MAT Prep Thread - 2nd November 2016

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    (Original post by lewman99)
    It's difficult to say for sure, but in general I've been giong under the assumption that each logical "step" is roughly 1 mark. So for a logic question (an introductory parts are normally 3 marks), splitting it up into different cases gets you 1 mark, evaluating each case gives you a mark in total, and coming to a conclusion gets you another mark.

    But honestly, it's probably not worth worrying too much about the exact mark when doing past papers. What's way more important is making sure that you've understood the question after looking through the marking sheme and perhaps trying it again,.

    Also, it's worth keeping in mind that the MAT has large amounts of discretion. Those marking your paper will see your working and the different ideas you've tried, even if they weren't correct. The MAT's only a part of your application, not the be all end all. There's no fixed cutoff for scores or anything like that - a great MAT score can make up for a not-so-great personal statement, and vice-versa.

    tl;dr: roughly 1 logical step/ark, but an exact score isn't as important as other exams.
    Personal statement can never make up for low marks in anything. It counts for literally nothing unless you started swearing in your PS.


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    For question 3 MAT 2010 (iv) why does the y=c line coincide with the second hump? It has to give exactly 5 solutions right? Is it because when it touches the line we count that as either 1/2 solutions?
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    (Original post by danielhx)
    For question 3 MAT 2010 (iv) why does the y=c line coincide with the second hump? It has to give exactly 5 solutions right? Is it because when it touches the line we count that as either 1/2 solutions?
    4 solutions - as 0 won't be a solution in this case. Touching the second hump is the only way to get 4 solutions.
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    (Original post by RichE)
    4 solutions - as 0 won't be a solution in this case. Touching the second hump is the only way to get 4 solutions.
    oh right! thank you but in general when it touches it can be considered as a repeated root hence 2 roots?
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    (Original post by danielhx)
    oh right! thank you but in general when it touches it can be considered as a repeated root hence 2 roots?
    Well that depends on how the question would be phrased. You might call such a case "2 solutions counting multiplicities".
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    (Original post by KloppOClock)
    How do you get started on spec a question 4ii
    ..
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    (Original post by KloppOClock)
    ..
    find the lengths of AR, BR, BP, CP, CQ and AQ first

    your aim is to do big triangle minus the three smaller outer ones
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    (Original post by DylanJ42)
    find the lengths of AR, BR, BP, CP, CQ and AQ first

    your aim is to do big triangle minus the three smaller outer ones
    ok thank you
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    (Original post by Moogle679)
    Awesome thank you, I really wasn't sure how it all worked as you mentioned it's really not specific at all. Does it show anywhere rough marks for longer questions?
    I haven't gone through many papers yet but in the 07 solutions it shows how many marks each part of the longer questions is worth:
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    Hi,
    Can someone please help me with 1H on the 2012 paper...
    Not really too sure how to do it, I've looked at the solutions but I still don't quite understand it... Can someone please explain it step by step? thanks!
    physicsmaths


    Also on Question 5 - MAT 2012, I got everything right except the last question where they ask what is (x 8k, y 8k)...
    I have no idea how you go from (X8, Y8) = (16,0) to (X8k, Y8k) = (16^k, 0)....
    I feel like I'm being stupid about this one ^ but I genuinely have no idea
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    Okay I am confused by this part of the question. From what I have understood, it is saying to prove the numbers in the bottom two rows and columns are the same in an n*n square if you apply R then C or C then R once. So in order to show this, you would have to show this holds for n= 1,2,3 and n>3 right? So why is this explanation good enough?

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    (Original post by KloppOClock)
    Okay I am confused by this part of the question. From what I have understood, it is saying to prove the numbers in the bottom two rows and columns are the same in an n*n square if you apply R then C or C then R once. So in order to show this, you would have to show this holds for n= 1,2,3 and n>3 right? So why is this explanation good enough?

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    I have found some of their explanations to not be the easiest to understand or as detailed as you'd possibly like.

    I understood it as this: R and C both only ever affect the top 2 rows or the left 2 columns. So therefore as long as n > 2 the right and bottom n-2 columns aren't affected by either R or C whichever order they are performed in. It's a tricky one to explain imo.
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    (Original post by Bruhh)
    I have found some of their explanations to not be the easiest to understand or as detailed as you'd possibly like.

    I understood it as this: R and C both only ever affect the top 2 rows or the left 2 columns. So therefore as long as n > 2 the right and bottom n-2 columns aren't affected by either R or C whichever order they are performed in. It's a tricky one to explain imo.
    i understand that for n>3 but for n=3, r then c and c then r change the top left corner of the final 2x2 square, so wouldnt you have to prove for n=3 as well as just stating that statement for n>3
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    (Original post by KloppOClock)
    i understand that for n>3 but for n=3, r then c and c then r change the top left corner of the final 2x2 square, so wouldnt you have to prove for n=3 as well as just stating that statement for n>3
    You don't need to prove anything for the question, it just says explain which is why its quite ambiguous as to how much detail is needed in the answer.

    Yeah the top left corner being changed is fine? It just has to be the same either way round you apply R and C, which it would be. If you look at the n=3 example they give, both the bottom row and right column remain unchanged whether you do it the way they've shown - R and then C or the other way C and then R.
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    (Original post by KloppOClock)
    i understand that for n>3 but for n=3, r then c and c then r change the top left corner of the final 2x2 square, so wouldnt you have to prove for n=3 as well as just stating that statement for n>3
    n=3 isn't a special case. It can be treated by the general argument. The effect of C, whether done first or second, is to do nothing on the right n-2 columns. So RC and CR both have the effect of just R on those columns.
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    (Original post by Bruhh)
    You don't need to prove anything for the question, it just says explain which is why its quite ambiguous as to how much detail is needed in the answer.

    Yeah the top left corner being changed is fine? It just has to be the same either way round you apply R and C, which it would be. If you look at the n=3 example they give, both the bottom row and right column remain unchanged whether you do it the way they've shown - R and then C or the other way C and then R.
    well yeah the bottom row and right column dont change, but it says the last two, not one. when n>3 thats easy to show as the affected rows and columns dont crossover when you apply r then c. but with n=3. after you apply r then c or c then r, the top left square of the remaining 2x2 square has been transformed twice, which surely you would then need to show that this transformation is the same regardless of order as well as just saying the initial statement?
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    (Original post by KloppOClock)
    well yeah the bottom row and right column dont change, but it says the last two, not one. when n>3 thats easy to show as the affected rows and columns dont crossover when you apply r then c. but with n=3. after you apply r then c or c then r, the top left square of the remaining 2x2 square has been transformed twice, which surely you would then need to show that this transformation is the same regardless of order as well as just saying the initial statement?
    Well it says 'the right n-2 columns' so when n is say 4 this will be the right 2 columns, but when n is 3 the n-2 columns is only the first column. I think it's just bad wording.
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    (Original post by KloppOClock)
    this is what im confused about
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    R and C effect the top row and left most column, you're showing it the other way round?
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    (Original post by Bruhh)
    R and C effect the top row and left most column, you're showing it the other way round?
    i deleted the post but i think youre right about the wording being all the rows and columns apart from the first two, makes a lot more sense
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    (Original post by KloppOClock)
    i deleted the post but i think youre right about the wording being all the rows and columns apart from the first two, makes a lot more sense
    Yeah you've made me have to think very hard about this, I just think it's a case of bad wording, I wouldn't stress yourself about it.
 
 
 
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