The Oxford Maths Admissions Test papers for 1996, 1997, 1998, 2000, 2003, 2004, and 2005, which are currently not available anywhere online, have now been obtained by a Freedom of Information request. The papers are thus attached to this post. Oxford said they don't have solutions in their possession, but the community will presumably be able to produce solutions collaboratively.

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 09082016 15:31

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 09082016 15:34
The 2005 paper is too big to attach so to download it, go to https://www.sendspace.com/file/i116d2

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 09082016 15:54
Thanks for this. Anyone up for solution threadding this?
Posted from TSR Mobile 
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 03092016 17:31
Thank you! Looks good.

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 05092016 10:22
Anyone done solving these papers?
Is it just me or the recent years' questions are harder ? 
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 06092016 03:43
(Original post by lightbulbmoment_)
Haven't finished going through all of these yet however the papers have definitely been getting harder in recent years. Last year's paper was imo quite a jump from the 2014 paper.
Can you post the solutions to some that you have done ? Would be awesome ~~ 
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 07092016 19:45
(Original post by lightbulbmoment_)
Not sure what others would say but my recommendation would be to do the 2015 paper earlier rather than leaving it to use as a mock. I felt it was quite different to previous years, and it would give you an indication to what the standard of this year's paper might be like.
And do you mean my solutions to the 2015 paper? The full solutions are up on the website but I don't mind posting mine too if you want.
There's also a specific MAT 2016 thread on here. There are some people on there that are a lot smarter than me
Oh I Will check that out, thank you. And I was talking about the 1990s solutions .. I don't think the website has got them.
Q.4 c of '97 was inaccessible for me.. would be great if you could help with that! 
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 03102016 15:15
Any of you guys up for some solution checking? I've done the 1996 and 1997 papers so far

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 03102016 16:30
(Original post by El TDP)
Oh I Will check that out, thank you. And I was talking about the 1990s solutions .. I don't think the website has got them.
Q.4 c of '97 was inaccessible for me.. would be great if you could help with that!
Label the pi/2 cos(x) curve as the Acurve, and the pi/2 cos(y) curve as the B curve.
You want to find:
Area_A = the area between the Acurve, the xaxis, and the lines x=0 and x=b, and subtract
Area_B = the area between the Bcurve, the xaxis and the lines x=0 and x=b.
I'm going to assume that you're supposed to know how to integrate cos x. This allows you to find Area_A.
The problem is that to find Area_B in the obvious way, you want a way of expressing the Bcurve in the form y = f(x), and you're going to end up needing to integrate arccos x, which I expect you're not supposed to know how to do.
So, instead, you need to find another way of finding Area_B.
Spoiler:ShowCan you find a region bounded only by straight lines and curveA that will have the same area as Area_B? 
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 03102016 17:59
(Original post by DFranklin)
Really could do with a diagram here, but...
Label the pi/2 cos(x) curve as the Acurve, and the pi/2 cos(y) curve as the B curve.
You want to find:
Area_A = the area between the Acurve, the xaxis, and the lines x=0 and x=b, and subtract
Area_B = the area between the Bcurve, the xaxis and the lines x=0 and x=b.
I'm going to assume that you're supposed to know how to integrate cos x. This allows you to find Area_A.
The problem is that to find Area_B in the obvious way, you want a way of expressing the Bcurve in the form y = f(x), and you're going to end up needing to integrate arccos x, which I expect you're not supposed to know how to do.
So, instead, you need to find another way of finding Area_B.
Spoiler:ShowCan you find a region bounded only by straight lines and curveA that will have the same area as Area_B? 
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 04102016 00:13
What did you guys get for multiple choice? Here's what I got:
1996:
a) i  Very standard  2 distinct roots therefore discriminant>0
b) iii  By testing the first few cases eliminate the remaining options
c) iv  Not sure whether this one is iii or iv but I think a 2x2 system can't have multiple solutions anyway (?)
d) iv  Using the double angle theorem and factorizing yields cosx(2sinx1)=0
e) ii  Not much to say here
f) i  A translation of i then a vertical reflection
g) ii  The nth root of 10000 approaches 1 as n becomes very large and positive
h) iii  Chain rule
j) iii  Fundamental theorem of calculus
k) iii  3(4/52)(48/51)(47/50) is approximately 0.2
1997:
a) iii  Solve or sub points in
b) i  f(2)<0 and f(1)>0
c) iii  Not sure about this one, really bad at combinatorics
d) ii  If 1<x<2, P becomes +()/()=+ therefore Q implies P, but solving P (by considering when 1x is positive/negative) doesn't give Q
e) iv  In the range, cosx ranges from 1 to 1, so cos(cosx) is positive throughout. This rules out ii and iii. i is wrong because for cosx=0 we require x=pi/2, which is greater than 1
f) ii  Not sure whether it's iv  as n approaches infinity, the (1)^n becomes insignificant
g) ii  The coefficient of x^n is (10Cn)/(2^n). Compare magnitudes
h) Not sure about this one but intuitively I think it's ii. Somebody explain?
j) iii  The nth line creates n more regions. As we have 1 region in the beginning the answer is 1+(the sum of the first n positive integers)
k) iv  Reducing the system to echelon form we find that the bottom row is all zeros 
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 04102016 07:33
(Original post by HapaxOromenon3)
It's not too difficult to integrate arccos(x) by parts, and since the MAT at that time did require C3 and C4 techniques as well as C1 and C2, it's possible that this was the intended method... 
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 04102016 07:44
(Original post by Insecure)
What did you guys get for multiple choice? Here's what I got:
1996:
a) i  Very standard  2 distinct roots therefore discriminant>0
b) iii  By testing the first few cases eliminate the remaining options
c) iv  Not sure whether this one is iii or iv but I think a 2x2 system can't have multiple solutions anyway (?)
d) iv  Using the double angle theorem and factorizing yields cosx(2sinx1)=0
e) ii  Not much to say here
f) i  A translation of i then a vertical reflection
g) ii  The nth root of 10000 approaches 1 as n becomes very large and positive
h) iii  Chain rule
j) iii  Fundamental theorem of calculus
k) iii  3(4/52)(48/51)(47/50) is approximately 0.2
1997:
a) iii  Solve or sub points in
b) i  f(2)<0 and f(1)>0
c) iii  Not sure about this one, really bad at combinatorics
d) ii  If 1<x<2, P becomes +()/()=+ therefore Q implies P, but solving P (by considering when 1x is positive/negative) doesn't give Q
e) iv  In the range, cosx ranges from 1 to 1, so cos(cosx) is positive throughout. This rules out ii and iii. i is wrong because for cosx=0 we require x=pi/2, which is greater than 1
f) ii  Not sure whether it's iv  as n approaches infinity, the (1)^n becomes insignificant
g) ii  The coefficient of x^n is (10Cn)/(2^n). Compare magnitudes
h) Not sure about this one but intuitively I think it's ii. Somebody explain?
j) iii  The nth line creates n more regions. As we have 1 region in the beginning the answer is 1+(the sum of the first n positive integers)
k) iv  Reducing the system to echelon form we find that the bottom row is all zeros
For 1997 question c), notice that the teams must consist of 2 people, 2 people, and 1 person. Thus the number of ways would be
5C2 * 3C2 * 1, except that this would then count e.g. (1, 2), (3, 4), (5) as different from (3, 4), (1, 2), (5), so we divide by 2 to give 15, I think.
1997 question f) is indeed 1, as you thought.
For 1997 question h), note that f(3x) gives the same points as f(x) between 1 and 2: it's a well known result that for a definite integral of f(x) with limits a and b, you can replace f(x) with f(a+bx) without changing the value of the integral. Now just multiply by 2 since we have 2*f(3x), to give the answer of 2 as you thought. 
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 04102016 08:47
(Original post by HapaxOromenon3)
1996 question j) should be f(x)f(0), i.e. option (i). Try it with an example function to realise this.
For 1997 question c), notice that the teams must consist of 2 people, 2 people, and 1 person. Thus the number of ways would be
5C2 * 3C2 * 1, except that this would then count e.g. (1, 2), (3, 4), (5) as different from (3, 4), (1, 2), (5), so we divide by 2 to give 15, I think.
1997 question f) is indeed 1, as you thought.
For 1997 question h), note that f(3x) gives the same points as f(x) between 1 and 2: it's a well known result that for a definite integral of f(x) with limits a and b, you can replace f(x) with f(a+bx) without changing the value of the integral. Now just multiply by 2 since we have 2*f(3x), to give the answer of 2 as you thought.
Could you explain 1996 j) a bit more? Also for 1997 c) the teams could also be 311 so we add 5C3=10 to 15, giving 25. 
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 04102016 13:53
(Original post by Insecure)
Hey, thanks.
Could you explain 1996 j) a bit more? Also for 1997 c) the teams could also be 311 so we add 5C3=10 to 15, giving 25. 
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 04102016 14:01
(Original post by HapaxOromenon3)
Actually I've thought about it a bit more and so I now think you're correct about 1996 j). 
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 04102016 14:04
What did you get for section B for the 1996 paper?
It's very weird as section B is VERY elementary (factorise x^2+x6). It seems implausible that 150 minutes is given to complete the entire paper whereas any able candidate would probably be able to finish the paper in about an hour.
I know the earlier papers were easier, but to this extent? 
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 04102016 19:38
(Original post by Insecure)
What did you get for section B for the 1996 paper?
It's very weird as section B is VERY elementary (factorise x^2+x6). It seems implausible that 150 minutes is given to complete the entire paper whereas any able candidate would probably be able to finish the paper in about an hour.
I know the earlier papers were easier, but to this extent? 
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 05102016 04:07
By the way, I think you're all over complicating the solution to 1997 4c).
The integral of the upper curve from 0 to b could easily shown to be (pi/2)sin(b). This region could be divided into three: the shaded region, the triangle like region, and a square. Call them A, B, C respectively.
We wish to find A, which is given by (pi/2)sin(b)BC. We could easily see that C=b^2, and B is simply the integral from b to pi/2 of the lower curve with respect to y. This is simply (pi/2)sin(y) with limits b and pi/2, so B=(pi/2)(pi/2)sin(b).
Combining all the above we have A=(pi/2)sin(b)(pi/2(pi/2)sin(b))b^2=pi(sinb)pi/2b^2 as required. 
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 05102016 15:30
Hello all.
This is important. You must register for the MAT by 15th October. Please make sure you do. Registration with UCAS does not automatically register you for the MAT.
Your school or college can register as a test centre. But this takes a few days, so don't leave it to the last minute.
Alternatively, you can register at an open test centre.
There are details at http://www.admissionstestingservice....wtoregister/.
Gavin
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Updated: October 7, 2016
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