FRA20: Remainders

Remainders, in a blog about fractions? Here's a miscellaneous collection of some of the many tasks for which there is not room to develop a weekly set. You might want to develop some variants yourself...

Task 20A: This looks like a standard equivalent fractions task, but it might provoke a short pause for thought. There is not a whole number multiplier that maps the numerator 3 onto the numerator 4, and that could then have been used to map the denominator 60 onto the missing denominator. However the task can be solved easily enough by using the whole number multiplier ×20 to map the numerator 4 onto the missing denominator.

We could of course use the fractional multiplier ×1⅓ to map 3 onto 4 and 60 onto the unknown denominator. Or we could divide the numerator and denominator by 3 to get the fraction 1/20 and then multiply the new numerator and denominator by 4.

Task 20B: In the first item, the fraction 4/45 is one third of 4/15, so its position is 1 cm from the left end of the rod. Some pupils might hastily think that 4/45 is three times as big as 4/15. Others might be thrown by the fact that the rod is not a whole number of centimeters long and that its length does not seem to have an obvious relationship to any of the other numbers that occur in the item. The faction 4/5 in the second item is more familiar than 4/45 so pupils should have a better idea of where it lies on the rod, namely quite close to the right end. In fact, being 3 times as large as 4/15, it lies at a mark 9 cm along the rod.

The rod looks to be about 11.2 cm long. It might be interesting to ask pupils what the precise length should be.

Task 20C: Here it is obvious that ¼ of Bar A is tinted yellow. Some pupils might see immediately that this is also true for Bar B. Others might get there after first determining that 3/12 of the bar is yellow. It turns out that Bar C is also a quarter yellow, even though we cannot assess the exact size of two of the yellow regions.

We can think of Bar C as being split into three regions, each of which is ¼ yellow. This means ¼ of the whole bar is tinted yellow.
We can think of this as an example of the distributive law:
¼ of X + ¼ of Y + ¼ of Z = ¼ of (X + Y + Z).

Task 20D: This is quite a simple task. The nice thing about it is that it can be solved in many different ways. For example, we could split the tinted shape into two triangles (with bases of 2 and 3 units and heights of 5 units, where the whole shape is a 5 by 5 square); or we could apply an area-of-trapezium formula to the existing shape; or we could shear the trapezium so that two corners are right-angled, and we could then cut-and-rejoin the shape to produce a 2½ units by 5 units rectangle.

Note that the task would still work if the whole shape were a rectangle or parallelogram rather than a square.

The variants, below, of the given shape are all half green. We can derive the fraction from the lengths of the parallel sides, namely (2+3)/(5+5).

Task 20E: This task can be solved in many different ways and at different levels of abstraction. Some pupils might count small squares: there are 15 tinted yellow out of a total of 60 small squares, so the desired fraction is 15/60. Others might derive these numbers by multiplying: there are 3×5 small yellow squares out of a total of 10×6 small squares, so the fraction is 3×5/10×6 = 15/60 = 1/4.

Others might work more abstractly (or maybe just more procedurally?) by multiplying the given fractions ⅚ and ³⁄₁₀. Note that this essentially gives us the same expression as 3×5/10×6, above, even though the thinking might be very different! For example, some pupils who multiply the fractions might be thinking in terms of scaling: we can transform the large blue rectangle into the yellow rectangle by scaling its height by the scale factor ⅚ and then scaling its width by the scale factor ³⁄₁₀. Or, pupils might adopt a similar-looking but more common static approach: we can form the yellow part of the large rectangle by first taking a ⅚ part of the rectangle and then a ³⁄₁₀ part of that: ³⁄₁₀ of ⅚ of 1 = 1/4.

A neat, geometric way of solving the task is to rotate the yellow region through 90˚, as shown below. We can see that the faction is a quarter.

The above rotation method works so neatly because of the way the task has been contrived. However, even where such a rotation does not produce an easily recognisable fraction, it nicely illustrates that a product a/u × b/v is equivalent to b/u × a/v. When u = v = 1, this is the same phenomenon as when we rotate an array showing the product of whole numbers a × b to demonstrate that a × b = b × a.

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FRA19: Dividing by three-sevenths

In this Week's task we again look at division as quotition (or measurment), but we also interpret it (in what some might regard as a rather laboured way!) as partition (or sharing). We compare the effect of dividing 10 by 3/7 using quotition and partition, and of dividing 4/5 by 3/7 using quotition and partition.

Task 19A: In this task pupils are asked to interpret the expression 10 ÷ 3/7 as "How many three-sevenths are there in 10?". This division is then broken down into three steps, starting with 10 ÷ 1/7.

The first step (How many 7ths are there in 10?') is fairly easy to construe, as can be seen from step 2: there are seven 7ths in 1, so there will be 10 times as many in 10, which gives us 10×7.

In the third step (How many three-sevenths are there in 10?) the divisor 3/7 is 3 times as large as the previous divisor, 1/7, so the number of times it fits into 10 will be one-third of the number of times 1/7 fits. So it fits 10×7÷3 times. 

Overall, it is fairly easy to see, but also to see why, we get the desired answer by multiplying 10 by the denominator of the divisor ³⁄₇, and by then dividing by the numerator. It follows, though this will be a step too far for some pupils, that we can rewrite our division 10÷³⁄₇ as 10×⁷⁄₃, where ⁷⁄₃ is the reciprocal (or multiplicative inverse) of ³⁄₇.

Task 19B: This task involves the same expression, 10 ÷ 3/7, as in the previous task, but this time we interpret the division as partition or sharing. The resulting statement, "10 is three-sevenths of a share" might seem rather awkward and forced, but the case can be made that it does make sense of a sort!Consider the division 10÷4, where the divisor is a whole number. We can think of this as fitting a story like '10 kg of rice are shared fairly between 4 people. How much does each person get?". This can be rephrased, a little awkwardly perhaps, as "10 kg is 4 shares, how much is 1 share?". Similarly, 10÷³⁄₇ can be phrased as "10 is ³⁄₇ of a share, how much is 1 share?".

Again, we can solve the task in stages. As with the measurement interpretation, a powerful first step is to consider one seventh. If 10 provides ³⁄₇ of a share, then ¹⁄₇ of a share will be one third of 10, which we can write as 10÷3. In turn, seven sevenths, or a whole share, will be 7 times that: 10÷3×7. 

Overall, it is fairly straightforward to see that we arrive at the answer by dividing the dividend 10 by the numerator of ³⁄₇ and then multiply by the denominator. So we can rewrite our division 10÷³⁄₇ as 10×⁷⁄₃, where ⁷⁄₃ is the reciprocal (or multiplicative inverse) of ³⁄₇. But again, this might be a step too far for some pupils and shouldn't be pushed too hard.

Task 19C: Here we revert to the quotitive or measurement interpretation of division, as in Task 19A, but this time the dividend is a fraction as well as the divisor.

We can again start with the simple and powerful idea that there are seven 7ths in 1; however, some pupils may well balk at the next step, namely that there will be ⅘ as many 7ths in ⅘, giving us ⅘×7. Previously we multiplied 10 by 7 rather than ⅘ by 7, which fits the common sense notion that 'multiplication makes bigger'; this time we can see that there will be fewer 7ths in our dividend ⅘ than in 1 - can this be achieved by multiplying?

As before, in the third step (How many three-sevenths are there in ⅘?) the divisor 3/7 is 3 times as large as the previous divisor, 1/7, so the number of times it fits into 10 will be one-third of the number of times 1/7 fits. So it fits 10×7÷3 times. 

Task 19D: We now interpret ⅘÷³⁄₇ as partition or sharing: "⅘ is ³⁄₇ of a share, how much is 1 share?". Here it is less easy to get a sense of what the answer might be, compared to Task 19B where the dividend was a whole number, 10. This might prevent some pupils from knowing what to do - even though we can carry out the same steps as before and, once started, they are probably just as easy to visualise and to discern how the numerator and denominator of ³⁄₇ come into play.

Task 19E: Here we summarise in a more formal way how we evaluated ⅘ ÷ ³⁄₇ in Tasks 19C and 19D.

For item 1, it will be interesting to see whether pupils give an explanation involving quotition or partition; or whether they provide a more formal argument such as this: 1÷⅟₇ = 7÷⁷⁄₇ = 7÷1 = 7.

In item 2, it might help to add an extra step, by rewriting ⅘×7 as ⅘×⁷⁄₁. Alternatively, in item 3 it might help to provide the interim expression ⅘×7÷3, though pupils might still find it difficult to accept that this can be written as ⅘×⁷⁄₃.

FRA18: Three-quarters divided by two-ninths

 

Here we look at ways of performing the division ¾ ÷ ²⁄₉. We adopt two approaches. One is to interpret and to visualise the division as quotition. The other is to transform fractions in ways that produce expressions that are easier to evaluate.

Task 18A: The number line is not essential for the items in this task and some pupils will probably ignore it. However the line can serve as a safety net (?!) for checking whether an answer makes sense.

The first item is straightforward.
In the second item, some pupils might be tempted into writing 2/3; however, this gives us a larger fraction than we started with, which doesn't make sense.
Some pupils might baulk at the third item, others might use a formal procedure or argue that the answer must be the reciprocal of the answer to the first item; of particular interest here is whether pupils can come up with a sensible estimate, perhaps based on a quotitive view of division (How many two-nineths are there in 2?) or a partitive view (If 2/9 of a share is 2, what is one share?).
The fourth item is similar to the third, though pupils who solve it might be puzzled by how we get an answer, 13½, that involves the fraction ½. Where did the half come from? Half of what?!

Task 18B: This is nudging pupils towards the idea of division as quotition:
"How many times does 2/9 go into 3/4?", or
"How many times does the brown line fit into the yellow line?".

To keep things focussed, the emphasis is on estimation rather than on finding the exact answer, though pupils might well go on to attempt that. We can see that the brown line fits more than 3 times but less than 4 times into the yellow line (or indeed more than 3¼ times and less than 3½ times). Note that we haven't numbered the scales on the number line, but that our unit, when it comes to the division, is the length of the brown line.

Digression: The task can be approached in other ways, of course. For example, one could imagine stretching the diagram containing the brown and yellow lines until the brown line is 1 unit long (below), using the scale factor ×⁹⁄₂. The yellow line would then be a little under 3½ units long (actually ¾ × ⁹⁄₂ = 3⅜). In effect, what we are doing here is changing both terms of the expression into something more convenient, by multiplying or dividing both fractions by the same number, on the principle that a/b = na/nb. We pursue this further in Task 18E.

We can also model the above stretch, with its scale factor of ×⁹⁄₂, using a double number line (DNL) such as the one below.

We could also set up a table like this and explore it on the basis that the relationship between the numbers in the columns (and rows) is multiplicative.

Task 18C: Here we work our way, step by step, to the division expression that we are ultimately interested in (¾ ÷ ²⁄₉). We start with a much simpler expression that it is hoped pupils can visualise, and then transform the division in further simple steps. Though these steps are simple in themselves, the situation might well become more abstract for pupils - how well do they cope?

The value of the expression in the first line is 9. (Why?) In the second step we halve that (4½). We then divide by 4 (1⅛). Finally we multiply by 3 (3⅜).

Task 18D: Again, we start with a (very) simple expression and move to our goal in simple steps.

The result of step 1 leaves ¾ unchanged. In step 2 we divide by something 9 times as small so the result will be 9 times as large (27/4 or 6¾). In step 3 we halve the current result (3⅜).

You might want to check whether pupils have understood the methods demonstrated in this and the previous task, by asking them to replicate the methods with a division similar to ¾ ÷ ²⁄₉.

Task 18E: Here we start with the desired division expression and transform it into equivalent expressions by multiplying or dividing both fractions in the expression by the same number. Identifying these transformations is quite challenging.

FRA17: Fration of a fraction of a fraction....

This week's set of tasks offers a continuation of last weeks's, but is probably more accessible. The individual fractions are simpler, but are repeated several times.

Task 17A: Here we make use of the classic 'diamond' method of halving the area of a square - by rotating a given square through 45˚ and shrinking it until it fits tightly inside the original square. The orange region is formed by halving the blue outer square and then halving and halving again. So its area is one eighth of the original square. The lemon region is formed by taking one third of the blue outer square, then taking a third of the resulting strip. So its area is one ninth of the original square.

Task 17B: Here we mix the operations. The area of the orange region is ½×½×⅓×⅓ of its blue outer square. The area of the lemon region is ⅓×⅓×½×½ of its blue outer square. So they are both ¹⁄₃₆ of the blue outer square.

Some pupils might form the above expressions to solve the task, while others solve the task in a step by step manner, by finding the area of successive smaller squares.

Task 17C: We can express the fractions as, say, ⅕ × ⅕ × ½ and ⅟₇ × ⅟₇ respectively. So the lemon region is slightly larger.

It might well appear as though the orange region is larger than the lemon region. In the diagram below we have rotated the region and placed it close to the lemon region. It is a tiny bit smaller!

Task 17D: This time the fractions are the same. We have ⅕ × ⅕ × ¼, say, and ⅟₁₀ × ⅟₁₀, say. So they are both ⅟₁₀₀.

Task 17E: We finish with something more challenging. We can see that the diagonals of the orange squares are 6 units, 5 units, 4 units and 3 units long, respectively. So their areas are half of 6², 5², 4² and 3² square units, respectively. This means that they cover 6/12, 5/12, 4/12 and 3/12 of their respective rectangles (or 1/2, 5/12, 1/3 and 1/4 - though this obscures the pattern in the set of fractions).

Expressed in general terms, the fraction is equal to h/12, where h is the height of the rectangle. Some pupils might have spotted this; other might need to be given time and encouragement! We can derive this algebraically:
the area of the 'diamond' is ½h²
and so the fraction is ½h²/6h = h/12.

FRA16: More of

The first four of these tasks are similar to those of the previous Week, but the diagrams are more curtailed and we don't explicitly mention scaling. A scaling interpretation fits here, but a more 'static' parts-of-a-whole approach would also work.

The first two tasks use the changing length of a bar to model the product of two fractions. When the changing lengths are viewed in terms of scaling, this provides a powerful and quite accessible interpretation of 'fraction of a fraction'. However, as with the corresponding tasks from last Week, it is relatively difficult to use the model to derive the precise value of the product. 

As with last Week, the next two tasks use area to model the product, where it is much easier to find the precise value of the result. On the other hand, if we want to encourage pupils to interpret the product as scaling, then they might find the idea of scaling in one direction and then another, as rather contrived.

Finally, as a coda, we model the fractions as points in the Cartesian plane.

Task 16A: Here 5/7 of a rod is (are?) shaded, and then 2/3 of that is shaded further. We can see that the resulting darker yellow region consists of about half the rod. How could we sub-divide the top and bottom scales on the rod so that the intervals on the top are the same size as those on the bottom?

In the diagram below, the scales along the top from A to B and along the bottom from D to E, have been made the same by subdividing each of the original 5 parts from A to B into 3 equal parts and by subdividing each of the original 3 parts from D to E into 5 equal parts: 5×3 = 3×5. This means 7×3 of these subdivisions would fit between A and C and so each of these is 1/21 of the rod.

The above argument is quite complex but it leads to a relatively simple observation: 7 and 3 are the denominators of the two fractions; their product, 21, gives the number of parts into which the rod has been subdivided.

Task 16B: Here we are multiplying the same two fractions as in Task 16A, but in the reverse order. Will the result be the same?

This time the scales along the top from A to B and along the bottom from D to E, have been made the same by subdividing each of the original 2 parts from A to B into 7 equal parts and by subdividing each of the original 7 parts from D to E into 2 equal parts: 2×7 = 7×2. This means 3×7 of these subdivisions would fit between A and C and so each of these is 1/21 of the rod. So  the product is 10/21.

We can see that the values of ⅔ of ⁵⁄₇ and ⁵⁄₇ of ⅔ are the same, by aligning the rods from Tasks 16A and 16B (below). Pupils will probably not be surprised by this, given their experience of multiplying whole numbers.  However, this doesn't mean that the diagram is particularly transparent. Why do the values turn out to be the same?

The diagram below models the much simpler products ½ of ⅓ and ⅓ of ½. Here it is somewhat easier to see why the products are the same and that their value is ⅙. However, the area model that is used to model a fractions product in the next task is much more transparent.

Task 16C: Here the pieces into which the shaded regions have been partitioned don't need to be partitioned further. Ten of these cover the darker yellow region, so all that remains is to determine their size by seeing how many are needed to cover the whole rectangle. It is relatively easy to see that this is 6 more than the ones shown in the diagram and that the total is given by the product of the fractions' denominators: 3×7 = 21. So ⅔ of ⁵⁄₇ = ¹⁰⁄₂₁.

Task 16D: Here the order of the fractions has been reversed again, but the result is the same.

In contrast to the diagrams for Tasks 16A and 16B involving a rod, in these later two tasks it is easy to see that the results are the same, and why they are the same. In both cases, the dark yellow region and the whole rectangle have been partitioned into the same number of equal parts, given by 2×5 = 5×2 = 10 and by 3×7 = 7×3 = 21, respectively.

Task 16E: This is much more sophisticated. Fractions y/x are represented by the coordinates of points in the Cartesian plane, and equivalent fractions by points on straight lines through the origin. So the fraction ⁵⁄₇ is represented by the point (7, 5) shown in red, and by the equivalent point (21, 15). Two thirds of this fraction is represented by the point (21, ⅔ of 15), or (21, 10) which represents the fraction 10/21.

Task 16E': This is an alternative version where the fraction ⅔ is plotted first. We again end up at the point (21, 10)

Here (below) we can see that the two graphs take us to the same place, the point (21, 10), which represents the fraction 10/21.

FRA15: Stretching

Here we further pursue the idea that multiplication can be thought of as scaling and apply this idea explicitly to expressions like 2⅔ × 2⅘ and ⅔ of ⅘.

Task 15A: Here scaling provides a vivid representation of multiplication, and one that is probably quite accessible to many pupils. However, many pupils will find it very difficult to evaluate the result precisely when the multipliers (or scale factors) are fractional rather than whole numbers.

For Aysha's modelling of 3×2, it is fairly easy to see that doubling the width of the letter A (and the square that surrounds it) and then trebling the new width, results in a letter (and rectangle) that is 6 times as wide as the original.

However, it is not so easy to find the precise value 2⅔ × 2⅘ using Aysha's approach. Before pupils try to do so, it is worth asking them to estimate the value based on the numbers in the expression and based on Aysha's diagram. The value will be greater than 2×2 = 4 and less than 3×3 = 9, and probably closer to the latter. The diagram suggest that the value is slightly less than 7½.

If we consider the scale above and the scale below Aysha’s scaled shape, the 1×2⅘ (or 14/5) mark on the top scale must coincide with the 1 (or 3/3) mark on the bottom scale. So if we split each 5th on the top scale into 3 equal parts, and split each 1/3 interval on the bottom scale into 14 equal parts, then the scales will have the same-size subdivisions: there will be 42 such subdivisions on the top scale between 0 and 2⅘, and the same number on the bottom scale between 0 and 1. These subdivisions will be 15ths (because there are 15 of these parts between 0 and 1 on the top scale). As there are eight 1/3 intervals along the bottom scale, each of which holds 14 of these subdivisions, a total of 8×14 of the subdivisions will fit along the width of the scaled figure and so its width is 8×14×⅟₁₅ = 112/15 = 7⁷⁄₁₅. These same numbers come into play when we evaluate 2⅔ × 2⅘ formally, by writing the mixed numbers as improper fractions:
2⅔ × 2⅘ = ⁸⁄₃ × ¹⁴⁄₅ = (8×14)/(3×5) = 112/15 = 7⁷⁄₁₅.

Task 15B: This tasks involves the same expressions as in Task 15A and again involves scaling, but this time the scaling occurs in directions that are at right angles. This means the results of the scaling are represented by an area which, it turns out, is relatively easy to evaluate.

For the product 3×2, the final scaled shape consists of 3 rows of 2 of Eric's original shape, making 6 such shapes altogether (below, left).
For the product 2⅔ × 2⅘, Eric's final shape can be partitioned into 8×14 small, equal parts (below, right). Fifteen of these parts cover Eric's original shape, so they represent 15ths. 8×14 15ths = 112 15ths = 7⁷⁄₁₅.

A bonus version for my grandson on his birthday.....

Task 15C: This again involves scaling, but this time we use 'of' to indicate 'multiply'. [We could have done this in the previous tasks too, for example by replacing '2⅔ × 2⅘' by '2⅔ lots of 2⅘', or even by '2⅔ of 2⅘' though this sounds a bit odd. Also, 'lots of' hints at repeated addition, rather than a single element being stretched.]

We can 'equalise' the scales above and below Aysha's final shape by splitting each interval in the top scale into 3 equal parts and each interval in the bottom scale into 4 equal parts. Fifteen of these new equal parts fit into the interval from 0 to 1 on the top scale, so each represents one fifteenth. And 2×4 = 8 of these 15ths fit along the width of the final shape, so it is ⁸⁄₁₅ units wide.
Formally, we can write
⅔ of ⅘ = (2×4)/(3×5) = ⁸⁄₁₅.

Task 15D: This involves the same product as in the previous task, but the result is represented by an area (as in Task 15B).

Eric's final shape can be partitioned in 2×4=8 equal shapes, as shown below. It is not too difficult to see (for example by extending the partition-lines) that 3×5=15 of these small shapes would fill Eric's original shape, so the value of ⅔ of ⅘ is (2×4)/(3×5) = ⁸⁄₁₅.

Task 15E: This is similar to Task 15D, but this time one of the multipliers (scale factors) is greater than 1 and the result is also greater than 1, so the result is a mixed number or improper fraction.

Eric's final shape can be partitioned into 16 times 4 small equal parts [or (2×7+2)×4 = 64 small equal parts]. A total of 7×5 = 35 such parts cover Eric's original shape,
so 2²⁄₇ of ⅘ = 64/35 units = 1²⁹⁄₃₅ units.

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FRA14: Splitting and shrinking

Here we continue to focus on the idea of fraction as quotient or division, in a context where we either split an object into equal parts, or where we scale or 'shrink' an object. Scaling provides a powerful way of moving from operating with natural numbers (where multiplication, for example, might be seen as repeated addition) to operating with rationals. However, this is not to suggest that as pupils become more familiar with scaling they will fully abandon earlier ideas or that such ideas will no longer be useful. 

Task 14A: The first diagram below can be thought of as representing the expression 3 ÷ 8, with the second representing (1 ÷ 8) × 3. Pupils who are familiar with the notion of fraction as division, for example through tasks such as those in Week 13, might feel that both expressions are obviously equivalent to 3/8, with nothing more to be said! 

Other pupils might argue that if we collect together the three blue 1/8 regions, as in the diagram below, left, then we can see that they align with the yellow region, so this too represents 3/8 of a 1m rod. 

A more powerful, purely logical argument, goes something like this:
split the yellow region into 3 equal parts, as in the diagram below, right;
do the same to the other seven adjacent regions (or imagine doing so!);
this means that 3×8 = 24 of these small equal parts will fit into the 3m rod;
and so 24÷3 = 8 of these parts will fit into a 1m rod, and so each part is 1/8 of a 1m rod;
and so the yellow region is 3/8 of a 1m rod, as are the blue regions altogether.

Task 14B: This is similar to the previous task. One can again think of the diagrams as representing 8 ÷ 3 and (1 ÷ 8) × 3, or perhaps 1/8 of 3 and 3/8 of 1. Some pupils will be quite comfortable with these expressions and the idea that they are all equivalent to 3/8. However, it is still worth supporting this with the a geometric argument of the sort given earlier, where the orange region is split into 3 equal parts and each part is shown to be 1/8 of a 1m rod.

It might be useful to present the task with diagrams where the orange and green regions are not aligned, as in the layout below. Here we can't so readily see that the two regions are the same and the idea that they might be, could well give some pupils pause for thought.

Task 14C: Here we introduce a shift in viewpoint. Instead of having a 3m rod (or in this case, a strip of elastic) that is cut into equal pieces (7 in this case), we use scaling (or shrinking) to form the smaller piece. The scale factor that maps the stretched strip onto the original strip is × ⅟₇.

Scaling provides a powerful way of interpreting multiplication and division when this involves rational numbers (fractions), especially when these numbers are smaller than one, when ideas such as repeated addition (for multiplication) and sharing (for division) no longer fit very well, and where ideas such as 'multiplication makes bigger' no longer apply. 

However, this does not mean that pupils will every fully abandon these earlier (and more primitive?) ideas as they become more familiar with scaling. There are many situations where these earlier ideas work perfectly well and others where they can coexist with scaling. Regarding the present task, pupils don't have to buy fully into this scaling viewpoint to be able to solve it. They could form a piece the same size as Yella's original strip by cutting a (rigid) version of the 3m strip into 7 equal pieces. This would give a piece of the desired size (3/7 of 1m), albeit with a design that looked slightly different:

Task 14D: Here we use scaling again, but this time the stretched 3m strip is partitioned into two parts, of length 2m and 1m. The scale factor that maps the stretched strip onto the original strip is × ⅟₅.

The original strip is one fifth of the stretched strip's length, so its total length will be 3/5 m and the green and yellow parts will be 2/5 m and 1/5 m respectively.

Task 14E: The scale factor that maps the stretched strip onto the original strip is × ⅟₃. As the stretched strip is more than 3m long, this means the original strip will be greater than 1m, or an improper fraction.

The original strip is one third of the stretched strip's length, so its total length will be 5/3 m (or 1⅔ m) and the red and green parts will be ⅔ m and 3/3 m (or 1m) respectively.