uponthyhorse
Badges: 15
Rep:
?
#1
Report Thread starter 2 years ago
#1
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
Question 6 ???
Question 7 by Lelanor
Question 8 by irishapplicant

Section B
Question 9 by uponthyhorse
Question 10 by Squiidsquid
Question 11 by Forecast
Question 12 by Integer123
Question 13 by Squiidsquid (beware, also links to answers for 14 and 19)
Question 14 by uponthyhorse + Squiidsquid (No method)
Question 15 by Forecast
Question 16 by Forecast
Question 17 by irishapplicant
Question 18 by uponthyhorse
Question 19 by uponthyhorse + Squiidsquid
Question 20 by TTTRRee

Solutions to Sample Paper 2
Section A
Question 1 by uponthyhorse
Question 2 by kishen111
Question 3 by uponthyhorse
Question 4 by southpacific
Question 5 by uponthyhorse
Question 6 by Sirlamb
Question 7 by Sirlamb
Question 8 by kishen111

Section B
Question 9 by Peractio
Question 10 by Sirlamb
Question 11 by LaiFuShi (and loads of other people, I just kept forgetting to put this on the original post! Sorry guys!)
Question 12 by mayanksharma318
Question 13 by Integer123
Question 14 by uponthyhorse
Question 15 by uponthyhorse
Question 16 by southpacific
Question 17 by Peractio
Question 19 by uponthyhorse + kishen111
Question 20 by uponthyhorse

Post solutions with spoiler blocks so anyone looking for an answer to another question doesn't read yours by accident, and please specify the paper and question number. Thanks!
4
reply
Forecast
Badges: 16
Rep:
?
#2
Report 2 years ago
#2
Sample Paper 1, Q1:

Spoiler:
Show



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}.


3
reply
Forecast
Badges: 16
Rep:
?
#3
Report 2 years ago
#3
Sample Paper 1, Q2:

Spoiler:
Show


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.

0
reply
Forecast
Badges: 16
Rep:
?
#4
Report 2 years ago
#4
Sample Paper 1, Q4:

Spoiler:
Show


\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.

0
reply
uponthyhorse
Badges: 15
Rep:
?
#5
Report Thread starter 2 years ago
#5
Sample Paper 1, Q3:

Spoiler:
Show

It's much easier to explain if I include a diagram
Name:  q3 sol.jpg
Views: 1150
Size:  467.7 KB
0
reply
uponthyhorse
Badges: 15
Rep:
?
#6
Report Thread starter 2 years ago
#6
Sample Paper 1, Q5 (Not sure about this one)

Spoiler:
Show



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.


0
reply
Forecast
Badges: 16
Rep:
?
#7
Report 2 years ago
#7
Sample Paper 1, Q11:

Spoiler:
Show


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.

Spoiler:
Show


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



0
reply
uponthyhorse
Badges: 15
Rep:
?
#8
Report Thread starter 2 years ago
#8
(Original post by Forecast)
Sample Paper 1, Q11:

Spoiler:
Show




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
1
reply
Forecast
Badges: 16
Rep:
?
#9
Report 2 years ago
#9
Sample Paper 1, Q15:

Spoiler:
Show





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.




1
reply
Forecast
Badges: 16
Rep:
?
#10
Report 2 years ago
#10
Sample Paper 1, Q16:

Spoiler:
Show


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.

0
reply
GimmeRice
Badges: 2
Rep:
?
#11
Report 2 years ago
#11
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???
0
reply
Forecast
Badges: 16
Rep:
?
#12
Report 2 years ago
#12
(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.
0
reply
kishen111
Badges: 8
Rep:
?
#13
Report 2 years ago
#13
Has anyone done questions from Sample Paper 2?
0
reply
uponthyhorse
Badges: 15
Rep:
?
#14
Report Thread starter 2 years ago
#14
(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?
0
reply
kishen111
Badges: 8
Rep:
?
#15
Report 2 years ago
#15
(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.
0
reply
fp1washard
Badges: 9
Rep:
?
#16
Report 2 years ago
#16
(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
0
reply
kishen111
Badges: 8
Rep:
?
#17
Report 2 years ago
#17
(Original post by fp1washard)
I got 8 for question 2 sample paper 2

My answer:

Spoiler:
Show


12 hmm...




Spoiler:
Show



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




Are these correct?
0
reply
uponthyhorse
Badges: 15
Rep:
?
#18
Report Thread starter 2 years ago
#18
(Original post by kishen111)
My answer:

Spoiler:
Show



12 hmm...





Spoiler:
Show




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:

Spoiler:
Show

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.
0
reply
uponthyhorse
Badges: 15
Rep:
?
#19
Report Thread starter 2 years ago
#19
(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

Spoiler:
Show

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.
0
reply
uponthyhorse
Badges: 15
Rep:
?
#20
Report Thread starter 2 years ago
#20
Sample Paper 2, Question 3:

Spoiler:
Show

\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
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

If you do not get the A-level grades you want this summer, what is your likely next step?

Take autumn exams (242)
47.08%
Take exams next summer (70)
13.62%
Change uni choice through clearing (111)
21.6%
Apply to uni next year instead (53)
10.31%
I'm not applying to university (38)
7.39%

Watched Threads

View All