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# AS Computer Science AQA Paper 2 14th June 2016

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1. (Original post by ShatnersBassoon)
No, I'm pretty sure you don't.

The general rule is: X.Y = (X'+Y''
For our situation, (A+(A+B)'', X = A and Y = A+B, so it simplifies to A.(A+B)
Why does the A+B not become A.B

I'm pretty sure I'm right as a couple of people coming out of the exam said zero but if you have solid proof let me know.

I've given hope on getting a good C.S grade after the first paper and now on this paper I dint do as well as I thought I did coming out of it.
2. (Original post by MrCoolVille)
I don't think you had to de morgan it. Was a case of expanding the brackets and simplifying!
Your statements are a non sequitur: there is often more than way to go about simplification. What answer did you get and how?
(Original post by Nemya_Nation)
Why does the A+B not become A.B.
Why would it?
3. (Original post by ShatnersBassoon)
I got A'.B as my final answer.
De Morgan's laws on the right bracket simplifies (A+(A+B)'' to A'.(A+B), which expands as A'.A + A'.B, which simplifies to [0 + A'.B = ] A'.B.

Then, we're left with (A'+B).(A'.B), or to rearrange it, A'.(A'+B).B.
Expand the left two multiplicands: (A'.A' + A'.B).B = (A' + A'.B).B
Expand the whole thing: A'.B + A'.B.B = A'.B + A'.B = A'.B.
I think I got that originally, however AQA have a tendency to utilise answers which are single characters, I should have suspected this may be different considering the scale of the question was 1.5x-2x that of a normal question that was stated within the specimen papers.

I hope that I will get a majority of the working out marks, considering that A'.B was in the working out if the answer is in-fact A'.B.

(Original post by ShatnersBassoon)
* One high level instruction can correspond to many low level statements, making high level languages faster to program in.
[Easier for humans to understand because]
* Meaningful identifiers
* Indentation
* Subroutines

* Low level programs sometimes execute faster because the instruction set of the program is very similar to that of the computer

I can't think of any others but I worry that I needed at least one more disadvantage.

... and we were simply taught that third generation languages are imperative and fourth generation are declarative; I could have described what 'declarative' meant but I'm pretty sure I'm not getting any marks for the first part of that question.
The advantages mirror some of what I stated, I however compared the ease of a high level language when considering the complexity of a lower level language.
I also detailed it further by stating that this would make large-scale programs much more manageable and document-able, especially when considering third parties.
I didn't specifically state your reasons for ease of understanding, seemed more of a paper 1 question.

For the disadvantages, I stated that lower level languages can be considered more efficient, as they directly communicate with the processor in terms of the instruction-set (Not as much need to translate)
I also put another point, but I can't remember off the top of my head, I will update this if another post makes me remember!
4. (Original post by ShatnersBassoon)
Your statements are a non sequitur: there is often more than way to go about simplification. What answer did you get and how?

Why would it?
De Morgans Law:

1.)Swap All operators you are applying de-morgans law to ergo

A+(A+B)' becomes A.(A.B)'

2.) Inverse

A'.(A.B) ---> A'.A.B cannot happen as you cannot have A and A' in same expression therefore equals zero.

This type of question came up in last years paper or the year before where the answer was also zero, hardly any one gets them as students tend to think that an Expression always must work lol.

Again if you find out otherwise, please let me know.
5. I got B
All you had to do was use De Morgan on the right side and then expand
The key was seeing that NOTA + A = 1
I think lolll
6. easy tiger what happened to breaking the bar before swapping the signs
7. (Original post by OturuDansay)
I got B
All you had to do was use De Morgan on the right side and then expand
The key was seeing that NOTA + A = 1
I think lolll
I got NotA And A = 0;
8. (Original post by Nemya_Nation)
I got NotA And A = 0;
yeah NOTA.A = 0

I did it your way in the exam but when it became NOTA.A.B to get the answer as 0 i didn't think that the marking added up to 4 marks at all but only 3. So I changed it back to what I had before lol
9. (Original post by Nemya_Nation)
I got NotA And A = 0;
should have gone to specsavers... answers !A.B
10. (Original post by Nemya_Nation)
De Morgans Law:

1.)Swap All operators you are applying de-morgans law to ergo

A+(A+B)' becomes A.(A.B)'

2.) Inverse

A'.(A.B) ---> A'.A.B cannot happen as you cannot have A and A' in same expression therefore equals zero.

This type of question came up in last years paper or the year before where the answer was also zero, hardly any one gets them as students tend to think that an Expression always must work lol.

Again if you find out otherwise, please let me know.
You've been taught an oversimplified version of the law. "Swap All operators" only works if there are exactly two operators. I suppose there are two operators here but the two operators are "A" and "A+B", not "A" and "B" as most of your classmates must have naively assumed.

To test it out:
The original thing is (A+(A+B)')'
My simplification is A'.(A+B)

If A = 0 and B = 0:
The original thing is (0+(0+0)')' = (0+0')' = (0+1)' = 1' = 0
My simplification is 0'.(0+0) = 1.0 = 0
Your simplification is 0'+(0.0) = 1+0 = 1

I could carry on but you get the gist. Your expression outputs different results to the original.
11. you read the question wrong
12. What was the answer for the number of bytes for the bitmap question?
13. (Original post by TechnoEngineer)
What was the answer for the number of bytes for the bitmap question?
64
14. (A+(A+B)')' using De morgan's law it becomes A'.(A+B) then distributing you get A'.A+A'.B which is 0+A'.B = A'.B
15. (Original post by OturuDansay)
64
Nope file size = resolution x colour depth
you had 3 colours so you need 2 bits to represent the 3 colours
00
01
10
11
so it was 16x16x2 =512 bits
16. (Original post by igotohaggerston)
Nope file size = resolution x colour depth
you had 3 colours so you need 2 bits to represent the 3 colours
00
01
10
11
so it was 16x16x2 =512 bits
It asked for bytes not bits. 512/8 == 64
17. For the assembly program i got
AND R3,R1,#1
CMP R3,#0
BNE ELSE:
MOV R2,#69
ELSE:
MOV R2,#79
END IF:
18. (Original post by OturuDansay)
It asked for bytes not bits. 512/8 == 64
OHH loool sorry i didnt read the question properly
19. (Original post by igotohaggerston)
For the assembly program i got
AND R3,R1,#1
CMP R3,#0
BNE ELSE:
MOV R2,#69
ELSE:
MOV R2,#79
END IF:
There needs to be an END IF after MOV R2, #69
20. (Original post by igotohaggerston)
OHH loool sorry i didnt read the question properly
Yeah I did a lot of that in the exam. On the Q to how to improve the processor I put multiple cores down and in the last minute, I re-read it and it had already said more cores in the Q
Silly AQA

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