What would be the next term in this sequence?                                                                                                  5.5,      5.75,      6,       6.25,      6.5,      6.75

What is the term to term rule in this sequence?                                                                                       9,      14,       19,      24

What would be the previous term in this sequence?   31,    36,    41,    46,    51

Look at the following sequence, which term below can not be part of that sequence?       29,  34,  39,  44,  49

If a represents the amount of apples. How could we represent double the amount of apples?

In this problem b represents the amount of bananas. There are 12 more apples than bananas. How could we represent the apples in terms of b?

Consolidating linear sequences and representing problems with algebra

﻿Hello, it's me Ms. Jones. And I'm ready to do some fun Maths with you today. In fact, I'm even more excited than usual today because today's lesson is all about problem solving and that's the reason we do Maths so we can solve problems in Maths, but even outside of Maths. I'm really, really looking forward to it. Anyway, to get us in the mood, I thought we'd start with another riddle. Today's riddle is, what begins with the letter E but only has one letter inside it? What begins with the letter E but only has one letter inside it? A little think. Okay, the answer is an envelope. An envelope begins with E and it has a letter inside it. Did you get that one? You're getting good at these, aren't you? Hopefully you can share that with your friends later on. Okay, now that we've got our brains warmed up, let's begin today's lesson. In today's lesson, we're going to be Reasoning and Problem Solving, yes. So we'll start off by looking at one type of problem together called function machine. We'll do an example together, and then you're going to have a go at a similar problem. Then we'll go through the solutions to that problem. After that we're going to look at a different type of task called Marbles. We'll discuss it together, and then you will go off to try and solve it. Once you've solved it or had to go at it, we're going to go through the solutions together. For this lesson, you will need a pencil and a piece of paper or something else to write with or write on. If you haven't got that yet, go and get it now and then come back. Okay, let's begin. Task one is called Function machine. Now you might have come across function machines before. And the premise is something goes in, something happens in the function machine, and then something comes out. Let's have a look as our problem. A Function machine creates a number sequence. Look at the table to try and identify the four operations at the function machine. Sorry, the operations that the function machine completes. Okay, so what's happening inside the machine? So something like one goes in and then three comes out. We need to decide what's happening here. Once we've done that, if the input is represented by the letter Y, write an algebraic expression for the function. So what would this be in terms of Y? Okay, before I write my algebraic expression, let's explore the function and see if we can work out what's happening from one to get to three. Well, you could think that one has been multiplied by three to get to three. That seems to work, but does it work for the other items in our sequence? Two multiplied by three is six. Hang on a minute, that's not a six. So this can't be our rule. We must have two different things happening here. Let's look for some patterns, so we can see in our output, we go from three to five to seven to nine. Each time we're adding on two. So we can see looking at our sequence that our output is very similar to the two times table. So our function must be something to do with multiplying by two. What is one multiplied by two, it's two, but that still isn't our output. So something else is happening here. Can you have a think about what else is happening? One multiplied by two is equal to two. Then if we add on one, we get three. So perhaps our function is times by two, add one. Let's see if that works for the others. Two times by two is equal to four, add one is equal to five, that works. Three times by two, add on one is equal to seven. So our rule is times by two, add one. Okay, pretty happy that we've worked that out. However, we haven't written it as an algebraic expression. So if we want to do that, we need to think about Y. So if Y is the input, we can represent the output as Y times two, I know I can write that as 2Y, add one. Our algebraic expression is 2Y, add one. If we take that as an example, if we had four in our answer, we would have two times four, I don't really need brackets coz multiplication takes priority. We add one and we get nine, same as what we've got here. Let's have a look at one more. This one is for you to have a try. A function machine creates a number sequence just like before, but this time we've got a different output. I want you to see if you can figure out what's happening in the function machine. And once you've done that, can you write that as an algebraic expression? If Y is the input, what would be the output? Go and have a go at your task now. Okay, let's go through this together. So looking at the pattern here, we've got eight, 13, 18. I can see that each one is five more. Did you spot that pattern? So this might have something to do with our five times table. Now I know to get from one to eight, we multiplied by eight, but that doesn't work all the way down. So two things must be happening. I'm going to use what I've noticed about the five times table to help me. So I think we are multiplying by five. However, one multiplied by five, is not eight. So what else is happening? We are timed by five and then are we adding or are we taking away? We're adding, so we timed by five, one times by five is equals five and then we add on one, two, three to get eight. The rule should have been, we multiply by five and we add three. Let's see if that works for the others. Two times five is 10, add three is 13. Three times five is 15, add three is 18. It works for them all. Now, how do we write that as an algebraic expression? Well if the input is Y, we know that the output will be five lots of Y, because we need to multiply by five, add three. That's what I think, let's check. 5Y plus three, great. Hopefully you managed to get there or find the rule. Don't worry if not. Hopefully my explanation has helped you to understand. Okay, we've got one more task to have a look at. Task 2 is called Marbles. I wonder what this could be about. What do you think it's going to be about, marbles? Sara has some marbles. Dev has twice as many marbles as Sara. Mia has the same amount as both Deb and Sara combined and an additional 30. Altogether they have 126 marbles. How many marbles does each child have? There's a lot of information here. So I think we need to use some sorts of representation to help us make sense of this problem. What different ways would you represent this problem? You might use a bar model. I know that Sara has an amount of marbles. I don't know what the amount is yet. I know that Dev though has twice the amount that Sara has. So I've made his bar twice as long, here. I've shown a little dotted line to show the halfway mark. Now I know that Mia has the same amount as both of these combined, which is three of these bars if you like. Looks like this, plus an additional 30. So this is what my bar model looks like now. I know another piece of information. I know that altogether, they have 126. Now I know that, can I use that information to help me work out the amount that each child has? How else could we represent this problem? We could use letters. We can say that Sara has an amount of marbles, which I've called M. That means Dev, who has twice as many, has 2M. Mia has the same as these combined, which is 3M, plus an additional 30. Can I use this information to help me work out the value of M? To work out Sara's and the value then of Dev's and Mia's amount of marbles. Okay, we've spent some time discussing this together. I'd like you to have a go at this problem. If you want to, you could use a bar model. You could use some algebraic expressions like these or another representation to help you work it out. If you want to, you can try and work it out two different ways and compare what's the same, and what's different and use that to double check that your answer's correct. Off you go to complete your task. When you're done, come back to the video and we'll go through the solutions. Okay, hopefully you've had a chance to complete your task. Let's go through it together. So as we already discussed, we could represent this problem with a bar model. We met that Sara had some marbles, Dev had twice as many, Mia had the same as those two combined plus an additional 30, and there are 126 all together. So using that information, how can we work it out? Well, if we take this 126, and first of all, let's think about what this amount might be by taking away 30. 126 take away 30 is equal to 96. Now, what I can do here is divide 96 by six, because I have one, two, three, four, five, six, equal parts. That way I can find out what one of these equal parts is? We get 16. Now I can see that now I know that Sara has 16 marbles. So I've worked out one part of my problem. Now, all I need to do is look what Dev had. Dev has twice as many as Sara. So he has doubled 16, which is 32. Mia has the same as Sara and Dev combined. So three lots of 16 or you can think of that as 16 plus 32. So Mia, we know has 48, but not just 48 because she also had an additional 30. So let's think about that all together. Mia has 48, plus an additional 30, which makes 78 all together. Sara has 16, Dev has 32, and Mia has 78. Let's have a look at this problem in a different representation. Now, if you remember, I also represented this problem using letters. Sarah has M, that means Dev has 2M, Mia has 3M. Now what I've done is that I know that all of this altogether makes 126. So I've written it out as a combined equation. M, which is Sara, plus Dev's which is 2M plus Mia's which is 3M plus 30, is all equal to 126. Now that I've done that, I can start to think about grouping my Ms together. So altogether here, I've got 6M added to 30, is equal to 126. Now, again, I can subtract that 30. Do you remember what we did there in our last representation to get 96 and to work out that one part was 16? Sara has 16, but I'm not finished coz I need to also think about what Dev and Mia had. Sara had 16, Dev has double that, 32, and Mia has 78, which is three lots of 16 added to 30. Think about both ways we represented these problems. What was similar and what was different? Did we end up with the same answer? Yes. It doesn't matter which way you represent it. As long as you're being accurate and using some sort of representation to make sense of it, if you need to. If you want to, you could always work out something in two different ways. And then that helps you to check that you're correct. That brings us to the ends of today's lesson. Before you go, make sure you do our Multiple Choice quiz, thank you.

If 3f + 8 = 20. What is the value of f?

Which expression would represent the following problem:  The cost of a photo album book is £1.50 for binding and printing with an additional cost of 25p per photograph.

a - 2b = 8. Which of the following pairs of numbers would work?

A function machine has the rule x = 2y + 1. Where y is the input and x is the output. If we put 4 into the function machine (y = 4) what would come out?

Reasoning and problem solving

In this lesson, we will explore two types of problem in depth, thinking carefully about how to represent each problem in order to help us find a solution.

Maths

Y6 Maths

Key Stage 2

Year 6

Miss

Calculation problems

Number

THEME

UNIT

Understanding which operations have equal priority

Order of operations

Consolidating order of operations

Linear number sequences

Expressing missing numbers algebraically

Expressing missing number problems algebraically

Finding solutions to problems with two variables

Reasoning and problem solving (Part 2)

True or False? 130 - 26 + 14 = 130 + 14 - 26

3 x 5 + 2 =

Solve the equation: 4 x (3 + 3) = ?

t + 7 = 20 What is the value of t?

Which expression would represent the following problem:  The cooking time for roast beef is 30 minutes for every kg plus an additional 15 minutes.

3c + 2d = 20. Which pair of numbers does does NOT work with this equation?

Lesson overview:Reasoning and problem solving

Calculation problems: