Numeracy fundamentals
This page explains important ideas about measurement and geometry. It also includes activities to help you practise and build your understanding as you go. If you would like more practice questions, especially if you are preparing for the Literacy and Numeracy Test for Initial Teacher Education (LANTITE), you should visit the LANTITE page. If you want to develop more advanced skills in geometry, you can visit the Geometry and trigonometry module.
As you work through the content, remember that there are often different ways to solve the same problem. In some cases, alternative methods are shown, but many examples only include one method. If you already know a different way to solve a problem, that is fine, as long as your method is correct. You do not need to change your approach. The most important thing is to use a method that makes sense to you.
In summary, this page explains how to:
- Convert units of measurement
- Draw diagrams to help solve measurement problems
- Solve problems involving time
- Use distance and time to calculate speed
- Understand and follow directions
- Calculate the perimeter, area and volume of simple shapes
Converting between units
The metric system is an international system of measurement that is based on the decimal (base 10) number system. This means measurements are organised in powers of ten, making them easier to convert.
The system uses two types of units: base units and derived units. The base units for three common types of measurement are as follows:
| Measurement | Unit | Abbreviation |
|---|---|---|
| Length | metre | m |
| Volume | litre | L |
| Weight | gram | g |
From these base units of measurement, other units are created to show smaller or larger amounts of the base unit. This is done by adding a prefix to the base unit to show how it has changed. The most commonly used prefixes are:
| Metric prefix | Relation to base unit | Abbreviation | Derived units |
|---|---|---|---|
| kilo | 1000 times bigger | k | kilometre (km), kilolitre (kL), kilogram (kg) |
| centi | 100 times smaller | c | centimetre (cm) |
| milli | 1000 times smaller | m | millimetre (mm), millilitre (mL), milligram (mg) |
When solving a problem that involves measurements, you need to make sure that all measurements of the same type are written in the same unit. If they are not, you will need to convert them into the same unit.
You can convert from one unit to another by using the relationships shown in the table above.
Example: Suppose you wish to tile the floor of a room that is \(3\textrm{m}\) wide with small tiles of width \(50\textrm{mm}\). How many tiles will you need to cover the width of the room?
Solution: To solve this you first need both measurements to be in the same unit. If you wish to have them both in metres, for example, you could convert the measurement that is in millimetres to metres by dividing by \(1000\):
\(50\textrm{mm} \div 1000 = 0.05\textrm{m}\)
Alternatively, if you wish to have them both in millimetres you could convert the measurement that is in metres by multiplying it by \(1000\):
\(3\textrm{m} \times 1000 = 3000\textrm{mm}\)
Once the measurements are both in the same unit, you can simply divide one by the other:
Method 1: \(3\textrm{m} \div 0.05\textrm{m} = 60\)
Method 2: \(3000\textrm{mm} \div 50\textrm{mm} = 60\)
Either way, \(60\) tiles are required.
Some other common equivalent measurements that you may need to convert between are:
- \(1\textrm{cm} = 10\textrm{mm}\)
- \(1\textrm{ metric tonne} = 1000\textrm{kg}\)
- \(1\textrm{ hectare} = 10000\textrm{m}^2\)
- \(1\textrm{ hectare} \approx 2.471\textrm{ acres}\) (an imperial measurement)
f you are unsure whether to divide or multiply when converting units, think about the size of each unit and whether your answer makes sense. For example, millimetres are much smaller than metres, so you need many millimetres to make a small number of metres.
You can also use your understanding of rates and proportions to check that your conversion is correct. If needed, refer to the Ratios, rates and proportions section.
Example: Suppose you are considering purchasing a property that is listed as being \(5\) acres. How big is the property in hectares?
Solution: If you wish to convert this to the metric unit of hectares, you have the following complete and incomplete rates:
\[\begin{aligned} 1 \textrm{ hectare} &= 2.471 \textrm{ acres} \\\ ? \textrm{ hectares} &= 5 \textrm{ acres} \end{aligned}\]Therefore, since \(5 \div 2.471 \approx 2.023\), it follows that \(5\textrm{ acres}\) is equivalent to \(2.023\textrm{ hectares}\).
Once you have worked through the examples above, have a go at the conversion problems in the activity below:
Drawing diagrams
Never underestimate how helpful a simple diagram can be when solving a measurement problem. Diagrams can support your thinking and make it easier to understand and solve the problem.
Example: Suppose you have a piece of cardboard measuring \(1.4\textrm{m}\) by \(1.2\textrm{m}\) and you want to cut it into squares of size \(25\textrm{cm} \times 25\textrm{cm}\) for a craft activity. You need \(23\) squares - will you be able to cut enough, or will you require some extra cardboard?
Solution: The first thing to do here is to convert the measurements to have them all in the same unit. Having all lengths in metres only requires one conversion, so doing this gives:
\(25\textrm{cm} \div 100 = 0.25\textrm{m}\)
Next, you might be tempted to find the area of the card and divide it by the area of the pieces you want to cut - but this doesn’t take into account the fact that you might have some unused strips of paper at the sides. A diagram illustrates this, and shows that instead you need to treat the length and width separately:
Doing this gives:
\(1.4 \div 0.25 = 5.6\), meaning \(5\) pieces can be cut from the length
\(1.2 \div 0.25 = 4.8\), meaning \(4\) pieces can be cut from the width
(Note you need to truncate - cut off - rather than round in both instances.)
Therefore \(5 \times 4 = 20\) pieces can be cut altogether, so you will need some extra cardboard.
Once you have worked through the example above, have a go at using diagrams to solve the problems in the activity below:
Time
Another idea you may need to use when working with measurements is time. Some common calculations with time include finding how long an event lasts and working out the difference between two times. There are a few different methods you can use to do this.
Example: Suppose you are timing a race, and the times for the first three competitors in minutes and seconds are \(23\):\(56\), \(24\):\(01\) and \(25\):\(12\). How much slower was the third placed competitor compared to the first?
Solution: There are a few different methods to find out the difference in minutes and seconds between \(23\):\(56\) (first) and \(25\):\(12\) (third).
For example, one approach is to convert both times to seconds by multiplying the minutes by 60 and adding the existing seconds, as follows:
\(23 \times 60 + 56 = 1436\textrm{ seconds}\), and \(25 \times 60 + 12 = 1512\textrm{ seconds}\)
You can then calculate the difference in seconds, which is:
\(1512 – 1436 = 76 \textrm{ seconds}\)
Finally, you can convert this difference back to minutes and seconds by dividing through by \(60\). When you do this the whole number part gives the number of minutes, while the decimal part can be multiplied by \(60\) again to give the number of seconds, as follows (note that the dot over the \(6\) indicates that the decimal is recurring - meaning it repeats forever):
\(76 \div 60 = 1.2\dot{6}\), and \(0.2\dot{6} \times 60 = 16\)
So the answer is \(1 \textrm{ minute}\) and \(16 \textrm{ seconds}\).
Another approach is to work it out in stages. For example, first of all you could calculate how many seconds are between \(23\):\(56\) and \(24\):\(00\), by subtracting \(56\) from \(60\) (i.e. \(60 – 56 = 4 \textrm{ seconds}\)). Then you could work out how many minutes are between \(24\):\(00\) and \(25\):\(00\) (i.e. \(25 – 24 = 1 \textrm{ minute}\)), and finally you could add on the extra \(12 \textrm{ seconds}\) from \(25\):\(12\) to give \(4 + 12 = 16 \textrm{ seconds}\).
So again the answer is \(1 \textrm{ minute}\) and \(16 \textrm{ seconds}\).
Once you have worked through the example above, have a go at the time problems in the following activity:
Speed
OOnce you understand how distance and time work, you can use them to calculate how fast something is travelling. This is called speed. You can calculate speed using the following formula:
\[\textrm{speed} = \frac{\textrm{distance}}{\textrm{time}}\]Example: Suppose you have just started a new job that is \(18\textrm{km}\) from your house, and you plan to ride your bike there. If you need to be at work at \(8\textrm{am}\) and you plan to leave your house at \(7.15\textrm{am}\), what will be the average speed you need to ride in kilometres per hour?
Solution: To calculate this, first note that you need to work out how much time you have in hours. Since you leave at \(7.15\textrm{am}\) and need to arrive by \(8\textrm{am}\) this gives a total time of \(45 \textrm{ minutes}\), and \(45 \div 60 = 0.75 \textrm{ hours}\).
Therefore your average speed will need to be: \(\frac{18}{0.75} = 24 \textrm{ km/hr}\)
Alternatively, \(\frac{18}{45} = 0.4 \textrm{ km/min}\) and \(0.4 \times 60 = 24 \textrm{ km/hr}\)
Once you have worked through the example above, have a go at the speed problem in the following activity:
Directions
Being able to read maps and understand directions is an important skill. There are two main types of directions you may need to use: the four cardinal directions (north, east, south, and west), and left and right directions.
When reading cardinal directions on a map, there is often an arrow that shows which way is north. If there is no arrow, you can usually assume that north is at the top of the map. Once you know where north is, you can work out the other directions. There are simple memory aids to help you remember the order of directions. For example, the phrase “Never Eat Soggy Weetbix” shows the clockwise order: North, East, South, West.
There are also directions between the main cardinal directions. These are called intermediate directions and sit halfway between them, as shown below.
For left and right directions, it can help to imagine yourself facing the same direction as you are ‘travelling’ on the map. You can also turn the map or your body to match this direction. This will make it easier to work out which way is left and right.
If you are unsure which side is left or right, you can hold both hands in front of you and make an ‘L’ shape with your thumb and fingers. The hand that forms a correct ‘L’ shape is your left hand.
Example: Suppose you are travelling south down Conway St and turn left on Allen St on the map shown. Will you make it to the shopping centre?
Solution: To figure this out, first note that north is at the top of the map, which means that south is at the bottom and that if you are travelling south down Conway St you are travelling ‘down’ the map. So try to imagine that when you get to Allen St you are coming at it from the ‘top’ side, facing downwards, and therefore that if you go left, you would go off the map and not towards the shopping centre.
Alternatively, you might like to try and remember that if you are travelling ‘down’ the map, the directions will be the opposite of what they would be if you were travelling ‘up’. So while it looks like the shopping centre is to the left of Conway St, actually you would need to turn right.
Once you have worked through the example above, have a go at the direction problems in the following activity:
Perimeter
The perimeter of a shape is the total distance around its outer edge. For example, the perimeter of a rectangle is found by adding together the lengths of all four sides. This can also be calculated by multiplying the length by two and the width by two, then adding these values together. As a formula this looks like:
\(\textrm{Perimeter of rectangle} = 2 \times \textrm{length} + 2 \times \textrm{width}\)
You may also need to find the perimeter of irregular shapes. In some cases, you will need to add up the lengths of all the individual sides. In other cases, there may be a quicker method based on the perimeter of one or more shapes that the irregular shape is made from or based on.
Example: What is the perimeter of the following shape?
Solution: One way of calculating the perimeter of this shape is to start by calculating the total perimeter of both the squares, which is \(12\textrm{cm} \times 4 + 12\textrm{cm} \times 4 = 96\textrm{cm}\)
Next, since both squares have been pushed together, there is a small part of the perimeter of each square that needs to be subtracted from this total. From the \(9\textrm{cm}\) distance on the diagram, we can work out that the distance that needs to be subtracted is two lots of \(3\textrm{cm}\) (since \(12\textrm{cm} - 9\textrm{cm} = 3\textrm{cm}\))
Therefore the perimeter of the irregular shape is \(96\textrm{cm} - 3\textrm{cm} - 3\textrm{cm} = 90\textrm{cm}\)
Once you have worked through the example above, have a go at the perimeter problems in the activity below:
Area
The area of a shape is the amount of space inside its boundary. The most common area you will need to calculate is the area of a rectangle. This is found by multiplying the length by the width:
\(\textrm{Area of rectangle} = \textrm{length} \times \textrm{width}\)
To calculate the area of a triangle, you multiply the base by the height and then divide the result by two. Dividing the result by two is the same as multiplying it by a half:
\(\textrm{Area of triangle} = \frac{1}{2} \times \textrm{base} \times \textrm{height}\)
Example: What is the area of the following triangle?
Solution: The area of this triangle is \(\frac{1}{2} \times 7\textrm{cm} \times 3\textrm{cm} = 10.5\textrm{cm}^2\)
Once you have worked through the example above, have a go at the area problems in the activity below:
Volume
The volume of a shape is the amount of three-dimensional (3D) space inside it. Volume can be measured in cubic units, such as cubic metres, cubic centimetres, or cubic millimetres. It can also be measured in litres, such as millilitres or kilolitres.
There is a relationship between these two types of measurements, which is shown below:
\(1 \textrm{litre} = 1000\textrm{cm}^3\)
Example: Suppose you have a rectangular fish tank that measures \(50\textrm{cm}\) long by \(30\textrm{cm}\) wide and \(40\textrm{cm}\) high. If you want to fill it three quarters of the way up, what would be the volume of water required in litres?
Solution: To do this you would first calculate the volume by multiplying the three measurements:
\(50\textrm{cm} \times 30\textrm{cm} \times 40\textrm{cm} = 60000\textrm{cm}^3\)
Next, three quarters of this volume is:
\(60000\textrm{cm}^3 \times \frac{3}{4} = 45000\textrm{cm}^3\)
Finally, you can convert to litres by dividing through by \(1000\):
\(45000\textrm{cm}^3 \div 1000 = 45 \textrm{ litres}\) required
Once you have worked through the example above, have a go at the volume problem in the activity below: