would be:

$10 \div 100 \times \$350 = \$35$ and then $\$350 + \$35 = \$385$

So the total cost of the computer is $\$385$

Example 2: Suppose you’d like to buy a TV, and you find one that has a $25\%$ off deal. If the TV was originally priced at $\$799$ how much will the sale price be?

Solution: One way of calculating this would be:

$100 - 25 = 75$ and then $75 \div 100 \times \$799 = \$599.25$

So the new price of the TV will be $\$599.25$

Alternatively, you may sometimes need to work out the original value before a percentage was applied, using the new value you already know. For example, you might need to find:

the original price of an item before a $20\%$ discount given the discounted price, or
the original bank balance before $3\%$ interest is added, given the current balance.

To do this, add the percentage to $100\%$ if it was an increase, or subtract the percentage from $100\%$ if it was a decrease. Then divide the new value by this percentage and multiply the result by $100$.

Example: Suppose you know that the cost including GST of a tablet is $\$451$ and you want to calculate the cost before $10\%$ GST was added.

Solution: To calculate this you would do:

$100 + 10 = 110$ and $\$451 \div 110 \times 100 = \$410$

So the original price of the tablet was $\$410$

Once you have worked through the examples above, have a go at the percentage problems in the following activity:

Calculating percentages without a calculator

While you can usually use a calculator to work out percentages with the formulas above, it is also important to understand how to estimate (approximate) percentages without a calculator. You can do this by combining the methods below. You can also round numbers to make the calculations easier when needed:

To calculate $1\%$ of a number: divide by $100$ by moving the decimal place two places to the left (For whole numbers, this is the same as removing two zeroes, if possible.)
To calculate $10\%$ of a number: divide by $10$ by moving the decimal place one place to the left (For whole numbers, this is the same as removing one zero, if possible.)
To calculate $25\%$ of a number: divide by $4$ (or divide by $2$ and $2$ again if that is easier)
To calculate $33\%$ of a number: divide by $3$
To calculate $50\%$ of a number: divide by $2$

Example: Approximate $21\%$ of $619300$

Solution: You could do this as follows:

$10\%$ of $619300$ is $61930 \approx 62000$ so $20\%$ of $619300 \approx 62000 + 62000 \approx 124000$

$1\%$ of $619300$ is $6193 \approx 6200$

So $21\%$ of $619300 \approx 124000 + 6200 \approx 130200$

Once you have worked through the example above, have a go at completing the percentage problems in the activity below without using a calculator:

Expressing one quantity as a percentage of another

Another way you may need to use percentages is to show one quantity as a percentage of another. To do this, divide the first quantity by the second quantity and then multiply by $100$

Example: Suppose that a student scored $34$ out of $40$ on a test, and you want to give them their result as a percentage.

Solution: To calculate what percentage $34$ is of $40$ you would do:

$34 \div 40 \times 100 = 85\%$

So the student scored $85\%$ on the test.

Finally, you may need to calculate a percentage change. For example, you might calculate the percentage change in a power bill from the previous amount to the current amount.

To do this:

First, subtract one value from the other to find the difference
Next, divide this difference by the original amount
Finally, multiply the result by $100$ to convert it to a percentage.

Example: Suppose a student obtains a mark of $72$ in a standardised test one year and a mark of $81$ the next. What is the percentage change in the student’s mark?

Solution: To calculate the percentage change from $72$ to $81$ you would do:

$81$ (new amount) $- 72$ (original amount) $= 9$

$9 \div 72$ (original amount) $\times 100 = 12.5\%$

So the student’s mark has increased by $12.5\%$

Once you have worked through the examples above, have a go at the percentage problems in the activity below:

Fractions

A fraction is part of a whole number. The bottom number (called the denominator) shows how many equal parts the whole is divided into. The top number (called the numerator) shows how many of those parts you have.

Understanding fractions is important for many everyday tasks, such as cooking, telling the time, reading maps, and working out discounts. It is also an important skill for other areas of maths, including ratio and proportion, algebra, geometry, and statistics.

You may often need to:

simplify fractions
convert them into equivalent fractions
add or subtract them
multiply or divede them.

You can use a calculator to do many of these calculations. However, it is also important to understand how to work them out by hand. This section explains how to do this.

First, there are some important things to know about fractions:

If the numerator is smaller than the denominator, the fraction is less than $1$. This is called a proper fraction. For example, $\frac{1}{4}$ is a proper fraction.
If the numerator is larger than the denominator, the fraction is greater than $1$. This is called an improper fraction. For example, $\frac{5}{4}$ is an improper fraction. These fractions can also be written as mixed numbers (a whole number and a fraction together) or as whole numbers, depending on the situation. For example, $\frac{5}{4}$ can also be written as $1\frac{1}{4}$ (because $5$ divided by $4$ equals $1$ with $1$ left over). Another example is $\frac{8}{4}$, which can be written as (2\) (because $8 \div 4 = 2$ with no remainder). Unless you are told otherwise, you should usually give answers as mixed numbers or whole numbers instead of improper fractions.
If the denominator of a fraction is $1$ the fraction is equal to the whole number in the numerator. For example $\frac{4}{1} = 4$
If the numerator and denominator are the same, the fraction is equal to $1$ and we usually write it as $1$. For example $\frac{4}{4} = 1$

4 glasses, one showing a quarter coloured in purple, one showing a half coloured in blue, one showing three quarters coloured in yellow, and one all coloured in red, showing one whole.

Simplifying fractions

Unless you are told otherwise, fractions should always be written in their simplest form. This means using the smallest possible numbers. You can change a fraction to its simplest form by following these steps:

Find the highest common factor (HCF) of the numerator and denominator. This is the largest number that divides exactly into both numbers. You may need to list all the factors of each number to find this, as there is no single method that always works.

Another option is to use any common factor (any number that divides both numbers exactly). If you do this, you will need to repeat the steps until no common factors remain.
When doing this, it is usually easiest to start with simple factors such as $2, 3, 5$ or $10$.

Divide both the numerator and denominator by the highest common factor or the chosen common factor. This will give you a new numerator and denominator.

If you used a factor that is not the highest, repeat the process until the fraction cannot be simplified any further.

Example: If $8$ out of $12$ participants who responded to a survey are female, what is this as a fraction in simplest form?

Solution: Following the steps above, we have:

The factors of $8$ are $1, 2, 4, 8$ and the factors of $12$ are $1, 2, 3, 4, 6, 12$
Since the largest number on both these lists is $4$ it means the highest common factor is $4$
Dividing the numerator of $8$ by this highest common factor gives $8 \div 4 = 2$ and dividing the denominator of $12$ by this highest common factor gives $12 \div 4 = 3$
So $\frac{8}{12}$ simplifies to $\frac{2}{3}$

Finding equivalent fractions

The fractions $\frac{8}{12}$ and $\frac{2}{3}$ from the previous example are equivalent fractions, which means they have the same value.

Often, you will find equivalent fractions by simplifying, as shown earlier. However, sometimes you may need to find an equivalent fraction with a given denominator (or less often, a given numerator). This is useful when you need to compare, add, or subtract fractions that have different denominators. You can find an equivalent fraction with a chosen denominator by following these steps (you can reverse the steps if you are working with a chosen numerator):

Divide the new denominator (from the incomplete fraction) by the denominator of the original fraction.
Multiply the numerator of the original fraction by the value you found in the first step.

If you prefer, you can do the division the other way around in the first step. If you do that, you would then divide in the second step instead of multiplying.

If you are unsure whether to multiply or divide, think about whether the new denominator (or numerator) should be larger or smaller than the original. This will help you decide.

Another method is to:

Multiply the know value in the incomplete fraction (for example, the denominator) by the “opposite” value in the original fraction (for example, the numerator),
The divide by the remaining value (for example, the denominator of the orginal fraction).

Example: If a student scores $\frac{32}{40}$ on a test, but the test only counts for $15$ marks of her final grade, what is her mark out of $15$ as an equivalent fraction?

Solution: Following the steps above, we have:

The denominator of the incomplete fraction is $15$ and the denominator of the complete fraction is $40$
So $15 \div 40$ gives $0.375$
The numerator of the complete fraction is $32$
So $32 \times 0.375$ gives $12$ and the equivalent fraction is $\frac{12}{15}$

Alternatively, you could multiply $15$ and $32$ (the ‘opposite’ values in the two fractions) and divide by the remaining value of $40$

Once you have worked through the examples above, have a go at the fraction problems in the activity below:

Adding and subtracting fractions

When adding or subtracting fractions, there are two main types of problems to consider.

The first type is when the fractions have like denominators (the same denominator). In this case, the process is simple. You add or subtract the numerators, and keep the denominator the same as in the original fractions.

Example: Nick eats $\frac{1}{8}$ of a pizza while his brothers Jo and Ryan each eat $\frac{3}{8}$ of the pizza. What fraction has been eaten altogether?

Solution: Adding the three fractions simply requires adding the numerators, as follows:

\[\frac{1}{8} + \frac{3}{8} + \frac{3}{8} = \frac{1 + 3 + 3}{8} = \frac{7}{8}\]

So $\frac{7}{8}$ of the pizza has been eaten altogether.

Adding or subtracting fractions with unlike denominators (different denominators) requires more steps. You must convert the fractions into equivalent fractions with the same denominator before you can add or subtract them. You can do this by following these steps:

Find the lowest common denominator of all the fractions This is the smallest number that each denominator can divide into exactly.
- If you cannot find the lowest common denominator, you can use the product of the denominators. This will always work, but it may not be the simplest option.
Convert each fraction into an equivalent fraction with the chosen denominator.
Add or subtract the fractions as usual once they all have the same denominator.
Simplify your final answer if needed.

Example: David has a big pile of assignments to mark, which he spreads over a number of days. He marks $\frac{1}{4}$ of the assignments on Monday, $\frac{1}{3}$ of the assignments on Tuesday and $\frac{2}{5}$ of the assignments on Wednesday. What fraction of the assignments has he marked over the three days?

Solution: Following the steps above, we have:

The three denominators in the problem are $4, 3$ and $5$ so the lowest common denominator is $60$ (which in this case is the same as the product of $4, 3$ and $5$ since $4 \times 3 \times 5 = 60$)
The fractions can be converted to equivalent fractions with denominators of $60$ as follows:
$\frac{1}{4} = \frac{15}{60}$ since $60 \div 4 = 15$ and $1 \times 15 = 15$
$\frac{1}{3} = \frac{20}{60}$ since $60 \div 3 = 20$ and $1 \times 20 = 20$
$\frac{2}{5} = \frac{24}{60}$ since $60 \div 5 = 12$ and $2 \times 12 = 24$
The sum now becomes $\frac{15}{60} + \frac{20}{60} + \frac{24}{60} = \frac{15 + 20 + 24}{60} = \frac{59}{60}$
This fraction cannot be simplified, so the fraction of assignments marked is $\frac{59}{60}$

Multiplying and dividing fractions

Sometimes you may need to find a fraction of a number, such as $\frac{2}{3}$ of another fraction, or $\frac{2}{3}$ of an integer. In both cases, you do this by multiplying the fraction by the number.

If the number is a fraction, multiply the two fractions by multiplying the numerators together and the denominators together to form a new fraction. Simplify the result if needed.
If the number is an integer, multiply the integer by the numerator of the fraction, and then divide by the denominator. You can do this in the order that is easiest for you to calculate.

Example: Suppose you are considering taking a job that is part-time, and you would be working $\frac{4}{5}$ of a normal load. If the full time salary is $\$60000$ what would your part-time salary be?

Solution: This requires you to calculate $\frac{4}{5}$ of $\$60000$ which is the same as multiplying $\frac{4}{5}$ and $\$60000$
You can do this in a number of ways (note it doesn’t matter which way around you write the numbers being multiplied, so do it in a way that makes sense to you). For example:

Method 1: $\$60000 \times \frac{4}{5} = \$60000 \times 4 \div 5 = \$240000 \div 5 = \$48000$

Method 2: $\$60000 \times \frac{4}{5} = \$60000 \div 5 \times 4 = \$12000 \times 4 = \$48000$

Method 3: $\frac{4}{5} \times \$60000 = 4 \div 5 \times \$60000 = 0.8 \times \$60000 = \$48000$

So your part-time salary would be $\$48000$

Furthermore, if you can multiply fractions, you are already halfway there when it comes to dividing fractions. Dividing by a fraction is the same as multiplying by its reciprocal. This means you swap the numerator and denominator of the fraction, and then multiply.

This works because multiplication and division are opposite operations.

Example: Suppose you have a piece of string that is $30$m long, and you want to cut it into lengths that are $\frac{3}{4}$m each. How many pieces will you end up with?

Solution: This requires you to divide $30$ by $\frac{3}{4}$ which is the same as multiplying $30$ by $\frac{4}{3}$
You can do this in any of the ways detailed above, for example:

$30 \times \frac{4}{3} = 30 \times 4 \div 3 = 120 \div 3 = 40$

So you would end up with $40$ pieces of string.

Once you have worked through the examples above, have a go at the fraction problems in the activity below:

Ratios, rates and proportions

A ratio is another way of comparing two or more numbers, similar to fractions and percentages. For example, if a class has $17$ male and $13$ female students, the ratio of males to females can be expressed as $17:13$ or as $17$ to $13$ or as $\frac{17}{13}$. The most important thing is to make sure the order of the items matches the order of the numbers.

Ratios compare values that use the same units, so the units are not included. We are only comparing the numbers. A rate, however, is a special type of ratio that compares two quantities with different units. When describing a rate, we use the word “per”. For example, the price of a food item might be $\$6$ per $2$ kg.

Like fractions, ratios are usually written in their simplest form, or lowest terms. For example, if there are $10$ male and $20$ female students then the ratio usually be written as $1:2$ not $10:20$

Rates are often written as unit rates, where the second number is one. For example, the cost of the same food item as a unit rate is $\$3$ per kg. You can simplify ratios and rates in the same way as fractions.

When two ratios or rates are equal, we say they are in proportion. For example, the ratios $1:2$ and $10:20$ are in proportion as they show the same relationship. The rates $\$6$ per $2$ kg and $\$3$ per kg are also in proportion.

Finding missing values in ratios or rates so they stay in proportion is a common task. You can do this in a similar way to finding missing values in equivalent fractions. The key idea is that all parts of the ratio or rate must be multiplied or divided by the same number to keep the relationship the same.

Example: Consider that an excursion is planned for a year two class of $30$ students. Given that there is a requirement for a ratio of one adult to every $6$ students, how many adults are required on the excursion?

Solution: To calculate this, first note that there is one complete ratio ($1:6$) and one incomplete ratio ($?:30$)

To find the missing value so that the ratios are in proportion, one way is to write the ratios on separate lines:

$1:6$
$?:30$

This way it is easy to see that you need to divide the two numbers that are above one another to find out the relationship between the ratios, and then apply this relationship to the other side of the ratios.

So in this case $30 \div 6 = 5$ meaning the second ratio is five times the first, and therefore that $1 \times 5 = 5$ adults are required:

Alternatively, you could multiply $1$ and $30$ (the ‘opposite’ values in the two ratios) and divide by the remaining value of $6$ which also gives $5$

Once you have worked through the example above, have a go at the ratio, rate and proportion problems in the activity below:

Averages

Sometimes you may need to calculate a simple average using a total and a number of people or items. To do this, divide the total by the number of people or items.

Example: Suppose that in a school reading challenge a class of $22$ students reads $99$ books. What was the average number of books read by each individual student?

Solution: To calculate the average books per student when $99$ books are read by $22$ students you would do:

\[99 ÷ 22 = 4.5\]

So the average number of books read per student is $4.5$ books.

Once you have worked through the example above, have a go at the average problems in the activity below:

Algebra

Having a basic understanding of algebra can help you solve problems that involve unknown values. This section explains how to write simple equations and how to substitute values to solve them. If you want to learn more about algebra, including how to rearrange equations, refer to the Algebra module.

Writing a simple algebraic equation involves using letters or symbols to represent unknown values. These are called variables. You then combine these variables with numbers, an equals sign, and mathematical operations to show the relationship between the values in the equation.

Example: Suppose you are a teacher planning a school incursion. You investigate prices and find an incursion you like which costs $\$5$ per student, plus an additional one-off booking fee of $\$50$ for the event. What is an equation that represents the cost of the incursion for different numbers of students?

Solution: In this case our unknown values are the number of students and the total cost. We could use the letter $s$ (for example) to represent the number of students, and the letter $t$ (for example) to represent the total cost. Two examples of equations to represent the total cost ($t$) of the incursion for $s$ students are then as follows:

$t = 5 \times s + 50$ $($or you can omit the $\times$ sign and just write $t = 5s + 50)$
$t = 50 + 5 \times s$ $($or you can omit the $\times$ sign and just write $t = 50 + 5s)$

In general, the purpose of writing algebraic equations is to solve them and find unknown values. You can solve equations with more than one variable, but this is more complex and requires the same number of equations as unknowns. This is not covered here.

Instead, we will focus on solving one equation with one unknown value. In particular, we will look at how to do this by substituting values for the other variables, so that only one variable remains. The goal is to have this variable by itself on one side of the equals sign.

Example: Consider the equation from the previous example. If there are $20$ students in the class, how much would the incursion cost?

Solution: You can solve to find the total cost by substituting (i.e. replacing) the $s$ with $20$ which gives:

\[t = 5 \times 20 + 50 = 100 + 50 = 150\]

So the equation becomes $t = 150$ and this tells us that the total cost of the incursion for $20$ students would be $\$150$

Alternatively, you may want to find how many students can attend an incursion with a set budget. To do this, you would need to rearrange the equation so that the $s$ is by itself on one side of the equals sign. For more information on rearranging equations, refer to the Algebra module.

Once you have worked through the examples above, complete the algebra problems in the following activity:

Are you a Curtin student who would like to work through a module and gain a certificate for learning more about key numerical skills? Access the Maths 4 University module.