MAT Prep Thread - 2nd November 2016

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    (Original post by Mystery.)
    Do you guys normally do all the questions or do like 2/3 from Q2-3 and try to secure all the marks?
    I'm not really sure what you mean. Generally, I do questions in the order of the booklet (for me that's 1,2,3,5 and 6), and where I can I try to do a whole question at once, because of how they progress. But if I'm really struggling I'll usually either skip the last part or a "show that" part (because I can still progress in the question with the given result) and come back to it later. It's kind of a dangerous game, though - the part you're skipping could be 1-2 marks, or it could be 4-6.

    If you're really struggling for time, then it's probably better to do the first few parts from all your questions rather than try and complete fewer questions, as earlier marks in a question tend to be easier. But really it's personal preference.
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    (Original post by Mystery.)
    Do you guys normally do all the questions or do like 2/3 from Q2-3 and try to secure all the marks?
    you should do all of them
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    I don't understand how they go about the long division in 2013 question 1, G?

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    Hi can someone please explain MAT 2007, Question 7d...

    Is there an alternative method to the one provided in the solutions?
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    (Original post by Someboady)
    Hi can someone please explain MAT 2007, Question 7d...

    Is there an alternative method to the one provided in the solutions?
    This is pretty similar to the given solution, but may help

    Assume we have N of the first cards and M of the second cards to make a match.

    Therefore 7N + M = N + 10M (on the first tile, we have N lots of 7 Xs on the top, and M lots of one X on the bottom; on the second tile, we have N lots of one X on the top and M lots of 10 Xs on the bottom - to make a match).

    We can rewrite this as 2N = 3M.

    Trialing we find N = 3 and M = 2, and get 2(3) = 3(2), which is correct.

    Therefore we have 3 + 2 = 5 tiles here.

    Seeing the working in bold clarifies that the only way of making this work on any occasion will be by using N as a multiple of 3 and M as a multiple of 2.

    Therefore N = 3k, M = 2k. 3k + 2k = 5k. Therefore we always have a multiple of 5.
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    2009, 1(f)

    How have they got the turning points to be x=0,1,2 in the worked solutions

    I've got x=-1,0, 3 through differentiating, equating to zero and then factorising
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    (Original post by jamestg)
    2009, 1(f)

    How have they got the turning points to be x=0,1,2 in the worked solutions

    I've got x=-1,0, 3 through differentiating, equating to zero and then factorising
    After differentiating y with respect to x, the given cubic can be factorised to = 12x(x−1)(x−3), the solutions of which are 0, 1 and 3 (I think they made a typo saying 2 because they use 3 later)

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    (Original post by some-student)
    After differentiating y with respect to x, the given cubic can be factorised to = 12x(x−1)(x−3), the solutions of which are 0, 1 and 3 (I think they made a typo saying 2 because they use 3 later)

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    Thank you!
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    someone help with q2 iii. iv 2010??
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    (Original post by Mystery.)
    someone help with q2 iii. iv 2010??
    For part (iii), consider the distance between the starting point, and the 4 points that are closest to it. What does this tell you about how the points are "consumed" as the circle increases in size?

    Part (iv)'s fairly similar, except you're trying to show the opposite of (iii). I've got some more detailed information below (try it without first). For part (iii):
    Spoiler:
    Show
    In part (iii), the point (0.5,0.5) is the same distance from each of the 4 closest points (0,0), (0,1), (1,0) and (1,1). Each of these points is the same distance from 4 other points, and so on. This means that, as r increases, points are added in "chunks" of 4, so it's a multiple of 4 for any value of r.

    And for part (iv):
    Spoiler:
    Show
    The point (sqrt(2), (sqrt(3)) is a different distance from each of the 4 closest points. Using similar logic to (iv), this means that you can expand the circle so that you can consume points 1 at a time, so the value of the function can be any positive integer.
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    (Original post by lewman99)
    For part (iii), consider the distance between the starting point, and the 4 points that are closest to it. What does this tell you about how the points are "consumed" as the circle increases in size?

    Part (iv)'s fairly similar, except you're trying to show the opposite of (iii). I've got some more detailed information below (try it without first). For part (iii):
    Spoiler:
    Show
    In part (iii), the point (0.5,0.5) is the same distance from each of the 4 closest points (0,0), (0,1), (1,0) and (1,1). Each of these points is the same distance from 4 other points, and so on. This means that, as r increases, points are added in "chunks" of 4, so it's a multiple of 4 for any value of r.
    And for part (iv):
    Spoiler:
    Show
    The point (sqrt(2), (sqrt(3)) is a different distance from each of the 4 closest points. Using similar logic to (iv), this means that you can expand the circle so that you can consume points 1 at a time, so the value of the function can be any positive integer.
    Thank you so much. Can I ask how are you preparing for this because I noticed you are a scottish applicant too. Have you learnt everything that wasn't covered in Higher for mat?
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    (Original post by Mystery.)
    Thank you so much. Can I ask how are you preparing for this because I noticed you are a scottish applicant too. Have you learnt everything that wasn't covered in Higher for mat?
    First, I learned the stuff that wasn't in the Higher syllabus for the MAT. Thankfully, quite a bit of that comes up early in the Advanced Higher when we were taught it (arithmetic/geometric series and the Binomial Thereom), but there's the odd thing that isn't in Advanced Higher at all (like the trapezium rule). Typically you can just look them up and get enough information, or use something like MyMaths if your school uses that. I'm still a bit rusty on the trapezium rule, but it's not a major part of the MAT and I've still got a bit of time.

    After that, I went through some of the resources for the MAT from the Reimann Zeta Club (look it up). They've got worksheets for various areas of the MAT (e.g Gemetry, Calculus and Logs Comparing Values), and a great PowerPoint going through multiple choice questions by topic. I started with the PowerPoint (there's a few formatting issues, but it's still mostly readable), then I used some worksheets to brush up on areas abetween past papers.

    Finally, I'm just working through past papers - roughly 1 a week (I'm skipping this week, but I'm doing last year's paper next week). Between papers, I'm looking through my work, looking at things I could improve on/make clearer, and going back to some of the worksheets to get some more practice.
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    (Original post by some-student)
    This is pretty similar to the given solution, but may help

    Assume we have N of the first cards and M of the second cards to make a match.

    Therefore 7N + M = N + 10M (on the first tile, we have N lots of 7 Xs on the top, and M lots of one X on the bottom; on the second tile, we have N lots of one X on the top and M lots of 10 Xs on the bottom - to make a match).

    We can rewrite this as 2N = 3M.

    Trialing we find N = 3 and M = 2, and get 2(3) = 3(2), which is correct.

    Therefore we have 3 + 2 = 5 tiles here.

    Seeing the working in bold clarifies that the only way of making this work on any occasion will be by using N as a multiple of 3 and M as a multiple of 2.

    Therefore N = 3k, M = 2k. 3k + 2k = 5k. Therefore we always have a multiple of 5.
    Thank you! much appreciated!
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    (Original post by lewman99)
    First, I learned the stuff that wasn't in the Higher syllabus for the MAT. Thankfully, quite a bit of that comes up early in the Advanced Higher when we were taught it (arithmetic/geometric series and the Binomial Thereom), but there's the odd thing that isn't in Advanced Higher at all (like the trapezium rule). Typically you can just look them up and get enough information, or use something like MyMaths if your school uses that. I'm still a bit rusty on the trapezium rule, but it's not a major part of the MAT and I've still got a bit of time.

    After that, I went through some of the resources for the MAT from the Reimann Zeta Club (look it up). They've got worksheets for various areas of the MAT (e.g Gemetry, Calculus and Logs Comparing Values), and a great PowerPoint going through multiple choice questions by topic. I started with the PowerPoint (there's a few formatting issues, but it's still mostly readable), then I used some worksheets to brush up on areas abetween past papers.

    Finally, I'm just working through past papers - roughly 1 a week (I'm skipping this week, but I'm doing last year's paper next week). Between papers, I'm looking through my work, looking at things I could improve on/make clearer, and going back to some of the worksheets to get some more practice.
    Yeah, I've been doing that too and I have improved in the multiple choice as the same- ish kind of questions come up but for the other questions sometimes I just don't even know where to begin. There's always one small trick and if you don't spot it then you are pretty much doomed for the whole question.

    Unfortunately, our school has decided to do the unit 3 first which is all calculus and none of that is in the mat so I have been studying sequences etc which I am finding quite easy in terms of the textbook questions but when it comes to the MAT i am not doing so well. Same goes with the trapezium rule, most mat questions seem to concern when the area is divided in some n parts. There was a fairly significant question in 2014 on the trapezium rule so I don't know if they'd bring it up again in the written questions. All the other questions on it are in the multiple choice i think.

    How many papers have you done and what scores are you getting if you don't mind me asking?
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    (Original post by Mystery.)
    Yeah, I've been doing that too and I have improved in the multiple choice as the same- ish kind of questions come up but for the other questions sometimes I just don't even know where to begin. There's always one small trick and if you don't spot it then you are pretty much doomed for the whole question.

    Unfortunately, our school has decided to do the unit 3 first which is all calculus and none of that is in the mat so I have been studying sequences etc which I am finding quite easy in terms of the textbook questions but when it comes to the MAT i am not doing so well. Same goes with the trapezium rule, most mat questions seem to concern when the area is divided in some n parts. There was a fairly significant question in 2014 on the trapezium rule so I don't know if they'd bring it up again in the written questions. All the other questions on it are in the multiple choice i think.

    How many papers have you done and what scores are you getting if you don't mind me asking?
    Yeah, I know a bunch of people doing things in different orders - some going in the order of the old AH (I think that's what we're doing), others in the new AH order, some doing all the calculus first... It's a real pain. And the textbook (the Maths in Action book) starts with proofs and number theory, again in a completely different order.

    For sequences, a lot of the question 5s use geometric series. For sequences, it's really about spotting and identifying where sequences can take place. Also, there's quite a few questions with 2 sequences "weaved" with each other (like 1,1,2,1/2,4,1/4...), and it asks you to find the sum of the first 2n term, in which case you'd just split it into 2 sequences and get the sum of each of them. I remember doing the 2014 question you're talking about, and generally with the trapezium rule questions it's about being comfortable with the formula, and also working with geometric series within the formula. Broadly, those questions can be split into 2 "types" - one based on flexibility with the formula, like the 2014 question, and questions based more about the idea of the trapezium rule (e.g the question tha asks how many strips you'd need for the estimate to be exact).

    Right now, I've finished all of the papers on the main website up to 2014, and I'm getting roughly 5-15 marks above the successful average each time (though it's not exact, and since I'm applying for Maths and CompSci the average might be slightly different).
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    I was a confident happy young man until I did 2013 and 2015 and realised I'm nothing but a little boy.
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    For question 3iv on 2009 I answered it differently but am wondering if it is still correct.

     \displaystyle \frac{3n-1}{(4n-1)} \leq An/(B+n)



1 is negligible for large values of n. Cancel the n's. Ignore B for large values of n as B is a constant.

Left with;

0.75 \leq A

    Would this still get the marks?
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    I've done a lot of practise and the multiple choice questions seem easier, but not questions 2-5. I usually can understand the method to get the answer, but finding what to do for the first step can sometimes be really difficult and I'm worried that in the exam I'll just have no idea where to start. I'm just gonna grind the papers during half term again and hopefully I'll feel better, but for the most part setting my eyes on imperial in stead of oxford now hahaha
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    (Original post by KloppOClock)
    For question 3iv on 2009 I answered it differently but am wondering if it is still correct.

     \displaystyle \frac{3n-1}{(4n-1)} \leq An/(B+n)



1 is negligible for large values of n. Cancel the n's. Ignore B for large values of n as B is a constant.

Left with;

0.75 \leq A

    Would this still get the marks?
    I'd say most likely yeah (it's how I originally did this part).
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    can someone help me understand 1G of 2014? I understand that the term required is nC4 i dont see how the coefficient is 4(nC4)
 
 
 
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