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    ABSOLUTELY NO DISCUSSING OF QUESTIONS EVEN AFTER THE INTERVIEWS HAVE FINISHED. Please remember people can get called back for reinterviews in January and there's every chance an applicant originally from a non CSAT college will have to take the test. It doesn't help any of us to discuss questions and Cambridge Academics do watch this thread... (...they're always watching... ). Best of luck!

    For anyone applying for CompSci this year, most colleges will be asking for the CSAT (Computer Science Admissions Test). Info can be found on the test here: http://www.cl.cam.ac.uk/admissions/u...missions-test/

    Thought this might be a good time to start a forum and perhaps discuss solutions to the sample questions on the site. It's non calculator by the way.

    Solutions:
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    Any ideas for ways to practice?
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    Of course give the sample questions on the site a good go... I can't do the first one! The 2nd and 4th I think I've done alright but there aren't any solutions... How have you found them??
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    This was my Q2 solution without use of calculator, did that whole permutations thing from S1... Had loads of long division and stuff on a separate sheet and had to work out factorials before hand so took a while (apparently you should try and do each question in about 10 mins but this probably took me 15...) Anybody get the same answer? And did anybody do it a simpler way?

    ERROR MADE: I stupidly said you could go up 11 steps 2 at a time... :/ :L Which added an extra incorrect combination. Inkblots has a better method.
    Attached Images
  1. File Type: pdf CSAT Q2.pdf (278.3 KB, 899 views)
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    My solution for question 2 involved doing it for 1 step then 2 steps then 3 and noticing it fell into a recognizable series.
    Spoiler:
    Show
    Fibonacci
    My answer at the end gave me
    Spoiler:
    Show
    144
    however so could have missed one or made a mistake in working out.
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    (Original post by Inkblots)
    My solution for question 2 involved doing it for 1 step then 2 steps then 3 and noticing it fell into a recognizable series.
    Spoiler:
    Show
    Fibonacci
    My answer at the end gave me
    Spoiler:
    Show
    144
    however so could have missed one or made a mistake in working out.
    UGH wish I'd seen that -_- :L I agree with your solution, I made a stupid error where I added a combination of taking only steps of 2 taking me to 145, however that's obviously not correct because it's an odd number of steps :/ :L Pretty funky how it links to Fibonacci
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    (Original post by Inkblots)
    My solution for question 2 involved doing it for 1 step then 2 steps then 3 and noticing it fell into a recognizable series.
    Spoiler:
    Show
    Fibonacci
    My answer at the end gave me
    Spoiler:
    Show
    144
    however so could have missed one or made a mistake in working out.
    Haha, I did the exact same thing Same answer too! I've done the two questions after, let me know what you get for them.
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    (Original post by CheetahCurtis)
    Haha, I did the exact same thing Same answer too! I've done the two questions after, let me know what you get for them.
    I didn't like my method for three, I have a feeling without a calculator it would be really inefficient. I got 9? and for 14 I got 670. You?
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    (Original post by Inkblots)
    I didn't like my method for three, I have a feeling without a calculator it would be really inefficient. I got 9? and for 14 I got 670. You?
    I also got 9 and 670. I'll post my methods (both done without calculator)
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    I'm just really glad they don't ask for a full solution, probably spend more time on the writing than the solving if that was the case.
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    Q3 and Q4 (or Q14 depending on where you're looking at it) posted here:
    Attached Images
  2. File Type: pdf Q3.pdf (228.4 KB, 644 views)
  3. File Type: pdf CSAT Q4.pdf (206.0 KB, 608 views)
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    (Original post by Inkblots)
    I didn't like my method for three, I have a feeling without a calculator it would be really inefficient. I got 9? and for 14 I got 670. You?
    Yep, same answers I'll post my solutions in spoilers (I didn't use a calculator, don't think you are meant to). Guy's don't open it if you don't want to know how it's done (and the answers).

    For question 3:
    Spoiler:
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    Answer: 9

    For the units digit, you are only concerned about adding the units digit from each (n!)^4 together. Every factorial from 5 onwards will end in 0 so the fourth power will also end in 0. So looking at factorials 1 to 4.
    1! = 1 therefore fourth power ends in 1
    2! = 2 therefore fourth power ends in 6 (2^4=16)
    3! = 6 therefore fourth power ends in 6
    4! = 24 therefore fourth power ends in 6
    1+6+6+6 = 19 so the answer must end in 9.

    That looks so much better on my scrap paper than on here
    For question 4:
    Spoiler:
    Show
    Answer: 670

    Entering into the function repeats the function odd, even, odd, even. For every 4 the function returns 2 + f(n-4). 1337 / 4 = 334 rem 1. If it's 334th multiple it will return 668 + f(1).
    f(1) returns 1 + f(-2)
    f(-2) returns 1

    668+1+1=670

    This also looks better on paper
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    (Original post by rohanpritchard)
    Q3 and Q4 (or Q14 depending on where you're looking at it) posted here:
    You beat me by 30 seconds and did the exact same as me :|
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    That was pretty much what I did for them, for some reason to me the B questions are much nicer than the A questions *shrug*. Anyone made any headway with 1 yet?
    Spoiler:
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    I got really fed up and said all of them would be 0 as it technically fits the solution but I doubt that's what they want.
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    (Original post by CheetahCurtis)
    You beat me by 30 seconds and did the exact same as me :|
    Pah Any idea on Q1?! It doesn't make sense to me how you can play around with one surd by just multiplying and then adding/subtracting to get a different surd... And when you square both sides you end up making a multiple of a surd equal to just a bunch of integer stuff... Might it just be a,b,c = 0?!
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    (Original post by rohanpritchard)
    Pah Any idea on Q1?! It doesn't make sense to me how you can play around with one surd by just multiplying and then adding/subtracting to get a different surd... And when you square both sides you end up making a multiple of a surd equal to just a bunch of integer stuff... Might it just be a,b,c = 0?!
    That is an annoying question, I have no idea. I made a quick program to try a, b, c from -100 to 100 (that's a lot of combinations - and I know we aren't meant to make programs for these but I couldn't resist) and it only came out with 0, 0, 0. So either that's the answer, I programmed it wrong, the question is wrong or one of the numbers is very large. Take your pick

    EDIT: Question 1 is different on the webpage to on the document, they must have changed one but not the other. Back to my program!
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    (Original post by CheetahCurtis)
    That is an annoying question, I have no idea. I made a quick program to try a, b, c from -100 to 100 (that's a lot of combinations - and I know we aren't meant to make programs for these but I couldn't resist) and it only came out with 0, 0, 0. So either that's the answer, I programmed it wrong, the question is wrong or one of the numbers is very large. Take your pick
    Pah I did the same thing!! (I went between -50 and 50). Agreed, nothing else. Then put it in WolframAlpha and they only came up with three zeros too... I think it's a terrible question if that's the answer... I mean c'mon... :L
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    It's a very unsatisfying answer . But hats off how many people would waste time trying to find the 'right solution' in the actual thing? That's sneaky.
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    (Original post by Inkblots)
    It's a very unsatisfying answer . But hats off how many people would waste time trying to find the 'right solution' in the actual thing? That's sneaky.
    I'm still convinced the question is wrong.
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    (Original post by Inkblots)
    That was pretty much what I did for them, for some reason to me the B questions are much nicer than the A questions *shrug*. Anyone made any headway with 1 yet?
    Spoiler:
    Show
    I got really fed up and said all of them would be 0 as it technically fits the solution but I doubt that's what they want.
    That's weird, I'm finding B questions to be trickier...
 
 
 

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