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Today's Computing Exam

John

Posted 28 May 2007 - 08:13 PM

It is 64MB

Amazoph

Posted 28 May 2007 - 08:13 PM

I got 64mb on that one myself :S

Most of the paper appeared to be pretty easy; a couple of trickier questions but they weren't too bad.

I did networking for third unit - hopefully I've done better in the exam than I've done on past papers ohmy.gif

Pete

Posted 28 May 2007 - 08:14 PM

Nope, that was my mistake. 64MB is the correct answer.

sunnyho

Posted 28 May 2007 - 08:34 PM

QUOTE(Pete @ May 28 2007, 09:14 PM) View Post
Nope, that was my mistake. 64MB is the correct answer.

I hate that question btw, it is a trick question, it said that the computer has 512Mb of ram...
but only 64Mb of ram is usable. weird....

verticalforce

Posted 28 May 2007 - 08:56 PM

Let say that for an A pass, I need 80%, how many marks can I afford to lose in the actual exam if I got like 54/60 in the coursework?

Michael V

Posted 28 May 2007 - 08:57 PM

What did everyone get for the size of the image from the camera? I can't remember the question number.

Was it 3000x2000x24 since the amount of colours was 2^24? Therefore 17.16MB?

John

Posted 28 May 2007 - 09:06 PM

QUOTE(verticalforce @ May 28 2007, 09:56 PM) View Post
Let say that for an A pass, I need 80%, how many marks can I afford to lose in the actual exam if I got like 54/60 in the coursework?


160 - 54 = What you need for an A

Pass mark will be more likely 68% to 74%

Yaz

Posted 28 May 2007 - 09:13 PM

QUOTE(Michael V @ May 28 2007, 09:57 PM) View Post
What did everyone get for the size of the image from the camera? I can't remember the question number.

Was it 3000x2000x24 since the amount of colours was 2^24? Therefore 17.16MB?


That's what I got too.
I thought of it as 3000*2000 gives the number of pixels then mulptiply by colour depth of 24 bits giving 17.166MB

Kiltox

Posted 28 May 2007 - 09:13 PM

I got 64mb the first time but it stated in the specifications that the system had 512mb of RAM.... what's the other 448mb doing ?biggrin.gif

Could be some kind of mix-up I guess.... or I've done that question wrongly now smile.gif

Meh.

ross_rosco

Posted 28 May 2007 - 09:27 PM

Yes I was confused on the camera question as well! I ended up with the 17.2 MB as well, but i only resorted to that because i did it the way i would have thought and ended up with over a terabyte for every photo! I was positive that if it said ...colours then you multiplied by that number (i.e. 16,777,216) and if it said ... bit colour then you multiplied by the smaller number. I was wrong!

Kiltox

Posted 28 May 2007 - 09:31 PM

QUOTE(ross_rosco @ May 28 2007, 10:27 PM) View Post
Yes I was confused on the camera question as well! I ended up with the 17.2 MB as well, but i only resorted to that because i did it the way i would have thought and ended up with over a terabyte for every photo! I was positive that if it said ...colours then you multiplied by that number (i.e. 16,777,216) and if it said ... bit colour then you multiplied by the smaller number. I was wrong!

17.2mb was what I got too (well, 17.17mb, but it's all good)

Pete

Posted 28 May 2007 - 10:05 PM

16,777,216 colours is 24 bits. You're expected to be able to recognise the number of bits that you would need to give values of 65536, 16777216, etc. The picture size should have came out at roughly 17.1-17.2MB, I can't remember exactly what.

Yaz

Posted 28 May 2007 - 10:34 PM

QUOTE(Pete @ May 28 2007, 11:05 PM) View Post
16,777,216 colours is 24 bits. You're expected to be able to recognise the number of bits that you would need to give values of 65536, 16777216, etc. The picture size should have came out at roughly 17.1-17.2MB, I can't remember exactly what.


It is good if you remember them but for people who don't remember you can use your calculators.
Using the log rules, the ln button can be used by inputting:

ln(number of colours)
ln2

in this case it would be:
ln16777216
ln2

the answer would be the power of 2 which gives the number of colours. Here it gives 24.
Hope that helps other ppl in the future! smile.gif

ross_rosco

Posted 29 May 2007 - 09:26 AM

log rules! - nice work haha. And as someone above said, we are expected to recognise these numbers as results of powers of two.

Hey - when drawing the mesh with five nodes, did anyone else notice the pentagram? ohmy.gif ohmy.gif

spooky...

sirjambob

Posted 29 May 2007 - 11:31 AM

"I got 64mb the first time but it stated in the specifications that the system had 512mb of RAM.... what's the other 448mb doing ?"

The specification stated 512 Mb (Megabits) of RAM, whereas the calculation gave you the answer in Megabytes. Really, 512 Mb is 64 MB so I'm guessing either of those answers would be right.

Why anyone would have a 3.9 GHz processor and only 64 MB of RAM is beyond me though...

John

Posted 29 May 2007 - 03:45 PM

QUOTE(sirjambob @ May 29 2007, 12:31 PM) View Post
"I got 64mb the first time but it stated in the specifications that the system had 512mb of RAM.... what's the other 448mb doing ?"

The specification stated 512 Mb (Megabits) of RAM, whereas the calculation gave you the answer in Megabytes. Really, 512 Mb is 64 MB so I'm guessing either of those answers would be right.

Why anyone would have a 3.9 GHz processor and only 64 MB of RAM is beyond me though...


That explains it then, and no one mentioned it was a lower case b tongue.gif

And come on, it's the SQA, they are numbskulls lol

Kiltox

Posted 29 May 2007 - 04:43 PM

Yeah because we all measure system RAM in Megabits.

That's just f*cking evil.

321

Posted 29 May 2007 - 05:59 PM

ohh well got it wrong.. haha

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