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# CPT 1 and 2 watch

1. The ASCII coding system uses seven bits to code a ccharacter. The character digits 0 to 9 are assigned the decimal number codes 48-57. An extra bit is used as a parity. A computer system uses the most significant bit as a parity bit for each bytee and works withh even parity.

i

what is bit pattern if the digits 37 are to be stored as characters

ii

explain how the parity bit is used by this computer system

can someone explain how to do this?
2. just woke up... all lovely jubbly.

had to come on computer to get my computing notes printed... yah, should've revised earlier methinks... whoops

was gonna revise last night but i just got so darned immersed in the local elections/couldn't be arsed

er... cpt1 is really easy. ums was 105/105 just by learning the past paper answers and writing as much as i could for each question (because lets face it, 1 and a half hours is waaaay to loong for that silly exam)

cpt2 today - gonna be a legendarious exam. just starting revision.... no problemo methinks. got about 2 to 3 hours to go.. sorted! and guten luck!

aurevoir!
3. ah ha, some computer geeks

Can someone pls explain the 2nd complement

I jst heard of it, I think Im meant to know it

4. (Original post by TheWolf)
The ASCII coding system uses seven bits to code a ccharacter. The character digits 0 to 9 are assigned the decimal number codes 48-57. An extra bit is used as a parity. A computer system uses the most significant bit as a parity bit for each bytee and works withh even parity.

i

what is bit pattern if the digits 37 are to be stored as characters

ii

explain how the parity bit is used by this computer system

anyone?

can someone explain how to do this?
5. (Original post by TheWolf)
The ASCII coding system uses seven bits to code a ccharacter. The character digits 0 to 9 are assigned the decimal number codes 48-57. An extra bit is used as a parity. A computer system uses the most significant bit as a parity bit for each bytee and works withh even parity.

i

what is bit pattern if the digits 37 are to be stored as characters

ii

explain how the parity bit is used by this computer system

can someone explain how to do this?
I looked at that, and thought "I've done that before" but I've forgotten it...

Are you sure the number 37 is right? If it was between 48 and 57 it'd make more sense to me.

I remember something about subsituting the numbers 0 - 9, so 0 is 48, 1 is 49 etc. but I really can't remember sorry

-=X=-
6. look it up at the mark schemes at aqa

that'll help explain it to you (probably)
7. as characters: 3 wil be 00000011 and the 7 is 00000111

but is it askin for 37 as a whole characyer(can this be dome, i dont kneo) If it can it will be 00010101

then for parity

if a system used even parity, the 8th bit (on the left) is made a 1 if the number of 1`'s is odd, and left as 0, if they add up to make an even num

Then the revieving comp ses, hmm..lets c..I use even parity

This has an even num of 1's, its correct data

Gud jb, wel dun!
8. Yeah, that's if it's just binary for 3 and then binary for 7 which I don't think it is. It's also asking for it to use 7 bits, not 14. Also, you've completely ignored the 48-57 range, so I'm assuming there's something else to it

I do remember doing it months ago...

-=X=-
9. or ye, i dit read the question (naughty me) jst jumped straight in

I'll have another luk

10. does any one know how to add binary

2nd complement i think
11. (Original post by austinmetro)
as characters: 3 wil be 00000011 and the 7 is 00000111

but is it askin for 37 as a whole characyer(can this be dome, i dont kneo) If it can it will be 00010101
how do you work that out in pure binary - itd be 00110111 right? for 37, annd here for ascii how did you calc it?
12. (Original post by TheWolf)
how do you work that out in pure binary - itd be 00110111 right? for 37, annd here for ascii how did you calc it?
sorry, i think I dun it completely wrong, waty paper is it
13. (Original post by austinmetro)
sorry, i think I dun it completely wrong, waty paper is it
0011 0011 1011 0111 is the answer - june 2003 paper
14. (Original post by TheWolf)
how do you work that out in pure binary - itd be 00110111 right? for 37, annd here for ascii how did you calc it?

I dont have a scooby anyomre

That number int in those limits, shud that matter!
15. (Original post by TheWolf)
The ASCII coding system uses seven bits to code a ccharacter. The character digits 0 to 9 are assigned the decimal number codes 48-57. An extra bit is used as a parity. A computer system uses the most significant bit as a parity bit for each bytee and works withh even parity.
i Split 37 up into 3 and 7.
3 will be given the code 51 (since 0=48, 1=49, 2=50) and similarly 7 will be given the code 55
51 in 7-bit binary is: 0 1 1 0 0 1 1
since even parity is used, there must be an even number of 1's...at the moment, there are 4 1's, so the parity bit is 0.....so 3 is coded as:
0 0 1 1 0 0 1 1

55 is: 0 1 1 0 1 1 1
for even parity, we must make the partiy bit a 1...so 7 is coded as:
1 0 1 1 0 1 1 1

so together, the code for 37 would be: 0 0 1 1 0 0 1 1 1 0 1 1 0 1 1 1

ii For even parity, the number of 1's is checked on sending and receving, to check that an even number of 1's have been sent. If it passes this check, then the data is assumed to be correct. vice-versa for odd parity.
16. (Original post by mockel)
i Split 37 up into 3 and 7.
3 will be given the code 51 (since 0=48, 1=49, 2=50) and similarly 7 will be given the code 55
51 in 7-bit binary is: 0 1 1 0 0 1 1
since even parity is used, there must be an even number of 1's...at the moment, there are 4 1's, so the parity bit is 0.....so 3 is coded as:
0 0 1 1 0 0 1 1

55 is: 0 1 1 0 1 1 1
for even parity, we must make the partiy bit a 1...so 7 is coded as:
1 0 1 1 0 1 1 1

so together, the code for 37 would be: 0 0 1 1 0 0 1 1 1 0 1 1 0 1 1 1

ii For even parity, the number of 1's is checked on sending and receving, to check that an even number of 1's have been sent. If it passes this check, then the data is assumed to be correct. vice-versa for odd parity.
thanks once again very useful - so to make even parity, -so if the number of 0s are odd, we add 1 moore at at the begining, if the number of 1s are odd, we add one more 1 at the beginning?
17. (Original post by TheWolf)
thanks once again very useful - so to make even parity, -so if the number of 0s are odd, we add 1 moore at at the begining, if the number of 1s are odd, we add one more 1 at the beginning?
yeah, so basically, even parity means that the total number of 1's is even (including the parity bit), and odd partiy would mean the total number of 1's must be odd (als including the parity bit).... and the parity bit goes at the beginning
18. (Original post by mockel)
yeah, so basically, even parity means that the total number of 1's is even (including the parity bit), and odd partiy would mean the total number of 1's must be odd (als including the parity bit).... and the parity bit goes at the beginning
ok so 1 byte = 8 bits, and 2^8 different combinations of 0 and 1s, therefore 2^8 different colours possible?
19. (Original post by TheWolf)
ok so 1 byte = 8 bits, and 2^8 different combinations of 0 and 1s, therefore 2^8 different colours possible?
i would think so...
20. Important: Parity bit goes at the front (left hand side) if it says "MSB" or Most Significant Bit, and at the end (the right hand side) if it says Least Significant bit. That's because the number on the right could only change the number by 1 if it was wrong, on the left it could change it by hundreds if the data is corrupt (depending on the size of the binary number)

010110101 < LSB
^
MSB

-=X=-

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