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PostPosted: Thu Sep 20, 2018 6:01 pm 
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Hi all

Would appreciate if anyone could help answer this auetion from the Bancrofts 2017 maths sample paper. My brain can’t seem to think of a solution. We have been looking at it for a long time.

A rectangular garden is surrounded by a path of a fixed width. The perimeter of the garden is 24 meters less than the distance along the outside edge of the path. What is the width of the path?


Picture is a rectangle with another smaller rectangle taken out of it, the outside being the path and the inner being the garden.

Thank you
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PostPosted: Thu Sep 20, 2018 6:04 pm 
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Image

the image


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PostPosted: Thu Sep 20, 2018 6:19 pm 
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I can think of a couple of ways.

Draw a line horizontally to the edge of the black shape from the white rectangle - the length of this is 2 path widths longer than the original shape - can you carry on from this?


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PostPosted: Fri Sep 21, 2018 9:53 am 
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what if you cut the lenth and width of the garden on the out line of path. what do you end up with?
that should be equal to 24.
if you can't find the answer i will let you know.


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PostPosted: Fri Sep 21, 2018 11:47 am 
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First, are you happy to visually see that it doesn’t matter if the garden is a square or a rectangle? If you are it makes the workings easier.

Assume that the outside length of the square path is a.
Assume that the width of the path is b.
The perimeter of the outside of the path is 4 x a
The perimeter of the inside of the path is 4 x (a-2b)
Therefore 4a -4(a-2b) = 24
Expand the brackets and solve for b
4a-4a+8b=24
8b=24
b=3
Path width =b=3


Last edited by Droftaw on Fri Sep 21, 2018 11:52 am, edited 1 time in total.

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PostPosted: Fri Sep 21, 2018 11:48 am 
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You absolutely do not need any algebra for this ...


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PostPosted: Fri Sep 21, 2018 12:34 pm 
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Ok, without algebra. By inspection ... the answer is 3 (joke).

Draw a grid (square or rectangle). Lightly shade in the outermost squares on that grid to look like the picture you posted. Imagine the shaded squares are paving slabs. You have a path which is one square (paving slab) wide. You could look at this two ways now.

1.Contrast the four corner paving slabs with the others slabs. Can you see that it must be these corner slabs that explain the difference in the perimeter of the inside and outside of the path? Each corner has two outside edges and no inside edges. In total the outside has 8 more edges than the inside. We know that this equals 24, therefore each individual edge must be 3, which is the width of the slab and the width of the path

2. Count the number of inside edges and deduct it from the number of outside edges. The answer is 8. The outside has 8 more edges than the inside. We know that this equals 24, therefore each individual edge must be 3, which is the width of the slab and the width of the path.

Whichever method used, the key point is that 8 x width of path = 24


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PostPosted: Fri Sep 21, 2018 12:38 pm 
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This is good example of problem that seems to be beyond KS2 curriculum.
If you know algebra, you will solve it as Droftaw suggested.
There is an easy way to solve it without algebra.


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PostPosted: Fri Sep 21, 2018 12:45 pm 
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Exactly - it is KS2 by inspecting the 'extra' bits at each corner.

There's also no need to assume we have a square - that would be marked wrong I think.


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PostPosted: Fri Sep 21, 2018 1:11 pm 
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Each outside edge of the path is 2 path widths longer than the equivalent outside edge of the garden.

There are 4 edges in a rectangle so the perimeter of the outside of the path is 8 path widths longer than the perimeter of the garden.

So Path width = 24/8 = 3 meters.


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