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    (Original post by lucybrooks17)
    for question 8iii i worked it out 2 ways and got two different answers - one of which was correct. What will they do about that will they give me any marks at all?
    What was 8iii again? And what answers did you put
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    (Original post by lucybrooks17)
    for question 8iii i worked it out 2 ways and got two different answers - one of which was correct. What will they do about that will they give me any marks at all?
    Perhaps. Usually if you give two answers either a 'list' principle will apply where they will award the first answer that naturally appears 1st in your working. If you put in full working for both methods, the examiner could take that two ways. You may benefit from ALL of the marks except the A mark(s) (for your final answer) or if the examiner isn't so lenient, he may dock you marks for 'hedging your bets' by equating two solutions.
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    (Original post by Charlieukharris)
    Is there an unofficial markschem
    In the original post, click 'show spoiler'. The questions aren't in the right order as I can't remember the order (I have said people can correct me but nobody has) but the answers are there.
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    use the discriminant b^2-4ac and no real roots so <0

    (-1)^2-(4x2x-3x-k)
    1+24-8k<0
    25<-8k
    -25/8>k
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    (Original post by Charlieukharris)
    Is there an unofficial markschem
    Not yet, although if you read the replies you can gather most of the answers that most people are interested in.
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    how many ums will 67 be
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    (Original post by ss1529)
    use the discriminant b^2-4ac and no real roots so <0

    (-1)^2-(4x2x-3x-k)
    1+24-8k<0
    25<-8k
    -25/8<k
    Actually k < -25/8 because the graph was a positive x2 function
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    (Original post by Madhutty)
    Did you get the scale factor (1/3) ? If so then that's probably going to be a mid-low A


    Posted from TSR Mobile

    Yeah I got SF 1/3,I'm hoping that it might still possible to get high 80s to low 90s on that C1, if the mark scheme is lenient.
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    (Original post by Rubberduckiller)
    how many ums will 67 be
    Most likely around ~87.
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    I spoke to Mr M, he is going to release his version of an unofficial mark scheme tomorrow. This is probably the most accurate we can get, since he will have the paper and he's a lot more trustworthy than the rest of us just saying what we got is right!
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    (Original post by Kadak)
    Posted from TSR Mobile

    Yeah I got SF 1/3,I'm hoping that it might still possible to get high 80s to low 90s.
    I'd estimate about 67-68 for 90
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    (Original post by Madhutty)
    Most likely around ~87.


    Posted from TSR Mobile

    I'm confused how you calculated this.Did you not divide 67 by 72 and times by 100 ?
    How did you calculate ums?
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    (Original post by Madhutty)
    Most likely around ~87.
    I strongly believe though that this C1 was one of the hardest in the last 3-4 years of exams. You can expect the UMS boundaries to be lower.
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    (Original post by chloe-jessica)
    I spoke to Mr M, he is going to release his version of an unofficial mark scheme tomorrow. This is probably the most accurate we can get, since he will have the paper and he's a lot more trustworthy than the rest of us just saying what we got is right!
    Who even is Mr M?
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    (Original post by Kadak)
    Posted from TSR Mobile

    I'm confused how you calculated this.Did you not divide 67 by 72 and times by 100 ?
    How did you calculate ums?
    UMS would depend on the grade boundaries that the board set. There is no 'set' way of calculating them. Each year the ratio of raw : UMS differs, so that is just an estimated value.
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    (Original post by gabby07)
    I strongly believe though that this C1 was one of the hardest in the last 3-4 years of exams. You can expect the UMS boundaries to be lower.
    I'd say it was on the same level as june 2013, which needed 61 for an A. it definitely won't be 65 for an A like last year
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    (Original post by ChoccyPhilly)
    Who even is Mr M?
    Maths god of TSR would be a reasonably accurate way to describe him.
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    (Original post by chloe-jessica)
    Maths god of TSR would be a reasonably accurate way to describe him.
    Oh, I knew that but I mean is he a teacher who just kindly posts the answers to us?
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    (Original post by Kadak)
    Posted from TSR Mobile

    I'm confused how you calculated this.Did you not divide 67 by 72 and times by 100 ?
    How did you calculate ums?
    These are all estimated, and based on my knowledge of previous paper difficulty + UMS boundaries. 61-62 is average difficulty, this was a fairly average paper so I'd say about 63 for 80 and 68 for 90.

    Don't take my word for it though! This isn't 100% certain.
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    (Original post by gabby07)
    I strongly believe though that this C1 was one of the hardest in the last 3-4 years of exams. You can expect the UMS boundaries to be lower.


    Posted from TSR Mobile

    Hopefully!
 
 
 
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