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Edexcel A Level Maths grade boundaries

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    I was wondering if anyone knew what the grade boundaries have been for the past few years on C1, C2 and M1 for Edexcel? I'm keen to know because I think I can scrape 80% on each paper but a friend told me that it wouldn't be enough to get me an A. I'd be grateful if anyone from this pool of experince could share some of their knowledge.

    Also, something strange has happened recently. I've forgotten how to factorise completely. I can't really explain it, I comfortably got an A in GCSE maths but now when I'm going over past papers I simply cannot factorise. Can someone tell me how to factorise more complex equations without trial and improvement?

    Cheers guys.
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    for C1 and C2, it probably wont be enough to get 80% for an A as boundaries for these exams tend to be really high (due to them being the easiest maths modules; many get 100%). As for M1, it all depends on how hard the paper is but 80% might just get you the A. Also, what kind of complex equations do you mean? an example would help.
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    Thanks for replying. They're not exactly complex they're just quadratic, it's GCSE material really but it means I keep dropping marks on questions I should be getting full marks on. Anyway, here's an example:

    15x^2+42x-9

    I just can't seem to work it out.
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    3x^2+14x+8 = (3x+2)(x+4)



becasue... we get... (3x+\alpha)(3x+\beta)/3



where \alpha * \beta = 8



and \alpha + \beta = 8*3
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    Consider the equation ax^2+bx+c=0 then if you look for numbers ac=xy and b=x+y and split your middle term using these new x and y it will be nicer.

    Your example:
    -15*9=-135 and 42 is their sum - obvious numbers satisfying this is 45 and -3.

    So 15x^2+45x-3x-9=0
    (5x-1)(3x+9)=0
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    (Original post by misternite)
    Thanks for replying. They're not exactly complex they're just quadratic, it's GCSE material really but it means I keep dropping marks on questions I should be getting full marks on. Anyway, here's an example:

    15x^2+42x-9

    I just can't seem to work it out.
    well use my technique so that (15x+\alpha)(15x+\beta)/15



where \alpha + \beta = 42



and \alpha * \beta = 15*-9
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    Can't you just use the quadratic equation :s:?

    Won't that be much easier?

    meaning: 15x^2 + 42x -9.
    Quadratic equation: -b +/- (b^2 -4ac) / 2a
    And substitute.
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    (Original post by Cynicism 101)
    Can't you just use the quadratic equation :s:?

    Won't that be much easier?

    meaning: 15x^2 + 42x -9.
    Quadratic equation: -b +/- (b^2 -4ac) / 2a
    And substitute.
    long and tedious way of doing it but it is a last resort.

    My method should work every single time however big the numbers.
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    (Original post by Cynicism 101)
    Won't that be much easier?
    That's debatable on a non-calc paper - I find the factorisation quicker than I square 42...


    Also, it's a technique necessary to be comfortable with to be able to solve other types of problems where we cannot explicitly use the formula because we have an unknown.
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    Oh, sorry, I tend to use quadratic formula automatically irrespective of whether it can be factorised....guess I'm just more used to it.
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    I use the "PAF" method. Quadratic is ax^{2}+bx+c=0. You split b into two integers with a product (P) of ac and a sum of b. You then factorise each half of the equation separately. Your example:

    15x^{2}+42x-9=0
    So we are looking for a product of -135 and a sum of 42... I get 45 and -3.
    15x^{2}+45x-3x-9=0
    Now factorise each half of the equation.
    15x(x+3)-3(x+3)=0
    Simplify.
    (15x-3)(x+3)=0
    Solve.
    x=\frac{1}{5}, x=-3
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    Why not just divide everything by 3 to get:

    5x^2 + 14x - 3

    Then you can use easily factorise or use the quadratic formula...
  13. Offline

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    (Original post by ShaolinTemple)
    Why not just divide everything by 3 to get:

    5x^2 + 14x - 3

    Then you can use easily factorise or use the quadratic formula...
    same thing since after you must mult by 3 to get the roots
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    No, the roots stay the same.
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    Here have a look at the Examiners' reports for the last few years for grade boundaries (Edexcel btw, you didnt say which board).
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    Sorry completely off subject now lol, Teacher told us that the boundries are:

    40/100>-E
    50/100>-D
    60/100>-C
    70/100>-B
    80/100>-A

    So if you can just scrape 80% then you'll get an A

    Well done as i can only just scape an E!
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    OP, this might help:

    http://www.edexcel.org.uk/VirtualCon...Boundaries.pdf
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    (Original post by Loz17)
    Sorry completely off subject now lol, Teacher told us that the boundries are:

    40/100>-E
    50/100>-D
    60/100>-C
    70/100>-B
    80/100>-A

    So if you can just scrape 80% then you'll get an A

    Well done as i can only just scape an E!
    Those are UMS boundaries, not raw mark boundaries.
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    Ums boundries? They were the boundries that my whole school used for all of our AS maths candadates grades so hopefully they should be right
  20. Offline

    (Original post by Loz17)
    Sorry completely off subject now lol, Teacher told us that the boundries are:

    40/100>-E
    50/100>-D
    60/100>-C
    70/100>-B
    80/100>-A

    So if you can just scrape 80% then you'll get an A

    Well done as i can only just scape an E!
    Those are UMS boundaries. Roughly speaking, 1% on the paper corresponds to 1 UMS, e.g. 60/75 is 80 UMS on average. However it varies a lot, look at the document I posted a couple of posts up, some papers you can get 100 UMS with about 67/75.

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