"very easy math"
It supposed to be easy lol
and well its suppose to give ppl a chance to earn the tickets way before the huge contest.

Why is 2 "D"
Other guys said that my answer is right
Please explain
and check your other answers aswell
since so far i haven't taken any tickets off for wrong answers since they were mostly corrected later on.

So you can post a correction if you want
If you are wrong, which you maybe.

Because I misread the question and didn't realise that the "-" outside the parenthesis was part of the expression.

Thanks for pointing it out. I'll correct my answer and mark the change.

crap, need more time...

1B correct
2C correct
3C by the laws of exponents
4B. factorize numerator to (w+3)(w+2) and denom to (w+3)(w-4)
5C. factorize denom to (b+2)(b-4) and set to zero.
6A by the laws of exponents
7C. factorize 1st numerator to (r+2)(r+5) and 2nd denom to (r-10)(r+5)
8D. factorize and cross-multiply to get (c+2)(c+1) / (c+7)(c+3). set denom factors to 0
9D. cross multiply
10A. cross multiply
11D. cross multiply
12C. lowest common multiple of 4 and 8
13B. lowest common multiple
14D. factorize denoms to (p+5)(p+2) and (p+2)(p+3)
15C. factorize denom to (w+5)(w-4), expand (w-3)(w-4).
16D. by inspection
17A. factorize denoms to (q+2)(q+3) and (q+1)(q+2)
18B. simplify the top to 4(t+2)(t-2).
19B. y=19/8
20C. a=19/2

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Enjoy

Hm... will have to think about this one.

Let's say I have a milkshake and you have a milkshake. I drink my milkshake, what will happen to your, yet untouched, milkshake?

Alternately:

Set the equation m/y = 2b/x up to y = mx + b form, show your work.

There are N pirates on a pirate ship trying to divide ten gold pieces among themselves. The pirate captain has the authority to propose a distribution, which the pirates all vote on. The vote passes if a majority is in favor or there is a tie, but if the vote does not pass, the captain is thrown overboard to the sharks, and the next most senior pirate makes his proposal. If his proposal fails, he too gets thrown overboard, and so on and so forth.

Pirates are rational self-interested folks who want to maximize their own share of the gold, while staying alive.

You are the captain. What is the proposal that maximizes your share of the gold while keeping you alive?


Milkshake: Nothing. It is untouched.

Equation: m/y = 2b/x | mx = 2by | y = mx/2b

For the pirates... depends on the amount of pirates there are since there's only 10 coins, so any amount larger than about 19 pirates will probably get the Cap'n thrown overboard (alternating the pirate he gives the coins to: Pirate 1 gets one, Pirate 2 doesn't, Pirate 3 does, etc.). So... 1 gold, or if there are fewer pirates (let's say less than 10 pirates), up to all 10 pieces (for 2 pirates) or 9 (for 3), and so on... I think.

*Edited out my first answer/response :D*

Not too hard for me as I know this riddle (yours is a bit modified but the idea is the same) :)

You need to light both ends of one fuse and one end of the other. The first one will burn out in 30 min. The other one is still burning so now we need to light its second end and call the start of the the 15 min time. It's been burning for 30 min so without lighting the other end it would burn for another 30 min. By lighting the other end we cut this time in half which gives us 15 min. The second fuse will burn out in exactly 15 min so we need to call the end of the 15 min time when it's burnt out completely.

Edit. My question:

You have 12 balls. The balls look exactly the same, but one of them has a different weight than the other 11 balls. You don't know if the ball is heavier or lighter than the other balls, all you know is that it has a different weight.
You have a traditional balance scale and you have to figure out which one of the balls has a different weight than the others with no more than 3 weighings.

You got the right idea, but I gave the question in terms of N so that solutions would be generalized, so you should express the solution formally in terms of N.

This is an information theory problem. The solution principle is to divide the set into 3 subsets, and weigh 2 of those subsets. If they weigh the same, the target is in the 3rd subset, and you reiterate the process within that subset to identify the target.

Flawless victory!