HSC Standard Mathematics Predictions / Thoughts (1 Viewer)

erucibon

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Did you guys get an email from NESA saying that they apologise for making it too hard and will consider that in scaling?
 

mmmm.

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oh I saw that too but I'm pretty sure it's just an email from a school's maths department like it says at the bottom to their students
Oh ok, i think that @erucibon might mean how they are going to change scaling

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L C

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why is it only on that website though? that doesn't seem right...
 

throwaway2318

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brent012

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Did you guys get an email from NESA saying that they apologise for making it too hard and will consider that in scaling?
The point of that email shown in this thread would be that if an exam is genuinely dificult and found challenging by most of the cohort, the scaling process will accomodate that. Not that any special scaling or aligning is being performed.
 

PC

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I used the table to find the total value of the first 20 days and add it to the next 10 (or 11?) days to get a total. Then i think i went to far by trying to turn it to a present value. I should get some working marks though.
I worked through this one on a spreadsheet. I know the answer is $34 486.50. But who knows how to get there using the PV table? It's doable for the Advanced guys with GPs, but this was in the Standard paper only.
 

DaBossMan

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I worked through this one on a spreadsheet. I know the answer is $34 486.50. But who knows how to get there using the PV table? It's doable for the Advanced guys with GPs, but this was in the Standard paper only.
Yeah I was actually wondering if that table was there to confuse us and wasn't even necessary. I doubt that is the case but I had no idea how to do that one.
 

quickoats

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I worked through this one on a spreadsheet. I know the answer is $34 486.50. But who knows how to get there using the PV table? It's doable for the Advanced guys with GPs, but this was in the Standard paper only.
That question did not require GPs. The PV table is to be interpreted to get the correct answer.
 

PC

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That question did not require GPs. The PV table is to be interpreted to get the correct answer.
Cracked it. Answer is $34486.

Short explanation.
Imagine if you were making withdrawals of $3000 for the whole 30 years.
PV = 3000 x 22.396 = $67188

But we're not. So now remember that the first 20 are only $1000, which is $2000 less.
For this difference, PV = 2000 x 16.351 = $32702

So total amount to invest = 67188 - 32702 = $34486.

But, seriously ...
 
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Riproot

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Why was that exam so hard. It was harder than all the practice paper I did. Can someone pls tell me what they got for the cricket question
χ = 22 - 0.525
= 21.475’C
ÿ = 684/20
= 34.2 chirps / 15s
(χ,ÿ) = (21.475 , 34.2)

Least squares regression line passes through these points is:
y = -10.6063 + bx
Hence
34.2 = -10.6063 + 21.475*b
44.8063 = 21.475*b
b = 2.086440046565774

Hence for point (19, y):
y = -10.6063 + 2.086440046565774*19
y = 29.03606088474971
Therefore
29 chirps in 15 seconds at 19’C
1603819654206.png1603819654243.png
 

PCNQQn2

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I worked through this one on a spreadsheet. I know the answer is $34 486.50.
I did exactly the same as you, and also had difficulty getting there with the PV table. I think in all the discussion about the crickets and the decagon that the complexity of this question has been lost a bit.

A teacher at my school came up with the following method, which is slightly different to yours and results in $34486.90, only 40c different from a spreadsheet model. The table is only good for 4 or 5 significant figures, so some errors are to be expected.

1. Consider the last 10 years. To withdraw $3000 per year you need to have 8.983*3000 = $26949 at the start of Year 21.
2. Now consider the first 20 years. To withdraw $1000 per year you need 16.351*1000 = $16351 at the start of Year 1. But you also need enough money at the start of Year 1 to cover the $26949 you need in 20 years' time. Let that be M. It compounds at 2% per year, so M*1.02^20=26949, so M=26949/(1.02^20)=18135.9.
3. Putting this together you need 16351+18135.9=$34486.90 at the start.
 
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DaBossMan

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I did exactly the same as you, and also had difficulty getting there with the PV table. I think in all the discussion about the crickets and the decagon that the complexity of this question has been lost a bit.

A teacher at my school came up with the following method, which is slightly different to yours and results in $34486.90, only 40c different from a spreadsheet model. The table is only good for 4 or 5 significant figures, so some errors are to be expected.

1. Consider the last 10 years. To withdraw $3000 per year you need to have 8.983*3000 = $26949 at the start of Year 21.
2. Now consider the first 20 years. To withdraw $1000 per year you need 16.351*1000 = $16351 at the start of Year 1. But you also need enough money at the start of Year 1 to cover the $26949 you need in 20 years' time. Let that be M. It compounds at 2% per year, so M*1.02^20=26949, so M=26949/(1.02^20)=18135.9.
3. Putting this together you need 16351+18135.9=$34486.90 at the start.
Did anyone on here that done the standard exam get the correct answer to that?
 

PC

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1. Consider the last 10 years. To withdraw $3000 per year you need to have 8.983*3000 = $26949 at the start of Year 21.
2. Now consider the first 20 years. To withdraw $1000 per year you need 16.351*1000 = $16351 at the start of Year 1. But you also need enough money at the start of Year 1 to cover the $26949 you need in 20 years' time. Let that be M. It compounds at 2% per year, so M*1.02^20=26949, so M=26949/(1.02^20)=18135.9.
3. Putting this together you need 16351+18135.9=$34486.90 at the start.
That's a pretty good solution too. I think that was where I was trying to go at first, but just stuffed up the compounding.
 

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