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CSAT For Computer Science Undergrad Admissions 2016 Entry

They are mostly pretty similar to the sample questions, so I would say that they're not particularly like either, but probably close to BMO. But it is less aimed at testing mathematical skills, and more aimed at testing general reason (in a mathematical context) if that makes sense.

Ok, thanks. I had a look at some BMO past papers and they seemed quite geometry-heavy. Were there many geometry questions on the CSAT? Geometry does not really play a big part in CS, does it?

Reply 321

sweeneyrod

15

Original post by NrGtic

Ok, thanks. I had a look at some BMO past papers and they seemed quite geometry-heavy. Were there many geometry questions on the CSAT? Geometry does not really play a big part in CS, does it?

No, you definitely don't need to learn geometry for it (or really any of the other BMO content). the CSAT isn't very similar to the BMO in content, more in style of questions. The UKMT Maths Challenge content is more similar, so it's probably only worth going onto the BMO if you find the SMC easy (in which case you will be in a pretty strong position for the CSAT).

Reply 322

NrGtic

Original post by sweeneyrod

No, you definitely don't need to learn geometry for it (or really any of the other BMO content). the CSAT isn't very similar to the BMO in content, more in style of questions. The UKMT Maths Challenge content is more similar, so it's probably only worth going onto the BMO if you find the SMC easy (in which case you will be in a pretty strong position for the CSAT).

By learning geometry, do you mean learning geometrical concepts required for BMO or doing geometry exercises?

Reply 323

sweeneyrod

15

Original post by NrGtic

By learning geometry, do you mean learning geometrical concepts required for BMO or doing geometry exercises?

Both.

Reply 324

NrGtic

Original post by sweeneyrod

Both.

Ideal. Were your interview questions similar to the CSAT questions or were they more similar to STEP or BMO?

Reply 325

sweeneyrod

15

Original post by NrGtic

Ideal. Were your interview questions similar to the CSAT questions or were they more similar to STEP or BMO?

My interview questions were pretty different to both. The first interview involved two logic problems -- if you look on the Oxf*rd site they have some fairly similar questions. The second interview had some surprisingly easy maths questions, more similar to AS Maths than anything else, and then a question based on a computer science topic. But I think interviews vary a lot between colleges (mine was at Churchill); I'm pretty sure most people get harder maths questions and some colleges do weird things like asking you to analyse an article about something computer sciency or do some non-compsci-related maths.

Reply 326

Forecast

16

The small "Sample Test" has now been replaced by a complete specimen paper, containing many questions from the 2015 test.

https://www.cl.cam.ac.uk/admissions/undergraduate/admissions-test/sample-test.pdf

Reply 327

Dobroslav

1

Did anyone get 2000/27 for the 1st question?
For Question 2 I got x+n^m (not sure about this one)
For Question 3 I think it is 1:2
Question 4 was on the previous examples (answer was 9)
Question 5, I don't understand the question, someone help me out, please
Question 6 I got when K is less than -4/3

(edited 7 years ago)

Reply 328

esfio

1

Original post by Dobroslav

Did anyone get 2000/27 for the 1st question?
For Question 2 I got x+n^m (not sure about this one)
For Question 3 I think it is 1:2
Question 4 was on the previous examples (answer was 9)
Question 5, I don't understand the question, someone help me out, please
Question 6 I got when K is less than -5/3

I got 2000/27 for the 1st question also, however I got "nm" for question 2

I'm not sure about question 2 either, but in your answer there's an "x", is that meant to be 0?

Reply 329

CharlieeS

Original post by esfio

I got 2000/27 for the 1st question also, however I got "nm" for question 2

I'm not sure about question 2 either, but in your answer there's an "x", is that meant to be 0?

I got 1000/27 for the first one somehow :/ Did you get a third polynomial equation that you search maximum value for?

By the way, how much did you get in the 19th? I recognised the pattern, but I am not sure how I should approach the solution, so I got +2 as my answer.

(edited 7 years ago)

Reply 330

Dobroslav

1

For question 1 I did x(10-2x)(10-2x), expand that and differentiate then make it equal to 0 and solve for X, this should give you the values of X for when it is a min or max. The graph goes to positive infinity (x*-2x*-2x) this means that the lowest value of the 2 values of X is the one that gives the maximum because of the shape of the graph. By subbing in the value into x(10-2x)(10-2x) you should get the answer 2000/27

Reply 331

CharlieeS

Original post by Dobroslav

For question 1 I did x(10-2x)(10-2x), expand that and differentiate then make it equal to 0 and solve for X, this should give you the values of X for when it is a min or max. The graph goes to positive infinity (x*-2x*-2x) this means that the lowest value of the 2 values of X is the one that gives the maximum because of the shape of the graph. By subbing in the value into x(10-2x)(10-2x) you should get the answer 2000/27

Yeah my bad I accidentally took x^2(10-2x), now I see my mistake, thanks

Reply 332

jpxw

5

Original post by Dobroslav

Did anyone get 2000/27 for the 1st question?
For Question 2 I got x+n^m (not sure about this one)
For Question 3 I think it is 1:2
Question 4 was on the previous examples (answer was 9)
Question 5, I don't understand the question, someone help me out, please
Question 6 I got when K is less than -5/3

Could you explain your workings for 3 and 6? I feel like I'm over complicating 3, and can't seem to do 6 at all for some reason

Reply 333

DanBing

1

Original post by ycry2

Hello,

I am successful applicant last year at Trinity College. I've been lurking on this thread last year and being able to see other people's solutions were very helpful for me! In response to new sample paper, I took the liberty of putting the solutions to the questions on my website:

http://ycry2.user.srcf.net/csat

I also typed up the solutions to the old questions:

http://ycry2.user.srcf.net/csat_old

This was done in a rush so obviously there will be errors: please tell me if you see any! Some of the questions I have done but not typed up yet: they will come soon. Good luck everyone!

Thank you so much for these solutions! I've been struggling with the sample questions so hopefully this will help clear them up!

Reply 334

Dobroslav

1

Original post by jpxw

Could you explain your workings for 3 and 6? I feel like I'm over complicating 3, and can't seem to do 6 at all for some reason

For Question 3, The line AB is the diameter of the circle, and the line BC will intercept the circle. Using circle theorem, the angle of APB will be 90. and let AP = y. This splits the isosceles triangle into 2 right-angled triangles sharing the same height.
We see that:
(BP)Cosθ=y
and
(CP)Cosθ=y

equate them and (BP)Cosθ=(CP)Cosθ, this means (BP=CP) meaning the ratio of the 2 lengths is 1:1, however we need BP:BC , therefore 1:2

6, The (k+1) etc are all constants, therefore we can say (k+1)=a, (3k+2)=b, 2k=c,

this leaves ax^4 -bx^2 -cx, we want to get a maximum, this means that d2y/dx2 has to be negative.

dy/dx = 4ax^3-2bx-c
d2y/dx2 = 12ax^2-2b

subbing x=-1, a and b back in leaves
12(k+1)-2(3k+2)<0
12k+12-6k-4<0
6k+8<0
6k<-8
k<-8/6
k<-4/3
Seems like I made an error in the first place oops, but this should be right haha

Reply 335

Dobroslav

1

I think the answer to question 9 is 125/216, anyone else get this?

Reply 336

Forecast

16

Original post by Dobroslav

I think the answer to question 9 is 125/216, anyone else get this?

Yup, just done it now. Really nice question, pretty sure this wasn't in the actual test last year which is a shame as I would probably have been able to get a good solution down in that case. I (foolishly) decided to spend most of my time attempting the more algorithmic type questions, which turned out to be really quite tricky - most of them aren't included in the new Sample Paper, which suggests that I wasn't the only one who found them too hard.

--

Anyone tried Q11? I think that the current wording renders the question impossible to do using school-level knowledge. As it currently reads, at the end of day

k=1

there is 1 cell,

k=2

there are 5 cells (4 produced during the day, plus the original cell),

k=3

there are 27.5 cells (22.5 produced that day, plus 5 from days 1 and 2), and so on. I think that the wording should be changed to either of the following alternatives (the subtle changes yield different answers in each case; the second one is likely the intended question):

An organism is born on day

k = 1

with

1

cells. During day

k = 2, 3, \dots

the organism has

\frac{k^2}{k-1}

times more cells than it had after day

k - 1

. Give a simplified expression for the number of cells after

n

days.

An organism is born on day

k = 1

with

1

cells. During day

k = 2, 3, \dots

the organism produces

\frac{k^2}{k-1}

times more cells than it produced on day

k - 1

. Give a simplified expression for the number of cells after

n

days.

(edited 7 years ago)

Reply 337

OverratedGman

4

For question 11:

I think the answer is ((k^2)/k-1)! from k to 2, but I think this is a partial solution but I'm not sure where to go from here

Reply 338

ILoveCS

4

Original post by Dobroslav

For Question 3, The line AB is the diameter of the circle, and the line BC will intercept the circle. Using circle theorem, the angle of APB will be 90. and let AP = y. This splits the isosceles triangle into 2 right-angled triangles sharing the same height.
We see that:
(BP)Cosθ=y
and
(CP)Cosθ=y

equate them and (BP)Cosθ=(CP)Cosθ, this means (BP=CP) meaning the ratio of the 2 lengths is 1:1, however we need BP:BC , therefore 1:2

6, The (k+1) etc are all constants, therefore we can say (k+1)=a, (3k+2)=b, 2k=c,

this leaves ax^4 -bx^2 -cx, we want to get a maximum, this means that d2y/dx2 has to be negative.

dy/dx = 4ax^3-2bx-c
d2y/dx2 = 12ax^2-2b

subbing x=-1, a and b back in leaves
12(k+1)-2(3k+2)<0
12k+12-6k-4<0
6k+8<0
6k<-8
k<-8/6
k<-4/3
Seems like I made an error in the first place oops, but this should be right haha

Hey Dobroslav,

Differentiating would seem like a good idea, but is it necessary/right?
Hint 1: try subbing in x=-1 first BEFORE you try to differentiate.
Hint 2: try subbing k=0, x=-1 and then k=1000, x=-1. Does that give you another hunch?

-R

Reply 339

ILoveCS

4

Original post by Dobroslav

I think the answer to question 9 is 125/216, anyone else get this?