The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Oxford MAT 2013/2014

Scroll to see replies

Reply 560
Avatar for cem101
cem101
11
stuck on 3i on 2010
have sinx<x<tanx
dont understand how they get from that to the answer help !
Reply 561
Avatar for IceKidd
IceKidd
i take 20 minutes on each of the longer questions. leave about 50 minutes for the multiple choice.

then have about 15 mins to go through any bits i missed (usually the very last part of a longer q)
Reply 562
Avatar for 31541
31541
2
Original post by souktik
Thanks for the clarification, I was under the impression that my solutions need to be singularly novel to earn full credit. I hope that I'll be able to explain everything logically. :smile:

What kind of MAT scores do accepted international applicants generally have? 80 minimum?

Also, are there any current Oxford math students here who would possibly consider reading my statement and telling me how bad it is? I know it's rather terrible, I just want the opinion of a current student. I really don't want to put it up online and ask for comments from everyone, though.


The Oxford website (the page which has the past papers) says the average score of successful applicants for each year.

I'm assuming you're hoping for a place in an Ivy too?
Original post by cem101
stuck on 3i on 2010
have sinx<x<tanx
dont understand how they get from that to the answer help !


No offense, but you do realize that there are answers to all tests online? There's nothing wrong with needing help, but once people start giving you tips, it's not exactly a realistic simulation of the actual test situation, so you might as well look those up.

If the online answers are not explanatory enough, ignore this comment. In that case I understand why you would come to TSR for help.
I asked earlier for help on question 5 (v) from 2007 earlier but I don't think anyone saw the post. I don't understand the answer given in the solutions, could someone please explain the answer?
Reply 565
Avatar for 31541
31541
2
Original post by SherlockHolmes
I asked earlier for help on question 5 (v) from 2007 earlier but I don't think anyone saw the post. I don't understand the answer given in the solutions, could someone please explain the answer?


What element don't you understand?
Reply 566
Avatar for 1298977
1298977
1
Original post by CD315
The Oxford website (the page which has the past papers) says the average score of successful applicants for each year.

I'm assuming you're hoping for a place in an Ivy too?


Hoping? Yeah. Expecting? No. :tongue:
Actually, I'm applying to a lot of American universities, my top choices being non-Ivy institutes where I don't stand any chance at all (MIT, Stanford, UChicago and Caltech). As for Oxford, my chances of being able to attend are really low even if I miraculously get accepted. I'm completely dependent on the Reach Scholarship if that happens, can't go if I don't get it.

Yes, I've seen those, I was just guessing that it'd be higher for international students.
Reply 567
Avatar for 31541
31541
2
Original post by souktik
Hoping? Yeah. Expecting? No. :tongue:
Actually, I'm applying to a lot of American universities, my top choices being non-Ivy institutes where I don't stand any chance at all (MIT, Stanford, UChicago and Caltech). As for Oxford, my chances of being able to attend are really low even if I miraculously get accepted. I'm completely dependent on the Reach Scholarship if that happens, can't go if I don't get it.

Yes, I've seen those, I was just guessing that it'd be higher for international students.


Awesome. I wish I had the extra curriculars to even apply to the US - would love to go.

I doubt it'd be much higher for internationals.
Original post by CD315
What element don't you understand?


Thanks for the response.

I don't understand what the solutions from the website is saying. They haven't used any words to explain what they're doing so I assume it is meant to be obvious.

I was hoping someone could explain how they work towards the solution.

I have attached the answer.
Question 5(v).png10.3KB
Original post by TheGoldenRatio
how is the answer to this DCapture.JPG



how did you know the answer was d?
This is in the 2006 paper which there isn't solutions for.
Reply 570
Avatar for 1298977
1298977
1
Original post by CD315
Awesome. I wish I had the extra curriculars to even apply to the US - would love to go.

I doubt it'd be much higher for internationals.


Oh, I don't have great extra curriculars either, and the ones I do have are academic ones. But I figured that it wouldn't hurt to try.
Reply 571
Avatar for 31541
31541
2
Original post by SherlockHolmes
Thanks for the response.

I don't understand what the solutions from the website is saying. They haven't used any words to explain what they're doing so I assume it is meant to be obvious.

I was hoping someone could explain how they work towards the solution.

I have attached the answer.


Using the definition of g(n) where n is even, you can do what you did in part iv.

So you have;

g(2l+2k)=1+g(2l−1+2k−1)=1+1+g(2l−2+2k−2)+...g(2^l+2^k) = 1 + g(2^{l-1}+2^{k-1}) = 1 + 1 + g(2^{l-2} + 2^{k-2}) + ...

Now, since l>k, we know that the second power of 2 will approach 0 before the first one. So we have;

k+g(2l−k+20)=k+g(2l−k+1)k + g(2^{l-k} + 2^0) = k + g(2^{l-k} + 1)

Since this is an odd term, you use the other definition for g(n) and combine this with your answer to part iv (subbing l-k instead of just k).
(edited 10 years ago)
Original post by CD315
Using the definition of g(n) where n is even, you can do what you did in part iv.

So you have;

g(2l+2k)=1+g(2l−1+2k−1)=1+1+g(2l−2+2k−2)+...g(2^l+2^k) = 1 + g(2^{l-1}+2^{k-1}) = 1 + 1 + g(2^{l-2} + 2^{k-2}) + ...

Now, since l>k, we know that the second power of 2 will approach 0 before the first one. So we have;

k+g(2l−k+20)=k+g(2l−k+1)k + g(2^{l-k} + 2^0) = k + g(2^{l-k} + 1)

Since this is an odd term, you use the other definition for g(n) and combine this with your answer to part iv (subbing l-k instead of just k).


Now I understand! Thank you very much.
Reply 573
Avatar for 31541
31541
2
Original post by SherlockHolmes
Now I understand! Thank you very much.


Awesome :smile:
I've done 1999 and 2002 from this website:
http://www.mathshelper.co.uk/oxb.htm
Although there aren't solutions. Would anyone else like to do them too and then we can compare our answers?
My answers for both papers:

Spoiler

Reply 575
Avatar for MEPS1996
MEPS1996
4
Anybody else find the MAT paper in 2010 extremely hard? I went from 85 percent to under 50 in this one. :frown:
Reply 576
Avatar for 31541
31541
2
Anyone absolutely bricking it? I feel like I'm not ready but I don't have anything else to do - I've done all the papers available to me twice..
Reply 577
Avatar for 1298977
1298977
1
Original post by MEPS1996
Anybody else find the MAT paper in 2010 extremely hard? I went from 85 percent to under 50 in this one. :frown:


That sounds scary, I'll take a look. Tell me, how are you grading yourself?
Reply 578
Avatar for 1298977
1298977
1
Original post by CD315
Anyone absolutely bricking it? I feel like I'm not ready but I don't have anything else to do - I've done all the papers available to me twice..


Quite the opposite, I've looked at just the last two papers. Now I'll check out 2010. :tongue:
Reply 579
Avatar for IceKidd
IceKidd
Original post by MEPS1996
Anybody else find the MAT paper in 2010 extremely hard? I went from 85 percent to under 50 in this one. :frown:


i found it hard...but still managed to complete as much. i got 84 in that one

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you