***JC 2014 Maths Paper Two - Mon. June 9th 2014 - all levels***

emersyn

The paper is up

yaEHya

Reckon I got 140 in that paper and 165 in paper one so I got a ****ing d

andyman

Higher paper done here. I got 7:8 for Question 10 also.

Liordi

andyman wrote: »

Higher paper done here. I got 7:8 for Question 10 also.

How did you find it?

thetalker

the hardest question was undoubtedly the one about questioning people at lidl about their favorite place to shop

Liordi

thetalker wrote: »

the hardest question was undoubtedly the one about questioning people at lidl about their favorite place to shop

Not a sign of Aldi. :rolleyes:

andyman

Liordi wrote: »

How did you find it?

Let r1 = 6cm and r2 = 3cm then apply the formula:

V = π (r1^2 + r1r2 + r2^2) h/3

Answer should leave you with (in terms of pi) 147π

Put that answer over the answer for part (i) and you get a ratio of .875:1 or 7:8

Liordi

andyman wrote: »

Let r1 = 6cm and r2 = 3cm then apply the formula:

V = π (r1^2 + r1r2 + r2^2) h/3

Answer should leave you with (in terms of pi) 147π

Put that answer over the answer for part (i) and you get a ratio of .875:1 or 7:8

I meant the paper in general.
Thanks, though

andyman

Liordi wrote: »

I meant the paper in general.
Thanks, though

It seemed very fair to me. I know a lot of people were afraid yesterday because of a nice paper 1 but it was nice to see they were just nervous.

Question two was very badly worded, which is an ongoing issue in Project Maths. I think the general idea was clear but a lot of students could easily have had to read it a couple of times.

Liordi

andyman wrote: »

It seemed very fair to me. I know a lot of people were afraid yesterday because of a nice paper 1 but it was nice to see they were just nervous.

Question two was very badly worded, which is an ongoing issue in Project Maths. I think the general idea was clear but a lot of students could easily have had to read it a couple of times.

Oh, are you a teacher or something?

andyman

Liordi wrote: »

Oh, are you a teacher or something?

Yes, just newly qualified.

Liordi

andyman wrote: »

Yes, just newly qualified.

Did you post the solutions online?

MangoMachine

andyman wrote: »

Yes, just newly qualified.

Hm if you could post a link with the solutions it would be brilliant!

MangoMachine

Answer to 8 (iii) anyone?

Liordi

MangoMachine wrote: »

Answer to 8 (iii) anyone?

x - 3y - 7 = 0

Emmie123

So happy with how paper 2 went! Hopefully that's pushed me to a high B

roisiny

DanielChester wrote: »

PS. That theorem of the circle was such a surprise, good thing I learned it the day before, I had doubts it would come up after seeing it in a previous exam paper.

Got up at 6:45 and learnt it this morning, even though my body was not impressed with six hours sleep

Kelly090

Can some one tell me how to do question 8? Thanks

andyman

Liordi wrote: »

x - 3y - 7 = 0

This.

andyman

Kelly090 wrote: »

Can some one tell me how to do question 8? Thanks

Which part in particular? Or all of it?

Liordi

Kelly090 wrote: »

Can some one tell me how to do question 8? Thanks

(i)
x - 3y - 6 = 0
Divide the whole line by three.
1/3x - y - 6 = 0.
Slope = 1/3.

OR

Let x = 0 let y = 0
0 - 3y - 6 = 0 x - 3(0) - 6 = 0
-3y = 6 x - 6 = 0
y = -2 x = 6
Point: (0, -2) Point: (6, 0)

y2 - y1

---

x2 - x1

0 - (-2)

---

6 - 0

2/6 = 1/3

(ii)
Just sub it in.
1 - 3(-2) - 6 = 0
1 + 6 - 6 = 0
1 =/= 0
=> The point (1, -2) isn't on the line.

(iii)
We know the point (1, -2) is on the line and as the line is parallel to the above line, m = 1/3.

y - y1 = m (x - x1)
y - (-2) = 1/3 (x - 1)
To neutralize the fraction; we multiply by 3.
3y + 6 = x - 1
x - 3y - 7 = 0.

ElmW13

roisiny wrote: »

Got up at 6:45 and learnt it this morning, even though my body was not impressed with six hours sleep

Same!!! For some weird reason I woke exactly an hour before my alarm was meant to go off... weird

Kelly090

Liordi wrote: »

(i)
x - 3y - 6 = 0
Divide the whole line by three.
1/3x - y - 6 = 0.
Slope = 1/3.

OR

Let x = 0 let y = 0
0 - 3y - 6 = 0 x - 3(0) - 6 = 0
-3y = 6 x - 6 = 0
y = -2 x = 6
Point: (0, -2) Point: (6, 0)

y2 - y1

---

x2 - x1

0 - (-2)

---

6 - 0

2/6 = 1/3

(ii)
Just sub it in.
1 - 3(-2) - 6 = 0
1 + 6 - 6 = 0
1 =/= 0
=> The point (1, -2) isn't on the line.

(iii)
We know the point (1, -2) is on the line and as the line is parallel to the above line, m = 1/3.

y - y1 = m (x - x1)
y - (-2) = 1/3 (x - 1)
To neutralize the fraction; we multiply by 3.
3y + 6 = x - 1
x - 3y - 7 = 0.

Thanks so much!!!:)

MangoMachine

Anyone know the marking scheme? Or at least have an idea?

MangoMachine

Can you get junior cert results online?

emersyn

MangoMachine wrote: »

Anyone know the marking scheme? Or at least have an idea?

To get a rough estimate multiply the time recommended for each question by 2, e.g. if it says to spend 10 minutes on a question it's generally about 20 marks. We won't know the exact marking scheme until September

Caral

I got 5.6m for the new height
I got 1:8 for the ratio
I got my angle to be 59 degrees.. Anyone else get these ... Did we actually have to use the frustrum formula? Its not on our course like...

Caral

Liordi wrote: »

Yeah.

No 6.403 was your distance AB

mayway

S_Hick12 wrote: »

Is 0.875:1 the same as 7:8?

Yes, it is.