This page takes you through some important concepts on the topics of number and algebra, with activities to consolidate your knowledge along the way. In particular, it covers how to do the following (use the drop-down menu above to jump to a different section as required):
For more sample questions on these concepts, in particular if you are preparing for the Literacy and Numeracy Test for Initial Teacher Education (LANTITE), you may like to visit the LANTITE page. Alternatively, if you would like to build up some more advanced skills on the topic of algebra, you may like to visit the Algebra module.
Finally, as you work through the content keep in mind that there are often different ways of solving the same problem. While in some cases alternative methods have been explained, for many examples just one method of working is provided. If you already know a different technique for solving the problem then that’s great (provided it is also correct, of course). Don’t feel you need to change your method - the important thing is that you use a technique that makes sense to you.
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.
Knowing what order to perform operations in is essential in order to obtain a correct answer for any problem involving multiple operations. Similarly, it is important to know how to use the order of operations to write expressions correctly so that others (including your calculator!) will evaluate them as you intended. The rule for this is known by a few different acronyms, one of which is BIMDAS. This stands for:
Example: Michelle the maths teacher would like to purchase a licence for some maths software for her students to use for a ten week term. The licence has a \($5\) registration fee per student, and then costs \($2\) a week per student. How much would it cost in total for Michelle’s class of \(27\) students?
Solution: One way to figure this out is to consider that for each student, it is going to cost \($5\) plus \($2\) per week for \(10\) weeks. So writing this as a formula, we have \(5 + 2 \times 10\) (note that no brackets are needed here, as BIMDAS tells us that the multiplication will happen first as required).
Next we need to multiply the cost per student by \(27\), since there are \(27\) students in the class. This requires brackets to make sure that the cost per student is calculated before it is multiplied by \(27\), so the final formula should be \(27 \times (5 + 2 \times 10)\).
Plugging this into a calculator (or you could always solve it without the aid of a calculator if wished, but this almost certainly wouldn’t be a requirement), and the answer for the total cost is \($675\) . Alternatively, another possible formula for the total cost is \(27 \times 5 + 27 \times 2 \times 10\). Check to confirm that this gives you the same answer.
Once you have worked through the example above, have a go at using BIMDAS to evaluate the problems in the following activity:
A decimal number is a number with two parts; the whole number part is the part to the left of the decimal point, while the part to the right is the fractional part.
Note that if the whole number part of a decimal number is \(0\), you would usually put \(0\) in front of the decimal point to avoid confusion. For example \(0.35\). On the other hand, if the fractional part of a decimal number is \(0\) you would usually omit the decimal point and write the number as a whole number. For example you would usually write \(2.0\) simply as \(2\). Being able to round decimal numbers is important as often it is impractical and unnecessary to state them in their entirety. Furthermore, in the case of irrational numbers like \(\pi\) (Pi) it is in fact impossible to state them in their entirety as there is an infinite number of decimal places without repetition.
In both cases the number of decimal places to round to may depend on the context (i.e. using two decimal places for monetary amounts, using three decimal places in a solution for which values in the question are given to three decimal places, or using common sense and rounding to the nearest whole number when calculating values such as numbers of people), or it may be stated (i.e. you may be asked to give a solution rounded to two decimal places).
Note however that when solving a problem involving multiple operations you should retain all accurate values as you work and not round until your final answer, in order to avoid an inaccurate solution.
When rounding a decimal number to a certain number of decimal places, or to the nearest whole number, you should follow these steps:
Find the relevant ‘round-off place’ in your number. In general, 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 simply the number to the left of the decimal point.
Determine the number that is one place to the right of your round-off place.
If this number is greater than or equal to \(5\), you should increase the number in the round-off place by \(1\) and remove all digits to the right of it. In other words, you should round up. On the other hand, if this number is less than \(5\) you need to round down. So in this case the number in the round-off place remains the same, and again you need to remove all digits to the right of 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 \(7\).
Once you have worked through the example above, have a go at the rounding problems in the following activity:
A percentage refers to the number of something ‘out of \(100\)’, and is shown by the symbol \(\%\). For example a tax rate of \(30\%\) indicates that you need to pay \($30\) out of every \($100\) you earn in tax. Some other facts about percentages are as follows:
You can calculate a percentage of a given quantity (including determining percentage increase and decreases), and you can also express one quantity as a percentage of another.
In certain situations, such as when finding a percentage increase or when expressing one quantity as a percentage of another, it is possible for the percentage to be more than \(100\%\).
Percentages are commonly used to analyse and compare data, and to express things such as sales discounts, commissions and interest rates.
Since a percentage represents an amount out of \(100\), to calculate a percentage of a given quantity you just need to divide the percentage in question by 100, and multiply by the quantity.
Example: Suppose that \(15\%\) of the students in a particular school are likely to need extra help with their numeracy each year. Given 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 \(360\):
\(15 \div 100 \times 360 = 54\)
Hence \(54\) students are likely to need extra help with their numeracy. Alternatively, 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.
As an extension of this, often you may wish to calculate the result of a percentage increase or decrease. For example, 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. You can find the new value once a percentage has been applied in one of two ways:
First calculate the specified percentage of the original value, as described previously, then either add (for an increase) or subtract (for a decrease) this value from the original value; or
First add (for an increase) or subtract (for a decrease) the specified percentage from \(100\%\), and then calculate this new percentage of the original value (as described previously).
Example: Suppose you want to find the total cost of a \($350\) compter, once \(10\%\) GST has been added.
Solution: To calculate this you could do either of the following:
\(10 \div 100 \times $350 = $35\), and \($350 + $35 = $385\); or
\(100 + 10 = 110\), and \(110 \div 100 \times $350 = $385\)
Either way, the total cost of the computer is \($385\).
Alternatively, sometimes you may be required to calculate an original value before a percentage was applied, given a new value. For example, calculating the original price of an item before a \(20\%\) discount given the new price, or calculating the original bank balance before \(3\%\) interest is added given the current balance.
You can do this by adding (for an increase) or subtracting (for a decrease) the specified percentage to or from \(100\%\), then dividing the new value by this and multiplying 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:
While you can generally use a calculator to evaluate percentages using the formulas above, it is also good to have an understanding of how to approximate percentages without a calculator. You can do this by performing combinations of the following as required, using rounding to aid in the calculations when needed:
To calculate \(1\%\) of a number: divide by \(100\), which you can do by moving the decimal place two places to the left (which is the same as removing two zeroes from a whole number if applicable)
To calculate \(10\%\) of a number: divide by \(10\), which you can do by moving the decimal place one place to the left (which is the same as removing one zero from a whole number if applicable)
To calculate \(25\%\) of a number: divide by \(4\) (or divide by \(2\) and \(2\) again if 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 following activity without the aid of a calculator:
Another way you might need to use percentages is to express one quantity as a percentage of another. To do this, you simply need to divide the first quantity by the second 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, sometimes you may wish to calculate a percentage change; for example the percentage change on a power bill from the previous bill amount to the current bill amount. To do this you should first subtract one value from the other to find the difference, then divide this by the original amount and 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 following activity:
A fraction is just part of a whole number. The bottom number of the fraction (the denominator) tells you how many parts the whole is divided into, while the top number (the numerator) tells you how many parts you have.
An understanding of fractions is required for many every day concepts, from cooking, to telling the time, reading maps, calculating discounts and more. Furthermore, knowledge of how to work with fractions is a vital skill required for many other mathematical concepts, from ratio and proportion to algebra, geometry and statistics.
Some of the things you may commonly be required to do with fractions are to simplify them, convert them to equivalent fractions, add or subtract them and multiply or divide with them. While many of these operations can be performed using the fractions feature on a calculator, it is also good to have an understanding of how to evaluate them manually - and this section details how. First though, some important things to note about fractions are as follows:
If the numerator of a fraction is smaller than the denominator, the fraction is less than \(1\) and is called a proper fraction. For example \(\frac{1}{4}\).
If the numerator of a fraction is bigger than the denominator, the fraction is greater than \(1\) and is called an improper fraction. For example \(\frac{5}{4}\). Improper fractions can also be expressed as mixed numbers (where a whole number and a fraction are together) or just as whole numbers as applicable. For example \(\frac{5}{4}\) can also be expressed as \(1\frac{1}{4}\) (since \(5\) divided by \(4\) gives \(1\) with \(1\) left over), while \(\frac{8}{4}\) can also be expressed as \(2\) (since \(8 \div 4 = 2\)). Unless otherwise stated, you would generally give answers as mixed or whole numbers rather than improper fractions.
If the denominator of a fraction is \(1\), the fraction is just equal to the whole number numerator. For example \(\frac{4}{1} = 4\).
If the numerator and denominator are the same, the fraction is equal to \(1\), and we generally write it as such. For example \(\frac{4}{4} = 1\).
Unless otherwise stated, fractions should always be given in their simplest form - that is, using the smallest numbers possible. You can convert a fraction to simplest form by following these steps:
Determine the highest common factor of the numerator and denominator (i.e. the largest number that divides into both a whole number of times). If you can’t find the highest common factor then at least determine a common factor of the numerator and denominator (i.e. any number that divides into both a whole number of times); it just means you will need to repeat these steps. Note that you can look for ‘easy’ factors such as \(2\), \(3\), \(5\) and \(10\), or you may need to list out all the factors of both the numerator and denominator in order to find the highest common factor (unfortunately there is no ‘magic’ way of determining this).
Divide both the numerator and denominator by this highest common factor or a factor to obtain a new numerator and denominator. If the latter is used, repeat as necessary until there are no more factors.
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 \(1\), \(2\), \(4\) and \(8\) and the factors of \(12\) are \(1\), \(2\), \(3\), \(4\), \(6\) and \(12\). 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 \(\frac{2}{3}\).
The fractions in the previous example (\(\frac{8}{12}\) and \(\frac{2}{3}\)) are equivalent fractions, which means that they are equal in value. While often you will determine an equivalent fraction by simplifying as above, other times you may need to find an equivalent fraction with a specified new denominator (or less frequently, a numerator) - perhaps for the purpose of comparing, adding or subtracting fractions with different denominators. You can determine an equivalent fraction with a specified denominator by following these steps (or vice versa for determining an equivalent fraction with a specified numerator):
Divide the denominator of your incomplete fraction by the denominator of your complete fraction.
Multiply the numerator of your complete fraction by the value obtained previously.
Note that you can always do the division in the first step the other way around if you prefer, in which case you will need to divide in the second step instead of multiplying (if you are ever in doubt about whether to multiply or divide in the second step, take note of the fractions and whether it would make sense for the new denominator, or numerator, to be bigger or smaller than for the original fraction).
Alternatively, you may want to work out the unknown value by multiplying the known value in the incomplete fraction (e.g. the denominator) by the ‘opposite’ value in the complete fraction (e.g. the numerator), and then dividing by the remaining value (e.g. the denominator of the complete 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 \(15\), while the denominator of the complete fraction is \(40\). So \(15 \div 40\) gives \(0.375\).
Multiplying the numerator of the complete fraction (\(32\)) by \(0.375\) gives \(12\), so equivalent fraction is \(\frac{12}{15}\).
Alternatively, you could multiply the ‘opposite’ values in the two fractions (\(15\) and \(32\)) and divide by the remaining value (\(40\)), which also gives \(12\) and hence an equivalent fraction of \(\frac{12}{15}\).
Once you have worked through the examples above, have a go at the fraction problems in the following activity:
When it comes to adding or subtracting fractions, there are two different types of problems you need to consider. The first is when the fractions have like denominators (i.e. the same denominators). When this is the case, adding or subtracting them is straightforward as you just add or subtract the numerators of the fractions, and keep the denominator of your answer the same as the original fractions.
Example: Nick eats \(1\) slice, or \(\frac{1}{8}\), 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, or different, denominators requires more work, as you need to convert the fractions to equivalent fractions with like denominators first before you can add or subtract them. You can do this by following these steps:
Determine the lowest common denominator of all fractions in the problem (i.e. the smallest number that all the denominators can divide into a whole number of times). If you can’t find this you can always just use the product of the denominators, which is always a common denominator but may not be the lowest.
Convert each fraction in the problem to an equivalent fraction with denominator as specified.
Once your fractions all have the same denominator, you can add or subtract them as usual.
Simplify your result if necessary.
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 \(\frac{59}{60}\)
Sometimes you may be required to find a fraction ‘of’ some number, such as \(\frac{2}{3}\) of another fraction, or \(\frac{2}{3}\) of an integer. Either way, this requires you to multiply the fraction by the number.
To do this when the number in question is also a fraction, you can multiply the two fractions together by simply multiplying the two numerators and the two denominators to make a new fraction (simplifying as required). On the other hand, if the number to multiply by is an integer then you can simply multiply the integer by the numerator of the fraction and divide by the denominator, in whichever order is easiest to calculate and makes sense to you.
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 with fractions then you are already halfway there (no pun intended!) when it comes to dividing with fractions, as dividing by a fraction with a certain numerator and denominator is the same as multiplying by the same fraction with the two values switched around (i.e. by the reciprocal). 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 following activity:
Like fractions and percentages, a ratio is a way of comparing two or more numbers. For example, if a class contains \(17\) male and \(13\) female students then the ratio of males to females can be expressed as \(17:13\), or as \(17\) to \(13\), or as \(\frac{17}{13}\) (either way, note that the crucial thing is to match up the order of the items being expressed in the ratio with their respective numbers).
Ratios such as this compare values that are in the same units, and therefore the units are excluded from the ratio as we are comparing the numbers only. A rate, on the other hand, is a special type of ratio that compares two quantities with different units of measurement. When describing a rate, the word ‘per’ is used between the two measurements; for example the price of a particular food stuff may be \($6\) per \(2\)kg.
Just as fractions are usually expressed in simplest form, ratios are usually expressed in lowest terms. For example, if there are \(10\) male and \(20\) female students then the ratio would typically be expressed as \(1:2\) rather than \(10:20\). Similarly, rates are typically expressed as unit rates, where the second term in the rate is one. For example, the cost of the same food stuff as a unit rate is \($3\) per kg. Note you can simplify ratios and rates like these in the same way as simplifying fractions.
When two ratios or rates are equivalent, we say that they are in proportion. For example, the ratios \(1:2\) and \(10:20\) you saw previously are in proportion as they represent the same relationship, and so too are the rates \($6\) per \(2\)kg and \($3\) per kg.
Finding unknown values to ensure ratios or rates are in proportion is a common requirement, and this can be done in much the same way as finding unknown values in equivalent fractions (as detailed in the Fractions section). The key thing to remember is that in order to be in proportion, the numbers in the ratios or rates need to be multiplied or divided by the same value.
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 the ‘opposite’ values in the two ratios (\(1\) and \(30\)) and divide by the remaining value (\(6\)), which also gives \(5\). So again \(5\) adults are required.
Once you have worked through the example above, have a go at the ratio, rate and proportion problems in the following activity:
Sometimes you may be required to calculate a simple average given a total and a number of people or things. To do this, you simply need to divide the total by the number of people or things.
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 following activity:
Having an understanding of some basic algebra is handy in order to solve problems involving unknown values. This section details how to write simple equations and how to substitute values in order to solve them. If you would like to learn more about algebra, including techniques for rearranging in order to solve equations, please refer to the Algebra module.
Writing a simple algebraic equation involves using letters or symbols to represent unknown values (called variables) and putting these together in an equation, complete with an equals sign and the relevant mathematical operations, to represent the relationship between them.
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 \($50\). 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 \(t = 50 + 5s\))
Generally, the purpose of writing algebraic equations is to solve them to find out unknown values. While this can be done for more than one variable at a time, note that this is more complex and requires the equivalent number of equations which is not covered here. Instead, we will focus on using one algebraic equation to find one unknown value, and in particular at how to do this by substituting values in for any other variables so that the only remaining variable is 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 \(t = 150\), which tells us that the total cost of the incursion for \(20\) students would be \($150\).
Alternatively, if you wanted to find out how many students could attend the incursion given a set budget, this would involve rearranging the equation so that the \(s\) was by itself on one side of the equals sign. For details on rearranging equations please refer to the Algebra module.
Once you have worked through the examples above, have a go at the algebra problems in the following activity: