This is Common Core State Standards Support Video for Mathematics. This is Grade 5 and it’s standard NF.7, and it’s actually three standards. It’s standard 7a, 7b, and 7c. Standard 7a says: Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. 7b says: Interpret division of a whole number by a unit fraction, and compute such quotients. And then 7c: Solve real-world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem.
Now it made sense to do all three of these at once because the only difference between a and b is which is a divisor and which is a dividend, you know, the whole number versus the unit fraction. And of course, while we’re looking at these two, it makes sense to also look at real-world contexts at the same time so that students get a feel for exactly what each of these would model. Now the typical context for division, and this is for whole numbers, and that’s what the kids are used to: one example would be where you would know the total and the number of groups and you’re looking for the size of the groups. Or vice versa, you would know the total and the size of the group and you’re trying to figure out how many groups that would involve.
So let’s look at 7a first. The most typical and easiest context for this scenario, where you have division of a unit fraction by a non-zero whole number would be if you knew the total and the groups and you’re looking for the size of the groups. Now groups is in quotation marks, because since we’re dealing with fractions, sometimes we are going to have to take that idea of groups with a grain of salt, because it might actually turn out to be something else; more than likely, better thought of as parts instead of groups.
So let’s take a scenario where let’s say we have 1/4 as the total, and we have three groups, and we’re looking for the size. Now keep in mind that the 1/4 deals with a whole of one, so here’s what we start off with. We’re dealing with 1/4, which is this here. Now we need to split that up into three parts, so it is just this chunk in green that we subdivide into three groups or three parts. Now students need to be keenly aware of what constitutes the whole in the fraction. So what we’re dealing with here is just this one small piece of that 1/4. So we’re dealing strictly with this. Well, let’s see, we have one shaded block out of, let’s see, we count those; we have a total of 12. So we have 1 out of 12 for our solution.
Now the problem is where does that 1/12 come from? We can figure out what happened with the visual, but it’s a little bit tougher to see what happens as far as the computation. So how do we get 1/12 out of that? So let’s investigate here and see what happens. We start off with ¼ and we end up with 1/12. What could have happened here? Well let’s see, we ended up with a 12 in the denominator, and we know that 4 times 3 is 12. So we had to have had a situation where we multiplied by 3 down here. So we had to have multiplied and on the top we had to have multiplied by 1. So somehow the 1/4 divided by 3 got converted to multiplication by 1/3 instead of 3 over 1, which is the reciprocal. So if you have students do this repeatedly with other types of contexts, with the visuals to go with it to help them out, they’ll see the pattern that what I’m ending up doing is changing this division problem to multiplying by the reciprocal.
Now some real-life contexts for this particular model would involve having to evenly distribute some quantity that’s a unit fraction, and the solution would be the size of each part or share. So the solutions can be obtained in context using visuals and counting, then expand that to a pattern activity where students would see that the result can be gotten by multiplication of the dividend by the reciprocal of the divisor. In these contexts, the numerator would be the result of 1 times 1, and then the multiplication of the two numbers occurs in the denominator. So for example, like what we did with the 1/4 divided by 3, that ended up as 1/4 times the reciprocal of 3, which of course is 1/3.
So let’s look at 7c where we look at some real-life contexts. So one possible scenario would be where we have an inheritance and 1/5 of that inheritance is going to be split evenly among four children. So what is each one’s share? Well let’s see, we have our visual where we have the whole inheritance, but it’s split up into five parts. But then that 1/5 is going to be what we actually are going to split up. So then we split that up into four parts, but we have to consider that one little green part with respect to the whole inheritance. So if we count all of these up, and if we do some splitting up of all the rest of these, we would end up with one piece out of 20. And again, if we do this where we had 1/5, and we ended up, divide by 4, that ended up being 1/5 times 1/4. So each of the children is going to get 1/20 of the whole inheritance.
Let’s look at the second scenario. You have three friends who are on a trip and they’re sharing the driving equally. If they’ve traveled 1/3 of the distance, what part of the total trip has each one driven so far? So again, very similar to what we did with the first example: our visual represents the whole trip split up into three parts, but then that 1/3 of the way that they’ve driven so far, we’ve split that up into three equal parts. So if we’re dealing with just one of those pieces, again we have to realize what that little piece is with respect to the whole trip. And again, if we subdivide the other two pieces, we’re going to have a total of nine. So each of the friends would have driven 1/9 of the total trip at this point. And again, the computation would be done just like we did with the first example.
Now let’s look at standard 7b where it’s division of a whole number by a unit fraction. For that type of model, the easiest context in all likelihood, is when you know the total and the constant size and you’re looking for the number of groups. So let’s say okay, we have a total of three, and the size of the groups is 1/4, and we’re looking for the number of groups. So symbolically that would be 3 divided by 1/4. So we draw three circles to represent our total of three. Now each one of those circles has to be divided into four parts. So now, again we have to take number of groups with a grain of salt. What is really happening here, the question being asked really is how many of these do I have in all of this?
So just using a little bit of common sense, the students could probably just look at this, well there are 12 of them. Okay, so again, students need to be aware of what the whole is, but now, basically, what happens is we’re dealing with all of this. Now students have to be keenly aware of what the whole is. But this is a little bit different situation because again, this is what’s happening. Basically how many of these are there in all of this? Which of course, the solution is 12. Now just like before, in a real-life context, here the solution can be obtained by counting. But then you expand that to a pattern activity where the students see the result is obtained by multiplication of the dividend by the reciprocal of the divisor. Now since the divisor is a unit fraction, you’ll actually end up multiplying by a whole number, because, for example, here my divisor is 1/4, the reciprocal is going to be the whole number of 4. So we end up with 12. Now a real-life context for this model would involve having to partition several wholes evenly, and then the solution will be the total number of those groups, or better yet, parts.
So now let’s take standard 7c, which deals with solving real problems, contextual problems, and apply it to 7b, which deals with division of a whole number by a unit fraction. So in this first problem, an investor has a five-acre parcel of land that will be subdivided into lots that are 1/4 acre each. How many of these quarter-acre lots will there be? So we have our visual, and these marks here indicate that our five-acre parcel has been split up into the five individual acres. But each one of these full acres is going to then be subdivided into quarter-acre plots, which means they to have four equal parts for each acre.
So we can tell already that this part here is going to be four of those quarter-acre plots. So students can already tell, well if I’ve got four here, I am going to have four here, and four here, and four here and four there. So I have four quarter-acre parcels in each one of these acres, and I’ve got 5 acres. So students intuitively can figure out that there’s going to be 20 of these quarter-acre lots. So looking at the computation, students have already seen the pattern that what’s going to end up happening here is that we have the five, and we’re going to end up multiplying the 5 times the reciprocal of the 1/4, which of course is just 4, which corresponds to the idea that we’re going to have 20 of these plots. So my solution in this case is going to be 20. And again, the 20 deals with having 20 of those quarter-acre plots.
Next problem, Shirley baked three pies and wants to share them with friends. If she wants to give each friend 1/2 of a pie, how many friends would get a share of the three pies? Well here we have our three pies, and we’ve cut each one of these in half. And it’s a pretty easy solution to just see, well, I’ve got 1, 2, 3, 4, 5, 6—so I’ve got six of these 1/2 pies. So Shirley will be able to share the pies with six friends, and again just like with the example here, they’ve already seen the pattern that they are going to take the 3 and multiply it times the reciprocal of 2 over 1 which would be simply 3 times 2 to give us our solution of 6 of these 1/2 pies.
One more example here: Let’s say you had $3.00. The question is how many quarters would that be? Well here’s $1.00 here, and then we have another $1.00 here, and another $1.00 here, and so, well, here’s 4 quarters, 4 quarters, 4 quarters. So that’s a total of 12 quarters. And again, same type of scenario: we can figure out in our heads that this is 12, but of course, you want the kids to do the actual computation where they change this to multiplying by the reciprocal. And then they will get their solution. And then, of course, you want to always make sure that the students use logic and reason it out to make sure that the solution makes sense. So here we have a solution of 12, and yes, that make sense where I’ve got 12 quarters in $3.00. And again notice that the quarters was the size of the set, and the 12 is actually the size of the groups, which in this case are actually the quarters.
Let’s go back to standard 7a. A little bit of a warning here; let’s look at the other scenario, the other context where students would know the total and the constant size, the equal size of each part and the number of groups or the number of parts. So let’s say we have the same numbers that we used before but in a slightly different context. So we have a total of 1/4, but the size is 3. Now I don’t know about you but I’m already starting to get a little bit confused, because wait a minute, I’ve got a total of 1/4 yet the size of the parts is 3. So that can get very confusing.
Now this 1/4 refers to just one circle, so this is what we’re dealing with. And then when I expand this to the other circles, again I don’t know about you, but I’m a little bit confused because I’ve got a total of 1/4 and I’ve got 1/12 here, because now I’m looking at all of this as a whole. So this is how many of these I have, out of this, is 1/12. And again, very confusing, so again, the warning here is don’t try to force fit a context into a model. Just go with what’s natural and what fits best for that particular context. So again, this particular situation here does not fit well into a situation where you have a unit fraction divided by a non-zero whole number.
Now let’s look at 7b, and let’s look at the other context where you would know the total and the number of groups and you’re looking for the size. Okay, I have a total, in this example, I have a total of three and the number of groups is 1/4. Okay, now wait a minute, I’ve got three. All right, I’ve got three items, but the number of groups is 1/4. So I’ve got 1/4 of a group. So this is just 1/4 of what I need. So then basically this is what happens. Okay, it makes a little bit more sense in that other scenario where, okay, but again it’s a whew! I don’t know about you, but it’s a little bit tough to visualize, you know, the size.
So here’s a better way of thinking. I mean this scenario will fit, but you’re going to have to adjust how you think of your knowns and your unknowns. Here’s a better way to think of this kind of scenario where this is actually a partial total, and this is actually the fractional part and you’re looking for the actual total. Now it makes sense. Okay, I’ve got three, which is my partial total, but that’s only 1/4 of the whole thing. So now that makes a lot more sense that I will need a total of 12 to be the actual total. So again, that’s a better way of thinking of this.
So there are some contextual problems that will fit this, and here’s an example, well a couple of examples. Let’s look at the first one. Okay, you’ve got a hiker that goes uphill from marker A to marker B and down hill from B to A of course. Now the hiker takes 15 minutes from B to A. So that takes 15 minutes and that’s only half the time it takes to go uphill, which makes sense. So how long does it take to go from A to B? Just using common sense, the students can see that well, it’s going to take 30 minutes because, you know, going downhill is only half the time of going uphill. But of course, when we set up the problem, you’re not starting off with a 15 and a 2, you’re starting off with 15 and 1/2. So again, it’s a matter of getting students used to how to set these up. But again, this type of context lends itself to the thinking where they can you pretty much go straight to what the process is going to involve. But it will make a lot of sense to them because, again, the conversion makes a lot of sense based on the context.
Let’s look at the second example. Julian has $10.00, but that’s only 1/3 of what he needs to buy a new game. So how much does he need to buy the game? Again students might want to jump the gun. They can already reason that, well, if $10.00 is only 1/3 of what he needs, he needs three times as much, so he is going to need $30.00. But again, it’s important that they set it up the way it’s supposed to be, because you’re only dealing with 10 and 1/3. But again students can very easily see from the context that, okay, I have $10.00. They already pretty much have figured out that the answer is $30.00. So what happens here is that again, he needs three times as much. So again, the idea of multiplying by the reciprocal here is pretty easy thinking, and they can make the connection.
So again, working problems in contexts like this really helps students to understand the rationale as to what really happens, you know, why it is that in this case that you end up multiplying by the reciprocal of the divisor. Hopefully these explanations for this standard will help, because again fractions, that’s a very difficult topic for students. And at this point, they’re just starting to get some experience with the operations. And when it comes to fractions, multiplying and dividing are the two toughest operations with these numbers. So again, make sure that you include contexts to where students can see why the computation happens as it does.