## > anonymous

### Application

Oxford

St John's

Mathematics

2008

offer made

### Applicant

A-levels

pre-qualification

home

N/A

Comprehensive School

yes (13 A*)

#### A-levels

(A at AS)

(A at AS)

(A at AS; predicted A; gained NA at A2)

(A at AS; predicted A; gained A at A2)

(A at AS; predicted A; gained NA at A2)

(A at AS; predicted A; gained NA at A2)

### Details about the offer

conditional

A in Mathematics (Further)

A, A grade in Physics or German

OR

II in STEP II/Merit in AEA

yes

N/A

grades pending/unknown

### Decisions about the application

Visited Oxford first, liked it.

I guessed on the college (I eliminated some colleges then fairly randomly picked between the remaining). I booked on St. John's open day fairly randomly and liked it a lot.

### Preparation

yes

One 'mock interview' with the head of sixth form. Absolutely nothing at all like the real thing.

Read some maths. Read some more maths? Enjoy maths.

### Interview

yes

Maths commons test. Wasn't so bad - the content is just C1 and C2. Just reiterating what's been said lots of times, but, know C1 and C2 really well. I'd advise reborrowing the textbooks and going through lots of times. Do the past/specimen papers on the oxford maths website.

no

no

Hard to say. Weren't "oh dear that was terrible", I didn't come away feeling amazingly confident. Just sort of 'meh'.

St Johns:

Draw a triangle, form inequality that the sum of any two sides is greater than or equal to the third side.

Now draw a quadrilateral, draw in diagonals.

Deduce that sum of the diagonals is greater than the sum of two opposite sides.

There are a collection of points on a plane. Join them together to make a circuit. Uncrossed circuits can be made by finding crossings, and simply uncrossing them. Repeat until uncrossed.

Deduce that there exists a shortest circuit - there are a finite number of points, hence there are a finite number of circuits. A finite set has a smallest member, hence there exists a shortest circuit. This circuit will not have any crossings, since the the length of the corresponding uncrossed circuit (created by uncrossing the crossing) would be longer (using result found above).

Mansfield:

Integration of some trigometric functions i.e. sin^2(x)cos^3(x)

When f(x+y) = f(x)f(y), prove f(0) = 1 where f is a non-zero, real valued function.

T-shirt and jeans. Why not? I never wear suits... suits make me feel awkward.

### Impressions

Fine. Two hot meals a day + a large breakfast. Can't complain.

### Final stage

ahhhh. bleh. Really horrible waiting for it.

### Looking back

Yes.

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