MONDAY: We start with a relatively simple task, where one can see that the result is 241, if one pauses to examine the given expression. Nonetheless it is likely that some pupils will perform the two multiplications shown in the expression and then find the difference. This is fine at this stage: a class discussion will soon reveal that there is a much simpler (and ultimately quicker) way!
TUESDAY: Like Monday's task, the numbers here are contrived to give a neat method of evaluating the given expression. We use the artefact of a conversation to focus attention on this method. However, you might want to present the task without the dialogue and instead alert pupils to the fact that there happens to be a neat solution - can they spot it?
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WEDNESDAY: Here we build on Tuesday's method, which is based on the property that multiplication is distributive over addition. But we add an extra twist....
It is worth emphasising that the tasks we present in this blog are only examples of what you might do. You might decide to give more than one task of a given kind, or to devise variants that differ to a lesser or greater extent from the original task.
For this task, some pupils might not spot, or be receptive to, the key idea that 3×114 is equal to 6×57. If so, you might want to give them more straightforward tasks, like Tuesday's, at least to begin with.
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THURSDAY: This task again rests on the distributive law, though the fact that it involves division rather than multiplication suggests that pupils may find it more demanding.
[Note 1: Streefland has a nice context for this, which he calls 'French division', where people are seated round a restaurant table and share pancakes as they arrive in batches, rather than waiting till they've all been cooked. Our numbers are rather large here, but sharing 48 pancakes and then 52 pancakes fairly among 4 people results in the same amount each as sharing the 100 pancakes in one go.]
[Note 2: while it is true that (A+B)÷C = A÷C + B÷C, it is not true that A÷(B+C) = A÷B + A÷C.]
Some pupils might evaluate the first expression by carrying out both divisions and adding the results (12+13). However this approach is far less straighforward for the second expression, which might prompt pupils to look for another approach. In turn they might spot that we can re-structure the expression as (49+51)÷4, ie as 100÷4.
You might want to make the task more demanding still (or challenge pupils to do so), eg by changing 49 and 51 to, say, 49.1 and 50.9. Another, variant is shown below. This is likely to be more challenging than the original task, as it seems less easy to visualise.
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FRIDAY: Some pupils might not be used to estimating and might thus find part a) quite challenging. What kind of answer, ie what level of accuracy is 'acceptable'? They might also find it difficult to envisage the effect of multiplying by the non-whole number 1.25, even if they have a procedure for doing so.
It will be interesting to see whether any pupils can come up with an argument of this sort:
1.25 is much nearer to 1 than to 2, so 1.25×32 will be much nearer 32 than 64.
Here are some methods for parts b) and c). What others do pupils come up with?
1 plus ¼ of 32 is 32 plus 8 = 40
1.25×32 = 1.25×4×8 = 5×8 = 40
1.25×32 = 1.25×8×4 = 10×4 = 40
1.25×32 = 0.125×320 = 1/8 of 320 = 40.
