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(i) (3 marks)

(ii) Two shaded circles on or above 1 and 3 with a line joining them (1 mark)

2. Biased die

(i) (2 marks)

(ii) (4 marks)

3. Factors and remainders

(i) Show a = -7 (2 marks)

(ii) (3 marks)

4. Constant acceleration

Distance = 130 metres and acceleration = 0.6 metres per second squared (4 marks)

5. Trigonometric equation

(i) Show trigonometric equation (2 marks)

(ii) Theta = 41.8 degs, 138.2 degs, 270 degs (4 marks)

6. Stationary point

(i) Show S.P. when x = 2 (4 marks)

(ii) Minimum (2 marks)

7. Yachtsman

(i) 3.11 km (3 marks)

(ii) Angle ABC = 61.6 degs and bearing = 152 degs (4 marks)

8. Integral

(i) Show value of integral = 2/3 (3 marks)

(ii) Part of the area is below the x axis and so would have a negative value (1 mark)

(iii) Area = 4 square units (3 marks)

9. Fairground

(i) 2 metres (1 mark)

(ii) 12 metres (2 marks)

(iii) 14 seconds (4 marks)

10. Circle

(i) Midpoint (4, 6) (1 mark)

(ii) (4 marks)

(iii) Show B is on circle (1 mark)

(iv) Show lines are perpendicular (3 marks)

(v) (0, 3) (3 marks)

(i) 2y = x + 6 (5 marks)

(ii) (3, 4.5) (3 marks)

(iii) Area = 125/12 (4 marks)

12. Highway Code

(i) Show formula (5 marks)

(ii) 38.75 feet (3 marks)

(iii) 23.2 mph (4 marks)

13. Chords

(i) a=12, b=6, c=1 (3 marks)

(ii) Show gradient of chord (3 marks)

(iii) (2 marks)

(iv) 12 (1 mark)

As h gets close to zero, the gradient becomes ever closer to 32. (3 marks)
2. First, of all, thanks a lot! However, just wondering whether you could have a copy of the paper to scan in - if so it'd be really helpful!
3. YES. OH GOD YESSS

But you bearing one is wrong. Degree is correct but bearing has to go clockwise from north, i.e. 180 degrees to get to south, 90 degrees more the get to east and then the original degree added to get the bearing.

4. (Original post by CheeseLord)
First, of all, thanks a lot! However, just wondering whether you could have a copy of the paper to scan in - if so it'd be really helpful!
It's not my property to scan. Under the terms of our agreement with OCR, I may only use past papers with my own students in my Centre.
5. (Original post by cyfer)
YES. OH GOD YESSS

But you bearing one is wrong. Degree is correct but bearing has to go clockwise from north, i.e. 180 degrees to get to south, 90 degrees more the get to east and then the original degree added to get the bearing.

Sorry my answer is correct. You were asked to find the bearing of B FROM C.
6. (Original post by Mr M)

4. Constant acceleration

Distance = 190 metres and acceleration = 0.6 metres per second squared (4 marks)

Is Distance not 130m..
I do not remember the figures but I am sure that the Distance/Answer was 130
7. (Original post by Mr M)

(i) (3 marks)

(ii) Two shaded circles on or above 1 and 3 with a line joining them (1 mark)

2. Biased die

(i) (2 marks)

(ii) (4 marks)

3. Factors and remainders

(i) Show a = -7 (2 marks)

(ii) (3 marks)

4. Constant acceleration

Distance = 190 metres and acceleration = 0.6 metres per second squared (4 marks)

5. Trigonometric equation

(i) Show trigonometric equation (2 marks)

(ii) Theta = 41.8 degs, 138.2 degs, 270 degs (4 marks)

6. Stationary point

(i) Show S.P. when x = 2 (4 marks)

(ii) Maximum (2 marks)

7. Yachtsman

(i) 3.11 km (3 marks)

(ii) Angle ABC = 61.6 degs and bearing = 152 degs (4 marks)

8. Integral

(i) Show value of integral = 2/3 (3 marks)

(ii) Part of the area is below the x axis and so would have a negative value (1 mark)

(iii) Area = 4 square units (3 marks)

9. Fairground

(i) 2 metres (1 mark)

(ii) 12 metres (2 marks)

(iii) 14 seconds (4 marks)

10. Circle

(i) Midpoint (4, 6) (1 mark)

(ii) (4 marks)

(iii) Show B is on circle (1 mark)

(iv) Show lines are perpendicular (3 marks)

(v) (0, 3) (3 marks)

(i) 2y = x + 6 (5 marks)

(ii) (3, 4.5) (3 marks)

(iii) Area = 125/12 (4 marks)

12. Highway Code

(i) Show formula (5 marks)

(ii) 38.75 feet (3 marks)

(iii) 23.2 mph (4 marks)

13. Chords

(i) a=12, b=6, c=1 (3 marks)

(ii) Show gradient of chord (3 marks)

(iii) (2 marks)

(iv) 12 (1 mark)

As h gets close to zero, the gradient becomes ever closer to 32. (3 marks)
Are you sure you didn't do this in a rush! I think it is a minimum point, bearing was around 118 degrees and distance was 130m? Can you please check these, though I might be wrong as well.
8. (Original post by metaltron)
Are you sure you didn't do this in a rush! I think it is a minimum point, bearing was around 118 degrees and distance was 130m? Can you please check these, though I might be wrong as well.
Well I got the Minimum and 130m
But I think the bearing is right
90 Degrees + Angle.. Corresponding etc.
9. (Original post by Mr M)
It's not my property to scan. Under the terms of our agreement with OCR, I may only use past papers with my own students in my Centre.
Oh right, that's no problem then
In that case, could I ask how you got the working out for question 4 and question 9 and 11? I suspect I completely stuffed those up, and am wondering how you did it... I remember distinctly looking at the distance question for Q4, and getting 130m as well (I even double-checked!), so a little confused and worried. I also got minimum for question 6, so wondering how you did it
10. (Original post by metaltron)
Are you sure you didn't do this in a rush! I think it is a minimum point, bearing was around 118 degrees and distance was 130m? Can you please check these, though I might be wrong as well.
It is a minimum (that is a typo) - I'll change it.

Bearing is right.
11. Hmmn I dont think all of those are right
12. Thank you SO much! But I swear it was minimum point.... :s
13. (Original post by Mr M)
It is a minimum (that is a typo) - I'll change it.

Bearing and distance is right.
Hmm... distance was using formula, s = ut + 0.5 at^2
Initial velocity and time were both 10. a was 0.6 and t was 10 again.

So 100 + 30 = 130m?

Could you please explain the bearing one to me, as I couldn't work out how to get it
14. (Original post by CheeseLord)
I remember distinctly looking at the distance question for Q4, and getting 130m as well (I even double-checked!), so a little confused and worried.
My arithmetic error - it is 130m!

Sorry!!! My excuse is I did do it in 25 mins.
15. Hi Mr M

s=(u+v)/2 x t
s=13x10
s=130

you could check this using
s=ut+0.5at^2
s=100+30
s=130
16. (Original post by Mr M)
My arithmetic error - it is 130m!

Sorry!!! My excuse is I did do it in 25 mins.
No Worries! More marks for me eh

This is great stuff though! I am hoping I have got a half decent mark
Was the calculus proof for the minimum point re differentiating dy/dx to get d2y/dx2 and putting the number in and as that was > 0 = Minimum?
17. I'm pretty sure it's 130 m :s
18. (Original post by Mr M)
My arithmetic error - it is 130m!

Sorry!!! My excuse is I did do it in 25 mins.
That's a relief.

Oh well about the bearings :P what is that? One or two marks (You get marks for method) out of 100?
19. (Original post by steviep14)
No Worries! More marks for me eh

This is great stuff though! I am hoping I have got a half decent mark
Was the calculus proof for the minimum point re differentiating dy/dx to get d2y/dx2 and putting the number in and as that was > 0 = Minimum?
You can do it that way - you would have got a value of 6.
20. Wait so was it a minimum or a maximum?

Also thank you very much for posting this+posting updates, much appreciated. Your pupils must love you.

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