Numbers and numerals

Axioms and arrays

Factors and multiples

Order of operations

Positive and negative numbers

Expressions, equations and inequalities

Angles

Classifying 2-D shapes

Constructing triangles and quadrilaterals

Coordinates

Area of 2-D shapes

Transforming 2-D figures

Prime factor decomposition

Conceptualising and comparing fractions

Manipulating and calculating with fractions

Ratio

Percentages

Different number systems

Sequences

Forming and solving equations

Forming and solving inequalities

Linear graphs

Accuracy and estimation

Algebra and problem solving

Ratio (8.6a)

Ratio, real life graphs, and rates of change

Direct and indirect proportion

Univariate data

Bivariate data

Famous maths problems

Angles and parallel lines

Angles in polygons

Bearings

Volume and surface area of prisms

Area review

Circles

FDP review (9.1a)

Probability

Sets and Venn Diagrams

Solving linear simultaneous equations algebraically

Solving linear simultaneous equations graphically

Angle review (9.5a)

Constructions, congruence, and loci

Pythagoras's theorem

Ratio review (9.7a)

Similarity and enlargement

Surds and trigonometry

Quadratic expressions (9.9)

Quadratic equations (9.10)

Indices and standard form

Growth and decay

Finance

Proportion problems

Percentages and statistics

Maths

Y8 Maths

Key Stage 3

Year 8

Ratio and proportion

THEME

UNIT

In this lesson, we will learn about different representations of ratio, including the relationship between fractions and ratio.

Representing ratio

﻿Hi, my name is Miss Kidd-Rossiter, I'm a maths teacher in Oak, and I'm going to be taking your lesson today on representing ratio. Take a minute now to clear away any distractions and find a quiet place to work if you can, okay? Are you ready? Let's get started. We're going to start today's lesson with the try this. In a moment, you'll see four students appear on the screen. They've been mixing green and white paint in different ways. Your job is to compare the shades that they've made. Pause the video now and have a go at this activity. When you're ready, resume the video. If you're struggling, it might help you to draw a diagram. Okay, so we going to have a look at the connect part of the lesson now. We going to start by comparing Antoni and Binh from the try this. So we know that Anthony has mixed his green and white paint in the ratio five to two. So that's five parts green and two parts white. We can represent this using a bar model, like the one on your screen. Similarly Binh has mixed her pain in the ratio six to two, where six parts are green and two parts are white. We could also show this using a bar model. Pause the video now and think about how you could use these bar models to compare the proportion of green paint in Antoni and Bihn's mixtures. So one way that I thought about doing it, was to put one on top of the other. When I do this, I can clearly see that Antoni's got less green paint in his mixture proportionally than Bihn has. Right, we going to have a look now comparing Bihn's paint mixture with Cala's paint mixture. We've already got on the screen one representation that we could use for Bihn. So we need to think about how we could represent Cala's. Three quarters of Cala's paint is green. So that means there are three parts of green and one part is white. So we could represent this in a bar model like this. Pause the video now and have a think about how you could compare these two bar models. Is there anything that you notice this time? Okay, so again, I'm going to compare mine by putting one on top of the other. So this time I've got Bihn's on the top and Cala's underneath. I can quite clearly see that the proportion of green paint in each of these mixtures is the same. Two parts of Bihn's mixture, is equal to one part of Cala's mixture. This means that six to two and three to one are equivalent ratios. We can represent this in a table also. So Cala's paint is mixed in the ratio three to one, which we know is equivalent to six to two. And we can find the constant of proportionality. So that's what we've multiplied the ratio three to one by, to get six to two. And in this case, we've multiplied by two. We have to remember that it's both parts of our ratio, both sides of our ratio that have multiplied by two. Okay now we're going to compare Bihn's paint mixture to Xavier's paint mixture. We've still got Bihn's representation on the screen. Xavier's paint mixture has three tenths of it being white. So that means that seven parts must be green, and three parts are white. So we could represent it in this way. Pause the video now and have a think about how we could compare these two bar models. Also, can you think about when a bar model might become a bit tricky to use. Brilliant, so again, I can compare these by putting one on top of the other like this. And I can see that six to two has more green paint proportionally than seven to three. I also asked you to think about the limitations of bar models. So when we might be able to not use a bar model to represent what we're wanting to see. If we get into having really big numbers, like 42 to seven or 127 to nine, it would be really tricky to draw those bar models. So you need to think about is there another way, that we could compare ratios? The one way that I've thought about doing it, is making one part of the ratio the same. So if we look here, we've got Bihn's ratio is six to two, Xavier's ratio is seven to three. Could I make one part of those ratios, either the green or the white the same? Pause the video now and think about how you could do that. So I'm going to do it. I'm making the white part of the ratio the same. But you could do it by making the green the same as well. They both work. So I know that six is a multiple of two and three. So I'm going to make the white paint in each ratio become six. So for Bihn's paint, this means that the constant of proportionality is three. So I've multiplied the white paint by three to get six here. So that means that I also need to multiply the green paint by three to create the equivalent ratio. So six multiplied by three gives me 18. For Xavier's paint my constant of proportionality this time is two. I've multiplied the three by two to get six. So that also means that I need to multiply the seven by two to create the equivalent ratio. So seven multiplied by two, we know is 14. So now we can compare these ratios because the white is the same in each case, six parts. We know that proportionally, this one has more green paint because 18 is a higher number than 14. Now you're going to apply what you've learned to the independent task. So pause the video and navigate to the worksheet, to have a go at the questions. We're now going to go through the answers to the independent task. Well done for persevering with it, some of the questions on there were quite tricky. So your answer to question one is five eights. Your answer to question two is five to three, and your answer to question three is five tens. And then extra well done if you remembered that you should simplify five tenths to one half. The question four, here are your answers. I'll let you have a look at those and I'm going to read them all out. So pause the video if you need to. And the question five, this was really tricky. There were loads of correct answers here. So well done if you managed to think about it in lots of different ways. One way that I thought about it, was that four to eight is an equivalent ratio to one to two. That would mean that the black part of my bar is one third of the whole bar. And the white part of the bar is two thirds of the whole bar. But as I said there's loads of different answers here that you could have come up with. So extra well done if you came up with loads of that. So the last thing that we going to do in today's lesson, is have a go at this explore task. Antoni has lots of tins of green and white paint. He is mixing up to seven tins together. How many different shades can he make? And what if he could mix eight tins or nine tins? Pause the video now and have a go at this task. Okay so hopefully you've had a really good go at this explore activity now, and I'm just going to talk you through one approach that I chose to use. You could have done it a different way and that's absolutely fine. So what I did was I listed all the possible ratios of tins of green paint to tins of white paint for seven tins. So I could have had one tin of green paint to one tin of white paint. I could have had one tin of green paint to two tins of white paint and so on. You'll notice that the parts of my ratio never exceed seven because I've only got seven tins. And then I thought, well, all of these options, different, so all of these different shades of green paint. Well I don't think they are in fact they're not, because I've got some equivalent ratios here. So if you look, I've got the ratio one to one, which is equivalent to the ratio two to two, and that is also equivalent to the ratio three to three. So there's several pairs or trios of equivalent ratios here. So again, I've got the ratio one to two that is equivalent to the ratio two to four. And you can continue to go through my answer if you like and find other equivalent ratios in there. That brings us to the end of today's lesson. So I hope you've had a really good learning experience and I look forward to seeing you again in the future, bye.

In this lesson, we will learn how ratio relates to enlargements of shapes.

Enlargement

﻿Hi, my name is Miss Kidd-Rossiter, and I'm going to be taking you through today's lesson on enlargement. Before we get started, make sure you've got no distractions. And if you're able to, you're in a nice quiet place, so you can fully concentrate on the work that we're going to do. Really excited to get started. If you need to pause the video now, then please do. If not, let's get going. We're starting today's lesson with a try this. Jasmin and Zaki are drawing enlargements of the triangle shown. All the triangles they draw, have integer side lengths. Jasmin says, "My new triangle has perimeter 33 centimetres." Zaki says, "The shorter side on my new triangle is six centimetres." Your job is to figure out what are the lengths of the triangles that they've drawn, and how many triangles with integer side lengths, can you draw that have a perimeter less than a hundred centimetres. Pause the video now and have a go at this task. If you're struggling, let me give you a couple of hints. The lines on this triangle tell us that their shape is isosceles, it's an isosceles triangle. That means that these two sides have the same length. So they're both four centimetres. The other thing you might be struggling with, is this word, integer. Integer just means whole number. So not a fraction, not a decimal, an integer is a whole number. So pause the video now and have another go if you were struggling. Brilliant work guys, I'm really impressed. Let's move on and have a think about these together. Let's look at Jasmin's statement first. She says, "My new triangle has perimeter 33 centimetres." What's the perimeter of the triangle that was given to you? Tell the screen now. Great, it was 11 centimetres. So I'm just using this p to represent perimeter today. So the perimeter was 11 centimetres. On Jasmin's triangle then, she's got a perimeter of 33 centimetres. What do you think we might have to multiply the side lengths by to get a perimeter of 33 centimetres? Tell the screen now. Excellent, if you said three, let's try it. So four centimetres times three is 12 centimetres. So we know that both of our equal sides are 12 centimetres, and three times three is nine centimetres. Let's check that that gives us a perimeter of 33 centimetres. 12 add 12 is 24, add nine is 33. So that's correct. This is the triangle that Jasmin was thinking of. Let's see what's similar. So the first shape, the one that was on the screen to begin with, had side lengths in a ratio, four to three. What is the ratio of side lengths in Jasmin's triangle? Tell the screen now. Brilliant, it's 12 to nine. Can you notice anything about these ratios? Zaki said that the shorter side on his new triangle is six centimetres. So let's have a go at drawing that. The shortest side, is it the base, or is it the pair of equal sides? Tell the screen now. Excellent, it's the base. So we know that the base of Zaki's triangle is six centimetres. What have I multiplied three by to get six centimetres? Tell the screen now. Excellent, two. So that means that I need to multiply the four centimetres by two as well, to get eight centimetres. As we did on the previous slide, we said that the ratio for the original shape was four to three. What's the ratio of side lengths in Zaki's shape? Tell the screen now. Excellent, eight to six. Do you notice anything about these ratios? Did you manage to answer the last question? I'm going to tell you that the answer is nine different triangles. If you didn't manage to find them all, maybe you want to pause the video now and have another go at that task. With the connect part of the lesson, you'll see the triangle from the try this on your screen. We already discussed that the side lengths, are in the ratio four to three, where four is the longer side, and three is the shorter side. I'm going to show you how I could represent this as a diagram. So each centimetre here is represented by a box, for the longer side. And for the shorter side, each centimetre is represented by the same size box. Here we can see quite clearly, that the shorter side is three quarters of the length of the longer side. Now let's think about the perimeter. What would the perimeter of this shape be? Tell the screen. Excellent, 11 centimetres. So what fraction of the perimeter is my shorter side? Tell the screen now. Excellent, it's three elevenths of the perimeter. We could write this in the ratio, eight to three, because the shorter side has three elevenths share of the perimeter. Now let's have a look at an enlargement of this triangle, and see whether we get the same thing. So what ratio are our sides in this time? Tell the screen now. Excellent, eight to six. And eight to six is an equivalent ratio of four to three. So if we write our four to three, and eight to six, we can find a constant of proportionality that we multiply both parts of our ratio by to get the new ratio. So what do I multiply four by to get eight? It's two. And what do I multiply three by to get six? It's also two. So our constant of proportionality for these ratios is two, and that's also the same as our scale factor of enlargement. Let's see what fraction of the longer side is the short side here. So it's six eighths of the longer side, and we know that six eights simplifies to three quarters. So again, we've got the same fraction as we had in the original shape. What about the perimeter of this one? What will it be? Tell the screen now. It's 22 centimetres, well done. What fraction of the perimeter is our shortest side? Tell the screen now. Excellent, it's six twentieth seconds. What do we know about this fraction? Excellent, it simplifies to three over 11. Again, we can see that our shorter side takes the same share of the perimeter. So what would our ratio be this time? It would be 16 to six, and we know that this is an equivalent ratio to eight to three, and it has the same constant of proportionality. What you're going to do now, is you're going to navigate to the independent task and you're going to apply what you've learned. When you're ready, check back in and we'll go through some of the answers. Well done for having a go over that independent task. Let's go through some of the answers now. So the ratio between the side lengths, AB and EF is four to 20, or you could have simplified that down to one to five. The constant of proportionality. Well, first of all, let's write our ratios alongside on ABCD is four to two. And on EFGH, is 20 to question mark. We don't know what that length is yet. What do I multiply four by to get 20? Well, I multiplied by five. So my constant of proportionality is five. Now I found my constant of proportionality. I can use it for part C to help me find FG. So FG is two multiplied by five, which is 10 centimetres. Don't forget your units. So I'm not going to go through this question in full, cause there's quite a lot of ratios that you have to find, but I will tell you whether the pairs of triangles are enlargements or not, and the constant of proportionality. So for A, are these triangles, enlargements of each other? Tell the screen now. Yes, they are, well done. So these ones are enlargements of each other and our constant of proportionality is four. So well done if you've got that, and well done for writing down the ratios too. Part B then. Are these pairs of triangles enlargements of each other? Tell the screen now. No, they're not, excellent. Because two pairs of sides have been enlarged by a scale factor of a half, but the final pair of side has not. If I was enlarging 10 centimetres by a scale factor of a half, it should give me five centimetres, not 5.5 centimetres. Part C then. Are these triangles enlargements of one another? Tell the screen now. Yes, excellent, well done they are. And our constant of proportionality here is 1.5. Or if you wrote it as a fraction three over two. And finally part D. Are these two triangles enlargements of each other. Tell the screen now. No, they're not. Well done. Two pairs of sides have been enlarged by a scale factor of 2.5, but the last side hasn't. If it had been, then it should be 8.75 centimetres, not nine centimetres. For this pair of pentagons, one is an enlargement of the other, state the ratio between the side lengths of the bases of the pentagons. So we know that that's four to 10, or you could have simplified that and written it as an equivalent ratio, two to five. Well done if you did that. Part B, state the constant of proportionality. So first of all, we need to write our ratio of our side lengths. So for our small pentagon, we've got the ratio four to 3.5 to x, and we can do the same with the larger pentagon, which is 10 to y to eight. So we need to find our constant of proportionality. So what do I multiply four by to get 10? What do I multiply four by to get 10. Tell the screen now. Excellent, 2.5. So our constant of proportionality is 2.5. So that's our answer to part B. Now let's use that to find our missing lengths. So 3.5 multiplied by 2.5 gives us y, which is 8.75 millimetres. Don't forget your units. And then this one's a little bit trickier because we're not multiplying this time, we're going to divide. So x times 2.5 gave me eight, so I can figure out what x is by doing eight divided by 2.5. What's the answer to that, tell the screen now. Excellent, it is 3.2 millimetres. So we're moving onto the explore task now. Zaki is using sticks to form shapes. He has lots of sets of the sticks that are shown on the right hand side of your screen. So he's got loads of those sticks. He's made a set of triangles that are enlargements of each other, so you can see them on your screen. What are the missing dimensions? Could he make another enlargement? And what other sets of shapes could he make? Pause the screen now, and have a go at this task. Well done for having a go at that task. It was quite tricky in places. What did you find then? What were the missing dimensions on this one, first of all? Excellent, four centimetres, well done. Because you could realise that the constant of proportionality between this smaller triangle and this larger triangle here was two. What about this larger triangle then. What's the base going to be here? Excellent, it's 4.5 centimetres, well done. Could he make any other enlargements? Tell me now the other enlargements that you made. One that I made, was one that looks like this. So I said that these were eight centimetres, my equal sides. What does that mean that my base is? Excellent, six centimetres. There were also lots of other sets of shapes that he could make. Well done for having to go at this explore task. I hope you can keep going with it for a bit, cause there's lots in there. So don't feel like you have to stop. That's it for today's lesson. I hope you've learned loads on enlargements. I've really, really enjoyed teaching you. If you've drawn lots of lovely enlargements and you'd like to share them with me, and please share your work with Oak National. If you'd like to, ask your parent or carer to share your work on Twitter, tagging @OakNational and #LearnwithOak. Don't forget to do the quiz for today's lesson, and hopefully I'll see you again in the future. Bye.

In this lesson, we will understand how to use double number lines to represent proportion problems.

Double number lines

﻿Hi, I'm Miss Kidd-Rossiter. I'm your maths teacher from home. And I'm going to be taking you through today's lesson on double number lines. It's going to build on the work that we've already done on ratio. Really excited to have you on board and to get started. Before we get going, can you please check that your distraction free, you're in a nice quiet space if you're able to be and you've got something to work with, a pen and paper is ideal, but if not a pencil and exercise book, something along those lines would be great. If you need to pause the video now to get prepared then please do, but if not, let's get going. So for today's try this activity, we've got a double number line on the screen and you're told that the markings are equally spaced. Your job is to figure out how many ways you could complete this double number line, using the number cards that you're given. If you're confident with this, pause the video now and have a go this activity. If you're a little bit unsure, just stay tuned and I'll give you a hint. So, if you need a little bit of help, that's absolutely fine. That's what I'm here for. So let's have a think about a times table. I'm going to go with the three times table. This box here, would represent our first multiple of three. So that is three. The missing one here would be our second multiple of three. Can you tell me what the second multiple of three is, please? Excellent, it's six. And then the third box here would represent the third multiple of three, which is nine. So I can fill that in. Now we're going to think about a different times table. So I'm going to go with the fours this time, on the bottom. So this box here is my first multiple of four, which is four. This box here, is my second multiple of four, which is eight. And then I could continue if I wanted to. So pause the video now and see if you can complete this number line in different ways. Well done everyone with that try this activity. We're not going to talk about it too much, because we're going to pull out the key features of a double number line, in the connect. And you can see on your screen that you've got the same activity. Can you tell me what you filled in on the top please? Excellent. And what did you fill in on the bottom? Excellent. Right, I'm going to give you one of my examples. Obviously, there's loads of them that you could have come up with. So I put in one here, three here, four here, and eight here. What do we notice as we go across the number line? What do we notice as we go across the number line? Pause the video now and think about that. Excellent. It is multiples, isn't it? So for the top here, we've got the multiples of one. Brilliant. On the bottom, we're still increasing by multiples, but of a different number this time, what are the multiples of this time? Four brilliant. So we got our first multiple four. Our second multiple of four. So you can clearly see that we've got something happening as we go across the number line. What about though, from here, to here? So from the top of the number line to the bottom of the number line. Well, to see this a bit better, we might need to fill in the missing numbers. So this one here, and this one here. Pause the video now and tell me what you think those missing numbers are. What did you get for this one? Two? Excellent. And what did you get for this one? 12. Excellent. Now I know that we didn't have a number card with 12 but we're just going to put it there to help us out with this explanation. So what do we notice about here from one to four? How would we go from one to four? We could add three, couldn't we? So that's an option. What else could we do? We could multiply by four, couldn't we? Is there any other way? Not sure. Let us see then. What about from here to here? How could we get from two to eight? We could add six, couldn't we? Or we could multiply by four. Interesting. Are you spotting a pattern? What about here going from three to 12. Well I could add nine couldn't I? Or I could multiply by four. So again, I've got a pattern here. So if we think about these multiplications to go from the top to the bottom, on each one of these, we're multiplying by four. And this is what's known as a constant of proportionality, so whatever we have on the top of the number line, in this case, we can multiply it by four to get what goes on the bottom of the number line. So this time, I'm going to have my multiples of two on the top. So I'm going to have two, four, six. And I'm going to have my multiples of three on the bottom. So three, six, nine and so on. What do I multiply two by to get three? What would I multiply four by to get six? What would I multiply six by to get nine? Pause the video now think about that. Excellent. It's 1.5 isn't it? So this time our constant of proportionality is 1.5. Well done, this also helps us if we were given a broken number line. So let's have a look at one more example together. So this time, I'm going to have my multiples of three on the top, so three, six, nine and it could continue. And I'm going to have my multiples of four on the bottom four, eight, 12. And it could continue. Pause the video now and work out the constant of proportionality please. Well done. This time, it's four over three, so we're going to write it as a fraction. So this time, our constant of proportionality is four over three. And this is the same going from any number on the top to any number on the bottom. So if we've had a broken number line here, we've broken our number line. We don't know how many equally spaced gaps there's been in between. And then we're told that 45 is here. How could we use our constant of proportionality to find the missing number at the bottom? Excellent, we just times don't we by our constant of proportionality. So we multiply by four thirds. And this time we get 60. So hopefully you've understood those double number lines. We're now going to apply what you've learned to the independent task. So pause the video now, navigate to the independent task, and then come back when you're ready to go through some answers. Well done for giving that independent task a go. There was some quite tricky questions in there. So I'm sure you've done a really good job of them. So given the start of these double number lines, what numbers would appear on the top and on the bottom? So what are we looking for here? Well, we know that going along the top are multiples of four. And going along the bottom are multiples of five. So we're looking for numbers that are both multiples of four and multiples of five. So we're looking for common multiples. What's the first common multiple of four and five? 20. Brilliant. So 20 would appear on both. What about the next one? Excellent. 40 is the next common multiple. We see a pattern here. What's our next common multiple of four and five? 60. Excellent. So now we can know that any number that's in the 20 times table, so any number that's a multiple of 20 will appear both on the top and the bottom of the number line. What about for seven and eight? What's our lowest common multiple of seven and eight? Excellent, it's 56. So 56 is our first common multiple. So we know that any multiple of 56 will appear on both the top and the bottom of our number line. So well done if you made those generalisations. We'll see here that I've given you the answers already. So if you need to pause the video and check your work. Excellent. Here are some more answers. So for question three, I've given you some of the answers already. For the second one, let's do this together. We know that two spaces on our bottom number line give us 9.6. So that means that one space must be half of 9.6, which is 4.8. So we've got the multiples of 4.8 on the bottom of our number line. On the top, then we've got the multiples of four, haven't we? So four, eight, 12. And that's how we figured it out. What would our constant of proportionality be here, from four to 4.8? Excellent, it's 1.2. Well done. And moving on to the Explore part of the lesson now. You've got a part of an number line here with equally spaced markings. Each Mark denotes an integer. So that's a whole number value. How many ways can you fill the boxes? Pause the video now and have a go at this task. If you're struggling, one way that you could fill in this box here is 24. So what are we increasing by from 24 to 36? Well, we're increasing by 12. Well done. And what's our constant of proportionality? What's our constant of proportionality here? Two. Well done. So that means to get from 36 to our missing box here, we will be multiplying by two. So this box here would be 72. That's one way of doing it. Pause the video now and have another go. Excellent work. Well done. So we've gone through one answer that you could have had. There are two other answers that you could have got. Can you tell me what they were? You could have had 27 and 64 here. Or you could have had 32 and 54. That's the end of today's lesson. So thank you so much for all your hard work on double number lines. I hope you've enjoyed the lesson. Don't forget to go and do the quiz now to show me what you've learned. And hopefully I'll see you again soon. Bye.

In this lesson, we will learn how to calculate proportion problems using a table.

The rule of four

﻿Hi, I'm Miss Kidd-Rossiter and I'm going to be taking you through today's lesson on the rule of four. Before we get started, can you please make sure that you're in a distraction free area, a quiet place if you're able to be, you've got something to write with, and that you're completely focused and ready to go. If you need to pause the video now to get anything sorted then please do, if not, let's get going. So for today's try this, you've got Cala and Xavier, and they're packing boxes of sweets for some affair. Each box has the same number of bags, and each bag has the same number of lemon and raspberry sweets. Cala says, the ratio of lemon to raspberry is three to four. Xavier says, there are 84 sweets in the box. Your job it is to find out how many bags could there be, in each box. Pause the video now and have a go at this task. If you need a couple of hints, you might want to think about some equivalent ratios to three to four, and you also might want to think about how many parts of the ratio there are in total, and whether the multiples of this number go into 84. Have a think about that resume the video when you're ready to go through it. How did you get on with this try this activity. I hope you had a really good go at it. We're going to talk about it now and then connect and try and pull out some of the important features here. So as you can see, Cala said that the ratio of lemon to raspberry is three to four. This could of course mean that we have three lemon sweets, and four raspberry sweets in a bag. If that's the case, how many sweets do we have in total in each bag? Tell me now, excellent. And we know that in a box, we have 84 sweets. So if we have seven sweets in each bag, and 84 sweets in total, how many bags do I need to put into my box? Tell me now, excellent, 12. So, if I have three lemon sweets in each bag, and I'm putting 12 bags into each box, how many lemon sweets will there be in total? Tell me now, excellent, 36. And if each bag contains four raspberry sweets, and we're putting 12 bags into the box, how many raspberry sweets will I have in total? Excellent, 48. So can we notice a relationship here, between three and 36, and four and 48 can you notice a relationship there? Excellent. Going across the table, we're multiplying by 12, aren't we? This is our constant of proportionality that we've seen before when we've written equivalent ratios, three to four is an equivalent ratio, as 36 to 48. And our constant of proportionality would be 12. What about the relationship going down the table? Is there's something I can multiply three by to get four, we know about our inverse operations. So we know that three times something equals four to find the something, we can do four divided by three which is four thirds, or 1.3 recurring. I prefer to leave it as a fraction. Let's just check whether this is the same 36 and 48. So 36 times something gives us 48. And to find our something, we use our inverse operation, 48 divided by 36, which does simplify to four thirds when we take out a common factor of 12 from the numerator and the denominator. So we have another multiplicative relationship going down the table, where we're multiplying by four thirds. Now of course, Cala's ratio, doesn't mean that we have to have exactly three, and exactly four sweets in the bag. We could have had an equivalent ratio that simplifies to three to four. Can you tell the screen now an equivalent ratio that simplifies to three to four? The one ratio that I came up with was 18 to 24. We can see that 18 to 24 is equivalent to three to four, what would our constant of proportionality be there? Tell me quickly, excellent six, So if I had 18 lemon sweets, and 24 raspberry sweets, how many sweets would I have all together in my bag? Tell me now, excellent, 42 and we know that we have 84 sweets in the box. So if I have 42 sweets in each bag, how many bags do I need to put into my box? Two, well done. I have 18 lemon sweets in each bag, and I'm putting two bags into each box so that means I will have 36 lemon sweets in the box, and if I have 24, raspberry sweets in each bag and I'm putting two bags in, that means I will have 48 raspberry sweets in each box. Do we notice anything here? Can you notice something compared to our last example? What's our constant of proportionality here? Going from 18 to 36 or 24 to 48. What was it? Excellent, it's two this time. And we already know that to go from 36 to 48, we would multiply by four thirds, let's just double check that this relationship is still true for 18 to 24. So 18 times something is 24, using our inverse operation, something is equal to 24 divided by 18, which is equal to four thirds when we take out a common factor of six from both the numerator and the denominator. So key point of today's lesson, is that when we have a ratio table like this, we have two multiplicative relationships. We've got the multiplicative relationship going across the table, which is our constant of proportionality, and then we also have a multiplicative relationship going down the table, which in this case is four thirds. You are now going to apply your learning from today's lesson to the independent task. So pause the video here, navigate to the independent task, and when you're ready to go through some answers, resume the video, see you soon. Well done for giving that independent task a go. There were some tricky questions in there. I'm not going to go through all of the answers in full we're just going to pick out the really interesting ones to talk about. So if at any point you need to pause the video to mark your work, then please feel free to. Question one, I filled in the grid for you, now we need to figure out what's the ratio of squash to water. So, squash to water we can see is 20 centilitres to 15 centilitres and if we wanted to, we could simplify this ratio to four to three. How much bigger is bottle two than bottle one? So what do we multiply 20 by to get 60? Or the same relationship here, what do we multiply 15 by to get 45? We can see that that is three times bigger. Question two, the answers are on your screen, so please mark those. Question three, you can see I've already done the working out for you, it should be on your screen. To find how high the photograph is, you first need to work out the scale factor of enlargement so that's what I've done here, and then you need to apply that, to the height of the photograph. We've already worked out scale factor of enlargement and then we can work out our ratio. Question four now this is a really nice question I like this one. Gemma says that the missing number here is 35, and Ed says the missing number is 19. Who do you agree with, and why? Tell the screen now, really good. I hope you have really good reasons for that. I'm going to tell you that I agree with Gemma. She's the person that I agree with here, because, I can see that going across here, my constant of proportionality from three to 15 is multiplied by five. So that would mean that I need to multiply by seven by five as well, to get the correct answer of 35. What did I do wrong? Excellent. He just added on 12, didn't he? 'Cos you can see that from three to 15. We add 12, and he's made the error here that I think it's an additive relationship when it's not it's a multiplicative relationship. Finally then, we're moving onto the explore task. So we've got Cala and Xavier again, they're arranging bunches of pink and blue flowers, they've given you some statements there. You need to figure out how many blue flowers could there be in each bunch. What you should think about the kind of process that we used in the try this activity, and trying to apply the same kind of thinking to this explore activity. Pause the video now and have a go at this task. That's it for today's lesson. So thank you so much for all your hard work. I hope you've learned loads. I've really enjoyed teaching you this lesson, please don't forget to take the quiz, to show what you've learned, and I hope to see you again soon. Bye.

I have 5 yellow counters and 7 green counters. What fraction of the counters are green?

I have a bag of pink and brown cubes. 5/11 of the cubes are pink. What is the ratio of pink cubes:brown cubes?

Purple paint is made by mixing blue paint and red paint in the ratio 4:3. How much blue paint is needed to mix with 24L of red paint?

Fill in the gap: the ratio between the side lengths AB and BC is 2 : ____.

Fill in the gap: The constant of _________ between the two triangles is 3.5.

What is the length of DE?

Fill in the gap: If two quantities are in the same ratio then there are _______ relationships I can use to find a missing value.

Lots of the same groups of yellow and red counters are put in a bag. Fill in the missing value in the ratio table.

Yellow and red counters are in the ratio 2 : 3. What fraction of the counters are yellow?

If a double number line starts with 4 on the top and 3 on the bottom, what will be the first number that appears both above and below the lines?

On a double number line, 4 is above the line and 4.8 is below the line. What is the constant of proportionality?

What is the constant of proportionality for this number line?

Unit Overview: Ratio (8.6a)

Lessons:

Representing ratio

Enlargement

Double number lines

The rule of four