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# MAT Prep Thread - 2nd November 2016 watch

1. (Original post by KloppOClock)
I was feeling sort of confident for this MAT exam... then I did the 2015 paper.
I fear I will experience this when i do the 2015 paper
2. (Original post by KloppOClock)
For 2014, question 4i, I managed to get the formula by using a C3 trig identity. Seen as this is meant to be based upon C1/C2, how else could you have got it, the mark scheme is not very helpful.

also i dont get this:
Spoiler:
Show
Note that F=AX/AC also. That means AX=kAC. Since AC is fixed, for each value of k, only one possibility of AX can be obtained. By applying cosine law two times, only one value of angle ABD can be obtained. Hence beta has a unique solution (pi/2-angle ABD/2) for every k
3. How many of you guys already did 2015 MAT? How's your marks?
4. How tf do they draw that for 2012 q5 ?!
5. (Original post by Mystery.)
How tf do they draw that for 2012 q5 ?!
Use the recursive definition of Pn+1 = PnLPnR.

We are given P1 drawn, so for P2 we turn left at the end of P1 and draw P1 again (in a different direction to the left turn). We now have P2 so for P3 we turn left at the end of P2 and draw P2 again. We now have P3 so for P4 we turn left at the end of P3 and draw P3 again.

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6. (Original post by 11234)
Help pls with q5 of 2002 paper

http://www.mathshelper.co.uk/Oxford%...est%202002.pdf
You have 6 steps which means that they all have to go forward.
There are 6 ways they can come to the second row then further 6 ways to the next row i.e 123,321 (last one goes first then 2 and first one goes last),231 etc
so 6x6=36 ways then
for b) i think its 0 ways as I when I draw it out It needs at least 8 steps but I maybe wrong
7. (Original post by some-student)
Use the recursive definition of Pn+1 = PnLPnR.

We are given P1 drawn, so for P2 we turn left at the end of P1 and draw P1 again (in a different direction to the left turn). We now have P2 so for P3 we turn left at the end of P2 and draw P2 again. We now have P3 so for P4 we turn left at the end of P3 and draw P3 again.

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cheers
8. Does anyone have any advice on tackling q5's because I am doing okay everywhere else but I always get q5 messed up
9. (Original post by LaserRanger)
How many of you guys already did 2015 MAT? How's your marks?
68 marks (Computer Science applicant)

Q2 and Q5 always mess me up, in the actual exam I'm just going to leave those 2 until the end so I can actually get some good marks on Q1, Q6 and Q7.
10. Help with Q4 2007 Please?
11. (Original post by Mystery.)
which part?
12. (Original post by KloppOClock)
which part?
iii

also, how does it go from p^2+q^2>=4pq to that?
Attached Images

13. (Original post by Mystery.)
iii

also, how does it go from p^2+q^2>=4pq to that?
where is that image from?

for part iii,
Spoiler:
Show
work out the area of one of those triangles, multiply it by two and take away the sector.
Here is how:
Spoiler:
Show
You know the base of the bottom triangle is the X coord of P take away one. You know the height of the triangle is 1. Base * Height * 0.5

(cot(x)+sec(x))*1/2

Times that by two.

Take away the sector which is 1/2 * r^2 * x which gives you pi/3

in order to know the angles just consider the triangles and quadrilateral separately as angles in a triangle ad up to pi and quads add upto 2pi.
14. (Original post by Quido)
68 marks (Computer Science applicant)

Q2 and Q5 always mess me up, in the actual exam I'm just going to leave those 2 until the end so I can actually get some good marks on Q1, Q6 and Q7.
I am also a Computer Science applicant - do you know what mark we should be aiming for as there is no indication on the Oxford website about questions 6 and 7?
15. (Original post by ComputeiT)
I am also a Computer Science applicant - do you know what mark we should be aiming for as there is no indication on the Oxford website about questions 6 and 7?
I dont really know, I have been aiming for as high as possible. My aim is to get around 75
16. (Original post by KloppOClock)
where is that image from?

for part iii,
Spoiler:
Show
work out the area of one of those triangles, multiply it by two and take away the sector.
Here is how:
Spoiler:
Show
You know the base of the bottom triangle is the X coord of P take away one. You know the height of the triangle is 1. Base * Height * 0.5

(cot(x)+sec(x))*1/2

Times that by two.

Take away the sector which is 1/2 * r^2 * x which gives you pi/3

in order to know the angles just consider the triangles and quadrilateral separately as angles in a triangle ad up to pi and quads add upto 2pi.
Its from 2008 Q4, and I actually meant part ii sorry but I still need part 3 too so thanks
17. (Original post by Mystery.)
Its from 2008 Q4, and I actually meant part ii sorry but I still need part 3 too so thanks
part two of 2007?

for the first part, if the top left angle is X and you do 90-X then it is the same angle as the bottom right so the areas would be equal as the areas would be symmetrical

areas A and B are equal as the angles are the same.
Work out the bottom left square and the rest of the circle and the entire triangle
Area of Big Triangle - (1x1 square on left) - (rest of the circle)

then divide by two.
18. (Original post by Quido)
I dont really know, I have been aiming for as high as possible. My aim is to get around 75
Yeah 75 is a good aim - multiple choice section, logic question 6 and most of question 7 are ok but I am struggling most with questions 2 and 5.
19. (Original post by ComputeiT)
Yeah 75 is a good aim - multiple choice section, logic question 6 and most of question 7 are ok but I am struggling most with questions 2 and 5.
Same with me, Q2 and Q5 are quite difficult. I've decided to do those last in the real exam so I can get marks everywhere else at least
20. (Original post by 11234)
Help pls with q5 of 2002 paper

http://www.mathshelper.co.uk/Oxford%...est%202002.pdf
(i) (a) All counters must be moved forward twice only to achieve the position.

We need to consider how we can pick a first counter, then a second counter and then a third counter, and how we can place the moves for each of them.

Assume we have a set of {1, 2, 3, 4, 5, 6} which is the move number.

This counter can choose its move from the set 6 choose 2 times = 15 times. We use n choose k here as moving on the 2nd and 5th term is the same as moving on the 5th and 2nd term.

As we have taken out two moves from the set, this counter can choose its move 4 choose 2 times = 6 times.

For the third counter, its moves will simply be the remaining moves.
This can happen one way.

15 * 6 * 1 = 90 ways.

As all columns perform the same moves, just in a different order, we do not need to worry about ordering the columns.

(b) There are no ways. We will have to move one counter back at least once, as we cannot move more than six times forward without a back move - and this will mean that not all counters can reach this position.

(c) We proceed as in (a). However, one column must make a backward move and then a forward move.
First we pick this column. There are three ways of doing this.

The first column will be 8 choose 4 = 70 ways. But we have two ways of the column moving: FBFF, FFBF = 2 times. 2 * 70 = 140 times.
The second column will be 4 choose 2 = 6 ways.
The third column is once again pre-determined.

We have 140 * 6 * 3 = 2520 ways.

------

For ii and iii you don't need reasons but I'll put them here to clarify my reasoning

Preliminary: consider a game. For a player to win, after their turn, the distances between each of their counters and the opponent's must sum to 0 - ie so that they cannot move. Call this Sd (I borrowed this idea).

(ii)

Sd now starts as 3. On white's move, Sd will always add or subtract 1, changing the parity to even. On black's move Sd will similarly change parity back to odd.

(a) Yes White must move a counter forward, assume the first. Now black can only move counters 2 and 3 forward, move 2 forward. White must now move 3 forward. Black can only move counter 2 back, and hence white moves 2 forward, which gives the position. Alternatively the given diagram has Sd of 0, and as Sd is even after white's move, it follows that this is possible.

(b) No Sd starts as 3.For black to win Sd = 0. On white's move, Sd will always add or subtract 1, changing the parity to even. On black's move Sd will similarly change parity back to odd and hence black cannot win as 0 is even.

(c) Yes see (a) and (b).

(d) No We have already established that black cannot win. For black to keep the game going forever, black must always be able to make a move after white's move. However we have established that white can make Sd = 0 and hence black cannot keep the game going forever.

(iii)

As above, Sd now starts as 8. On white's move, Sd will always add or subtract 1, changing the parity to odd. On black's move Sd will similarly change parity back to even.

(a) No For this position, Sd must be 0 after white's move - but it is always odd after white's move.

(b) Yes black can win as 0 is even.

(c) No We have already established that white cannot win. For white to keep the game going forever, white must always be able to make a move after black's move. However we have established that black can make Sd = 0 and hence white cannot keep the game going forever.

(d) Yes we have already established in (b) that black can win.

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