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MAT Prep Thread - 2nd November 2016

Reply 420

alfmeister

2

Original post by KloppOClock

I was feeling sort of confident for this MAT exam... then I did the 2015 paper. :frown:

I fear I will experience this when i do the 2015 paper :s-smilie:

Reply 421

LaserRanger

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

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

Reply 422

LaserRanger

2

How many of you guys already did 2015 MAT? How's your marks?

Reply 423

Mystery.

21

How tf do they draw that for 2012 q5 ?!

Reply 424

some-student

7

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.

Posted from TSR Mobile

Reply 425

Mystery.

21

Original post by 11234

Help pls with q5 of 2002 paper

http://www.mathshelper.co.uk/Oxford%20Admissions%20Test%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

(edited 7 years ago)

Reply 426

Mystery.

21

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.

Posted from TSR Mobile

cheers

Reply 427

Mystery.

21

Does anyone have any advice on tackling q5's because I am doing okay everywhere else but I always get q5 messed up

Reply 428

Quido

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.

(edited 7 years ago)

Reply 429

Mystery.

21

Help with Q4 2007 Please?

Reply 430

KloppOClock

14

Original post by Mystery.

Help with Q4 2007 Please?

which part?

Reply 431

Mystery.

21

Original post by KloppOClock

which part?

iii

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

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

work out the area of one of those triangles, multiply it by two and take away the sector.
Here is how:

Spoiler

(edited 7 years ago)

Reply 433

ComputeiT

1

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?

Reply 434

Quido

9

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

Reply 435

Mystery.

21

Original post by KloppOClock

where is that image from?

for part iii,

Spoiler

work out the area of one of those triangles, multiply it by two and take away the sector.
Here is how:

Spoiler

Its from 2008 Q4, and I actually meant part ii sorry but I still need part 3 too so thanks

Reply 436

KloppOClock

14

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.

(edited 7 years ago)

Reply 437

ComputeiT

1

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.

Reply 438

Quido

9

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 :tongue:

Reply 439

some-student

7

Original post by 11234

Help pls with q5 of 2002 paper

http://www.mathshelper.co.uk/Oxford%20Admissions%20Test%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 S_d(I borrowed this idea).

(ii)

S_d now starts as 3. On white's move, S_d will always add or subtract 1, changing the parity to even. On black's move S_d 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 S_d of 0, and as S_d is even after white's move, it follows that this is possible.

(b) No S_d starts as 3. For black to win S_d= 0. On white's move, S_d will always add or subtract 1, changing the parity to even. On black's move S_d 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 S_d = 0 and hence black cannot keep the game going forever.

(iii)

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

(a) No For this position, S_d 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 S_d = 0 and hence white cannot keep the game going forever.

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

MAT Prep Thread - 2nd November 2016

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