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    Hello all, I haven't seen a thread on this anywhere (not sure why) during my travels so I thought I'd make one. Discussing the answers and solutions might help since it seems Cambridge haven't provided any answers to the sample questions.

    What is the CSAT?
    Information on the Computer Science Admissions Test can be found on the Cambridge Website. The Cambridge CST department has also started the CSAT practice platform to aid CSAT preparation based on the sample papers. Also see the official FAQ.

    CSAT Sample Papers
    These are the sample papers provided by Cambridge, based on some real questions. There have been reports that the real exam last year was noticeably harder than these but it should be a good start.
    Solutions to Sample Paper 1
    Section A
    Question 1 by Forecast
    Question 2 by Forecast
    Question 3 by uponthyhorse
    Question 4 by Forecast
    Question 5 by uponthyhorse
    [url="https://www.thestudentroom.co.uk/showthread.php?t=4949576
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    Sample Paper 1, Q1:

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    V = x(10-2x)^2 = 100x - 40x^2 + 4x^3 and so \dfrac{\mathrm dV}{\mathrm dx} = 100 - 80x + 12x^2. To maximise V, set \dfrac{\mathrm dV}{\mathrm dx} = 0 and solve for x, giving x = 5 and x = \frac{5}{3}. Clearly x = 5 gives the minimum volume of 0, and thus the maximum volume that can be achieved is \frac{5}{3} (10-\frac{10}{3})^2 = \frac{2000}{27}.


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    Sample Paper 1, Q2:

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    Repeatedly applying g to x gives h_n(x) = g^n(x) = x + n. So, h^m_n(x) = x + \underbrace{n + n + ... + n}_{m \text{ times}} = x + nm. So h^m_n(0) = nm.

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    Sample Paper 1, Q4:

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    \displaystyle \sum_{n=1}^{1337} (n!)^4 = 1^4 + 2^4 + 6^4 + 24^4 + 120^4 + .... Note that all terms after 24^4 contain a factor of 10 and thus their units digits are 0. The units digit of the first four terms are 1, 6, 6, 6 respectively. Thus the units digit of the sum is 19 \mod 10 = 9.

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    Sample Paper 1, Q3:

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    It's much easier to explain if I include a diagram
    Name:  q3 sol.jpg
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    Sample Paper 1, Q5 (Not sure about this one)

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    max(0, min(a_2, b_2) - max(a_1, b_1)) I can probably go into more detail about why I think it's this if anyone wants to know but an analysis of the various possibilities for the overlaps of two intervals will yield the first formula, min(a_2, b_2) - max(a_1, b_1). The outer formula is there to ensure that if there is no overlap, then we output 0 as the first formula will give a negative number.


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    Sample Paper 1, Q11:

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    On day k, the organism produces \dfrac{1^2\cdot2^2\cdot3^2\cdots k^2}{2\cdot3\cdots (k-1)} new cells, which simplifies to give k!k. The total number of cells produced by the end of day n is therefore \displaystyle \sum_{k=1}^n k!k.

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    \displaystyle \sum_{k=1}^n k!k = \sum_{k=1}^n ( (k+1)! - k! ) = (n+1)! - 1, using the method of differences.



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    (Original post by Forecast)
    Sample Paper 1, Q11:

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    Arghh I managed to crack the first half of it but couldn't simplify it further like you went on to do within the second spoiler tag. I've never been taught the method of differences but I feel like I could have gotten there by some other route. I'll add it to the OP
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    Sample Paper 1, Q15:

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    Note that 30 = 2\times3\times5 and n^5 - n = n(n^4-1) = n(n^2-1)(n^2+1) = n(n-1)(n+1)(n^2+1). The factors of n, (n-1) and (n+1) ensure that n^5-n will always be divisible by 2 and 3. If n is one more than a multiple of 5 then (n-1) will be divisible by 5; if n is two more than a multiple of 5 then n^2 + 1 = (5k+2)^2 + 1 = 5(5k^2 + 2k + 1), and thus n^2 + 1 will be divisible by 5; if n is three more than a multiple of 5 then n^2 + 1 = (5k+3)^2 + 1 = 5(5k^2 + 6k + 2), and again n^2 + 1 will be divisible by 5; if n is four more than a multiple of 5 then (n+1) will be divisible by 5. So n^5-n always has factors of 2, 3 and 5, and thus 30 does divide n^5-n for all positive integers n.




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    Sample Paper 1, Q16:

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    Let A be the initial value of the investment. After n years, the total value of the investment is A\left(1 + \frac{r}{100}\right)^n. For the investment to have doubled, 2A = A\left(1 + \frac{r}{100}\right)^n \Rightarrow 2 = \left(1 + \frac{r}{100}\right)^n \Rightarrow \ln 2 = n \ln \left(1+\frac{r}{100}\right). Using the definition from Q7, \ln 2 = n\left(\frac{r}{100} - \frac{r^2}{2 \times 100^2} + \dots \right). If r is sufficiently small, we can ignore the r^2 term and all subsequent terms, giving n \approx \frac{100\ln2}{r} \approx \frac{69.3}{r} and so x = 69.3.


    Further reading: https://en.wikipedia.org/wiki/Rule_of_72

    Additional note: in an earlier version of Sample Paper 1 (plus the actual 2015 test I believe), the value of \ln2 was given. I'm not sure why it was removed.

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    For question 1, by Forecast... Isn't it should be 10-2x instead of 10-x? because we are cutting out 4 squares from the card....If we do, for one width and length we must subtract 2 x from the length and width each???
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    (Original post by GimmeRice)
    For question 1, by Forecast... Isn't it should be 10-2x instead of 10-x? because we are cutting out 4 squares from the card....If we do, for one width and length we must subtract 2 x from the length and width each???
    You're right. Amended.
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    Has anyone done questions from Sample Paper 2?
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    (Original post by kishen111)
    Has anyone done questions from Sample Paper 2?
    Yes I'll try and type them up, are there any questions in particular you wanted to discuss?
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    (Original post by uponthyhorse)
    Yes I'll try and type them up, are there any questions in particular you wanted to discuss?
    1, 2 and 3 of Section A please. I've done 2 and 3, but I don't even understand question 1.
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    (Original post by kishen111)
    1, 2 and 3 of Section A please. I've done 2 and 3, but I don't even understand question 1.
    I got 8 for question 2 sample paper 2
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    (Original post by fp1washard)
    I got 8 for question 2 sample paper 2

    My answer:

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    12 hmm...




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    CEDAB,CADEB,AECDB,EACDB,ACEDB,EC ADB,ACDEB,ECDAB,CDEAB,CDAEB,CAED B,CEADB




    Are these correct?
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    (Original post by kishen111)
    My answer:

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    12 hmm...





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    CEDAB,CADEB,AECDB,EACDB,ACEDB,EC ADB,ACDEB,ECDAB,CDEAB,CDAEB,CAED B,CEADB





    Are these correct?
    (Original post by fp1washard)
    I got 8 for question 2 sample paper 2
    I am agreeing with kishen here for that question, those are the combinations I came up with as well.

    For a more in depth explanation:

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    B always has to be at the end as all other scores are bigger. Only A, C and E can be at the front. Notice that even though C > D, it could be the case that C > E > D, or even C > A > E > D. There are 12 combinations as listed by kishen.
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    (Original post by kishen111)
    1, 2 and 3 of Section A please. I've done 2 and 3, but I don't even understand question 1.
    I'm not sure I understand Question 1 either but here's what I interpreted it as

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    The function f(x) = min t^2 for all real x where t < x plots the smallest squared value below a certain x value. Where x is negative, t^2 > x^2 and hence the smallest squared value would be x^2. So for x < 0, the function takes the shape of the regular x^2 parabola. For x > 0, the smallest t^2 is just where t = 0. So for x > 0 the curve is always 0.
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    Sample Paper 2, Question 3:

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    \frac{15}{11} is just over 1 and hence a = 1 . Subtract this from \frac{15}{11} to get \frac{4}{11}.

    With the remaining fractions, create a common denominator repeatedly to obtain

    \frac{1}{ \frac{bcd + b + d}{cd + 1} } = \frac{4}{11}

    \frac{cd + 1}{bcd + b + d} = \frac{4}{11}

    Equate the numerators and the denominators to get :

    cd = 3 so c = 3 or 1 or d = 1 or 3.
    bcd + b + d = 11
    3b + b + d = 11
    4b + d = 11

    d = 3 gives b = 2, so the solution is a = 1, b = 2, c = 1, d = 3
 
 
 
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