Here's two brainteasers for you:
How can three men and one woman have mutually safe heterosexual intercourse with just two condoms ? Assume that no condom can break or leak, and that you cannot wash a used one.
2. A rock is dropped from the top of the Empire State Bldg. At what speed does it hit the ground, and how long does it take it to get there ?
The answers will follow -at the end of the day-.
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does that ho have any cold sores?
and does she like to bite?
Like that.
Also this, fr. Cookies w/ Milken
Please, first one is from heard on the street, the second one is basic physics.
Maybe I'm reading the question incorrectly, but: 2 men use the condoms, 1 doesn't. Assuming the woman has nothing, the men are mutually safe.
Hits the ground in approximately 9-10 seconds travelling at ~100m/s.
op go diaf
(but yeah this question has been asked before and I won't spoil it)
Troll Answer: It hits the ground at a speed of zero. It's moving before impact and after impact, however the moment the rock impacts the ground the ground pushes back with equal force. It takes approximately no time to get there (there being the speed on impact) as the rock starts at a speed of zero.
Real Answer: Probably having to do with the terminal velocity of the rock and how long it takes to reach that. Wouldn't you need to know the weight of the rock? Me no don't do good science stuff.
old one...
//www.wallstreetoasis.com/forums/dirty-trading-puzzle
and StryfeDSP you really should have listened in sex-ed
Given the possibility the bitch has oral herpes, I can think of at least 4 other orifices that can get the job done.
All you need is a rubber hose, and a large kimono dragon.
I got the solution to #1 while talking with a friend over dinner...dunno what that means for me >.>
First question I had was how broadly we were defining heterosexual sex here, but I figured that got way more behavioral than the interviewer would be looking for...so assuming each man just has penile-vaginal intercourse with the woman, here's the solution, assuming no leakage and Latex erosion, and assuming condoms are handled gingerly to prevent hand-to-hand infection somehow (basically all of the crazy scenarios are out):
Guy #1: Doublebags it with two condoms. Finishes. Takes off the outside condom and passes it to Guy #2. Guy #2: The outside is potentially contaminated with the woman's STD's, but the inside is completely safe. Uses only the outside one, same orientation. Finishes. Passes it to Guy #3. Guy #3: The outside and inside of the outside condom is infected, potentially. The only solution left is to take Guy #1's inside condom, turn it inside out, and thus place this inside condom inside of the outside one, and finishes. All sides of the condoms are now infected, but all participants are satiated.
Make the answer a bit less or a bit more formal sounding depending on the place to fit the culture, and good to go.
2 haven't thought about yet, will edit soon.
EDIT: Might be a weak answer to #2, but if it's dropped from the top of the Empire State, wouldn't it just hit the top of it? If you mean right off, then I guess you might factor in terminal velocity and do a quick physics calc, but that seems too nontrivial. Then again, don't know the context of the firm.
BONUS EDIT: Thought approach to #1: Fixing Condom A, and Condom B, then since there are two sides to each and four participants, we know that having fixed these condoms that each participant must touch one side only and one condom only. Arbitrarily choose Condom B outside to be the woman's side. Then one man gets the inside, and the other two get Condom A. Since woman's side was already fixed, this immediately necessitates the doublebagging in all but one case. Finally it's very quick to see how the order that must occur.
Yes, that's the answer.
Was answered by ytnifny
If you want to detail, you can say: Ignoring air resistance,
We need to know how high Empire State is. From memory, about 100 floors/ each fl. about 4 meters high- so that's 400 meters high.
Using the simple equation of motion, v^2 = u^2 + 2gD, where v=final velocity, u=initial (zero), D = distance = 400 m, g (gravity)=9.8 m/s^2 ~ approx 10 m/s^2
v^2= 210400 = 8000 m^2/s^2, v= approx. 90 m/s
v= u + g*t, so t ~ 9 s.
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