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General discussion (archive)Win some gogs, by doing some very easy math.(164 posts)(164 posts)
(164 posts)
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This is my favourite topic
Posted July 30, 2011
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stonebro: I'll participate when the contest starts demanding some real math.

Come on, elevate it to freshman calculus, at least!
"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.
Posted July 30, 2011
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rooshandark8: Since you usally reply asap, just letting you know if you don't know already,
That Round 6 is up.
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Barefoot_Monkey: Thanks. I appreciate that.

1. right (B)
2. wrong (D) - by inspection
3. C - by inspection
4. B - (w + 2)(w + 3) / (w - 4)(w + 3)
5. C - denominator = (b-4)(b+2)
6. A - by inspection
7. C - (r+5)(r+2)(r-10)3 / 3(r-10)(r+5)
8. C - demoninator = (c+7)(c-2)(c-5)(c+3)
9. D - by inspection
10. A - by inspection
11. D - (r+5)(r-2) / (r+7)(r-2)(r+7)(r-3)
12. C - (3/4) = (6/8)
13. B - by inspection
14. D - denominators are (p+2)(p+5) and (p+2)(p+3)
15. C - w^2 + w - 20 = (w-4)(w+5), so multiply by (w-4)
16. D - by inspection
17. A - denominators are (q+2)(q+3) and (q+2)(q+1), so numerator q(q+1)+(q+3)
18. B - 3(t - 2)(t+2) / (t-2)
19. B - 0 = 8y - 16 - 3
20. C - 5a -10 = 3a + 9 --> 2a = 19
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.
Posted July 30, 2011
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rooshandark8: Why is 2 "D"
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.
Posted August 01, 2011
crap, need more time...
Posted August 02, 2011
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
Posted August 02, 2011
Tickets Awarded to all who posted a answer
Bonus Round
Reward: 10 tickets
No questions
Here are the rules
Come up with the HARDEST question you can
and basically wait for a answer

To get reward must do the following
Come up with 1x questions
Answer 3x questions
Dance around in a circle singing your national anthem
Buy me a game any game :P
Go swiming in space
Eat chicken
And say
GOG FTW
and in a whisper rooshandark8 is awesome
Enjoy
Posted August 02, 2011
Hm... will have to think about this one.
Posted August 02, 2011
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.
Posted August 02, 2011
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?


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Garugo: 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.
Milkshake: Nothing. It is untouched.

Equation: m/y = 2b/x | mx = 2by | y = mx/2b
Post edited August 02, 2011 by oasis789
Posted August 02, 2011
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oasis789: *snip*
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*
Post edited August 02, 2011 by Garugo
Posted August 02, 2011
Here's a fun problem. For some contrived reason you will soon need to measure 15 minutes of time. When you are ready you must call out when the 15 minutes begins and then call out when it ends, give or take 30 seconds. The only tools you have are 2 long pieces of fuse and your trusty lighter. The fuses do not burn at a steady rate (parts of it burn quickly and parts burn slowly), but despite this each is guaranteed to burn for exactly 1 hour in total.

How do you go about doing this?

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Garugo: 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.
I have the milkshake, therefore I will drink it. I wouldn't want to leave it long enough for you to get hold of it first. :P

For your other question, we can start by establishing that neither y nor x is 0 because then the given equality would not hold.

Knowing this, we can multiply both sides by mx, revealing mx = 2by.

From here, y = (m/2b)x for nonzero values of b. For b = 0 all we can say is m = 0, x != 0, y != 0
Posted August 02, 2011
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Barefoot_Monkey: Here's a fun problem. For some contrived reason you will soon need to measure 15 minutes of time. When you are ready you must call out when the 15 minutes begins and then call out when it ends, give or take 30 seconds. The only tools you have are 2 long pieces of fuse and your trusty lighter. The fuses do not burn at a steady rate (parts of it burn quickly and parts burn slowly), but despite this each is guaranteed to burn for exactly 1 hour in total.

How do you go about doing this?
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.
Post edited August 02, 2011 by nagytow
Posted August 03, 2011
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oasis789: *snip*
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Garugo: 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*
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.
Posted August 03, 2011
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Barefoot_Monkey: Here's a fun problem. For some contrived reason you will soon need to measure 15 minutes of time. When you are ready you must call out when the 15 minutes begins and then call out when it ends, give or take 30 seconds. The only tools you have are 2 long pieces of fuse and your trusty lighter. The fuses do not burn at a steady rate (parts of it burn quickly and parts burn slowly), but despite this each is guaranteed to burn for exactly 1 hour in total.

How do you go about doing this?
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nagytow: 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.
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.
Posted August 03, 2011
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nagytow: 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.
Flawless victory!
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This is my favourite topic
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General discussion (archive)Win some gogs, by doing some very easy math.(164 posts)(164 posts)
(164 posts)