***2015 LC Maths Paper 2 - Higher Level - June 8th***

DarraghF197

Hon the Dubs wrote: »

actually anyone get the smallest value of n to be 7 ?

Did you get something like 8.5 to be n? I thought it would rounded up considering that would be the smallest value of n but still less than the 0.0001.

randylonghorn

The_N4sir wrote: »

http://www.irishtimes.com/news/education/leaving-cert-maths-paper-two-ordinary-level-students-relieved-1.2241690

"John Brennan, a teacher at the Ballinteer Institute and at projectmaths.com"
"Eamonn Toland of TheMathsTutor.ie "
"Jean Kelly, a maths teacher at the Institute of Education"
"Aidan Roantree, senior maths tutor at the Institute of Education"

Interesting.

So does the IT think that only those who charge students to teach them maths (or are part of a business which does) have a valid opinion, or are these the people who immediately contact the IT to tell them their opinion, thus saving their journalists work?

Hon the Dubs

DarraghF197 wrote: »

Did you get something like 8.5 to be n? I thought it would rounded up considering that would be the smallest value of n but still less than the 0.0001.

nah i got 6.something
but yeh it would have been rounded up

skippy1977

Fiona G wrote: »

How do we know it's symmetrical? I know it's probably obvious but I just can't see a way to be sure :P

Glad I assumed that anyway if it's right. The only answers I couldn't get out in that paper was proving that p(n+1) = 0.2p(n) thing and then I just made an educated guess at the smallest value of n in the last part of that q, I said 2. Can't believe how well it went overall though

The quadrilaterals share all the same characteristics. The radii of both circles are the same, they share the common line AB. If QP is a tangent then the lines QB and AP are perpendicular and so make the same angles with the bottom quadrilateral as HA and KB make with the top one. Good you got it anyway!

Bah_Humbug

Hon the Dubs wrote: »

actually anyone get the smallest value of n to be 7 ?

I got 7 too! Pretty confident with it, though the questions leading up to it were an absolute headwreck

Luke Armstrong

randylonghorn wrote: »

"John Brennan, a teacher at the Ballinteer Institute and at projectmaths.com"
"Eamonn Toland of TheMathsTutor.ie "
"Jean Kelly, a maths teacher at the Institute of Education"
"Aidan Roantree, senior maths tutor at the Institute of Education"

Interesting.

So does the IT think that only those who charge students to teach them maths (or are part of a business which does) have a valid opinion, or are these the people who immediately contact the IT to tell them their opinion, thus saving their journalists work?

Well they are known to be the top maths teacher's in Ireland so I guess their opinion is worth quite a lot.

Magnate

Luke Armstrong wrote: »

Well they are known to be the top maths teacher's in Ireland so I guess their opinion is worth quite a lot.

Controversial... :P

Fiona G

skippy1977 wrote: »

The quadrilaterals share all the same characteristics. The radii of both circles are the same, they share the common line AB. If QP is a tangent then the lines QB and AP are perpendicular and so make the same angles with the bottom quadrilateral as HA and KB make with the top one. Good you got it anyway!

Thanks for the explanation, I always doubt myself doing that stuff

randylonghorn

Luke Armstrong wrote: »

Well they are known to be the top maths teacher's in Ireland so I guess their opinion is worth quite a lot.

By who?

Themselves and their PR people?

I've no doubt they're good at what they do, otherwise they wouldn't still be in business, but there are lots of excellent maths teachers out there who don't have PR machines (or want them).

DarraghF197

Hon the Dubs wrote: »

nah i got 6.something
but yeh it would have been rounded up

Yeah at least I was close to (what seems to be) the right answer, which I'm happy with. I expected it to be a lot more tbh, into the teens or so. Still, happy out with maths, moving on briskly!

Luke Armstrong

randylonghorn wrote: »

By who?

Themselves and their PR people?

I've no doubt they're good at what they do, otherwise they wouldn't still be in business, but there are lots of excellent maths teachers out there who don't have PR machines (or want them).

By the public, not saying they are the best teachers just known to be the best

randylonghorn

Luke Armstrong wrote: »

By the public, not saying they are the best teachers just known to be the best

Ah, so you agree that it's all about the bass hype.

Magnate

Kremin

Magnate wrote: »

https://www.youtube.com/watch?v=eKmRkS1os7k


Anyway I don't actually know how or where I went wrong with that question with the angle bisector and find k... but honestly I don't really care anymore. I know paper 2 was worse than paper 1 but not 40+ marks worse so I still like my odds :P.

oktplz

If anyone stumbles across solutions be sure to put a link up :P

theRB

If anyone is planning on doing out the whole thing again, would you have a look at doing the question with (10, k) and the bisector by using the fact that the point of intersection is (4,0) and then using tan to get the slope? I did this and got 0.68 and -52.2, slightly different to the answers the other way got but I really don't think I did anything wrong.

skippy1977

theRB wrote: »

If anyone is planning on doing out the whole thing again, would you have a look at doing the question with (10, k) and the bisector by using the fact that the point of intersection is (4,0) and then using tan to get the slope? I did this and got 0.68 and -52.2, slightly different to the answers the other way got but I really don't think I did anything wrong.

The idea is right and the numbers are very close. I'd say the error is just down to rounding off with your use of Tan? I'll do it out after and post what I get.

dalta5billion

Was kicking myself when I realised that if it's the bisector of the angle between the two lines, that means it must also go through the point of intersection of the other two lines.

The problem with this damn project maths is that going the long way round and not seeing the shortcut can be disastrous for your timing. I was off drawing horrendous diagrams to figure out what angles to add to each other.

Also as an applied maths student, I think I was unfairly advantaged in that golf ball projectile question. We knew instantly to let the height=0 to find the time of landing, and "horizontal speed" is a term that would confuse others.

Hon the Dubs

dalta5billion wrote: »

Was kicking myself when I realised that if it's the bisector of the angle between the two lines, that means it must also go through the point of intersection of the other two lines.

The problem with this damn project maths is that going the long way round and not seeing the shortcut can be disastrous for your timing. I was off drawing horrendous diagrams to figure out what angles to add to each other.

Also as an applied maths student, I think I was unfairly advantaged in that golf ball projectile question. We knew instantly to let the height=0 to find the time of landing, and "horizontal speed" is a term that would confuse others.

Was it not let the height equal -8? Cause it landed 8 below the start?

skippy1977

Hon the Dubs wrote: »

Was it not let the height equal -8? Cause it landed 8 below the start?

No see the formula takes no notice of the hill. The ball landed on the x-axis..regardless of where it was hit from and so you let the formula equal 0.

theRB

skippy1977 wrote: »

The idea is right and the numbers are very close. I'd say the error is just down to rounding off with your use of Tan? I'll do it out after and post what I get.

Thanks, I'd imagine so too. Definitely got the angles right, I'd say I rounded tan to one decimal place because I did out the line formula instead of just multiplying it by 6 and didn't want it to be too messy. Ought to only lose a mark or two.

Hon the Dubs

skippy1977 wrote: »

No see the formula takes no notice of the hill. The ball landed on the x-axis..regardless of where it was hit from and so you let the formula equal 0.

ah didn't notice that. Will that be all the marks gone for that question now do you think?

Troxck

dalta5billion wrote: »

Was kicking myself when I realised that if it's the bisector of the angle between the two lines, that means it must also go through the point of intersection of the other two lines.

The problem with this damn project maths is that going the long way round and not seeing the shortcut can be disastrous for your timing. I was off drawing horrendous diagrams to figure out what angles to add to each other.

Also as an applied maths student, I think I was unfairly advantaged in that golf ball projectile question. We knew instantly to let the height=0 to find the time of landing, and "horizontal speed" is a term that would confuse others.

I suppose studying Physics gave me an advantage there too? In fairness even if you studied neither Applied Maths or Physics, it might have taken only a bit more of effort to realise?

skippy1977

Hon the Dubs wrote: »

ah didn't notice that. Will that be all the marks gone for that question now do you think?

Oh god no, you'll get most of them. A lot of people won't have let the formula equal anything. If it's worth 10 I'd say you'll get 7. Method is the key!

Hon the Dubs

skippy1977 wrote: »

Oh god no, you'll get most of them. A lot of people won't have let the formula equal anything. If it's worth 10 I'd say you'll get 7. Method is the key!

Ah that's great news thanks so much! Actually my angle still worked out to be 3
Hope today went well for you

randylonghorn

Hon the Dubs wrote: »

Hope today went well for you

I'd say he might have passed it with a bit of luck! :pac:

Kremin

Troxck wrote: »

I suppose studying Physics gave me an advantage there too? In fairness even if you studied neither Applied Maths or Physics, it might have taken only a bit more of effort to realise?

Ye thats what i was thinking, I didn't even realise anyone who did applied maths would be at an advantage. Like, someone said s=ut would help but cmon, it travels for 4 seconds at 32 m/s, that's jc science at best.

Troxck

Kremin wrote: »

Ye thats what i was thinking, I didn't even realise anyone who did applied maths would be at an advantage. Like, someone said s=ut would help but cmon, it travels for 4 seconds at 32 m/s, that's jc science at best.

I remember doing questions similar to that in class when we did functions before so, no real AM/Physics advantage there!

skippy1977

theRB wrote: »

If anyone is planning on doing out the whole thing again, would you have a look at doing the question with (10, k) and the bisector by using the fact that the point of intersection is (4,0) and then using tan to get the slope? I did this and got 0.68 and -52.2, slightly different to the answers the other way got but I really don't think I did anything wrong.

Did a bit of an exploration of a method using slopes rather than perpendicular and it becomes a bit unwieldy!!

Definitely works but requires a bit more thinking...and there is less accuracy due to having to round decimals...now this answer not fully fleshed out but does go most of the way to show how to get one of the correct answers 0.75...interestingly when I took the (negative version of tan) I got the 13.8...that some others mentioned (but which isn't a valid solution).

I have the method I would use underneath.

DarraghF197

Oh my God, I'm such an idiot for not thinking about using the perpendicular distance formula! Did it the long way with the tans, and see where I went wrong. I did alpha being +/-29.74* rather than use the unit circle. Bit of a shambles!