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UKMT Senior Individual Maths Challenge 2016 - Preparation and Tips watch

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    (Original post by qeyoo)
    Yes, you can get max 25+14x4= 81, last year it was 50 for a bronze

    Posted from TSR Mobile
    DO u guys think boundaries will be up or slightly down on last year
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    (Original post by mathspro2580)
    do u guys think boundaries will be up or slightly down on last year
    i pray they go down
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    (Original post by HapaxOromenon3)
    Yep I'll send you a PM in a minute.
    Me as well, please.

    (Original post by louisforrest)
    when are the official answers released?
    Tomorrow 9 am onwards.
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    Anyone explains question 19? Such a weird question........

    BTW I didn't do that one and made a silly mistake on the last question, so 116 hopefully.
    • Thread Starter
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    Can anyone guess what I got overall? I feel I got mid-70s (but I'm Year 10, so ok I guess...?)
    1B 2D 3D 4B 5A 6C 7A 8B 9C 10B 11C 12C 13 14D 15 16D 17D 18E 19D 20A 21 22C 23D 24 25D
    Misread half the questions! (Scale, set with M, .......!!)
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    (Original post by d010534)
    Guys,
    I answered 14 questions and now I'm worried. Can I at least get a Bronze??
    It obviously depends on how many you got right; if you got full marks, that's 14*4 + 25 = 81 marks, which would have been one mark off a gold last year (and people are saying this paper was harder). So I think a reasonable estimate (everyone's likely to have got 1/2 questions wrong) would be that you got silver, although gold or bronze are both possible.

    Remember that the people posting on this thread are going to be some of the most enthusiastic/dedicated people who took the test, and therefore probably some of the smartest, so just because everyone's saying they answered more questions doesn't mean that's normal. And remember, a bronze/silver is still a respectable achievement.
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    (Original post by sqrt(e/m)=c)
    How did you get that? It works, but I just want to know how, for the future.
    It wasn't a great way of doing it, but I just tried each answer in the 'X' box one by one, then found the average of X and 25 to find the box second from the end.

    Then I called the box to the left of X 'Y', then tested whether each one worked or not.

    For example, with answer A:15, I found it would give '20' in the box second from the right. Then I called the 3rd box Y, and so the 2nd was (10+Y)/2, and then you can see that Y=(((10+Y)/2)+15)/2

    So: 2Y=((10+Y)/2)+15
    So: 2Y-15=(10+Y)/2
    So: 4Y-30=10+Y
    So: 3Y=40
    So: Y=40/3, and we can assume that these should be integers, so 15 does not work.

    Next: 2Y=((10+Y)/2)+17
    So: 2Y-17=(10+Y)/2
    So: 4Y-34=10+Y
    So: 3Y=44
    So: Y=44/3

    Finally: 2Y=((10+Y)/2)+19
    So: 2Y-19=(10+Y)/2
    So: 4Y-38=10+Y
    So: 3Y=48
    So: Y=48/3=16 and so this must be the correct answer.

    I only used trial and error because I could see that X would have to be odd immediately, because otherwise the one second from the right would be a fraction (which seemed unlikely).

    A better way to do it would be to call the box to the left of X 'Y' (as I did above), and NOT substitute in the answers till the end (which was what I did).
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    Question 24 was fierce, I ended up in a web of inverse and compound tan functions...answer came out though. Got 120 according to general TSR agreement. Anyone know of a BMO1/2 thread going at the moment?
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    (Original post by Mathspro2580)
    DO u guys think boundaries will be up or slightly down on last year
    Hoping they go down (by quite a lot but doubt it)

    Lowest they ever were was 2002 but I think it is being too optimistic to hope for it to happen again...
    Bronze: 47
    Silver: 57
    Gold: 70
    BMO1: 82
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    Do you guys think getting a silver is respectable if u want to do physics at a good university ( i really want to go to imperial)
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    (Original post by surina16)
    Hoping they go down (by quite a lot but doubt it)

    Lowest they ever were was 2002 but I think it is being too optimistic to hope for it to happen again...
    Bronze: 47
    Silver: 57
    Gold: 70
    BMO1: 82
    Where do you get these grade boundaries from?
    And I doubt they'll ever go that low again lol, 82 for BMO1 lmao.
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    (Original post by WhiteScythe)
    These are spoof answers -barely any are correct!
    Agreed!
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    Given the mark scheme after school today (got 85/125) so almost certainly gold and possible into the kangaroo challenge
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    (Original post by sqrt(e/m)=c)
    Can anyone guess what I got overall? I feel I got mid-70s (but I'm Year 10, so ok I guess...?)
    1B 2D 3D 4B 5A 6C 7A 8B 9C 10B 11C 12C 13 14D 15 16D 17D 18E 19D 20A 21 22C 23D 24 25D
    Misread half the questions! (Scale, set with M, .......!!)
    I would say 83, well done!
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    (Original post by Chittesh14)
    Where do you get these grade boundaries from?
    And I doubt they'll ever go that low again lol, 82 for BMO1 lmao.
    stalked the UKMT twitter and saw they tweeted a picture of a spreadsheet of previous grade boundaries
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    (Original post by surina16)
    Hoping they go down (by quite a lot but doubt it)

    Lowest they ever were was 2002 but I think it is being too optimistic to hope for it to happen again...
    Bronze: 47
    Silver: 57
    Gold: 70
    BMO1: 82
    Where do you get the past thresholds from. I will score 77 does that qualify me for a gold?
    • Thread Starter
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    What's good for a Year 10? Came in looking for a BMO, but I'll be lucky if I get a gold
    (Misread half the questions)
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    (Original post by DavidBick)
    Question 24 was fierce, I ended up in a web of inverse and compound tan functions...answer came out though. Got 120 according to general TSR agreement. Anyone know of a BMO1/2 thread going at the moment?
    Here's a nice way to do it: Set up a coordinate system with the origin as the bottom left corner of the square, and let the side length of the square be 2, so that the midpoint of the bottom side is (1,0). Thus the circle is x^2+(y-2)^2=4, and the tangent line is y=m(x-1), where m is the gradient. Hence substitute to give x^2+(mx-m-2)^2=4 -> (1+m^2)x^2 - 2m(m+2)x + m^2 + 4m = 0. Now as the line is tangent, the quadratic should have only one solution (to give just one point of intersection), so using the discriminant, 4m^2(m+2)^2 - 4(1+m^2)(m^2+4m)=0 and by expanding and cancelling, this becomes 3m^2-4m=0, so as m is clearly non-zero, m = 4/3. Hence the ratio is 4:3.
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    (Original post by Chittesh14)
    Me as well, please.



    Tomorrow 9 am onwards.
    Chittesh14, I have PMed you now.
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    Everyone reading this, plz check out the 2002 SMC and just be glad we weren't sitting it then. It is insane compared to this
 
 
 
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