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.
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 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?
