Week 12: Hidden fractions

This is essentially about fractions, though we don't mention them by name! We use the number line to focus on the measurement and ratio aspect of fraction rather than part of a whole.

Task 12A: This gives us an opportunity to see what strategies children use to estimate parts of a line. We use a line with a large number of divisions to discourage children from simply counting (though they can do so to check their estimates) and to convey the idea that we don't expect the estimates to be spot on. We have chosen numbers where the use of benchmarks can be helpful. For example, the 18 mark will be just under halfway along the line segment, 12 will be just over a quarter (or under one third), 25 is over one half but less than three quarters, say. Some children might spot that 25 is 5/8 of the way along and, similarly, that 30 is three quarters along the 0 to 40 segment.

Task 12B: We don't require anything precise here, beyond the fact that the 13 mark is more than halfway along its line segment (actually quite close to 2/3) and that 23 is slightly less than halfway along its segment - so 13 is nearer the right-hand end of its segment. Item 2 attaches a story to the same numbers - do children appreciate that it makes sense to use ratios to compare the two types of bean rather than the differences in the number of germinating and non-germinating beans? In other words, do they appreciate that it is more appropriate to use multiplicative rather than additive reasoning?

Some children might express the relations using fractions or, more likely perhaps, decimals or percentages. If it seems appropriate, you might want to discuss the idea that comparing 13-out-of-20 and 23-out-of-50 is essentially about division: 13÷20 is greater than 23÷50.

Task 12C: Again, we don't expect a very precise answer here (the red arrow is actually pointing at 35.2), but do children at least appreciate that the red arrow is pointing more than half way along the line segment?

Some children might argue that since 11 is 1 more than half of 20, the red arrow will be pointing at 1 more than half of 64 (that is, 33). This is quite a sophisticated answer, though it involves additive rather than multiplicative reasoning. The gap between the red arrow and the halfway mark, 32, will actually be just over 3 times 1 (since 64 is just over 3 times 20).

Task 12D: The 3 mark is further along the 0 to 19 line segment than along the 0 to 20 line segment: 3 out of 19 is greater than 3 out of 20. However, some children will think the opposite, especially if they are focussed on 'parts of a whole': 3 out of 19 involves fewer parts so will be smaller.

 

Task 12E: The red arrow is pointing at the 15 mark. How close do children get to this? Do they at least chose a number below 20 (half of 40)?

Some children will recognise that we are in effect dealing with the equivalent fractions 3/8 and 15/40. And/or they might see that numbers on the bottom line are 5 times the corresponding numbers on the top line. Some children might think of splitting the top line into 8 equal parts, perhaps by halving, halving and halving again. Doing the same on the bottom line reveals the numbers 20, 10 and then 15.

Week 11: Another SAT interlude

We look at some more SAT items taken from Paper 1 of the Key Stage 2 National Curriculum Mathematics tests from 2025. As before, we ask children to solve a given SAT item (in any way they choose) and then look at some variants that explore some of the mathematical relationships engendered by the item.

Task 11A: The SAT item here is Question 8 from the 2025 Paper 1 test. The most obvious way to solve it is to evaluate 12×3 and then multiply the result by 10, which we can express like this:  (12×3)×10. Variant 1 makes this explicit. Variant 2 brings out the fact that we could also solve the item by evaluating 3×10 and then multiplying by 12: 12×(3×10). This works because multiplication obeys the associative law: (A✯B)✯C = A✯(B✯C). Addition obeys this law too.

Multiplication is also commutative, so (12×10)×3 would be another way to solve the item, which is a variant that children might well come up with.

Variant 3 is perhaps more challenging. Where did the 24 come from?! One explanation would be to say we have doubled 12 and halved 10. Here is a more formal way of explaining it:
12×3×10 = 12×3×2×5= 12×2×3×5 = (12×2)×(3×5) = 24×15.

A systematic way to generate more variants like Variant 3 is to write the original expression as a product of prime factors: 12×3×10 = 2×2×3×3×2×5.
The factors can then be reordered and combined in a variety of ways.

Task 11B: The SAT item here is Q9 from the 2025 Paper 1 test. The missing term in the given equation is 54. Variants 1 and 2 have each been formed by adding or subtracting a given amount to/from the right hand side of the equation and to/from the first term on the left hand side. So in each case the second term on the left hand side of the equation remains the same: 54.

Variant 3 is rather different: 326 and 380 have both been halved. What happens to 54?

Task 11C: There is a very simple rule that can be used to solve this SAT item (Q11): 'multiply the numerators, multiply the denominators', or 'multiply tops and bottoms'! The aim of the task is to give some meaning to what is going on, by using a number line to model the fractions. This requires some quite careful thought:
we can think of 5/6 as being 5/6 of the distance from 0 to 1; however, the fraction 2/5 is operating on this 5/6 distance, not on the 0 to 1 distance. The diagram shows that there are 5 equal intervals from 0 to 5/6. 2/5 of these 5 intervals is 2 intervals, which is 2/6 of the way from 0 to 1.

The distance that is half of the distance from 0 to 1, can be though of as 3 of the 6 given intervals from 0 to 1; this is also 3 of the 5 intervals from 0 to 5/6 which is 3/5 of the distance from 0 to 5/6:
3/5 of 5/6 = 1/2.

Task 11D: The SAT item here (Q14) is testing children's knowledge of the long multiplication algorithm. Michael Gove and Dominic Cummings had to know this algorithm when they were at school; in 2014 they declared that our young children should have to know it too. The algorithm rests on the fact that multiplication is distributive over addition. So instead of trying to multiply 614 by 32 in one go, we split 32 into 30+2 and multiply by 30 and by 2 separately. Formally, we write this: 614×32 = 614×(30 + 2) = 614×30 + 614×2.

Variant 1 is a slightly longer form of the standard algorithm: it shows explicitly that we are multiplying by 30 and by 2. How aware are children that this is what the algorithm is actually doing?!

Variant 2 underlines the fact that we are splitting the multiplier 32 into smaller parts. We normally do this by splitting the number into its 10s component and 1s component, but we can spit it in other ways, for example as here: 32 = 20+12. We should get the same answer!

Variant 3 uses a different idea which takes us back to the associative law:
32 = 4×8 and so 614×32 = 614×4×8 = (614×4)×8.

Task 11E: This SAT item (Q15) is rather neat, though I don't know the rationale for including it in the Paper 1 test. Are children expected to use the 'bus stop' algorithm? (4 into 2 doesn't go; 4 into 20 goes 5, remainder 0; 4 into 0 is 0; 4 into 0 is 0; answer 500.) Or do they think of 2000 as 100×20 as in Variant 2; or do they visualise 2000 split into 4 equal parts as in Variant 3?