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# CPT 1 and 2 watch

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1. Hi could someone please tell me when the AQA computing exams are for CPT 1 and 2. AS levels.
Thanks
2. (Original post by mark001282)
Hi could someone please tell me when the AQA computing exams are for CPT 1 and 2. AS levels.
Thanks
afternoon - and do you know how to do the algorithms and the programming bit of the past papers?
3. (Original post by TheWolf)
afternoon - and do you know how to do the algorithms and the programming bit of the past papers?
what do we need to know?
4. (Original post by mockel)
what do we need to know?
algorithms etc eg http://www.aqa.org.uk/qual/gceasa/qp...W-QP-Jun02.pdf

question 8 i dont know how to do it
5. Thanks, hmm i'm not sure about algorithms and programming, it was never my strongpoint in computing. All I can suggest is look at the past papers and then look at the answers. Thats what i'm gonna do
6. (Original post by TheWolf)
algorithms etc eg http://www.aqa.org.uk/qual/gceasa/qp...W-QP-Jun02.pdf

question 8 i dont know how to do it
okay...i'll try to go through it step by step:

First x is set to 5
Then Index is set to 0
Now in the loop, they say the value of 'y' is the remainder when you divide 'x' by 2. So starting right from the beginning, x=5, so y = 1 (remainder from 5/2).
The next step is to find the new value of 'x'. 'x' is given by how many times 2 can be divided into 'x'. since x=5, the new value of x=2 (2 goes into 5 twice).
Next line is to add 1 to Index...so Index is now 1
Next line is to set the Result[Index] as the value of y. Since the current Index=1, and y is currently 1, Result[1] = 1
Next line is UNTIL x=0, so because x isn't 0, you have to repeat the process

2nd time:
y = xMOD2, so y = 2MOD2....y=0
x = xDIV2, so x = 2DIV2....x=1
Index = 1+1...Index = 2
Result[Index] = y....Result[2] = 0
Is x=0? No, so you have to repeat

3rd time:
y = xMOD2, so y = 1MOD2. since 2 doesn't divide into 1 any times, the remainder of the division is 1...y=1
x = xDIV2, so x = 1DIV2. since 2 doesn't divide into 1 any times, x=0
Index = 2+1....Index=3
Result[Index] = y......Result[3] = 1
Is x=0? Yes, so you can stop

I hope this is right (and it makes sense)
7. (Original post by mockel)
I hope this is right (and it makes sense)
yayyyy!! just checked the mark scheme and it's right
tell me if you have any problems understanding it....
8. (Original post by mockel)
okay...i'll try to go through it step by step:

First x is set to 5
Then Index is set to 0
Now in the loop, they say the value of 'y' is the remainder when you divide 'x' by 2. So starting right from the beginning, x=5, so y = 1 (remainder from 5/2).
The next step is to find the new value of 'x'. 'x' is given by how many times 2 can be divided into 'x'. since x=5, the new value of x=2 (2 goes into 5 twice).
Next line is to add 1 to Index...so Index is now 1
Next line is to set the Result[Index] as the value of y. Since the current Index=1, and y is currently 1, Result[1] = 1
Next line is UNTIL x=0, so because x isn't 0, you have to repeat the process

2nd time:
y = xMOD2, so y = 2MOD2....y=0
x = xDIV2, so x = 2DIV2....x=1
Index = 1+1...Index = 2
Result[Index] = y....Result[2] = 0
Is x=0? No, so you have to repeat

3rd time:
y = xMOD2, so y = 1MOD2. since 2 doesn't divide into 1 any times, the remainder of the division is 1...y=1
x = xDIV2, so x = 1DIV2. since 2 doesn't divide into 1 any times, x=0
Index = 2+1....Index=3
Result[Index] = y......Result[3] = 1
Is x=0? Yes, so you can stop

I hope this is right (and it makes sense)
thanks! finally i think imgetting somewhere what does mod mean by teh way?
9. (Original post by TheWolf)
thanks! finally i think imgetting somewhere what does mod mean by teh way?
it explains it in the question.....if you had 11MOD3, the answer would be 2, because you want to find the remainder when 11 is divided by 3, i.e. 11/3 = 3 rem 2
10. (Original post by mockel)
it explains it in the question.....if you had 11MOD3, the answer would be 2, because you want to find the remainder when 11 is divided by 3, i.e. 11/3 = 3 rem 2
ahh ok thanks! do you mind having a look at those! i dont think i get them either

http://www.uk-learning.net/t47222.html
11. (Original post by TheWolf)
ahh ok thanks! do you mind having a look at those! i dont think i get them either

http://www.uk-learning.net/t47222.html
Question 7 june 2003 http://www.aqa.org.uk/qual/gceasa/qp...W-QP-Jun03.pdf

a)
(i) Var S1: String
(ii) If S1 = S2 Then.......End If
(iii) For Ptr := 1 To 3 Do......End For

b) copy and concat are both functions, becasue they return a value
print is a procedure, since no value is returned, it is just a command

c) First main step is setting S1 to "PAT" and S2 to ""
Next line means you have to start at Ptr=1, and every time you come back to it, Ptr must increment by 1. So firstly, Ptr starts at 1.
L is given by Copy(S1, Ptr). This means take the nth letter of S1 where n is the current value of Ptr. So Ptr=1, so they want the first letter of S1 to be extracted, which is a "P"...so L = P
S2 is given by joining what is currently in L ("P") with what is currently in S2 (nothing), so S2="P" (remember, the order is important)
Now we must repeat, until Ptr becomes 3, so....

Ptr=1+1, so Ptr=2
L is now given by the second letter of S1, so L="A"
S2 is now given by L and S2 joined together, so S2 = "AP"
Now repeat once more....

Ptr = 3
L is now the 3rd letter of S1, so L="T"
S2 is L and S2 joined together, so S2 = "TAP"
Now since Ptr=3, we can come out of the loop, and they ask the question: "Is S1=S2". If it is, then you have to write 'True'...if it isn't then you have to write 'Fasle' Since S1="PAT" and S2="TAP", then you must write 'False'
12. ok ill have a look at it now, thanks in advance! ill give u rep ta
13. (Original post by TheWolf)
ok ill have a look at it now, thanks in advance! ill give u rep ta
no problem.....helped me remember a few things as well
14. (Original post by mockel)
no problem.....helped me remember a few things as well
im revising everything in the syllabus right now, actually learning it lol, but it should be ok, you coming on tommorow morning?
15. (Original post by TheWolf)
im revising everything in the syllabus right now, actually learning it lol, but it should be ok, you coming on tommorow morning?
umm....probably. i haven't finished revising yet.
got 89% and 94% for CPT1 and 2 respectively in the mocks, so i've just been slacking since then when it's come to computing....i've been putting in so much hard work for P3, i haven't really had the time for computing
16. (Original post by mockel)
umm....probably. i haven't finished revising yet.
got 89% and 94% for CPT1 and 2 respectively in the mocks, so i've just been slacking since then when it's come to computing....i've been putting in so much hard work for P3, i haven't really had the time for computing
me = worse cus im doing half and as and half as next year, i couldnt be arsed to revise as ive got 4 other as to do, so im cramming it in these past 2 days..
17. (Original post by TheWolf)
me = worse cus im doing half and as and half as next year, i couldnt be arsed to revise as ive got 4 other as to do, so im cramming it in these past 2 days..
oh, okay....i'm dropping it after as. just can't be bothered with it, and i need to concentrate on my other subjetcs.
i'm off to bed know, good luck with your revision ....i'll probably see you tomorrow morning, bye
18. (Original post by mockel)
oh, okay....i'm dropping it after as. just can't be bothered with it, and i need to concentrate on my other subjetcs.
i'm off to bed know, good luck with your revision ....i'll probably see you tomorrow morning, bye
ok see ya ! good luck 2 u 2
19. How is every1 reatin there chances 2morrow

Or u not speculatin
20. Honestly, CPT1 is not hard. I did no work for it, and still got 92/105 of the UMS points or whatever they are. I've got CPT2 today - I did CPT1 in January. The algorithms are like the only hard bit in the paper, and you might not get them.

What do I need to know for CPT2?

-=X=-

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