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MAT Prep Thread - 2nd November 2016 watch

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    (Original post by Zacken)
    No, a typical MAT prep-er won't know enough content to attempt the vast majority of STEP questions and will struggle overly much on the few questions that are accessible. It's not worth it, MAT is easy enough - you don't need so much preparation.
    Okay, thank you for the reply
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    (Original post by Zacken)
    No, a typical MAT prep-er won't know enough content to attempt the vast majority of STEP questions and will struggle overly much on the few questions that are accessible. It's not worth it, MAT is easy enough - you don't need so much preparation.
    Arghhhhh I hate how easy you make it seem :P

    Wish I was a fraction as good as you are!

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    Just got 57 on 2015 as my mock, not very impressed... Mostly it was that I messed up the multiple choice I think, normally I get 32-36, and I got 24 this time.

    Don't really know what to do now other than collapse into a ball of disappointment.

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    Nature probably isn't pleased with me because of the amount sheets I've used up for three days' work. I need to plant a few trees, ASAP.

    I am really hoping for the exam to pass smoothly.
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    (Original post by Zacken)
    Yuck, no. Geometry is shite, unless you mean some of the actual geomety like algebraic geometry.
    .......rly? Geometry is bae. cmon euclidean geometry is kinda fun now and again. I cant stand combinatorics though....
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    (Original post by Zacken)
    No, a typical MAT prep-er won't know enough content to attempt the vast majority of STEP questions and will struggle overly much on the few questions that are accessible. It's not worth it, MAT is easy enough - you don't need so much preparation.
    Would you mind explaining 2013 q1 (I)?

    Thanks!
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    Would someone please explain how you do 2013 Q5 iii and iv?

    Thanks!

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    (Original post by RuairiMorrissey)
    Would you mind explaining 2013 q1 (I)?

    Thanks!
    We work out the number of functions that have the value of -1.

    Imagine it as a 'tree'

    3

    6 7

    and so on

    We find how many times this tree splits until we get a maximum of F(100).

    On each row, the number doubles each time - and on the nth line, we have 2^(n - 1) numbers. Taking the rightmost part, we find the highest value is 63. This is on the fifth line, of which there are 2^4 = 16 numbers.

    We need to also find the numbers that are less than or equal to fifty - the sixth line will contain numbers stemming from these. We find the numbers we want on the fifth line are 48, 49 and 50. These therefore produce five numbers (96 to 100).

    Our total number is 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 5 = 1 + 2 + 4 + 8 + 16 + 5 = 36.

    Therefore our answer is 100 - 2 * 36 = 28 (b)
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    How many marks would I get for the question 6 of 2006? http://www.mathshelper.co.uk/Oxford%...est%202006.pdf

    6(i) If the message on the silver box is true, then the message on the gold box is false and hence the prize would be in the gold box. If the message on the silver box is false, then the message on the gold box must be false (if it was true, exactly one would be true) and hence either way the prize is in the gold box.

    6(ii) If the message on the lead box is false, at least two messages would be true, and then the prize would be in both the silver and gold box - this is a contradiction. Hence the message is true. Therefore the messages on both the gold and silver boxes must be false (as the lead message is true) and hence he should choose the lead box.

    6(iii) If the message on the lead box is true, the dagger is in the silver box. He should choose either gold or lead.

    If the message on the lead box is false we have that at most one message is false, and as the lead message is false - both gold and silver messages are true and hence he should choose lead or silver.

    The only consistent option is lead and hence he should choose the lead box.
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    (Original post by molly221)
    Would someone please explain how you do 2013 Q5 iii and iv?

    Thanks!

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    iii) from part a we know that from 0 to 99, there are n+1 numbers with digit sum n.

    If we now put a 1 on front of everything, we essentially reduce the digit sum for the last two numbers by 1 (i.e. 1 + second digit + third digit = n, so second digit + third digit = n - 1)

    We saw in part iii that n-1 will have n solutions.

    Continuing this; numbers in 200 have n-1 solutions for n
    numbers in 300 have n-2 etc.

    I'll leave you to work out the full sum : P



    iv) Use the formula from part iii, removing the solutions from 1-499
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    How do you draw the graph for Q4(i) in 2013 paper? Name:  Screen Shot 2016-10-26 at 15.24.34.png
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    Thanks
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    (Original post by GrungeGirl)
    How do you draw the graph for Q4(i) in 2013 paper? Name:  Screen Shot 2016-10-26 at 15.24.34.png
Views: 128
Size:  23.3 KB

    Thanks
    I started with polynomial long division to get 1 + 2a/a-x. Then it should look like the graph of 1/x but moved about cos of transformations. I think the asymotote is then at x=a and y=1
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    (Original post by some-student)
    How many marks would I get for the question 6 of 2006? http://www.mathshelper.co.uk/Oxford%...est%202006.pdf

    6(i) If the message on the silver box is true, then the message on the gold box is false and hence the prize would be in the gold box. If the message on the silver box is false, then the message on the gold box must be false (if it was true, exactly one would be true) and hence either way the prize is in the gold box.

    6(ii) If the message on the lead box is false, at least two messages would be true, and then the prize would be in both the silver and gold box - this is a contradiction. Hence the message is true. Therefore the messages on both the gold and silver boxes must be false (as the lead message is true) and hence he should choose the lead box.

    6(iii) If the message on the lead box is true, the dagger is in the silver box. He should choose either gold or lead.

    If the message on the lead box is false we have that at most one message is false, and as the lead message is false - both gold and silver messages are true and hence he should choose lead or silver.

    The only consistent option is lead and hence he should choose the lead box.
    I did this paper a few days ago and although I couldn't find a mark scheme, for question 6 I got exactly the same answers as you with exactly the same reasoning. So we would have got the same score... which is hopefully 15/15.

    If you did the rest of the paper, do you mind checking to see if your answers were the same as mine? Question 5 was a 'show that', but for questions 1, 2 and 3, I'm not sure what, if anything, I got wrong.

    Spoiler:
    Show

    Question 1:
    A: d
    B: c
    C: b
    D: a
    E: a
    F: b
    G: b
    H: c
    I: a
    J: c

    Question 2:
    (i) x = -1, x = 0
    (ii) |y| < (2√3)/3 or in another form, -2/√3 < y < 2/√3
    (iii) y = (2√3)/3 and x = (-√3)/3

    Question 3:
    (i) A graph, which hits the x axis at y = 0, 1, 2 and has turning points at (1-√3, -2/3√3) and (1+√3, 2/3√3)
    (ii) k = -2/3√3 (or k = (-2√3)/9 if you rationalised)
    (iii) X1 = 1; g(1) = 0.25
    (iv) X2 = 2
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    (Original post by some-student)
    We work out the number of functions that have the value of -1.

    Imagine it as a 'tree'

    3

    6 7

    and so on

    We find how many times this tree splits until we get a maximum of F(100).

    On each row, the number doubles each time - and on the nth line, we have 2^(n - 1) numbers. Taking the rightmost part, we find the highest value is 63. This is on the fifth line, of which there are 2^4 = 16 numbers.

    We need to also find the numbers that are less than or equal to fifty - the sixth line will contain numbers stemming from these. We find the numbers we want on the fifth line are 48, 49 and 50. These therefore produce five numbers (96 to 100).

    Our total number is 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 5 = 1 + 2 + 4 + 8 + 16 + 5 = 36.

    Therefore our answer is 100 - 2 * 36 = 28 (b)
    Understood it now, thanks.
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    (Original post by GrungeGirl)
    How do you draw the graph for Q4(i) in 2013 paper? Name:  Screen Shot 2016-10-26 at 15.24.34.png
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Size:  23.3 KB

    Thanks
    Start by thinking about what happens when x=0, x is approximately -a and approximately a
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    anyone have any tips on how to do questions which have a general style?
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    (Original post by Zacken)
    Integrate both sides of the given identity from -1 to 1, remember that the existing integral in that identity is just a number, you might find it helpful to call it u = \int_{-1}^1 f(t) \, \mathrm{d}t and then integrate both sides like so:

    \int_{-1}^1 6 \, \mathrm{d}x + \int_{-1}^{1} f(x) \, \mathrm{d}x = 2\int_{-1}^1 f(-x) \, \mathrm{d}x + u\int_{-1}^1 3x^2 \, \mathrm{d}x

    which then simplifies down to 12 + u = 2u + 2u where u is the answer you want.
    I don't fully understand how you got that last term?
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    (Original post by Zacken)
    Yuck, no. Geometry is shite, unless you mean some of the actual geomety like algebraic geometry.



    okay, integrate x dx between -1 and 1, what number do you get?
    Integrate t dt between -1 and 1, what number do you get?

    The t and x inside the integral don't matter, they're called dummy variables as long as the integral is a fefinite one with limits, then the letter doesn't matter since the integral just represents a number anyway.

    It's a bit (actually, almost exactly) why in summation notation, it doesn't matter what letter you choose. The sum from k=1 to 100 of k is the same as the sum from m=1 to 100 of m. The letters are just placeholders.

    This has tripped up many a student in the past.
    yes that's the type I meant. Like this question :https://www.imperial.ac.uk/media/imp...tions-2010.pdf
    Q3 I
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    Can someone help with Q4 (v) 2009?
    Dont understand what it means by 'unique value of a'.
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    Can someone please explain to how to do this? It is from 2011, Q3.
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