if you think that you only lost 5 marks, i dont think you have the time come here to place your 171 posts!Originally posted by iptteacherssuck
i reckon ive lost max mark of 5 marks..
ive lost 3 from desk checking and
2 prolly from silly mistakes !!!!!
)))
What Did U Think ?? (2 Viewers)
- Thread starter iptteacherssuck
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iptteacherssuck
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haha but mmost of my posts are spam and paying out of iptteachers
I thought it was fairly straightforward except for Question 23(a) *looks utterly confused* What the hell was that all about? I think I managed to fudge perhaps only 3 or 4 marks from that part, tops (I suck at interpreting other people's algorithms!). The rest I thought was okay though - overall I thought it was easier than last year's, even though there were so many algorithms to do. I'm not going to predict my mark - I always get it wrong.
I'm just so glad it's over!
I'm just so glad it's over!
Well..... i reckon it was as hard as last year's one in some ways... like this year's one was easier to understand, but more algorithms and other crap to write...
I ran out of time... and left out some questions
For q23 a iv, i spotted one of two errors but because i ran out of time, i didn't write a desk check.... i reckon it should've been worth more than 3 marks if they wanted us to do a deskcheck and identify the errors. BTW the error i found is on line 29. If NumberTicketsSold is greater than EventTicketsAvailable, then EventTicketsAvailbe will end up with negative value.... Is this what everyone else got?
The sort algorithm was a bit*h too in q23, and the algorithm in q25 c iii was harder (i think). Did anyone get 13 as the reminder in q25 c ii? And what did everyone else write for q25 b?
Overall... it was a hard exam... looking at max raw mark of mid 70s (that's very very generous in marking) ... but at least we're happy it's all over.
I ran out of time... and left out some questions
For q23 a iv, i spotted one of two errors but because i ran out of time, i didn't write a desk check.... i reckon it should've been worth more than 3 marks if they wanted us to do a deskcheck and identify the errors. BTW the error i found is on line 29. If NumberTicketsSold is greater than EventTicketsAvailable, then EventTicketsAvailbe will end up with negative value.... Is this what everyone else got?
The sort algorithm was a bit*h too in q23, and the algorithm in q25 c iii was harder (i think). Did anyone get 13 as the reminder in q25 c ii? And what did everyone else write for q25 b?
Overall... it was a hard exam... looking at max raw mark of mid 70s (that's very very generous in marking) ...
but at least we're happy it's all over. In the Q23 algorithm I just read the events into the array in sorted order, since they were already sorted by event number in the file - do you reckon that's okay? *hopes so* Or maybe I just interpreted the question wrong.
I got 0 as the remainder in 25(c)(iii), because wasn't the sum divided by 13?
I just wrote about how f.p. takes up more bytes and requires more complicated calculations than integers but it's more precise or something (made it up on the spot... I hardly had time to study for the option topic!).
Exam was alrite
Damn i was really glad i didn't just rely on SDD knowledge, i found the exam much easier than last years, did all the questions without much trouble and im hoping to get band 6 but more realisticly it might be a band 5.
Damn i was really glad i didn't just rely on SDD knowledge, i found the exam much easier than last years, did all the questions without much trouble and im hoping to get band 6 but more realisticly it might be a band 5.
man u must b pretty good if u found it easier than last year.
i found the short answer questions okish, but the thing that killed me was time! i had to think and construct 3 algorithms and find errors in another one within 3hrs under the pressure of exam
conditions.
bah :\
i found the short answer questions okish, but the thing that killed me was time! i had to think and construct 3 algorithms and find errors in another one within 3hrs under the pressure of exam
conditions.
bah :\
Well i actually tutor this subject to other students since the teachers that teach it are almost always wrong. So i guess what really helped me was all the variety of questions/problems ive come across and just practice of doping em over and over.
The thing with algorithms is u gotta spend like 5 mins planning, drawing a basic outline with ure main progs and waht modules u gonna have and then try to actually draw it or write it all up in under 15 mins.
It was quite a lenghty exam, only had 5 mins to spare.
The thing with algorithms is u gotta spend like 5 mins planning, drawing a basic outline with ure main progs and waht modules u gonna have and then try to actually draw it or write it all up in under 15 mins.
It was quite a lenghty exam, only had 5 mins to spare.
yeah, i'm with everyone else, 3 algorithms is way too much for the time we got, especially since it finished at 5! it was just question 23a) and i won't get full marks for that 6mark algorithm in 23b)! and i now wish i did the evolution option, not hardware, the algorithm wasn't too bad, but once again, i won't get full marks for it. all i can say is at least there wasn't a circuit you had to create like last years!
It has to scale better than last years doesn't it? i mean for a hard test (and i know it doesn't theoretically scale on difficulty) but everyone here is pretty smart, and still complaining! i can't see how so few people were put in the band5/band6
It has to scale better than last years doesn't it? i mean for a hard test (and i know it doesn't theoretically scale on difficulty) but everyone here is pretty smart, and still complaining! i can't see how so few people were put in the band5/band6
Okay... let's redo q25 a i and ii again step by step:
The packets are: 0101100101 0110100111
Rules: the 1st digit and the last digit are start and end bits.
the 2nd digit is the direction X and Y.
In X direction: 1 is right, 0 is left
In Y direction: 1 is up, 0 is down
i. Spliting the digits away they become:
0 (start bit) 1 (right) 0110010 (mm) 1 (end bit)
0110010 = 50 base 10
therefore x direction: right 50 mm
similarly with the 2nd packet:
0 (start bit) 1 (up) 1010011 (mm) 1 (end bit)
1010011 = 83 base 10
therefore y direction: up 83 mm
ii. as for the checksum for the data above, the checksum bits are the sum of the two packets... In the exam i didn't include the direction into the calculation. I did 83 + 50 = 133 / 13 and get a reminder of 3 (sorry when i said i got 13 as the remainder).
If you included the directions into the addition, i.e:
10110010 + 11010011 then you should get 0 as the remainder.
The question is now... were u suppose to include them or not?
The packets are: 0101100101 0110100111
Rules: the 1st digit and the last digit are start and end bits.
the 2nd digit is the direction X and Y.
In X direction: 1 is right, 0 is left
In Y direction: 1 is up, 0 is down
i. Spliting the digits away they become:
0 (start bit) 1 (right) 0110010 (mm) 1 (end bit)
0110010 = 50 base 10
therefore x direction: right 50 mm
similarly with the 2nd packet:
0 (start bit) 1 (up) 1010011 (mm) 1 (end bit)
1010011 = 83 base 10
therefore y direction: up 83 mm
ii. as for the checksum for the data above, the checksum bits are the sum of the two packets... In the exam i didn't include the direction into the calculation. I did 83 + 50 = 133 / 13 and get a reminder of 3 (sorry when i said i got 13 as the remainder).
If you included the directions into the addition, i.e:
10110010 + 11010011 then you should get 0 as the remainder.
The question is now... were u suppose to include them or not?
yes u do include those bits becoz at the end of the 1st paragraph it specifies "eight data bits" and the quetions sez "the data bits from the two packets are added"
i did include them and i got 389(the total of bits) / 13 = 29 r 12
can sum1 confirm??
i did include them and i got 389(the total of bits) / 13 = 29 r 12
can sum1 confirm??
idontknow
New Member
i think more people will get into band 6 this year as they did a review this year of last years marking and found SDD band 6 to be set too high. So this year band 6 might be in the range of 85 - 100 depending on how people go. Also they are allowed to be more flexible in the marking scheme this year my teacher said (he marked Q21 last year) so that if people interpreted the question differently or came up with something not in the scheme but still correct then they will give marks, unlike last year.
freaken hell!!!!!!!
such a stupid paper, i dind't even finish the damn paper! i left 11 marks out in question 23,...and i make mistakes at other places.
what the hell was those multiple chioce questions, they suuuckkk!!!!!
ddddddddaaaaaaaaammmmmmmmmmnnnnnnnnnnnnn!!!!!!!!
such a stupid paper, i dind't even finish the damn paper! i left 11 marks out in question 23,...and i make mistakes at other places.
what the hell was those multiple chioce questions, they suuuckkk!!!!!
ddddddddaaaaaaaaammmmmmmmmmnnnnnnnnnnnnn!!!!!!!!
I got remainder 3 in that one
What pisses me off is how much easier the evolution of programming option topic was compared to the hardware one
There wasn't enough time and I had about 45 seconds to do the desk check (thankfully it was only 3 marks)
Most of the theory was alrite but the algorithms were rather hard
What pisses me off is how much easier the evolution of programming option topic was compared to the hardware one
There wasn't enough time and I had about 45 seconds to do the desk check (thankfully it was only 3 marks)
Most of the theory was alrite but the algorithms were rather hard