FRA05: Common parts

Here we continue with the notion of fraction as measurement. We compare two rulers split into different size intervals and consider whether we can split the intervals into equal 'mini-parts' that are the same size on the two rulers. In effect we are looking at common denominators.

Task 05A: Some pupils might be able to solve this by using ¼ of ⅕ = 1/20, or ⅕ ÷ 4 = 1/20. A more grounded approach is to consider what would happen if Andy splits all his 5ths into 4 equal parts: there would be 5 lots of 4 mini-parts in an inch, so each is one 1/20 of an inch. Similarly, Beth would have 3×7 = 21 mini-parts in an inch, so each is 1/21 of an inch. So Andy's min-parts are slightly wider than Beth's.

[Note: We could have modified the task to show all of Andy's 5ths split into 4, which would have allowed pupils just to count how many mini-parts there are in an inch. The virtue of the current form of the task is that it requires pupils to find ways of structuring the task themselves (unless they are permanently stuck!) and to think more deeply about the relationships involved - it can also us give us more insight into pupil's thinking.]

Task 05B: There are 6×4 = 24 of Amy's mini-parts to an inch. Ben could have the same number if he split each of his 8ths into 24÷8 = 3 equal parts.

Task 05C: This might be more challenging for some pupils, but it should broaden pupils' understanding, even if they are already conversant with common denominators.

If Alma splits all here quarter intervals in 2, 3, 4, 5, 6, 7, 8, .... equal parts, she would get 8, 12, 16, 20, 24, 28, 32, .... equal parts in an inch. Similarly, Bosco could end up with 20, 30, 40, 50, .... equal parts to an inch.
So they could end up with the same width mini-parts if Alma splits a quarter into 5 equal parts and Bosco splits a tenth into 2 equal parts (4×5= 20 = 10×2). Or they could split their respective intervals into 10 and 4 equal parts (4×10= 40 = 10×4), or 15 and 6 equal parts, or 20 and 8, etc. In each case, the resulting number of min-parts in an inch is a multiple of 4 and 10.

Task 05D: Pupils might find this a lot more challenging, though it might click if they are given time!

It turns out that 6 of the 'red intervals' will fit exactly into one of Alan's 5ths, and that 5 of these will fit exactly into one of Beth's 6ths. (Can pupils explain why?)

Thirty of the red intervals would cover an inch (5×6 = 30 = 6×5), so the distance between the original two red lines is one 30th of an inch.

We can fit 4 red intervals between the green lines, so they are 4/30 of an inch apart (or 2/15).  

Task 05E: This is slightly more complex than Task 05D. It might help pupils consolidate their understanding and perhaps broaden it.

 

We can split one of Aysha's 5ths into 7 equal parts, exactly 5 of which will cover one of Bob's 7ths. 35 of these parts will cover an inch, and two of them cover the 'red interval'. So the distance between the two red lines is 2/35 of an inch.

We can 'see' (!) that there are 5+5+5+4 = 19 of the small intervals between the green lines, so the distance between the green lines is 19/35 of an inch (which is just over half an inch).