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A repeating pattern is a set of multiple identical groups of different symbols, items or shapes that are copied in the same order each time.
Shape patterns are repeating combinations of
Children are expected to order and arrange combinations of mathematical objects in patterns and sequences as part of the English Year 2 national curriculum.
It is important to learn how to complete repeating patterns because Mathematics is full of patterns and rules. By identifying missing shapes or symbols in sequences, children begin to build the logic required to solve more complex sequences. It may also help them to pick up rules and patterns in any aspect of Mathematics that they learn in the future.
Here is another example of a repeating shape pattern formed by blue then red circles.
We can see that we have blue, red, blue, red, blue, red and so on.
Every blue always has a red following it. Every red has a blue following it.
The complete pattern can be formed simply by taking one blue and then one red and repeating these two. After each red we start again with another blue, red.
In this next example we have two blue counters followed by two red counters.
We always have two reds following two blues.
We can form the pattern by repeating two blues and two reds.
In this example, we have the pattern of two blues then a red repeated.
In this example above, the red is not as common as the blues. It marks the end of the repeating part of blue, blue, red.
We can easily see that blue, blue, red repeats after each red.
To find the rule for completing patterns and sequences we can use the following tips:
When introducing sequences and patterns to children early on, most sequences will be fairly simple with only two or three different repeating symbols.
Here is an example of a shape pattern formed by: triangle, triangle, cross.
We can construct the whole pattern by repeating triangle, triangle, cross.
We eventually have one more space remaining at the end.
We know that we repeat triangle, triangle, cross.
After a cross, we return to the first shape in the sequence, which is a triangle.
Here is an example of a repeating sequence with a missing shape at the end. We need to find the missing shape to complete the pattern.
The first symbol is a blue right arrow. We then have two yellow left arrows.
We look for other blue right arrows and we can see that two yellow left arrows always follow a blue right arrow.
The pattern blue right arrow, yellow left arrow, yellow left arrow repeats to make the sequence.
After two yellow arrows, we are back to the beginning, with one right blue arrow.
Here is another example of a shape pattern with a missing symbol.
We need to decide how to complete the pattern.
We can see that we have pairs of arrows in the pattern.
We have two blue right arrows followed by two yellow left arrows. These pairs of arrows alternate throughout the pattern.
Before the missing symbol, we have just one yellow left arrow. The missing shape must be another yellow left arrow.
We now have two blue right arrows followed by two yellow left arrows repeating.
Here is another example of a repeating pattern with missing shape.
We can use the first symbol in the pattern to help us. We have a red pentagon.
Following this red pentagon, we have a blue right arrow and then a yellow down arrow.
We look for other red pentagons in the pattern and see if they are also followed by a blue right arrow and a yellow down arrow.
We can see that this pattern repeats.
The missing shape after the blue right arrow is a yellow down arrow.
In this repeating shape pattern example, we have two missing shapes.
Again we can look at the first symbol and then look at which symbols follow this until we get back to the first symbol again.
We begin with a purple down arrow and following this we have a pink up arrow and a blue right arrow. We then get back to the first shape in our sequence.
We can fill in our pattern from the next purple down arrow. After the purple down arrow we should have a pink up arrow followed by a right blue arrow.
We already have a pink up arrow and then a blank space, so after the pink up arrow we put in a right blue arrow.
We are then back to starting the pattern again, using a purple down arrow.
The pattern of purple down arrow, pink up arrow and blue right arrow repeats throughout the sequence.
Now try our lesson on Roman Numerals 1 to 10 where we learn how to write the numbers 1 to 10 in Roman numerals.
Correspondence problems are also known as integer scaling problems. In the English national curriculum for Year 4, we are asked to ‘solve harder correspondence problems such as n objects are connected to m objects’.
Correspondence problems involve multiplying or dividing groups of amounts that are in given ratios.
Some examples of correspondence problems would be:
These problems involve working out how many groups, packets or boxes we have in total and then using this to work out the missing number needed.
To solve correspondence problems, first work out the total number of groups using division and secondly, multiply to find the missing amount.
In the example below we have three bags of marbles each containing 1 blue and 3 red marbles. We want to know how many blue marbles this is in total.
We already know that we have three bags in total, so we do not need to work this out.
We have 3 groups and each group contains 1 blue marble.
We have 3 lots of 1 blue marble. We have 3 × 1 = 3.
We wanted to know how many blue marbles there are and so we multiplied the number of blue marbles in each bag by the total number of bags.
We multiply the number of the required item in each group by the number of groups.
In this next example we still have the same 3 bags of marbles, however this time we need to find the total number of red marbles.
We multiply the number of the required item in each group by the number of groups.
The required item is red marbles and so we multiply the number of red marbles in each bag by the total number of bags.
In both of these examples, we already knew the total number of groups. We knew that we had 3 bags.
In the next scaling problem we have an example where we first need to calculate the number of groups.
We buy bags of marbles that contain 1 blue and 3 red. However, we need 30 red marbles in total and are asked how many blue marbles we would have if we did so.
The first step is to find the number of bags required to give us 30 red marbles.
We can think of this as 30 red marbles shared equally into groups of 3 because there are 3 red marbles in each group.
In Mathematics, we use the divide operation to share equally.
We have 30 red marbles shared equally into groups of 3.
This is 30 ÷ 3 = 10. There are 10 bags of marbles.
We know that if we had 10 bags, each containing 3 red marbles then this would give us 30 red marbles in total.
Now that we know that we have 10 bags, we can find how many blue marbles this would be. There is 1 blue marble in each bag.
10 lots of 1 is 10 and so, there would be 10 blue marbles if we bought 30 reds.
Alternatively, we could see that in the given ratio, we have 3 reds to 1 blue. There are three times more red marbles than blue and so, we can just divide the amount of red marbles by 3 to find the number of blue.
If we had 30 red, then dividing this by 3 gives us 10 blue marbles.
In this next example we have the same bag containing 1 blue and 3 red marbles.
We know we have 12 red marbles and are asked how many blue marbles we must have.
Again we can divide 12 red marbles by 3 to find that we have 4 bags in total.
In 4 bags, there are 4 lots of 1 blue marble, which gives 4 blue marbles in total.
Alternatively we know that no matter how many red marbles there are, we can divide this by 3 to find the number of blue marbles.
12 ÷ 3 = 4.
Here is an example of a scaling problem with a different ratio of marbles.
We have 2 purple to 3 green marbles in each bag. We have 6 greens in total and need to find out how many bags there are so that we can find how many purple marbles we have.
To find the number of bags we divide the number of green marbles in total by the number of green marbles in each bag.
6 ÷ 3 = 2 and so, we have 2 bags.
Each bag contains 2 purple marbles and so, we have 2 bags each containing 2 purple marbles. 2 lots of 2.
2 × 2 = 4 and so, we have 4 purple marbles if we have 6 green.
In the next correspondence problem, we have the same bag of marbles but are asked how many green marbles we have if we have 10 purple marbles.
We first find the number of bags by dividing the 10 purple marbles by the number of purple marbles in each bag.
There are 2 purple marbles in every bag.
10 ÷ 2 = 5 and so, there are 5 bags in total.
Now that we know that we have 5 bags, we can multiply this by the number of green marbles in each bag, which is 3.
5 × 3 = 15. We have 15 green marbles in total.
In this example we have packets of pencils, each containing 2 blue and 4 green.
We are asked how many green pencils we have if we buy 6 packets.
Since we already know that we have 6 packets, we simply multiply 6 by the number of green pencils in each packet.
There are 4 green pencils in each packet and so, we have 6 lots of 4.
6 × 4 = 24. We have 24 green pencils in 6 packets.
In this next correspondence problem, we have 40 green pencils in total and are asked how many blue pencils we have.
40 ÷ 4 = 10 and so there are ten packets in total.
In ten packets, there are 20 green pencils since there are 2 in each packet.
Alternatively we can solve this scaling problem using the ratio of pencils in each packet.
In each packet there are 2 blue and 4 green. 2 is half of 4. There are half as many blue pencils as there are green pencils.
If we have 40 green pencils, then we have half this number of blue pencils.
Half of 40 is 20.
In this next problem we know that we have 8 blue pencils. There are 2 blue pencils in each packet and so we divide 8 by 2 to find the number of packets required.
8 ÷ 2 = 4 and so there must be 4 packets.
We are asked how many pencils there are in total of both colours.
There are 2 blue and 4 green in each packet. We have 2 + 4 = 6 pencils in each packet.
We have 4 packets, each containing 6 pencils. 4 lots of 6.
4 × 6 = 24 and so, we have 24 pencils in 6 packets.
In this next scaling problem, we have a box of chocolates containing 2 strawberry, 1 dark, 3 mint and 2 caramel.
We buy 6 of these boxes and are asked how many mint chocolates this would be.
We have 6 lots of 3 mint chocolates.
6 × 3 = 18 and so, we have 18 mint chocolates in 6 boxes.
In this correspondence problem, we have 14 strawberry chocolates and are asked how many chocolates we have in total of all flavours.
We need to work out the number of boxes and then multiply this amount by the number of chocolates in each box.
There are 2 strawberry chocolates in each box, so to work out the total number of boxes, we divide 14 by 2.
14 ÷ 2 = 7 and so we have 7 boxes. 7 boxes, each containing 2 strawberry chocolates must give us the 14 strawberry chocolates required.
We now calculate the number of chocolates in total in 7 boxes.
The number of chocolates in each box is the sum of the strawberry, dark, mint and caramel flavours.
In each box there are 2 + 1 + 3 + 2 chocolates. There are 8 chocolates in each box.
We have 7 boxes, each containing 8 chocolates. 7 lots of 8.
7 × 8 = 56 and so there are 56 chocolates in seven boxes.
Now try our lesson on Addition and Subtraction Word Problem Keywords where we learn how to solve word problems.
The larger the numerator (on top), the larger the fraction.
Like fractions are fractions that have the same denominator, which means the same number at the bottom of the fraction. Like fractions are also known as similar fractions or are simply called fractions with the same denominator.
If fractions have different denominators then they are unlike fractions.
The denominator is the number at the bottom of a fraction, below the dividing line.
Any set of fractions that have the same number as their denominator are like fractions.
For example, here are like fractions with the same denominator of 3.
We have 1 / 3 , 2 / 3 and 3 / 3 .
Here is another example of similar fractions with a denominator of 5.
We have 2 / 5 , 3 / 5 and 4 / 5 .
These fractions are like fractions because they all have the same denominator on the bottom of 5.
Here is a fraction wall which shows examples of like fractions. Each row of the wall contains a new set of similar fractions.
We can see rows of halves, thirds, quarters, fifths, sixths and sevenths.
To compare like fractions use the values of the numerators. The larger the numerator, the larger the fraction. The smaller the numerator, the smaller the fraction.
Here is an example of comparing fractions with the same denominator of 3.
We start with one whole and divide it into three equal parts called thirds.
Since the denominator is ‘3’, we divide the whole shape into 3 equal parts.
The numerator tells us how many of these parts we have.
The fraction of 3 / 3 has a numerator on top of 3. We have three thirds.
The fraction of 2 / 3 has a numerator on top of 2. We have two thirds.
The fraction of 1 / 3 has a numerator on top of 1. We have one third.
We can see that the more parts we have, the larger the fraction we have.
The numerator on top of the fraction tells us how many parts we have. Therefore the larger the numerator, the larger the fraction when comparing like fractions.
3 / 3 has the largest numerator and so is the largest fraction.
1 / 3 has the smallest numerator and so is the smallest fraction.
2 / 3 is in between the other two fractions shown.
Here is another example of comparing like fractions.
We are comparing fractions with the same denominator of 5.
To compare like fractions we compare the size of their numerators. The largest fraction has the largest numerator and the smallest fraction has the smallest numerator.
4 / 5 has the largest numerator and so is the largest fraction.
2 / 5 has the smallest numerator and so is the smallest fraction.
The like fractions are put in order from smallest to largest as 2 / 5 , 3 / 5 and 4 / 5 .
To order fractions with the same denominator, order them using their numerator on top of the fraction.
If all of the fractions have the same denominator on the bottom, then simply look at the numerators on top.
In this example of ordering similar fractions we will order them in ascending order.
Ascending order means from smallest to largest.
We have 2 / 4 , 4 / 4 , 1 / 4 and 3 / 4 .
The fractions are all similar fractions because they all have the same denominator of 4. To put them in order, we put them in order of their numerators on top.
Looking at the numerators alone, the ascending order is 1, 2, 3 and 4.
Therefore the ascending order of these like fractions is 1 / 4 , 2 / 4 , 3 / 4 and 4 / 4 .
They are arranged from smallest to largest.
1 / 4 is the smallest fraction and 4 / 4 is the largest fraction.
Here is another example of ordering like fractions from smallest to largest.
We have 3 / 8 , 7 / 8 , 5 / 8 and 2 / 8 .
We will arrange the fractions in ascending order from smallest to largest by ordering them using their numerators.
Considering only the numerators, we have an order from smallest to largest of 2, 3, 5 and 7.
Therefore the order of the like fractions from smallest to largest is 2 / 8 , 3 / 8 , 5 / 8 and 7 / 8 .
2 / 8 is the smallest fraction and 7 / 8 is the largest fraction.
Now try our lesson on Comparing Unit Fractions where we learn about unit fractions.
Then multiply the numerators and the denominators separately.
2 / 3 to make it 3 / 2 .
To divide fractions, it is very important to know how to multiply fractions first. We then turn the fraction division into a fraction multiplication so that we can work it out.
The method for dividing fractions is to keep the first fraction the same, turn the division into a multiplication and flip the fraction you are dividing by.
The rules for dividing fractions by fractions are shown in these steps:
Division is the opposite of multiplication.
To divide by a fraction, we can turn the division into a multiplication if we flip the fraction we are dividing by upside down.
The fraction of 3 / 4 means 3 ÷ 4.
The opposite of this is 4 ÷ 3, which can be written as the fraction 4 / 3 .
÷ 3 / 4 is the same as × 4 / 3 .
We can see that we simply flip the fraction that we are dividing by.
Here is an example of 2 / 5 ÷ 3 / 4 .
When teaching dividing fractions, it is useful to write the division calculation as a multiplication below the original.
We keep the first fraction the same and so, we write 2 / 5 below.
We turn the division into a multiplication.
Because we did this, we flip the fraction we are dividing by. 3 / 4 becomes 4 / 3 .
The fraction division is now written as a multiplication and so we can multiply to work out our answer.
We multiply the two numerators, which are the numbers on the top. These give us the numerator on the top of our answer.
2 × 4 = 8
We then multiply the denominators, which are the numbers on the bottom. These give us the denominator on the bottom of our answer.
5 × 3 = 15
2 / 5 ÷ 3 / 4 = 8 / 15
In this next example of dividing fractions, we have 3 / 10 ÷ 2 / 3 .
We use the method for dividing fractions of keep, multiply and flip..
We keep the first fraction of 3 / 10 .
We turn the division sign into a multiplication.
Because we did this, we flip the second fraction. 2 / 3 becomes 3 / 2 .
We have turned the division of 3 / 10 ÷ 2 / 3 into the multiplication of the fractions 3 / 10 × 3 / 2 .
Multiplying the numerators we have 3 × 3 = 9.
Multiplying the denominators we have 10 × 2 = 20.
3 / 10 ÷ 2 / 3 = 9 / 20 .
In this next dividing fractions example we have 3 / 8 ÷ 1 / 2 .
We keep 3 / 8 .
We multiply.
We flip the fraction 1 / 2 to make 2 / 1 .
We can multiply the numerators and denominators to get 6 / 8 .
When doing any fraction calculations, we try and simplify our answers wherever possible.
6 / 8 is made of the digits 6 and 8, which are both even.
We can divide both 6 and 8 by two to simplify the fraction.
6 ÷ 2 = 3 and 8 ÷ 2 = 4.
6 / 8 can be simplified to 3 / 4 .
3 / 8 ÷ 1 / 2 = 3 / 4 .
Try to simplify all fractions that can be simplified when they are written as the final answer.
We can notice a trick to help us do this division more easily.
If we can divide the numerator of the first fraction by the numerator of the second fraction exactly and we can divide the denominator of the first fraction by the denominator of the second fraction exactly, then we can do this to get our answer.
We see that in the original division question of 3 / 8 ÷ 1 / 2 , we can see that 3 ÷ 1 = 3 and 8 ÷ 2 = 4.
The division on the top of the fractions gives us 3 and the division on the bottom of the fractions gives us 4. The answer is the same as we got to using the keep, multiply and flip method earlier. We get 3 / 4 .
Simply dividing the numerators and the denominators is much easier, however we can only do this with some questions where there is an exact division on both the top and bottom of the fraction. If we do not have this, then we use the method of keep, multiply and flip.
In this next example we have 4 / 15 ÷ 2 / 3 .
We keep the first fraction of 4 / 15 the same.
We turn the division into a multiplication.
We flip the second fraction of 2 / 3 to produce 3 / 2 .
We can now multiply the tops and bottoms of the two fractions separately to obtain the final answer.
4 × 3 = 12 and 15 × 2 = 30
The answer to 4 / 15 ÷ 2 / 3 is 12 / 30 .
Again in this example, we can see that this final answer can be simplified.
Both 12 and 30 are in the 6 times table and so, both numbers can be divided by 6.
12 ÷ 6 = 2 and 30 ÷ 6 = 5.
12 / 30 can be simplified to 2 / 5 .
Again in the original question of 4 / 15 ÷ 2 / 3 , we can see that we have an exact division on the tops of the two fractions and an exact division on the bottoms.
Looking only at the numerators of the two fractions:
4 ÷ 2 = 2
and looking only at the denominators of the two fractions:
15 ÷ 3 = 5
4 / 15 ÷ 2 / 3 = 2 / 5 . This is the same result that we obtained earlier.
Now try our lesson on Converting Fractions to Percentages where we learn how to write a fraction as a percentage.
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The number of shaded parts is written on the top of the fraction.
If an amount is divided into 2 parts, these parts are called halves.
If an amount is divided into 3 parts, these are called thirds.
If an amount is divided into 4 parts, these are called quarters.
If an amount is divided into 5 parts, these are called fifths.
Here is a poster showing the names of some common fractions.
In the example below, the rectangle is divided into 2 equal parts, which are called halves.
To write one half as a fraction, we write a number above a dividing line with another number below the dividing line.
We first count how many parts are shaded in. We have 1 part shaded in and so we can write a ‘1’.
We put our dividing line below this number.
Then we count how many parts the shape has been divided into in total. We have 2 parts.
We write a 2 below the dividing line.
We have 1 / 2 . The shape is divided into two equal parts, directly down the centre.
Here is another example of one half.
This triangle is divided into two equal parts down the centre.
We have one half shown shaded in.
One half is written as 1 / 2 . The line in the fraction means to divide, so 1 / 2 means one whole divided by two.
One third is an amount divided into 3 equal parts.
This rectangle is divided into 3 equal parts and so, it is divided into thirds.
We have one whole divided into 3, which can be written as 1 / 3 .
We have one part shaded in and so we write a 1 above the dividing line. We have three parts in total so we write a 3 below the dividing line.
In this example we have one third of a circle.
We have one out of three equal parts shaded in.
The fraction of one third can fit into the whole shape three times.
One quarter is when something is divided into 4 equal parts.
In the example below, we have one out of four equal parts.
One divided by 4 can be written as 1 / 4 .
We have one out of the four parts shown.
Two quarters is two out of four equal parts. Two quarters is the same as one half.
In the example below, two parts are shaded in. So we write a 2 above the dividing line.
There are 4 parts in total and so, we write a 4 below the dividing line.
We have two of the four quarters.
Two out of four can be written as 2 / 4 .
We can see that if we remove one of the dividing lines in the shape, we have a rectangle split into 2 equal parts.
2 / 4 is the same size as 1 / 2 .
Two quarters is the same as one half.
In the example below we have 2 parts shaded. We write a 2.
There are 4 parts in total and so, we write a 4 below the dividing line.
We have the fraction of 2 / 4 .
We know that two quarters is the same as one half. We can see that we have half of the shape shaded in if we double it. When we add another two squares, the whole shape is filled in.
Two quarters make one half.
One fifth is an amount is divided into five equal parts.
In the example below, the pentagon is divided into five equal parts.
We have 1 part shaded and so we write a 1 above the dividing line of the fraction.
There are five parts in total and so we write a 5 below the dividing line.
One fifth is written as a fraction as 1 / 5 .
Here is another example of one fifth.
The circle below is divided into five equal parts.
One fifth of the circle is shaded in.
Now try our lesson on Writing Fractions of Geometric Shapes where we learn how to find further fractions of amounts.
We can repeat them up to 3 times to add them.
Here is a chart showing all of the roman numerals to 100.
The roman numerals to 100 are written using the 5 numerals of I, V, X, L and C.
Each of these numerals has the following values:
There are only two digits that are created by subtracting. These are 4 and 9.
We can add the same numeral up to three times but not more than this.
For example, the numeral I can be written twice as II, which equals 2.
We can make 3 with three I numerals: III.
However we cannot write 4 as IIII because we have used the same numeral more than three times. Instead the digit 4 is one of our special digits that we write as a subtraction.
We subtract 1 from 5 to make 4. We write four as IV.
Similarly we can add to five to make 6, 7 and 8.
Six is VI, which is 5 + 1.
Seven is VII, which is 5 + 1 + 1.
Eight is VIII, which is 5 + 1 + 1 + 1. Eight requires three of the same numeral, I.
We cannot write 9 as VIIII because this requires four of the numeral I. We can only use three at once.
9 is our second special digit that is made by subtracting.
9 is 1 before 10, written as IX.
Once we know the rules for Roman numerals 1 to 10, we can continue to use these to make larger numerals.
To write Roman numerals, partition the number into its tens and units. First write the numerals for the tens digit and then write the numerals for the units digit.
We can combine the numerals of I, V, X, L and C to create all numbers up to 100.
For example, we can partition 11 into 10 + 1.
10 is X and 1 is I and so, 11 is written as XI. Ten plus one.
12 is 10 + 2. We write this as XII.
XII means X + I + I, which means 10 + 1 + 1.
Here are the Roman numerals from 1 to 30.
We can see that once we know the Roman numerals from 1 to 10, we can create the numerals to 30 by writing them after the appropriate tens digit numeral.
To write the numbers 11 to 20 in Roman numerals, write the Roman numerals 1 to 10 after the numeral ‘X’.
For example, 18 is made from 10 + 8.
We can write the numeral X for 10 and then follow this with VIII, which is 8.
18 is written as XVIII, which means 10 + 5 + 1 + 1 + 1.
Our numerals decrease in size from 10 to 5 to 1, so we add them. We have stuck to the rule of only using three of the same numeral at once.
To write the Roman numerals 21 to 30, we write the Roman numerals 1 to 10 after two X numerals.
XX is worth 10 + 10, which is 20. We write XX for twenty and then add on the Roman numerals between 1 and 10 afterwards to make the numbers from 21 to 30.
For example, 21 is partitioned into 20 + 1.
20 is two tens, written as XX.
We need to add 1 more to 20 to make 21.
21 is written as XXI, which means 10 + 10 + 1.
We can see in the list of Roman numerals to 30 below, that the numerals 1 to 9 are shown in red. There is a clear pattern of these numerals repeating from column to column. We simply put an extra X numeral in front every time that we move to the next column and add ten.
For example, we can start with 5, move right one column to get to 15 and then move right one more column to get to 25.
We start at V for five. We add ten by writing an X numeral in front. XV is 15. Another X is written in front to make XXV, 25.
We can continue this pattern as we look at the Roman numerals 31 to 60.
In the first column, we have the numerals to 40.
We have the Roman numerals 1 to 9 following three X numerals.
XXX is 30, which means 10 + 10 + 10. It is okay in Roman numerals to use the same numeral up to three times.
38 is made by partitioning 38 into 30 + 8.
We know that 30 is XXX and 8 is VIII.
38 is written as XXXVIII, which means 10 + 10 + 10 + 5 + 1 + 1 + 1.
We cannot write 40 as four tens. XXXX is using four of the same Roman numeral, which is not allowed.
Like the number 4 is written as IV, 1 before 5, we need to write 40 as 10 before 50.
10 before 50 is XL. X is 10 and L is 50.
40 is XL and then 3 is III.
43 is written as XLIII in Roman numerals.
XLIII is XL + III, which is 40 + 3.
Be careful not to think of XLIII as 10 + 50 + 1 + 1 + 1.
X (10) is a smaller value than L (50) and we do not write smaller numerals before larger numerals to add them. If a smaller numeral is written before a larger numeral, then subtract the smaller numeral from the larger numeral.
XL is 50 – 10, which is 40.
It is easiest to remember that 4, 40, 9 and 90 are created in this way. Both 4 and 9 are one less than the multiples of five, 5 and 10.
We can see all numbers in the 40 column start with XL.
50 is written as L in Roman numerals. 50 has its own Roman numeral just like 5 does.
All numbers in the fifties start with L and are followed by the Roman numerals 1 to 9.
60 is made by 50 + 10, written as LX.
All numbers in the sixties column start with LX.
For example, 69 is written as 60 + 9.
60 is LX and 9 is IX.
69 is written as LXIX in Roman numerals. LX + IX means 60 + 9.
We can continue to count on from 50 to make 70 and 80.
70 is two tens more than 50, which is written as LXX. LXX means 50 + 10 + 10.
80 is three tens more than 50, which is LXXX. LXXX means 50 + 10 + 10 + 10.
All numbers in the seventies start with LXX and all numbers in the eighties start with LXXX.
For example, 84 is written as 80 + 4.
We start with 80 and add on the numerals for 4.
80 is LXXX, 50 + 10 + 10 + 10.
4 is IV, 1 before 5.
84 is written as LXXXIV, which means L + X + X + X + IV.
Remember to look out for the special digits of 4 and 9, which are created by subtracting 1 from 5 or 10 respectively.
To write any larger numbers, we need our next Roman numeral.
100 is written as C in Roman numerals.
We cannot write 90 as LXXXX because this uses four of the same numeral, X.
Like 9 is written as 1 before 10, 90 s written as 10 before 100.
90 is written as XC in Roman numerals. XC is 10 before 100. XC means 100 – 10.
We subtract 10 from 100, because when we have a smaller numeral in front of a larger numeral, we subtract the smaller numeral from the larger numeral.
All of the numbers in the nineties are written beginning with XC.
For example, 92 can be partitioned into 90 + 2.
90 is XC and 2 is II.
92 is written as XCII in Roman numerals.
XCII means XC + II, 90 plus 2.
We still continue to use the Roman numerals 1 to 9 with each new column.
Roman numerals are often used in modern times to number more formal or significant events.
For example, some major events are numbered with Roman numerals, such as the World Wars I and II, along with the Olympic games.
Kings, queens and popes are often numbered with Roman numerals, such as Queen Elizabeth II and King Henry VIII. Some people named after others in their family may use these numerals too.
Old buildings and monuments often have significant dates carved on them in Roman numerals.
Book volumes or chapters, along with movie copyrights often use Roman numerals as it can appear more formal.
Roman numerals are not used frequently today primarily because they take up too much space and are difficult to use in mathematical calculations.
For example, 88 is much shorter than LXXXVIII.
Also by using our number system separately to the letters of the alphabet, it is easier to identify which is a number and which is a letter. Algebra is a branch of Mathematics that uses the letters of the alphabet and this would be very difficult if we still used Roman numerals today.
Where are Roman Numerals used Today?
Roman numerals are most commonly seen today on clock or watch faces. The numerals to 12 make up the 12 hours on the clock face. The digit 4 is often represented as ‘IIII’ on a clock face, instead of the usual ‘IV’. This is to distinguish it more clearly from other numerals such as VI for 6.
Now try our lesson on Negative Numbers on a Number Line where we learn about negative numbers.
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If there are no keywords, use the verb to decide the context.
Verbs that mean ‘removing’ or ‘comparing’ tell us to subtract.
To solve an addition or subtraction word problem, the first step is to decide whether it is an addition or subtraction operation.
It is important to read the entire worded problem when deciding if it is an addition or subtraction.
In general, if the numbers are being combined to make a total, then we have an addition word problem.
If the numbers are being removed, then we have a subtraction.
If the numbers are being compared, then we have a subtraction.
Keywords do not guarantee whether the word problem is an addition or subtraction, however they can be a big hint.
Here is a list of some common addition and subtraction keywords.
Any of these keywords found in a word problem will most likely indicate whether it is an addition or subtraction.
Many word problems do not contain typical keywords. Look at the verbs in the word problem to decide if we are combining or removing the numbers.
In the example below, “James has 6 apples and Jessica has 4 apples.”
We are asked, “How many do they have in total?”
In this example we have the addition keyword of ‘total’.
‘Total’ is a common addition keyword seen in worded problems. It gives us a clue that we are performing an addition. We just need to read the question carefully to decide if the two amounts are being combined.
We are combining the two amounts of apples.
Drawing a picture can sometimes be a helpful way to visualise a word problem.
We will add the two numbers given in the word problem.
6 and 4 are number bonds to 10. It is helpful to look for common number facts.
6 + 4 = 10
There are 10 apples in total.
Here is another example of a worded problem.
“Olivia has 8 toys and Henry has 2 toys”, we are asked, “How many toys do they have in total?”
Again, we have the keyword ‘total’, which tells us to combine the two amounts.
We write down the two numbers involved in the worded problem and then add them.
8 and 2 are another number bond to ten.
8 + 2 = 10
They have ten toys in total.
In this next example, “Sophia has 14 cherries and Alex has 5”. We are asked, “How many do they have altogether?”
Altogether is a common keyword we might see in word problems.
Altogether means to combine the values to make a total.
We read the full question again to check and we do need to combine the two values.
We have an addition word problem.
We add 14 and 5.
Using our number facts to 10, we know that 4 + 5 = 9.
Therefore 14 + 5 = 19 because 14 is ten larger than 4.
Alternatively, we know that from our number bonds to 20, 15 + 5 = 20.
14 is one less than 15 and so, 14 + 5 is one less than 15 + 5.
14 + 5 is one less than 20, which gives us 19.
It can help to draw a picture of the word problem if you are struggling to understand it. It can also help to use manipulatives to act out the situation for young children.
Manipulatives are physical items that we can hold, such as wooden blocks, cubes and counters. We can use a counter or cube for each cherry in this question.
In this next example we have, “Chloe has 10 yoyos and gives 5 to her friend.” We are asked, “How many does she have left?”
There are none of the typical addition and subtraction keywords in this question.
In many word questions, we will not have familiar keywords to tell us what to do.
Instead, we look at the verb in the question. The verb is the action word or ‘doing word’.
If the verb implies that we are combining the amounts, then we may have an addition.
If the verb implies that we are removing an amount, then we may have a subtraction.
This is not a fixed rule because word problems can be written in different ways but it can be a big clue.
The verb ‘gives’ implies that we are removing the 5 yoyos. Chloe is giving them away, or taking them away.
We have a subtraction word problem.
10 – 5 = 5 and so, there are 5 yoyos left.
The word ‘left’ also is a common word in word problems. It means what do we have after a subtraction has taken place and so, it gives us another clue that this is a subtraction.
The word ‘remaining’ is often seen instead of the word ‘left’ and it is a very similar clue.
Again, if using manipulatives, we could have started with 10 cubes and taken away 5.
In this word problem example we are told, “Michael has 13 bananas and last week he ate 6.” We are asked, “How many are remaining?”
Again there are no typical keywords in this word problem so we look at the verb and try to understand the context.
The verb is ‘ate’. This implies that we remove some of the bananas by eating them. We are subtracting the 6 bananas from the original 13.
The word ‘remaining’ also implies that we have a subtraction. It means ‘what is left after doing a subtraction?’
We start with 13 bananas and subtract 6 of them.
13 – 6 = 7
There are 7 bananas remaining.
Now try our lesson on Vertical Column Subtraction without Regrouping where we learn how to do vertical column subtraction.
