MAT Prep Thread - 2nd November 2016

TRY this, draw that circle, the interior including boundary is the points that satisfy it.
Now draw line x+y=k

Thank you, did you just try coordinates to see? like (-1,0) (1,0)

Also is the answer A to this :

People get in with really low scores but no way its the average.
Check this official info out.
https://www.whatdotheyknow.com/request/mat_scores_mathematics_degrees
So people who sat the 2014 paper for 2015 entry.
Average score of all Imperial applicants:51
Average score of successful: 66
So nkt 20 marks lower.
My friend got a 3A* offer this year for getting 48 in mat.


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wait so doesn't that make the average score of Imperial applicants higher than Oxford ones? trippy

Would I get any marks for 2015 Q2 (v) for this?
To be a cube number, $k^3 + 2k^2 + 2k + 1 = (k + a)^3$
Clearly $a = 1$ and hence for $k^3 + 2k^2 + 2k + 1$ to be a cubed number, it must be equal to $(k + 1)^3$

However, $(k + 1)^3 = k^3 + 3k^2 + 3k + 1 \neq k^3 + 2k^2 + 2k + 1$ and hence $k^3 + 2k^2 + 2k + 1$ is not a cube number.

Also, can someone explain why the same Q2 part (iv) uses $3^{2015} - 2^{2015} = (3^5 - 2^5)(\dots)$ when we established in (i) that $a^{n + 1} - b^{n + 1} = (a - b)(\dots)$ - $n = 2014$?

Thanks

Not perhaps that surprising. You can't apply to Oxford and Cambridge both, so there will be a lot of Imperial applicants who are Cambridge, rather than Oxford applicants. Also Imperial don't interview - so it's perfectly possible to get an offer from Oxford with a moderate MAT score and good interviews. Imperial are more reliant on the test.

PS Also looking now at the data, the average across all Imperial applicants was higher than Oxford's but the average amongst successful applicants was quite a bit lower.

Seems fine for full marks for part V, maybe state that it's because the expansion ends in +1?

And there's no reason not to use the expansion of part i, so they did.

Seems fine for full marks for part V, maybe state that it's because the expansion ends in +1?

And there's no reason not to use the expansion of part i, so they did.

is the answer a ?

What was the answer to this?

There is no mark scheme.

Anyone know the answer to this ?

Anyone know the answer to this ?

There is no mark scheme.

Anyone know the answer to this ?

(this is probably wrong because other people got a different answer, oh well)
I extended AP and BQ out of the circle until they touched (lets say at point 'S' and formed a kite shape. From there i said that angle RPS= pi - x (because angles on a straight line = pi) and from that i said that angle RQS= pi - (pi - x)= x (opposite angles in a kite add to pi). Finally, since RQB and RQS are on a straight line, they equal pi too, Therefore my answer is 'x + y = pi'.

Can someone please check if this is wrong or right please? I feel like ive definitely made a mistake here

Can anyone go through the 2005 multiple choice answers with me please?

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