Lesson video

Hello, I'm Miss Miah, and I'm so excited to be a part of your learning journey today.

I hope you enjoy this lesson as much as I do.

In this lesson, you'll be using known facts from the 2, 5 and 6 times tables to solve problems involving the 7 times table.

Your keywords are on the screen now, and I'd like you to repeat them after me.

Multiple.

Partition.

Distributive law.

Array.

Fantastic, let's learn what these keywords mean.

Now, a multiple is the result of multiplying a number by another whole number.

Partitioning is the act of splitting an object or value down into smaller parts.

The distributive law says that multiplying a number by a group of numbers added together is the same as doing each multiplication separately.

An array is the layout of items, arranged in rows and/or columns.

This lesson is all about using our known facts from the 2, 5 and 6 times tables to solve problems involving the 7 times table.

Now this lesson is made of two lesson cycles.

Our first lesson cycle is using known facts from 2, 5, and 6 times tables, we then move on to solving problems using what we know.

Now, using what we already know from the 2, 5, and 6 times tables can make solving problems with the 7 times tables easier.

And in this lesson, I'm going to teach you how to break down the 7 times tables into smaller and familiar steps, so you can solve problems faster and with more confidence.

It's basically like using shortcuts to get to your answer quicker.

Let's get started.

In this lesson, you'll meet Jacob and Andeep.

Jacob and Andeep look at a multiplication equation with a missing number.

8 times 7 is equal to, hmm? Andeep says, "We can use knowledge of our 2, 5, and 6 times tables to help us." Jacob says, "Knowing how to find adjacent multiples of 7 can also help us." Jacob uses a part-part-whole model to partition the factor 7.

and you can see that he's done that here.

So he says, "I have partitioned 7 counters into 5 and 2." Andeep says, "I'll represent this model using equations." So here we are, so 5 plus 2 is equal to 7, and we also know that 7 subtract 2 is equal to 5, and 7 subtract 5 is equal to 2.

Now Jacob says, "Each counter is worth 1, which I can label." "I'll rewrite the expressions and equations too." So here we are, and we can see how 7 times 1 has been partitioned into 5 times 1 and 2 times 1.

So we can represent this as 5 times 1 add 2 times 1 is equal to 7 times 1.

"I can say 7 groups of 1 is equal to 5 groups of 1 plus 2 groups of 1." "What if I change the value of the counters to 2 instead?" And we've done that now.

Well, Andeep says, "Then I'd need to change the expressions and equation again." So this time, it's 5 times 2, add 2 times 2, which is equal to 7 times 2.

Hmm, Jacob says, "What if I change the value of the counters to 5 instead?" And Andeep says the same.

"Here's the expressions and equation amended again! 5 times 5 add 2 times 5 is equal to 7 times 5." "Last one, what about if they were each worth 8?" "Here's the expressions and equation once more.

5 times 8, add 2 times 8 is equal to 7 times 8." "This looks very familiar!" "It's the equation we started with!" So Jacob and Andeep partition 7 into 5 and 2 because they know their 2 and 5 times tables.

7 is equal to 5 plus 2, so 7 times 8 is equal to 5 times 8 plus 2 times 8.

"I think there are more ways of interpreting this part-part-whole model.

It shows other equations." What do you think? "Well, 5 is equal to 7 subtract 2, so 5 times 8 is equal to 7 times 8 subtract 2 times 8.

2 is equal to 7 subtract 5, so 2 times 8 is equal to 7 times 8 subtract 5 times 8." So by using the distributive law, Jacob and Andeep were able to solve the problem.

Over to you.

I'd like you to complete the part-part-whole model and the equation to represent 7 times 6.

You can pause the video here and click Play when you're ready to rejoin us.

So what did you get? Well, you should have got 5 times 6 and 2 times 6 as your parts, and then 5 times 6 add 2 times 6 is equal to 7 times 6.

Jacob and Andeep partition 7 groups of 8 in two ways.

So on the left, we've got one way, and on the right, we've got another way, have a look.

What's the same, what's different, hmm? Well Andeep says, "The factor of 7 has been partitioned in different ways.

The structure of the calculation is different for each." "The 7th multiple is the total of the 5th and 2nd multiple.

The 7th multiple is the total of the 6th and 1st multiple.

The value of each counter is the same in each model." Oh, this time we've got three part-part-whole models here.

What's the same, what's different? "So the factor of 7 has been partitioned in the same way this time, so the structure of the calculation is the same for each." "One of the equations is adding and the others are subtracting." Jacob and Andeep look at the multiplication equation with a missing number.

So here's another example, and this time, we've got a number line, which shows multiples of 7, and then after the multiple of 35, there's a blank box.

Now Jacob says, "I know my 5 times tables, so know this one." Hmm, Andeep says, "I'm not sure, but I also know that 5 times 7 is equal to 35." I wonder how they can use that to calculate what 6 times 7 is.

Ah, Jacob says, "That's how I know it! I add on one more group of 7." "Yes, it's an adjacent multiple, so must have a difference of 7." So that means 6 times 7 is equal to 42.

Now Andeep says, "We worked out the fifth multiple of 7, or 5 groups of 7." So that's 5 times 7.

"And then we added on another group of 7 to get the 6th multiple.

So we can say that 6 times 7 is equal to 5 times 6 add 7, which is equal to 42." Now Jacob and Andeep solve another equation by using their knowledge of 2, 5, and 6 times tables.

So on the screen here, we've got 4 times 7 is equal to something.

We've got a number line, and we can see that they've subtracted 7, so they're finding the adjacent multiple of 7 to 35.

Now Andeep says, "I know that the 5th multiple of 7 is 35, so the 4th multiple will be 7 less.

So we can represent the equation like this.

4 times 7 is equal to 5 times 7, and we're subtracting 7, and that's equal to 28." Now Jacob says, "Remember, adjacent multiples of a number have a difference equal to that number.

We can easily calculate multiples of a number by adding on or subtracting that number from an adjacent multiple." Back to you.

I'd like you to complete the missing numbers in the equation using your knowledge of the 2, 5, and 6 times tables.

You can pause video here and click Play when you're ready to rejoin us.

So what did you get? Well, you may have got 7 times 8 is equal to 5 times 8, add 2 times 8, or we can apply the law of commutativity here, you may have also got 7 times 8 is equal to 2 times 8 add 5 times 8.

Let's move on.

So this is the main task for this lesson cycle.

For Question 1, you're going to complete the part-part-whole models and stem sentences, using your knowledge of the 2, 5, and 6 times tables.

So you've got two there, A and B, and then you're going to fill in the blanks.

For Question 2, you're going to complete the part-part-whole model using the equations.

So again, you've got two questions there, A and B.

And for Question 3, you'll need a set of cards from 1 to 12.

You're going to pick a card and write an equation to multiply by 7.

Partition one of the factors using the distributed law and represent it as a part-part-whole model and an equation.

You can pause the video here, and click Play when you're ready to rejoin us, off you go, good luck.

So what did you get? Well, for Question 1 this is what you should have got.

So let's look at 1A.

We can see that 7 has been partitioned into 6 and 1, so we know that 7 times 7 can also be partitioning to 6 times 7 and 1 times 7.

6 is equal to 7 minus 1, so 6 times 7 is equal to 7 times 7 minus 1 times 7, and for 1B, so 7 is equal to 6 plus 1, so 7 times 7 is equal to 1 times 7 plus 6 times 7.

This is what you should have got for Question 2.

We can see 9 times 7 has been partitioned into 6 times 9 and 1 times 9, so 6 times 9 add 1 times 9 is equal to 7 times 9, and for B, 7 times 11 has been partitioned into 5 times 11 and 2 times 11, 5 times 11 add 2 times 11 is equal to 7 times 11.

For Question 3, you may have got something like this.

So Jacob picked 6, and this is how he represented 6 times 7.

So he partitioned 6 times 7 as 5 times 6, and 2 times 6.

Onto the second lesson cycle.

So Oak Academy is organising a basketball competition across Year 4.

How many players will there be in total? Hmm, Andeep says, "We can't answer this yet! We don't have enough information." Jacob asks, "What else do we need to know?" Well, what information do you think is missing? Well, there will be 7 teams of 7 children competing.

Now we can see an array.

So Andeep says, "We need to calculate seven groups of seven, so it's multiplication." "Hmm, I don't know 7 times 7, So I'll use the distributive law." Andeep says, "I'll draw it as an array to help us visualise it." Wow, that's a big array! Let's partition, quick." So 7 times 7 is equal to something, we don't know, this is what the boys are trying to figure out, and Andeep has visualised this using an array.

So Jacob says, "I'm going to partition the 7 into 6 and 1." So this can be represented like this, the 7 has been partitioned into 6 and 1, so now we've got 6 times 7 add 1 times 7.

Jacob says, "I know that 6 times 7 is equal to 42, and 1 times 7 is equal to 7.

So 42 add 7 is equal to 49." Moving on, Oak Academy also has a netball tournament.

Netball teams have 7 players.

How many players will there be in teams? So again, here, Andeep has partitioned 7 into 1 and 6.

So we could partition 7 into 1 and 6, which will give us an answer of 42.

But Andeep says, "He's going to partition the 7 into 2 and 5." Now, he's got 6 times 2, add 6 times 5.

He says, "He knows that 6 times 2 is 12, and 6 times 5 is 30, so he can find the product from this." And that's 42, because 12 add 30 is 42.

Netball teams have 7 players in each team.

How many players will there be in 9 teams? Jacob says, "I think there's a way of using subtraction." So we know that 10 times 7 is 70, so 10 times 7, subtract 1 times 7, would be 70 minus 7, which is 63.

"I see, you change the factor 9 to 10 because that's easier to work with.

Then you subtracted one group of 7 because you had one group too many." Back to you, which equation does this representation show? Is it A, 5 times 9 add 2 times 9 is equal to 63, B, 5 times 7 add 2 times 7 is equal to 49, or C, 9 times 7 add 9 times 5 is equal to 108? You can pause the video here and click Play when you're ready to rejoin us.

So what did you get? Well, you should have got A, and this is because this shows 7 times 9, the 7 has been partitioned into 5 and 2, so by reeding 5 and 2, we end up with 7, and then our factors are 9 and 7, and the product of 9 and 7 is 63.

Moving on, Andeep is trying to solve this problem.

Guinea cards come in packets of six.

Lucas buys seven packets of cards.

How many cards are there altogether? Andeep says, 'You can factorised one of the factors and apply the distributive law to help you." Jacob says, "I think there's a way of using addition." And we're calculating what the total of 7 packets of 6 cards are.

Now, we've got two methods here.

Have a look whose method was more efficient? Andeep says, "He prefers the first method because he can quickly calculate the 6th multiple of 6." And I agree with him because we know that six 6ths is 36 or 6 times 6 is 36.

And Jacob says, "I like the last method because I'm confident in my 2 and 5 times tables, and by finding the 2nd and 5th multiple of 7, I can add these products together." So there's no right or wrong answer, it really does depend on what times table you are more confident in to find the answer.

"So both our preferred methods partitioned different factors, so efficiency is partly personal choice." Over to you, which of these methods is more efficient for finding the product of 7 and 9.

Explain your reasoning.

You can pause the video here and click Play when you're ready to rejoin us.

So which method did you prefer? Now, Andeep says, "I find this subtraction more efficient because I know my 10s really well, and that's the starting point." And Jacob says, "I think that addition equation is more efficient because I'm more confident adding." So again, there's no right or wrong answer, it really depends on what you're more confident with.

Onto the main task for this lesson cycle.

So for Question 1, you're going to use your knowledge of the 2s, 5s, and 6s were appropriate, to solve the following problems. So for 1A, you're going to calculate 7 times 4, 1B, you're going to calculate 7 times 12, 1C, 7 times 6 and 1D, 7 times 11.

For Question 2, you're going to complete the following word problems using your knowledge of 2, 5, and 6 times table.

So for 2A, Andeep loves Squishy Star sweets, which come in packets of 6 sweets.

He has 6 packets of Squishy Stars and is given another packet.

How many Squishy Stars does he have altogether? 2B, Jacob's mum does 6 sets of 7-minute exercises in one day.

If she adds on another set, for how many minutes does she exercise in one day? 2C, Andeep fills 5 baskets with 7 oranges each.

He gives one basket of oranges away.

How many oranges does he have in total? You could pause the video here, off you go, good luck, and click Play when you are ready to rejoin us.

So how did you do? For Question 1, this is what you should have got.

So 7 times 4 is equal to 28, and we know that 5 times 4 add 2 times 4 is equal to 7 times 4, so the 7 has been partitioned into 5 and 2.

Then we know that 5 times 4 is equal to 20, 2 times 4 is equal to 8, 20 add 8 is 28.

For Question B, this is what you should have got.

7 times 12 is equal to 84.

The 7 was partitioned into 6 and 1, so you should have got 6 times 12, add 1 times 12 is equal to 7 times 12.

6 times 12 is equal to 72, 1 times 12 is equal to 12, so 72 add 12 is equal to 84.

For Question C, you should have got 42.

That's by partitioning 6 into 1 and 5.

And for Question D, you should have got 77, and that's because you know that 10 times 7 add 1 times 7 is equal to 11 times 7, so 70 add 1 is equal to 77.

Well done if you manage to get that correct, you are showing that you can use the distributive law, as well as your knowledge of the 2s, 5s, and 6s to help you answer the problems. Now for Question 2 , this is what you should have got.

So you know that Squishy Star sweets come in packets of 6 sweets.

So if Andeep already has 6 packets of 6 sweets, that's 6 times 6, is then given another packet of sweets, so that's another 1 packet of 6 sweets, adding 1 times 6, which is equal to 7 times 6.

So 7 times 6 is equal to 42, Andeep has 42 sweets altogether.

Now if you know that 6 times 6 is 36, you could have added another 6 to that to get the answer, which is 42.

For 2B, you should have got 49 minutes of exercise, and for 2C, you should have got 28 oranges altogether.

Well done if you manage to get all those questions correct, I'm super proud of you.

We've made it to the end of the lesson, let's summarise our learning.

In this lesson, you learned how to use your known facts from the 2, 5, and 6 times tables to solve problems involving the 7 times tables.

You should now be able to use the distributive law and partition 7 to help you multiply by 7.

And you should know that if 7 is equal to 5 plus 2, then 7 groups of 6 is equal to 5 groups of 6, plus 2 groups of 6.

You should now also understand that 7 times a number and 6 times the same number are adjacent multiples.

Thank you so much for joining me in this lesson, and I look forward to seeing you in the next one, bye!.