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junior college 1 | H1 Maths
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Then based on the ratio, the length is 5/4 x cm.
Let the number of enclosures be n.
Total area of all enclosures
= n(length x width of 1 enclosure)
= n(5/4 x cm )(x cm)
= 5/4 nx² cm²
So 5/4 nx² = 1440 ①
The fencing includes the partitions between each enclosure and the outline of the entire structure.
For each enclosure, there are 2 widths. So there are 2n widths worth of fencing to be counted in total. 2n widths means the fencing used for all the widths is 2nx cm.
However, some partitions (lengths) are shared/common. We cannot just count two for each enclosure as that would result in over counting.
How to count them :
The right length of each enclosure is the left length of the next one, except the last enclosure(rightmost one), whose right length is unshared.
So we don't need to count in the right lengths. We just count the left length for each enclosure.
There are n left lengths for n enclosures. adding the last right length that is unshared, there are a total (n + 1) lengths to be counted in our fencing.
An easier way to see this : (from diagram)
When there's 1 enclosure, there are 2 lengthwise fences.
Where's there's 2 enclosures, there are 3.
When there's 3 enclosures, there are 4.
So number of lengthwise fences for n enclosures = (n + 1)
The fencing used for these
= (n+1)(5/4 x) cm
Total fencing = ( 2nx + (n + 1)(5/4 x) )cm
= ¼x(13n + 5) cm
so ¼x(13n + 5) = 327 ②
5/4 nx² = 1440 ①
¼x(13n + 5) = 327 ②
From ①,
5/4 nx² = 1440
nx² = 1440 x 4/5 = 1152
n = 1152/x² ③
Sub ③ into ②,
¼x(13(1152/x²) + 5) = 327
x(13(1152/x²) + 5 ) = 1308
14976/x + 5x = 1308
Multiply throughout by x,
14976 + 5x² = 1308x
5x² - 1308x + 14976 = 0
(5x - 1248)(x - 12) = 0
5x = 1248 or x = 12
x = 1248/5
= 249.6
(N.A, x must be a positive integer. We can't have partial enclosures.
(Not easy to get this factorisation, so you can opt to use quadratic formula or completing the square instead. Or use calculator's quadratic function to check first then do working
n = 1152/(12²) = 8
8 enclosures need to be built.