* Maths paper 2 (Ordinary) tips / discussion / aftermath * (1 thread only please)

Aaron547

Anybody got any tips??

LedZeppelin

Yeah im worried bout paper two after todays excuse for an exam..

Mysteriouschic

This has loads of good tips for studying for maths .

http://www.rte.ie/exams/maths.html

lisabisa

I left my exam papers in school, gonna print off some tomorrow to study, but I was wondering how much marks are going for the vectors question? Like the option in general? Is it the same as a normal question?

Gav77

lisabisa wrote: »

I left my exam papers in school, gonna print off some tomorrow to study, but I was wondering how much marks are going for the vectors question? Like the option in general? Is it the same as a normal question?

yes

._.

Anybody planning on doing the Geometry question? (Q4)

GoldFour4

lisabisa wrote: »

I left my exam papers in school, gonna print off some tomorrow to study, but I was wondering how much marks are going for the vectors question? Like the option in general? Is it the same as a normal question?

yup 50 marks

Dbstf

Which of the 4 options would be the easiest?
Further Geometry
Vectors
Further Sequences and Series
Linear Inequalities and Linear Programming

martyllia

Dbstf wrote: »

Which of the 4 options would be the easiest?
Further Geometry
Vectors
Further Sequences and Series
Linear Inequalities and Linear Programming

we've done only vectors! it's not bad...bit simmilar to q 4 on paper 1!

Planemo

Which of the 4 options would be the easiest?
Further Geometry
Vectors
Further Sequences and Series
Linear Inequalities and Linear Programming

Linear programming definitely. Did vectors last year and couldn't wrap my head around it at all. With LP I have a chance of actually passing!

Mysteriouschic

Dbstf wrote: »

Which of the 4 options would be the easiest?
Further Geometry
Vectors
Further Sequences and Series
Linear Inequalities and Linear Programming

I think linear programming. I was awful at it before I didn't even know how to start it as I hadn't learnt it properly. I've only learnt how to do that question last week , after getting my teacher to go over it. It's quite simple once you get the hang of it. The questions are usually similar .

C__

Can someone help me. I am currently doing some maths paper 2 study I am doing 2009 q6 c(iv) and i am having some trouble with the last question.
It states:3 boys and 2 girls are seated in a row as a group. in how many different ways can the group be seated if - the 2 girls must be seated beside each other,
I checked the marking schemes and the answer was 4! x 2! = 48
Can someone please explain why it is 4! x 2! ??????

C__

And one final question on Linear programming when you are doing out the part where you need to find which combination of items is the best for profit and stuff how do you know which points to use?

PJelly

C__ wrote: »

Can someone help me. I am currently doing some maths paper 2 study I am doing 2009 q6 c(iv) and i am having some trouble with the last question.
It states:3 boys and 2 girls are seated in a row as a group. in how many different ways can the group be seated if - the 2 girls must be seated beside each other,
I checked the marking schemes and the answer was 4! x 2! = 48
Can someone please explain why it is 4! x 2! ??????

Five people ok? If two MUST BE beside each other. Take them as one person. So there's only 4 left. Which gives 4!
But remember, those two people can be sitting in different ways. It can Be John on the left and Mary on the right, or vice versa. Because they can switch places, its multiplied by 2!

Say if there were 6 people. And 3 had to be beside each other. Take the three as one group. So there's only 4 people (3 by themselves, and the group) which can be arranged in 3! ways. But the group of people itself can be arranged in 3! ways too.
So you get 4! x 3!
Geddit?

C__

PJelly wrote: »

Five people ok? If two MUST BE beside each other. Take them as one person. So there's only 4 left. Which gives 4!
But remember, those two people can be sitting in different ways. It can Be John on the left and Mary on the right, or vice versa. Because they can switch places, its multiplied by 2!

Say if there were 6 people. And 3 had to be beside each other. Take the three as one group. So there's only 4 people (3 by themselves, and the group) which can be arranged in 3! ways. But the group of people itself can be arranged in 3! ways too.
So you get 3! x 3!
Geddit?

Awh i thank you sir.
I only started probability last week and this was the part i kinda had trouble with but i get it now thanks

Planemo

And one final question on Linear programming when you are doing out the part where you need to find which combination of items is the best for profit and stuff how do you know which points to use?

you have to do out a table of all 4 points from the graph and multiply the x value by each x point and the y value by each y point. add them together and the value that ends up being the highest is the profit.

eg: 90 of type x and 100 of type y; (0,0) (50,0) (0,40) (20,30)
90 x 0= 0 100 x 0= 0 x+y=0
90 x 50= 4500 100 x 0= 0 x+y= 4500
90 x 0= 0 100 x 40= 4000 x+y= 4000
90x 20= 1800 100 x 30= 3000 x+y= 4800
highest is 4800 so profit is 4800 euro with 20 of type x and 30 of type y needed.

C__

hawaiichair wrote: »

you have to do out a table of all 4 points from the graph and multiply the x value by each x point and the y value by each y point. add them together and the value that ends up being the highest is the profit.

eg: 90 of type x and 100 of type y; (0,0) (50,0) (0,40) (20,30)
90 x 0= 0 100 x 0= 0 x+y=0
90 x 50= 4500 100 x 0= 0 x+y= 4500
90 x 0= 0 100 x 40= 4000 x+y= 4000
90x 20= 1800 100 x 30= 3000 x+y= 4800
highest is 4800 so profit is 4800 euro with 20 of type x and 30 of type y needed.

I get this part but the part i am on about is how do you know which points to choose from your graph?
i was doing one tonight and i didnt know which points to choose.

Ficheall

There'll be a region bounded by the x and y axes and the lines you've drawn in. The corner points (usually 4) of this region are the points you take.

C__

Ficheall wrote: »

There'll be a region bounded by the x and y axes and the lines you've drawn in. The corner points (usually 4) of this region are the points you take.

Ok thanks

Iceboy

Feel confidant on Q1,4 and 5 atm

need to know 3 more by monday

galwayman17..

i need to get around 270 marks outa 300 in this paer after the disaster on friday!

dear god i hope i get it! paper 2 was always gonna be my better paper but paper 1 was worse than i expected. anyone else feel this way?

daex

C__ wrote: »

Can someone help me. I am currently doing some maths paper 2 study I am doing 2009 q6 c(iv) and i am having some trouble with the last question.
It states:3 boys and 2 girls are seated in a row as a group. in how many different ways can the group be seated if - the 2 girls must be seated beside each other,
I checked the marking schemes and the answer was 4! x 2! = 48
Can someone please explain why it is 4! x 2! ??????

BBBGGG

B B B GG - consider the girls as one.

now there's four places. .. x .. x .. x .. (four places)
4 x 3 x 2 x 1= 24
then because the two girls can be seated in two different places
you multiple 24 x2=24

Mysteriouschic

galwayman17.. wrote: »

i need to get around 270 marks outa 300 in this paer after the disaster on friday!

dear god i hope i get it! paper 2 was always gonna be my better paper but paper 1 was worse than i expected. anyone else feel this way?

I was expecting to to able to do question at least 4 full questions I didn't even get one and failed that paper. . I hope they won't give us a hard paper 2. Could you imagine if simpson rule didn't come up. I need to get 6 full questions to pass now.

r0cks0l1dd

I asked this on another thread but I don't know what page it was on :P

So, since it doesn't specify to us to use 3.14 or 22/7 for Pi, can I just put down the symbol for Pi on my answer and use the Pi button on my calculator? I usually do this.

Also, does anyone know if the Unit Circle for Trigonometry is worth knowing, I don't understand it. And do you think the Compound Angle formulas will come up? Haven't came up yet as far as I know.

I have to attempt all every part in this. I decided not to do the second Algebra question on Paper 1 to do Q6 and I kind of messed that up.

r0cks0l1dd

[/B]

Mysteriouschic wrote: »

I was expecting to to able to do question at least 4 full questions I didn't even get one and failed that paper. . I hope they won't give us a hard paper 2. Could you imagine if simpson rule didn't come up. I need to get 6 full questions to pass now.

If they do that, especially after no first principles, there will be war....:D

Aaron547

Will there be formulas i cant find in the log tables?

How do i fin a perimeter of a square when and there i a circle in the square with the radius of 9cm?

also the above question is 2010 p2 q1 a

._.

Leaving out Geometry and Trigonometry, it'll be fine

MathsManiac

Aaron547 wrote: »

Will there be formulas i cant find in the log tables?

How do i fin a perimeter of a square when and there i a circle in the square with the radius of 9cm?

also the above question is 2010 p2 q1 a

The only reason the circle is there in that question is to see if you can figure out that the distance across the square is twice 9, which is 18.

The perimeter of any shape is just the distance around the edge. A square whose sides ar 18cm long has perimeter 18+18+18+18 = 72cm.

A square is just a special kind of rectangle, so its area is just (length X width) = 18 x 18 = 324 cm^2.

r0cks0l1dd

Aaron547 wrote: »

Will there be formulas i cant find in the log tables?

How do i fin a perimeter of a square when and there i a circle in the square with the radius of 9cm?

also the above question is 2010 p2 q1 a

Well the one side of the square is double the radius the circle and a square has 4 equal sides, it would be 18 x 4=72

Answers and solutions for all questions since 1996 (no solutions for 2011 yet, just answers). http://www.studentxpress.ie/ordinary/ordarith.htm

And I don't think there are any formulas you need that aren't in the tables.

J_E

r0cks0l1dd wrote: »

I asked this on another thread but I don't know what page it was on :P

So, since it doesn't specify to us to use 3.14 or 22/7 for Pi, can I just put down the symbol for Pi on my answer and use the Pi button on my calculator? I usually do this.

What I usually do for those shape questions is treat Pi like an x and work around it until I get something like 1537.50pi cm^3 for example. I'll post an example if you're confused.

BmxNoob

Any help with this question ? :

p(3, 0) is a point.
t and s are two distinct points on the y-axis and |pt| = |ps| = 5.
(i) Find the co-ordinates of t and the co-ordinates of s.

MathsManiac

BmxNoob wrote: »

Any help with this question ? :

p(3, 0) is a point.
t and s are two distinct points on the y-axis and |pt| = |ps| = 5.
(i) Find the co-ordinates of t and the co-ordinates of s.

Draw a diagram. Mark in what you know, and use Pythagoras' theorem.

mc3ac

just done that today, you have two right angle triangles and the 5 is the hypotenuse - the only way to understand it is to draw it out first then use pythagoras to find out co-ordinates

nitromaster

For probability...if you mess it up..and can't figure out it's 4 x 3 x 2 or whatever...can you just do it out manually and still get the marks?

BmxNoob

Still cant understand that question ... I tried drawing the triangle but the only point i have is (3,0) and the value 5 but i dont know what that is ... Can anyone help ?

Ficheall

BmxNoob wrote: »

Any help with this question ? :

p(3, 0) is a point.
t and s are two distinct points on the y-axis and |pt| = |ps| = 5.
(i) Find the co-ordinates of t and the co-ordinates of s.

Well, (3,0) is a point on the x-axis.
The other two points are on the y-axis from the question.

Draw it, it should be much clearer. You know the length between |pt| and (3,0) is 5. The distance between the origin and (3,0) is 3.
So by Pythagoras theorem the distance of the other side of the triangle is 4. This is directly above the origin (as it is on the y-axis), so the point is (0,4).
Correspondingly the other point is (0,-4).

Drawing it is the simplest option to understand it, I suspect.

SDTimeout

Anybody think something mad like no Simpsons Rule could happen after the 1st Principals incident ?

kt.alice

Is anyone else doing the theorems for Q4-b)?

Just wondering are you supposed to do the diagram in pencil? Cause I always did em in pen, but just realised this could be an issue. :rolleyes:

BmxNoob

Ficheall wrote: »

Well, (3,0) is a point on the x-axis.
The other two points are on the y-axis from the question.

Draw it, it should be much clearer. You know the length between |pt| and (3,0) is 5. The distance between the origin and (3,0) is 3.
So by Pythagoras theorem the distance of the other side of the triangle is 4. This is directly above the origin (as it is on the y-axis), so the point is (0,4).
Correspondingly the other point is (0,-4).

Drawing it is the simplest option to understand it, I suspect.

Thanks for trying to help but i still dont understand it ... Is there any website i can get the instructions to the question ?

I cant understand how im meant to draw the diagram ... I have the point (3,0) but i dont know how to get the coords of p and s ... Sorry for the trouble

Ave Sodalis

Does anyone here do Project Maths?

r0cks0l1dd

SDTimeout wrote: »

Anybody think something mad like no Simpsons Rule could happen after the 1st Principals incident ?

Was first principles on every year besides this year?

I checked my book and the only thing they could really ask in place of Simpsons Rule is Area/Volume Irregular Prisms but I'm sure that it will be there

r0cks0l1dd

sup_dude wrote: »

Does anyone here do Project Maths?

Download the one second from the top. Podcast about PM from Countdown to 806

http://www.rte.ie/radio1/podcast/podcast_countdownto806.xml

J_E

r0cks0l1dd wrote: »

Was first principles on every year besides this year?

I checked my book and the only thing they could really ask in place of Simpsons Rule is Area/Volume Irregular Prisms but I'm sure that it will be there

Simpsons Rule has been on every year from at least 1998 so I really doubt they'd do something that drastic. I have a feeling they could do either an awkward probability or awkward linear programming though.

Ficheall

BmxNoob wrote: »

Thanks for trying to help but i still dont understand it ... Is there any website i can get the instructions to the question ?

I cant understand how im meant to draw the diagram ... I have the point (3,0) but i dont know how to get the coords of p and s ... Sorry for the trouble

You know that they're on the y-axis. It's given in the question.
All you have to figure out is how far up/down on the y-axis they are.

Look
> Draw your x-axis and y-axis

> Mark the point (3,0). It's on the x-axis.

> Your other two points are on the y-axis, given from the question. K?
We don't know where on the y-axis yet though.

> Given that they are the same distance from the point (3,0), which is on the x-axis, the points ps and pt must be symmetric about the origin, though you don't know how far up and down the y-axis they are.

> You can find out though! Focus only on one half of the triangle, say the half above the x-axis. 5 is the length of the hypotenuse as given in the question (the distance between (3,0) and ps. The distance between (3,0) and the origin is 3. By Pythagoras' theorem, the length of the other side of the triangle must be 4.

> So the point is on the y-axis, as given in the question, and is 4 away from the origin, ie. (0,4).
> The same applies for the triangle underneath the x-axis, and the point turns out to be (0,-4).

Any better?

BmxNoob

Ficheall wrote: »

Well, (3,0) is a point on the x-axis.
The other two points are on the y-axis from the question.

Draw it, it should be much clearer. You know the length between |pt| and (3,0) is 5. The distance between the origin and (3,0) is 3.
So by Pythagoras theorem the distance of the other side of the triangle is 4. This is directly above the origin (as it is on the y-axis), so the point is (0,4).
Correspondingly the other point is (0,-4).

Drawing it is the simplest option to understand it, I suspect.

I've got it now Ficheall thanks ... I thought it was a difficult question !

r0cks0l1dd

Cydoniac wrote: »

Simpsons Rule has been on every year from at least 1998 so I really doubt they'd do something that drastic. I have a feeling they could do either an awkward probability or awkward linear programming though.

Well we didn't do probability (we were but our teacher was out for over 3 weeks once and the sub kept doing algebra so we didn't have enough time out of time) so that doesn't really affect me but the linear programming will be worrying if they do make it a bit odd, same with Trigonometry. I'm locked at 6 questions with 1,2,3,5,7 and 11 so I hope they don't mix it up to much.Oddly enough I find the (a) and Linear programming harder than (b).

SDTimeout

I don't think it won't be there, just preparing myself. I was really hoping 1st principals came up in a question 6 the other day as it's usually my weakest of all 6 , 7 and 8. Easy 20 marks and all that but shockingly it was probably the one i felt most confident answering during the test.

FlawedGenius

Quick question: How do you know when to use Pythagorass Thereom and what is the formula again??

soccymonster

Gonna do so bad tomorrow. And I went from honours into a special class made for honours drop outs. And I seemed to be the only person who hadn't covered vectors in that class so I don't understand them at all for the optional question. FML. Hopefully my paper 1 will save me from the trainwreck that'll occur tomorrow.

r0cks0l1dd

soccymonster wrote: »

Gonna do so bad tomorrow. And I went from honours into a special class made for honours drop outs. And I seemed to be the only person who hadn't covered vectors in that class so I don't understand them at all for the optional question. FML. Hopefully my paper 1 will save me from the trainwreck that'll occur tomorrow.

If you still have your book or anything, try to look over and understand Linear Programming Q11

Ficheall

FlawedGenius wrote: »

Quick question: How do you know when to use Pythagorass Thereom and what is the formula again??

If you have a right-angled triangle and know the length of two of its sides, you can use Pythagoras' theorem to find the length of the third side.

PT: The length of the hypotenuse (the longest side) squared = the sum of the squares of the lengths of the other two sides.

ie. if the sides are a,b,h, where h is the hypotenuse, then
h^2=a^2+b^2