Hello! I'm new to the forum.
I've just recently started to prepare for this test. I was ill for the past few months and have only recently got better... I'm extremely worried now!! The past papers seem impossible!!!
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Oxford MAT 2013/2014 watch

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 02112013 15:04

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 02112013 15:29
(Original post by jadoreétudier)
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:Show1999
a i
b iii
c i couldn't get this one
d iv
e iii
f i
g ii
h iii
j i
k couldn't get this one either! my answer was 4*49*48/(52*51*50) = 0.07 which isn't an option
2. a) (x+3)(x2)
b) a is greater than or equal to 1/4.
x=  1 plus or minus the square root of (1+4a) all over 2
(just the quadratic formula)
c) when x=1, the whole thing equals zero, no matter what b is.. you get 1  b  1 + b
b=1/4 which I've said is the only value for b
3. I don't think I knew what I was doing in this question, especially c.
a) y=x+3
b) subbing in x=1, gives you y=2, no matter the value of a. just like question 2
y= 2 + 1/a  x/a
c) x=1 is one line. I haven't been able to come up with a general formula for the equations of//// the lines, but I think y=(4root3)x + root 3 2, might be one of them
4. Quite confused with this question.
a) for both integrals I got 2/3
b)
c) I got zero, but it says the area should be measured positively and I don't think zero counts as positive. Maybe because I've done top curve minus the bottom curve in my integration I'm showing that I'm measuring it positively?
d) I don't know! It's difficult to see graphically that the difference in areas above and below the x axis will be the same as it is for the quadratic.
5. a) 16*15*14*13= 43680
b)4^4 = 256
c) 4*3*2*1=24
2002
a a
b a
c c
d b
e d
f b
g a
h d
i c
j a
Question 2 appeared in a later paper, so I have the answers for that. 3 didn't work out well for me. 4:Attachment 250205
5. i) i haven't written down an answer, for a guess I'll go for 18
ii) a yes b no c yes d no
iii) a yes b yes c no d yes
For the 1999 paper:
Q1 I got the same as you did except
iii for c), ii for e) and for k) I did: 3*((4*48*47)/(52*51*50)) which is approx= 0.2 Not sure about this though
Q2 Same as you
Q3 The same for part a)
For b) I got the equation "y=x+1" since the slope of the line would be m=1 and by using the coordinates (1,2)
c) I didn't quite get the question I started using the change in x over change in y to represent the slope and got a messy equation, so it's probably wrong...
Q4
a) it is 2/3 for both
c) The total area is 2/4. What I did was to split the area in two part and measure each positively (from 1 to 0 and from 1 to 0), otherwise the areas cancel out to give 0. The top curve minus the bottom curve is right.
d) I'm not sure for this part either
Maybe we had to spot that both integrals over the interval was the same (2/3) and that the area bounded between them was also the same above and below the xaxis(that's why they cancelled out each other) and arrive to some conclusion from there
I didn't really get into question 5 yet, but I suppose that probability is not part of the syllabus for our test on Wednesday? 
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 02112013 15:35
(Original post by MathGirl)
Hello! I'm new to the forum.
I've just recently started to prepare for this test. I was ill for the past few months and have only recently got better... I'm extremely worried now!! The past papers seem impossible!!!
I guess the only thing that we can do right now is to practice a bit more and try our best on that day 
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 704
 02112013 16:06
(Original post by souktik)
Okay, for part (i), this is how I counted
To get to the given position, each bead must move forward 2 steps net. So the number of forward moves is 6+k and the number of backward moves is k. Total is 6+2k, an odd number. This immediately shows us that the answer to b is 0. As for a, one can notice that each way to reach that configuration corresponds uniquely to an arrangement of 2 A's, 2 B's and 2 C's. (If we move the left bead we write A, middle bead  B, right bead  C.) So the number of ways is 6C2 times 4C2 times 2C2, or 90. For part c, two of the beads are moved 2 times forward, the other bead is moved forward thrice and backward once. There are 3C1 ways to determine the bead that moves 3F, 1B. It can move FFBF or FBFF, but not BFFF or FFFB, as it can't go backward from its initial position or move forward 3 steps. (Here, F=forward, B=backward.) Now for the arrangement. 8C4 ways to pick the times when we move the bead that moves 3F,1B. The 4 remaining moves can be assigned to the two other beads in 4C2 times 2C2 ways. Total number of ways = 3.2.8C4.4C2 = 2520. I'm not sure if I got the same answer last time, might have messed up the calculations. Please tell me if I'm undercounting or overcounting, somehow, or if I've made mistakes in my calculation.
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 02112013 16:29
Q1 A (on 2002)
differentiate to find has turning points at: a+root(a^2+3) all over 3
since a<root(a^2+3) always
then it will have one turning point negative x and one turning point positive x
consider the graph of positive cubic with yintercept (0,2)
will have 1 sol. only when x>0
Q1 G (2002)
consider a equilateral triangle slice of your hexagon. By setting side length 2, the line down the centre(the radius of the inner circle) is of length root3
so we have radius1=2 and radius2=root3
area1=4*PI area2=3*PI
therefore ratio 4:3
I could easily have made a mistake but I hope this helps 
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 706
 02112013 16:52
(Original post by Yezi_L)
Could you please explain Q1 A and G ? Is there a particular formula to get the answer that I've missed out(Original post by sun_tzu)
Q1 A (on 2002)
differentiate to find has turning points at: a+root(a^2+3) all over 3
since a<root(a^2+3) always
then it will have one turning point negative x and one turning point positive x
consider the graph of positive cubic with yintercept (0,2)
will have 1 sol. only when x>0
Q1 G (2002)
consider a equilateral triangle slice of your hexagon. By setting side length 2, the line down the centre(the radius of the inner circle) is of length root3
so we have radius1=2 and radius2=root3
area1=4*PI area2=3*PI
therefore ratio 4:3
I could easily have made a mistake but I hope this helps
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 02112013 16:53
I'd really appreciate it if someone could explain that part to me.
2012 1.H
Sent from my GTN7100 using Tapatalk 4 
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 02112013 17:04
(Original post by seohyun)
I'd really appreciate it if someone could explain that part to me.
2012 1.H
Sent from my GTN7100 using Tapatalk 4
This is a lousy explanation, I understand. Tell me if I need to clarify anything.
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 02112013 17:20
(Original post by souktik)
You're looking at the integral from ZERO to x, not from pi to x. For x greater than pi, that integral would be equal to the integral from 0 to pi, plus the integral from pi to x. The integral from 0 to pi is, say, X. X is clearly positive. The integral from pi to x is Y, suppose. Y is clearly negative. Now look at the graph. As long as x hasn't reached 2pi, the lower negative area is less than the full positive area between 0 and pi. So the sum of the areas of the positive and negative sections remains positive. At x=2pi, the positive and negative areas become equal and cancel out, giving a total of 0.
This is a lousy explanation, I understand. Tell me if I need to clarify anything.
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Sent from my GTN7100 using Tapatalk 4 
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 02112013 17:27
Posted from TSR Mobile 
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 02112013 17:28
once you realise sin(sin(x)) looks pretty much the same as sin(x) with the same xintercepts its pretty clear
(Original post by souktik)
Yeah, I did the very same things, if I remember correctly. Hey, sun_tzu, what's your score range like? 90ish?
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I don't know whether to wish bad luck or good luck because surely if everybody else does well then relatively my performance will be worse xD 
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 712
 02112013 17:36
(Original post by sun_tzu)
once you realise sin(sin(x)) looks pretty much the same as sin(x) with the same xintercepts its pretty clear
yeah usually, I found it pretty daunting at first actually but came to realise that the questions always have the same themes pop up
I don't know whether to wish bad luck or good luck because surely if everybody else does well then relatively my performance will be worse xD
Hey, does anyone know what the highest score is usually like? Is it always 100, or usually in the high 90's?
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 02112013 17:39
(Original post by sun_tzu)
Q1 A (on 2002)
differentiate to find has turning points at: a+root(a^2+3) all over 3
since a<root(a^2+3) always
then it will have one turning point negative x and one turning point positive x
consider the graph of positive cubic with yintercept (0,2)
will have 1 sol. only when x>0
Q1 G (2002)
consider a equilateral triangle slice of your hexagon. By setting side length 2, the line down the centre(the radius of the inner circle) is of length root3
so we have radius1=2 and radius2=root3
area1=4*PI area2=3*PI
therefore ratio 4:3
I could easily have made a mistake but I hope this helps
Did you arrive at this conclusion because:
a+root(a^2+3) would give a positive turning point and
aroot(a^2+3) would give a negative one?
...and this would allow you to draw the graph of a positive cubic(inc=>dec.>inc) intersecting the yaxis at (0,2) as it decreases?
Thanks a lot for your help! Now everything's much clearer (I feel stupid for drawing that triangle but not spotting its relation with the radius of the smaller circle) 
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 714
 02112013 17:48
(Original post by sun_tzu)
yeah usually, I found it pretty daunting at first actually but came to realise that the questions always have the same themes pop up 
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 02112013 17:56
(Original post by TSR561)
I've found this too. Stuff like functions and a bit of complex trig always always seems come up.
Which is supposed to make it fairer for student with different backgrounds. Although there is still quite a big difference  those who have a relatively large amount of help available school or those who have private tutoring, compared to those who haven't even covered everything found in the MAT syllabus at school. (My school is an example...we do a weird selection of things but miss out others ) 
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 716
 02112013 18:04
(Original post by souktik)
If you're from the UK and score 90+, I can't think of any plausible scenario in which you DON'T get into Oxford.
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(Original post by Yezi_L)
Did you arrive at this conclusion because:
a+root(a^2+3) would give a positive turning point and
aroot(a^2+3) would give a negative one?
...and this would allow you to draw the graph of a positive cubic(inc=>dec.>inc) intersecting the yaxis at (0,2) as it decreases?
Thanks a lot for your help! Now everything's much clearer (I feel stupid for drawing that triangle but not spotting its relation with the radius of the smaller circle) 
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 717
 02112013 18:21
(Original post by sun_tzu)
haha thanks but how about the scenario where I'm applying to cambridge and imperial not oxford xD
Then there's the scenario where you'll be getting offers from both Have you been practicing with STEP questions instead of the MAT ones? Btw, which college in Cambridge did you apply to if it's alright to ask?
yes you've got it exactly, glad to have helped. By considering the best way to explain it, it actually makes it clearer in my mind so I understand it better too 
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 02112013 18:29
(Original post by Yezi_L)
Then there's the scenario where you'll be getting offers from both Have you been practicing with STEP questions instead of the MAT ones? Btw, which college in Cambridge did you apply to if it's alright to ask?
my current strategy is to do all 8 MAT papers before wednesday
and trinity college
you applying for cambridge too? 
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 719
 02112013 18:39
(Original post by sun_tzu)
haha thanks but how about the scenario where I'm applying to cambridge and imperial not oxford xD
Best of luck, I'm sure you'll get in everywhere! 
TheFuture001
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 02112013 19:09
Hey I was wondering whether my solution makes sense to Q1 H on the second specimen paper http://www.mathshelper.co.uk/Oxford%...20Test%202.pdf
What I basically did was bring the quadratic from the right to the LHS, and allowed c = 2 + (x^2 + 1)^10, which I then used the discriminant on. I ended up with 4 4(x^2 + 1)^10 < 0, thus there are no real solutions. Wanted to know whether it was mathematically correct to do this, claiming the term with the x and power 10 to be part of the constant.
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