In this week's tasks we are given a simple arithmetic expression involving an unknown number and are asked to guess the number's value which we then test, usually with a calculator. The aim is to encourage children to interpret expressions and to estimate, rather than rush, unthinkingly, into performing calculations.
[NOTE: The calculator has dangers as well as strengths: each time children reach for their
calculator, ask them to justify the key sequence they are planning to
enter; discourage them from entering sequence after sequence with very
little thought.]
Task 04A: Here children are asked to search for a number whose square is 70. To start with, and to keep things simple, we at first restrict the search to whole numbers (Method 1) before moving on to simple decimals, ie decimals involving just one decimal place (Method 2). Finally children are asked to calculate the number (as accurately as their calculator will allow) by using the calculator's square root key.
The suggestion that children use a calculator in Method 1 can be ignored for the given version of the task. However, it can be useful should you decide to replace 70 with something much larger, as in the examples below, where it could be a distraction to calculate G x G 'by hand' (unless you are keen for this to become the focus of the task!).
There is plenty of scope for investigating this situation further by replacing the 80 or the 3, or both, by smaller or larger numbers. You might encourage children to come up with some variants, or you might ask questions like this:
What will happen to G if we replace 80 by something larger?
Task 04D: The first part of this task is fairly easy, if children know their 8 times table! 8×12 = 96, which is only 3 away from 99, so 8×12 is closer to 99 than 8×13. So, as with earlier tasks, we don't necessarily need a calculator for Method 1, unless we vary the task by changing the given numbers. In Method 3 we can calculate G using G = 99÷8. Do children spot this?
Task 04E: The given equation here might look daunting at first, but a nice feature of the task is that the arithmetic is fairly simple - for most of the process it just involves doubling and subtracting with whole numbers. On the other hand, using 'trial and improvement' with an equation where the unknown appears on both sides can be challenging: it is not always easy to interpret the result of a trial - should I try a bigger or smaller number next?
[Note: Many of the equations that children meet at school can be solved quite easily by trial and improvement or by other informal methods*. For example an equation like 14/(e+2) = 2 can quite easily be solved using the cover-up method: 14 divided by something equals 2; ah, 7; something plus 2 equals 7; ah, 5. That is why equations with the unknown on both sides, even 'simple' ones like 3e + 5 = e + 13 are particularly useful for teaching formal methods. The given equation G = 100–2G can be solved formally in just a few steps: adding 2G to both sides gives 3G=100; dividing both sides by 3 gives G=100/3 = 33⅓.
* I wrote about this in a paper called 'Quantitative and formal methods for solving equations', in the journal Mathematics in School (Volume 12, 5, November 1983)]
It is possible that some children might spot that G must be 1/3 of 100, perhaps by visualising an image like this:
It is worth noting the generative power of a model of this sort. Even if a child doesn't hit upon an image like the one above straight away, it might well emerge from an attempt to produce one. Imagine that a child makes an initial guess of G=20 and makes a sketch like this of the result:
Some children might arrive at a similar conclusion by formally transforming the equation G = 100 – 2G by adding 2G to both sides.
You might want to try other variants of the given equation, such as G = 100 – 3G. In the variant below it is relatively easy to spot what fraction G is of 100, namely a half:
