Week 11: Here we look at the structure of basic equations (or number sentences) of the form A✭?=B or ?✭A=B, where ✭ is one of the four arithmetical operations +, –, × or ÷. We ask pupils to find the missing number with a calculator, by keying-in the given numbers in such a way that the missing number appears as a result in the display. This means in effect transforming the given equation into the form ?=A✩B or ?=B✩A, where ✩ is one of the four operations again. We don't expect pupils to perform such transformations in a formal way, but, rather, to use their developing number sense with regard to the four operations. Pupils are likely to find this easier for equations where the given numbers, and the solution, are whole numbers and relatively small. It then becomes interesting to see how readily pupils transfer this insight to more 'opaque' equations, where it is difficult to find the missing number in any other way.
Though pupils may more readily be able to transform simple ('transparent') equations, they may not see the
need for the
calculator-based 'transformation' approach with such equations, since
they may have informal methods for solving them, such as inspection
or trial and improvment (with or without a calculator). If this is the
case, celebrate the informal methods, but also encourage pupils to find the missing number by using the calculator in
the desired way. The informal solutions
can then serve as a check.
The operations + and x are commutative and it is likely that pupils will generally find the equations involving these operations easier to solve in the desired, calculator way, than those involving – and ÷.
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Monday: Part a) can be solved fairly easily by adding on, with or without a calculator. But what about part b) ?!
We can estimate the size of the missing number in part b) by adding on - it's about 60,000! But it would be extremely tedious to find its exact value this way. On the other hand, we can find it directly (and more or less instantly if we use a calculator) by performing the calculation 893774 – 829177. This switch in perspective may not come easily with these numbers, so it is worth returning to part a) to try to elicit the switch there, where it is more easily 'visualised'.
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Tuesday: Parts a) and b) can be solved fairly easily without a calculator, but it is worth finding a calculator method that can then help with parts c) and d).
Subtraction is not commutative, so parts a) and b) involve different missing numbers, as do parts c) and d). Pupils might well solve part a) by seeing it as 92 + ? = 126 and thus 'adding on' (92+
8+
26 = 126); it might be more of a challenge to see it as ? = 126–92, which would help with part c). It is possibly easier to write part b) in the desired form, ? = 92+126, which would help with part d).
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Wednesday: Part a) is straightforward and can be done without a
calculator. What does the answer tell us about the answer to part b)?
The answer to part b) will be less than the answer to part a), so less than 3. If pupils spot this, some will say the answer is 'impossible' since the next (whole) number less than 3 is 2 and 32×2 doesn't get us to 90. Other pupils might now set about finding the missing number using trial and improvement, by feeding expressions like 32×2.9 into the calculator. The chances of finding the actual value, 2.8125, in this way are fairly remote, which is a strong incentive for finding our desired calculator method. As before, it might help pupils in this search to look again at part a).
Part c) is there to underline the fact that multiplication is commutative.
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Thursday: Here we continue with tasks involving "32×?", but where the solution hovers around 1. Pupils might be able to spot (or home in on) the unknown number in part a) but that becomes increasingly dificult in the other two parts.
Some pupils might be able to find the number in part a) by trial and improvement, or by a grounded argument like '40 = 32 plus a quarter of 32', leading to 1.25. This becomes harder in part b) [38 = 32 plus 3/16 of 32] and also very challenging to express as a decimal. It is much 'simpler' to use the inverse of multiplication here (? = 38÷32) but pupils might find this challenging too, as it is not easy to visualise with the given numbers. Part c) will involve a further challenge for some pupils in that we are multiplying by a number less than 1.
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Friday: These tasks involve division rather than multiplication, so it will be interesting to see whether this makes it more or less challenging for pupils to 'visualise' the operations and transform the equations.
The solution to part a) is a small whole number (7), so pupils might well be able to find it by trial and improvement. Some pupils might transform the task into 91 = ?×13 and then use trial and improvement or some kind of 'inspection' approach.
Part b) underlines the fact that division is not commutative. Pupils should be encouraged to 'visualise' the division (for example, alsong the lines of '91 people share some apples, and get 13 apples each'). This can help pupils see that the number is going to be 'large' (much bigger than 91!) and perhaps even that it will be 91×13 ('91 lots of 13 apples').
Part c) is related neatly to part a) in that 52 is 4 times 13, and so the solution is a quarter of the solution to part a). Some pupils might spot this and use it as their initial method or as a check to the calculator method, ? = 91÷52. Pupils who don't spot this might noetheless notice that 52 is relatively close to 91 (more than halfway...) and so the solution will be less that 2, which provides another useful check.
