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AQA Core 4 - Monday 10th June 2013 (AM) - Official Thread Watch

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    Wish I was taking this exam in 2006...
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    Horrible paper I missed a few questions because I ran out of time. And the first question just put me off. I made the last question up...and I didn't use the quadratic formula so I must have thought it factorised. -_-

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    Pretty sure I got the A* the paper was quite hard but I guess my revision came to use

    Some of the ans I got were

    C was 256 - 2pi sq. Hence it was 6m to the nearest metre
    One was (-50,0)
    There was a 32 in the cartesian equation
    Sinx = 196 and 344
    Vector was 5 1 2
    I write some others later
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    (Original post by Anime Lover)
    I'm not sure, I needed 87 on this paper to get an A* overall, and I need a D on my M2 paper to get the A* provided that this averages an A*. I'm pretty sure I'll get an A overall but idk if it'll be any higher than that.
    To get A*, you need to get 90UMS in C3 and C4. All other papers are irrelevant. Even if you get 100UMS in every other paper, you still MUST get 90UMS in C3 and C4 for the A*.
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    (Original post by mulac1)
    how many marks do you think dh/dt - 1.3cos(t/12) would get? one for dh/dt and another for 1.3 - or am I being hopeful?
    I was hoping that, but seeing the working out for value of k, it might be 1 mark for what you did and 2 marks for finding k. Could be wrong though.
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    Sorry I made a mistake on here ..

    for k I did 2pi/12 which is just pi/6
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    (Original post by Gary)
    For Coordinates of Q with normal w/e did people get (-50,0)?
    the gradient of tangent was -4. Gradient of normal was 1/4. The equation of normal was y-12=1/4x-2. Where it cuts the y axis is zero, so x=-46.
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    Wait I may not have missed out that 3 marker. Was finding the coordinates the question before finding the cartesian equation? And x and y were in parametric form? I need to see the paper to remember if I did it.
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    (Original post by stuart_aitken)
    To get A*, you need to get 90UMS in C3 and C4. All other papers are irrelevant. Even if you get 100UMS in every other paper, you still MUST get 90UMS in C3 and C4 for the A*.
    Isn't it that you need to average 90 UMS in C3 and C4, So for example,

    C3 - 95
    C4 - 85

    = A*
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    (Original post by Jamie T)
    Wish I was taking this exam in 2006...
    Agreed. They had it so easy.


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    The year 12's in my school are not too bad I suppose...
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    (Original post by nothepreacher)
    the gradient of tangent was -4. Gradient of normal was 1/4. The equation of normal was y-12=1/4x-2. Where it cuts the y axis is zero, so x=-46.
    The x coordinate was -2 so it was y-12=1/4x+2
    hence x =-50
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    (Original post by nothepreacher)
    the gradient of tangent was -4. Gradient of normal was 1/4. The equation of normal was y-12=1/4x-2. Where it cuts the y axis is zero, so x=-46.
    I thought the equation was y-12 = 1/4x + 2

    so x=50.

    The point that we were putting into the normal was (-2, 12) wasn't it?

    x = -50 I should say
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    (Original post by ItsLewis)
    *your
    Such a student room nerd answer
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    (Original post by Qwob)
    Wait I may not have missed out that 3 marker. Was finding the coordinates the question before finding the cartesian equation? And x and y were in parametric form? I need to see the paper to remember if I did it.
    if you're the person who was getting emosh over dropping 15 marks, don't worry! Boundaries were 53 for an A in january 2013. I genuinely think It'll be lower than that.
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    Well I think I did OK but I incorrectly evaluated the function in 8b and got x=2.28.
    Some answers:
    Trig polynomial:
    \sin( \theta )=\dfrac{3- \sqrt( 17 )}{4}
    \theta = 196^{\circ},\, 344^{\circ}
    Question 7:
    \dfrac{\rm{d}h}{\rm{d}t}= 1.3 \, \cos \left( \dfrac{\pi}{6} \, t \right)
    Question 8:
    a
    \dfrac{4 \, t \sin\left(\frac{1}{4} \, \pi t\right)}{\pi} + \dfrac{16 \, \cos\left(\frac{1}{4} \, \pi t\right)}{\pi^{2}}\ \ (\, + C\, )
    b
    16 \, {\rm x}^{2} = \dfrac{4 \, t \sin\left(\frac{1}{4} \, \pi t\right)}{\pi} + \dfrac{16 \, \cos\left(\frac{1}{4} \, \pi t\right)}{\pi^{2}} - \dfrac{16}{\pi^{2}} + 256
    x  = 3.65 \quad (=3.64621 \ldots)
    (x=4.2 comes from using degrees other answers, I have no idea)
    I may do a few more later.
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    Easily the hardest C4 Paper AQA has ever sat.

    I can just about remember my final answer - x = 365cm. Note that doing your calculations in degrees rather than radians would result in an answer greater than 4m - I know, because I made that mistake and frantically corrected it with three nanoseconds to go.

    Here's hoping for low grade boundaries :P
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    Definitely the hardest paper I've ever seen, it was as hard as FP3! I resat this for a A*, looks like i got the vector questions slightly wrong, and missed out question 7! Not impressed
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    (Original post by ItsAnnaYay)
    Horrible paper I missed a few questions because I ran out of time. And the first question just put me off. I made the last question up...

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    I know right, I hope the grade boundaries are low this time.....
    like 60 for an A or something......
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    (Original post by Barcelona'99)
    Isn't it that you need to average 90 UMS in C3 and C4, So for example,

    C3 - 95
    C4 - 85

    = A*
    Hopefully! But I thought it was 90UMS in each. Meh... my chances of it are scuppered anyway
 
 
 
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