That's a very good idea. Maybe you should keep a note of all of the worked solutions too, not just the correct answers, that way people can learn from others, and also it's easier to challenge incorrect answers (or demonstrate their accuracy)

Section 2 question 16... I agree with you that it seems odd. I did the following: 1.) write out the numerator having an index of 3, and the denominator with an index of -4 = sqrt[(2x10^3+0.8x10^3)/(4x10^-4 + 3000x10^-4)] 2.) simplify Step 1 = sqrt[(2+0.8)x10^3/(4+3000)x10^-4] 3.) bring the 10^-4 up to the numerator (and thus switch the sign, so the numerator is now to the power of 7) = sqrt(2.8x10^7/3004) 4.) bring 2^6 outside of the root (so it becomes 1000) = sqrt(28/3004) x 1000 5.) bring a factor of 4 outside of the root (so it becomes 2) = sqrt(7/3004) x 2000 This is simplified as much as possible. 3004 is not divisible by 7, and the square root of 7 is irrational. I get exactly the same answer as you

Ah, I see what the problem is now - it's the quality of the photocopying I (and so did A-Parsons it seems) read the second term in the denominator as 3x10^-1 (and therefore rewritten it as 3000x10^-4), but you've got it as 3x10^-4 So the "-4" has photocopied to look like a "-1"

I agree with you completely for questions 2 and 6, but could u elucidate your working for this one? I got A I think, but I might be mistaken... I also take maths in french so the whole "the cube has unit length sides" might have confused me... anyways I'd be glad to understand. Thankyou Crims

With regards to S2Q16, yes it was bad scanning - I looked at the original paper earlier, and the denominator was in the order -4, not -1. That clears it up, and I can write an explanation fine now. :L And don't worry about jumping around all over the place, I can fill them in wherever you want - order doesn't matter.

thanks for putting this up, i feel slightly better knowing that i did get some questions right, I could only remember the ones i got wrong

Entropy, I hope you don't mind, but because it was a difficult one to write out in words, I have posted a link to your workings for S2Q16 (which are beautiful, in all their lilac glory, by the way. :L). However, would you mind rewriting it in darker ink, so it is easier to read (if less pretty)? Oh, I've also used your one for S2Q8. Thanks for this, everyone who is contributing.

I think you meant rewriting S2Q16. (And lol, I apologized for the lilac-ness. That's the only pen I have in front of me right now!!) I'm too lazy to rewrite the entire thing, but I played with the picture a little bit and adjusted the saturation. I don't think this one needs a diagram. And besides, I'm not that great with probability, but C is the answer I chose as well.

LESLIE!!! STOP CORRECTING ME!!! Yes, that is what I meant. Gr. And thanks, it is far more understandable now. And also, why are you not on MSN/AIM, yet you are replying on here? Have you blocked me? :L

If someone asks "Why are you calling Entropy 'Leslie'?", I may be tempted to post that photo on here... Though I suppose I can't, as you would probably go mental 'cos you are afraid of paedophilles/ stalkers, and stuff like that... :L Bummer. I will juust have to explain it if anyone asks. And as for the horrible probability one, I think all the maths ones need a pretty drawing, but now you've explained it a bit, I might be able to do my own pretty drawing for it... I'll try in a bit.

Don't you DARE post that picture up! I will personally have to come to UK and kill you then (regardless of whether I get any interviews). As for the pretty drawing... well you know, you can ... try. Oh, Chwirkytheappleboy, can you please take a shot at S1Q3?

I feel like such a tard now... But thx that did help now I know what I did wrong... It's unit side so it's 1 but since it's half cuz the base is vertex etc blablbla u divide that side by two and then the maths logically follows thx again guys. I'll try to contribute this weekend when I have a bit more time^^ Goodnight guys Crims

Oh yeah, that one's actually pretty easy if you want to do it by a "brute force" method. I guess there must be a more efficient and logical way of doing it, but I chose a method which definitely gets the correct answer even if it is a bit slow. Basically because all of the charts except the correct one have exactly two bars wrong, then you know that once you've got the "correct" chart, then when you compare the others to it, they'll all have exactly two bars which look different. So just pick a random chart, assume it's "correct", and compare all the other charts to it. Then you can start eliminating incorrect answers. So say we pick an initial guess. Chart A, for example. We assume this is our correct position, but then we notice that when we compare it to Chart C, bars B, C, and E are different. So since 3 bars are "incorrect", Chart A cannot actually be the correct answer. Carry on, and eventually Chart D seems to be the correct answer; -> Chart A.) Bars B & C are different -> Chart B.) Bars B & E are different -> Chart C.) Bars C & E are different -> Chart E.) Bars A & D are different

Damn, now I have to rewrite that answer in <5 lines... :L I think D was what I put, and I did the same method as you, though less random. I started from chart A, bar 1, then worked my way through. :L