Discussion and Suggested Solutions - Mathematics 2017

[immagonow]

I got 89/100 for this exam and I was ranked 51/128 in my school, what band would I be sitting in?

[jamonwindeyer]

Hey Jamon,

Just wondering how accurate ur predictions for raw mark cutofffs for b6s are? How successful have u been in prev yrs? Just dont wanna get my hopes up too high lol.

Thanks

No idea, I don't track my success rate for these predictions, ahaha - But I think most people have found this exam tougher than those in recent years, you can use that idea and this database to come to your own conclusions

I got 89/100 for this exam and I was ranked 51/128 in my school, what band would I be sitting in?

Your exam mark will align to a B6, so it depends on the performance of your cohort! See this guide!

[Abhi]

Just wondering how scaling works, say i get an 83 rawmark, is that aligned to high 80s and then scaled once again?

[mariamoussa99]

honestly I felt like it was easier than many of the past years' exams. I think you only found it hard because you went the very long, hard way in some questions. in question 15 a ii you  could have just found sina and sinb as trigonometric ration since the triangles were right angled. in question 16 c i you could have used the parallel intercept theorem.

[liya1234]

Just curious about alignment - so if raw low 90's aligns to mid/high 90s, what would raw high 90s (97/98) align to?

[Sahana2]

for the kangaroo question, i got 870 because i used the unrounded answer of k . i gave the final answer in 2sf. do you think i will still get the mark?

I DID THAT ASWELLL

[Sahana2]

HEY JAMON
Just wanted to know if I'm sitting at a 90 raw and a rank of 5/52. Am i eligible for a B6.

[Annie657]

Thankyou so much for the solutions!

So if I got 91 raw, is there any way my cohort would be able to bring me down from a b6? I just have a really bad rank (8/13) cause I got 77% in trials :/ they all found the exam hard so I am worried :/

[avalansh7]

Hey Jamon,
So I've been doing some research and discussion with our teachers, and apparently a GP must have a common ratio that is a fixed value not equal to 0. So r cannot be zero. The only exception to this is the special GP, where a=0, r=0. i.e. 0,0,0,0,0. But if 'a' is any non-zero number to form a GP 2, 0, 0, 0, even though a limiting sum of 2 is reached, a GP doesn't exist, as Term2/Term1=0/2=0 but Term 3/2=0/0=undefined. So the answer should actually be 0<a<4, where a is not equal to 2. That probably explains why it is 3 marks. 1 for find Sinfinite --> 2=a/(1-r), therefore a=2(1-r). The other for graphing a vs r, or using another method to find the range of a (0<a<4), and the final for considering that the r=0 case doesn't make a GP. Sorry to everyone who got this q wrong (i did too), but its probably a q to set apart the really high mark students lobbying for state rank.

[Thebarman]

That was a bloody tough exam. When it came to proving that sinA = sinB when L' = 0, I went about it in a different way than shown in the working on page one (i.e. I worked out sin A and Sin B separately and then more or less subbed it in), but I couldn't find the value of x from it. I immediately lost those 3 marks then :/ Overall, I'm not too confident on this one, but I knew that when going in. At least I've got a backup subject.

How many booklets did everyone use in the end? I used 14

[Zainbow]

That was a bloody tough exam. When it came to proving that sinA = sinB when L' = 0, I went about it in a different way than shown in the working on page one (i.e. I worked out sin A and Sin B separately and then more or less subbed it in), but I couldn't find the value of x from it. I immediately lost those 3 marks then :/ Overall, I'm not too confident on this one, but I knew that when going in. At least I've got a backup subject.

How many booklets did everyone use in the end? I used 14

Your concerns regarding the exam are similar to mine  I think I used an entire booklet to try working out x in the sinA = sinB question, but I couldn't get it either. Now that I think about it, there were plenty of easier ways to answer the question lol but I can't go back now obviously.

As for booklets, I used up all the personalised ones plus 9 extras

[jamonwindeyer]

Just wondering how scaling works, say i get an 83 rawmark, is that aligned to high 80s and then scaled once again?

Yep, so it is aligned to a Band cut off then scaled by UAC, here is the guide for anyone interested

honestly I felt like it was easier than many of the past years' exams. I think you only found it hard because you went the very long, hard way in some questions. in question 15 a ii you  could have just found sina and sinb as trigonometric ration since the triangles were right angled. in question 16 c i you could have used the parallel intercept theorem.

I think I also made my life tough doing the exam in less than half the prescribed time I totally agree that my approaches in Q16 could have been more efficient, but I personally still found the exam quite difficult in comparison to say, 2014 (my years exam). I think it was tougher than last years, on the whole, as well.

Important to know this is my perspective - Like, what I would consider 'difficult' is different to what you would consider difficult. But in talking to people broadly, I feel like people found this tougher than last years, I might even go find the 2016 Exam Discussion thread and compare a little later

I mean, go you! If you've found the exam easy that bodes really well for your score - Good job

Just curious about alignment - so if raw low 90's aligns to mid/high 90s, what would raw high 90s (97/98) align to?

Still high 90's, there isn't much room to move!!

HEY JAMON
Just wanted to know if I'm sitting at a 90 raw and a rank of 5/52. Am i eligible for a B6.

Yep, if your cohort performed well I reckon you are in with a good shot

That was a bloody tough exam. When it came to proving that sinA = sinB when L' = 0, I went about it in a different way than shown in the working on page one (i.e. I worked out sin A and Sin B separately and then more or less subbed it in), but I couldn't find the value of x from it. I immediately lost those 3 marks then :/ Overall, I'm not too confident on this one, but I knew that when going in. At least I've got a backup subject.

I reckon you'll get marks, because that would definitely have been the expected approach, rather than my brute force method

[avalansh7]

Your concerns regarding the exam are similar to mine  I think I used an entire booklet to try working out x in the sinA = sinB question, but I couldn't get it either. Now that I think about it, there were plenty of easier ways to answer the question lol but I can't go back now obviously.

As for booklets, I used up all the personalised ones plus 9 extras

For the sina=sinb, you should first differentiate to find dL/dx. From that you get an expression of dL/dx= (x)/root(x^2+25) -(9-x)/root(-(9-x)^2).

If you sub in dL/dx=0, x/root(x^2+25)=(9-x)/root((9-x)^+2+49) which if you look at the diagram is sina=sinb.

This is because the LHS and RHS of the above expression [x/root(x^2+25)=(9-x)/root((9-x)^2+49)] is sina=sinb in the triangles, using normal trig ratios and part (i).

Thus for the min distance travelled, when dL/dx=0, sina=sinb. (i.e. you don't need to find the actual value of x at this stage), but can use it to calculate x in (iii).

[jamonwindeyer]

Hey Jamon,
So I've been doing some research and discussion with our teachers, and apparently a GP must have a common ratio that is a fixed value not equal to 0. So r cannot be zero. The only exception to this is the special GP, where a=0, r=0. i.e. 0,0,0,0,0. But if 'a' is any non-zero number to form a GP 2, 0, 0, 0, even though a limiting sum of 2 is reached, a GP doesn't exist, as Term2/Term1=0/2=0 but Term 3/2=0/0=undefined. So the answer should actually be 0<a<4, where a is not equal to 2. That probably explains why it is 3 marks. 1 for find Sinfinite --> 2=a/(1-r), therefore a=2(1-r). The other for graphing a vs r, or using another method to find the range of a (0<a<4), and the final for considering that the r=0 case doesn't make a GP. Sorry to everyone who got this q wrong (i did too), but its probably a q to set apart the really high mark students lobbying for state rank.

Hmm, very interesting! Thank you for the thoughts!

I definitely agree broadly, but I'm not 100% sure I agree in the context of the question. So while the way you've described a ratio poses issues for \(r=0\), realistically a common ratio of zero is okay (I can multiply by zero to get my next term, no problem), and it is within the \(-1<r<1\) range that NESA normally expects/mandates in this question type. We aren't taught \(|r|<1,r\neq0\), we are just taught \(|r|<1\).

So while enforcing \(r\neq0\) would probably be more mathematically correct, essentially, we are debating a semantic, and I don't know if NESA will even care about that semantic. High chance you get paid the full marks whether you considered it or not.

I wouldn't be surprised either way - I'm very torn, I guess we'll have to wait and see what they do!

[Anything]

Hmm, very interesting! Thank you for the thoughts!

I definitely agree broadly, but I'm not 100% sure I agree in the context of the question. So while the way you've described a ratio poses issues for \(r=0\), realistically a common ratio of zero is okay (I can multiply by zero to get my next term, no problem), and it is within the \(-1<r<1\) range that NESA normally expects/mandates in this question type. We aren't taught \(|r|<1,r\neq0\), we are just taught \(|r|<1\).

So while enforcing \(r\neq0\) would probably be more mathematically correct, essentially, we are debating a semantic, and I don't know if NESA will even care about that semantic. High chance you get paid the full marks whether you considered it or not.

I wouldn't be surprised either way - I'm very torn, I guess we'll have to wait and see what they do!

Just adding on, conceptually wouldn't a limiting sum = 2 be approaching 2 when the number of terms approach infinite instead of being exactly equal to 2? (in the case of r = 0 and a = 2)