# Thread: S1/17 WAM Predictions

1. ## Re: S1/17 WAM Predictions

Originally Posted by Zoinked
So you're saying no combination of numbers can be divided by three to give a decimal place of .25 or .75? lol.
yeh

2. ## Re: S1/17 WAM Predictions

Originally Posted by Zoinked
So you're saying no combination of numbers can be divided by three to give a decimal place of .25 or .75? lol.
They have to be whole numbers...

3. ## Re: S1/17 WAM Predictions

Originally Posted by Zoinked
So you're saying no combination of numbers can be divided by three to give a decimal place of .25 or .75? lol.
The total of n integers will have a remainder of either 0, 1, 2, 3, …, or (n-1) upon division by n. This implies that the possible fractional parts for the average of n whole numbers are just: 0, 1/n, 2/n, 3/n, …, (n-1)/n.

E.g. If 1 subject: only can have decimal of 0.

If 2 subjects: only can have decimal of 0 or 0.5 (i.e. 0 or 1/2).

If 3 subjects: only can have decimal of 0, 0.333..., or 0.666..., (i.e., 0, 1/3, or 2/3).

If 4 subjects: only can have decimal of 0, 0.25, 0.5, or 0.75 (i.e. 0, 1/4, 2/4, or 3/4).

Etc.

Furthermore, if 0 < f < 1 and the smallest positive integer for which f occurs as a possible fractional part in the average is n0, then f can occur as a fractional part in the average of N subjects (N a positive integer) if and only if N is a (positive) multiple of n0 (easy exercise).

So for instance, a fractional part of 0.25 in the average means the no. of subjects used in the average must have been some multiple of 4, since 4 is the smallest positive integer for which 0.25 can occur as a fractional part.

(Assumptions: weights in the average are equal and subject marks are whole numbers. If one or more of these assumptions is violated, then other decimal values may be possible.)

4. ## Re: S1/17 WAM Predictions

Originally Posted by InteGrand
The total of n integers will have a remainder of either 0, 1, …, (n-1) upon division by n. This implies that the possible fractionals parts for the average of n whole numbers are just: 0, 1/n, 2/n, 3/n, …, (n-1)/n.

E.g. If 1 subject: only can have decimal of 0.

If 2 subjects: only can have decimal of 0 or 0.5 (i.e. 0 or 1/2).

If 3 subjects: only can have decimal of 0, 0.333..., or 0.666..., (i.e., 0, 1/3, or 2/3).

If 4 subjects: only can have decimal or 0, 0.25, 0.5, or 0.75 (i.e. 0, 1/4, 2/4, or 3/4).

Etc.

So for instance, a fractional part of 0.25 in the average means the no. of subjects used in the average must have been 4 or a multiple of 4.

(Assumptions: weights in the average are equal and subject marks are whole numbers. If one or more of these assumptions is violated, then other decimal values are possible.)
Legend

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5. ## Re: S1/17 WAM Predictions

Originally Posted by matchalolz
Lol is this infs? I think I flopped that shit :')
I think it's been 2 law subjects (based on speaking with my law mates) that've been added so far for me rather than 1 law + INFS1602.

6. ## Re: S1/17 WAM Predictions

Actually, the WAM reflected on the stream declaration still isn't fully revised right? Like, it still may not include every courses you have taken in this semester correct?
That WAM is sending me some vibe that I might fail ECON1203... -_-

7. ## Re: S1/17 WAM Predictions

Originally Posted by BHLee
Actually, the WAM reflected on the stream declaration still isn't fully revised right? Like, it still may not include every courses you have taken in this semester correct?
That WAM is sending me some vibe that I might fail ECON1203... -_-
You can figure this out with some calculating but yes, for most people it is likely that not all courses have been added yet.

8. ## Re: S1/17 WAM Predictions

Originally Posted by BHLee
Actually, the WAM reflected on the stream declaration still isn't fully revised right? Like, it still may not include every courses you have taken in this semester correct?
That WAM is sending me some vibe that I might fail ECON1203... -_-
As RoT explains it may not be fully revised but if you go back a page I think we have some evidence that ECON1203 has been entered.

9. ## Re: S1/17 WAM Predictions

any ideas about MATH2011, MATH2501 and MATH2801? My WAM has not changed and I find it highly unlikely i failed anything.

10. ## Re: S1/17 WAM Predictions

Originally Posted by InteGrand
The total of n integers will have a remainder of either 0, 1, …, (n-1) upon division by n. This implies that the possible fractionals parts for the average of n whole numbers are just: 0, 1/n, 2/n, 3/n, …, (n-1)/n.

E.g. If 1 subject: only can have decimal of 0.

If 2 subjects: only can have decimal of 0 or 0.5 (i.e. 0 or 1/2).

If 3 subjects: only can have decimal of 0, 0.333..., or 0.666..., (i.e., 0, 1/3, or 2/3).

If 4 subjects: only can have decimal or 0, 0.25, 0.5, or 0.75 (i.e. 0, 1/4, 2/4, or 3/4).

Etc.

Furthermore, if 0 < f < 1 and the smallest positive integer for which f occurs as a possible fractional part in the average is n0, then f can occur as a fractional part in the average of N subjects (N a positive integer) if and only if N is a (positive) multiple of n0 (easy exercise).

So for instance, a fractional part of 0.25 in the average means the no. of subjects used in the average must have been some multiple of 4, since 4 is the smallest positive integer for which 0.25 can occur as a fractional part.

(Assumptions: weights in the average are equal and subject marks are whole numbers. If one or more of these assumptions is violated, then other decimal values may be possible.)
Lol didnt realise that you could only get a whole number

11. ## Re: S1/17 WAM Predictions

Originally Posted by Zoinked
Lol didnt realise that you could only get a whole number
I'm not sure how it works at MQ, but at UNSW whole number marks is the norm and an effort will be made to round marks up where necessary. This may be different at MQ due to the use of the GPA and a focus on whether or not you got HDs/DNs/CRs instead of a number, but you should keep this into consideration when judging WAMs.

12. ## Re: S1/17 WAM Predictions

Ok guys would love some help here. Old wam was 73.958 with 24 subjects counted. New wam is (using wam glitch) 73.462. Anyone know how many subjects have been added from this semester for me so far?

13. ## Re: S1/17 WAM Predictions

Originally Posted by Randox
Ok guys would love some help here. Old wam was 73.958 with 24 subjects counted. New wam is (using wam glitch) 73.462. Anyone know how many subjects have been added from this semester for me so far?
I will make the assumptions that all subject marks are whole numbers and all subjects are equally weighted. (If these assumptions don't both hold, you would have to provide us with what assumptions to use I suppose.)

The fractional part of the new Average is 0.462 (at least to three decimal places). This implies the total no. of subjects N (which is at least 25) cannot be 25, since the possible fractional parts then are precisely of the form k/25 (k an integer from 0 to 24), which would make the fractional part be a multiple of 1/25 = 0.04, and hence have a third decimal place of 0.

Note that 12/26 = 0.46153..., which is 0.462 to three decimal places, so 26 is a candidate for the total no. of possible subjects used.

Play around with your calculator on other numbers like 27 and 28 and see what the fractional parts around 0.4 are like for those numbers (so numbers of the form k/27 or k/28 that just less than 14/27 or 14/28), and see if any are 0.462 to three decimal places. If none are, 26 is probably the total no. of subjects used in the new Average.

14. ## Re: S1/17 WAM Predictions

y'all just wait until results come out

if u put this much effort into studyin y'all wouldn't have to worry about analysing ur damn WAM

15. ## Re: S1/17 WAM Predictions

Originally Posted by InteGrand
The total of n integers will have a remainder of either 0, 1, 2, 3, …, or (n-1) upon division by n. This implies that the possible fractional parts for the average of n whole numbers are just: 0, 1/n, 2/n, 3/n, …, (n-1)/n.

E.g. If 1 subject: only can have decimal of 0.

If 2 subjects: only can have decimal of 0 or 0.5 (i.e. 0 or 1/2).

If 3 subjects: only can have decimal of 0, 0.333..., or 0.666..., (i.e., 0, 1/3, or 2/3).

If 4 subjects: only can have decimal of 0, 0.25, 0.5, or 0.75 (i.e. 0, 1/4, 2/4, or 3/4).

Etc.

Furthermore, if 0 < f < 1 and the smallest positive integer for which f occurs as a possible fractional part in the average is n0, then f can occur as a fractional part in the average of N subjects (N a positive integer) if and only if N is a (positive) multiple of n0 (easy exercise).

So for instance, a fractional part of 0.25 in the average means the no. of subjects used in the average must have been some multiple of 4, since 4 is the smallest positive integer for which 0.25 can occur as a fractional part.

(Assumptions: weights in the average are equal and subject marks are whole numbers. If one or more of these assumptions is violated, then other decimal values may be possible.)
Lol I think only people on BOS would have debate over this lol

16. ## Re: S1/17 WAM Predictions

Originally Posted by InteGrand
I will make the assumptions that all subject marks are whole numbers and all subjects are equally weighted. (If these assumptions don't both hold, you would have to provide us with what assumptions to use I suppose.)

The fractional part of the new Average is 0.462 (at least to three decimal places). This implies the total no. of subjects N (which is at least 25) cannot be 25, since the possible fractional parts then are precisely of the form k/25 (k an integer from 0 to 24), which would make the fractional part be a multiple of 1/25 = 0.04, and hence have a third decimal place of 0.

Note that 12/26 = 0.46153..., which is 0.462 to three decimal places, so 26 is a candidate for the total no. of possible subjects used.

Play around with your calculator on other numbers like 27 and 28 and see what the fractional parts around 0.4 are like for those numbers (so numbers of the form k/27 or k/28 that just less than 14/27 or 14/28), and see if any are 0.462 to three decimal places. If none are, 26 is probably the total no. of subjects used in the new Average.
Thanks for this. I think it's 26. So 2 counted

17. ## Re: S1/17 WAM Predictions

Originally Posted by Queenroot
y'all just wait until results come out

if u put this much effort into studyin y'all wouldn't have to worry about analysing ur damn WAM
I study my WAM so that I can get a better WAM.

18. ## Re: S1/17 WAM Predictions

Maths results will be out tomorrow, I think.

19. ## Re: S1/17 WAM Predictions

Originally Posted by 4025808
Maths results will be out tomorrow, I think.
Fark, i hope.

20. ## Re: S1/17 WAM Predictions

Originally Posted by 4025808
Maths results will be out tomorrow, I think.
Good. I'm getting impatient as hell.
Originally Posted by Queenroot
y'all just wait until results come out

if u put this much effort into studyin y'all wouldn't have to worry about analysing ur damn WAM
Can't say I didn't try to.

21. ## Re: S1/17 WAM Predictions

Originally Posted by 4025808
Maths results will be out tomorrow, I think.
Do you check MATH results on the math mark page?

22. ## Re: S1/17 WAM Predictions

Originally Posted by noctua
Do you check MATH results on the math mark page?
Yes those are your preliminary marks. Most of the time they don't change

23. ## Re: S1/17 WAM Predictions

ACTLs been uploaded to stream declaration

24. ## Re: S1/17 WAM Predictions

I have a number now. I did ENGG1000, ENGG1811, PHYS1121, MATH1131. Not sure which

Edit: I am 90% sure it's ENGG1811

25. ## Re: S1/17 WAM Predictions

MATH is currently being logged in now I believe

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