A friend of mine had an interview a couple weeks ago with Google Inc. He provided me a list of just some of the questions he was asked. I’ve added a few more from others I have talked to who had interviews with the internet giant, Google, as well. See if you can answer them. Many are open ended with several right answers, therefore I did not provide the answers.
1. How many golf balls can fit in a school bus?
2. You are shrunk to the height of a nickel and your mass is proportionally reduced so as to maintain your original density. You are then thrown into an empty glass blender. The blades will start moving in 60 seconds. What do you do?
3. How much should you charge to wash all the windows in Seattle?
4. How would you find out if a machine’s stack grows up or down in memory?
5. Explain a database in three sentences to your eight-year-old nephew.
6. How many times a day does a clock’s hands overlap?
7. You have to get from point A to point B. You don’t know if you can get there. What would you do?
8. Imagine you have a closet full of shirts. It’s very hard to find a shirt. So what can you do to organize your shirts for easy retrieval?
9. Every man in a village of 100 married couples has cheated on his wife. Every wife in the village instantly knows when a man other than her husband has cheated, but does not know when her own husband has. The village has a law that does not allow for adultery. Any wife who can prove that her husband is unfaithful must kill him that very day. The women of the village would never disobey this law. One day, the queen of the village visits and announces that at least one husband has been unfaithful. What happens?
10. In a country in which people only want boys, every family continues to have children until they have a boy. if they have a girl, they have another child. if they have a boy, they stop. what is the proportion of boys to girls in the country?
11. If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?
12. If you look at a clock and the time is 3:15, what is the angle between the hour and the minute hands? (The answer to this is not zero!)
13. Four people need to cross a rickety rope bridge to get back to their camp at night. Unfortunately, they only have one flashlight and it only has enough light left for seventeen minutes. The bridge is too dangerous to cross without a flashlight, and it�s only strong enough to support two people at any given time. Each of the campers walks at a different speed. One can cross the bridge in 1 minute, another in 2 minutes, the third in 5 minutes, and the slow poke takes 10 minutes to cross. How do the campers make it across in 17 minutes?
14. You are at a party with a friend and 10 people are present including you and the friend. your friend makes you a wager that for every person you find that has the same birthday as you, you get $1; for every person he finds that does not have the same birthday as you, he gets $2. would you accept the wager?
15. How many piano tuners are there in the entire world?
16. You have eight balls all of the same size. 7 of them weigh the same, and one of them weighs slightly more. How can you find the ball that is heavier by using a balance and only two weighings?
17. You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How should he allocate the gold in order to maximize his share but live to enjoy it? (Hint: One pirate ends up with 98 percent of the gold.)
Do you still think you have what it takes to work for Google?
(64 votes, average: 4.11 out of 5)
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Could you just put the answers after?
Now that I think about..I am sure I could just find the answers using google xD
These are great. I’ve figured out a number of them but would still like the answers to them all. Did your friend have to answer them on the spot?
Here are answers that I come up with. Let me know if you come up with something I don’t have
1) School bus = 50′ x 10′ x 8′ = 4000 ft^3
Golf ball = 1.5 in ^ 3 = 1.5/1728 ft^3
4000/(1.5/1728) = approximately 4.6 million
2) Not worry because I would fall in between the blades and the plastic at the bottom
3) Hmm… let’s assume there are 10,000 (1E4) windows on the average metropolitan seattle building and there are about 500 buildings…. thus about 5E7 (50 million) windows in seattle…. I would probably charge about 1$ a piece so 50 million dollars
4) … no idea
5) It keeps numbers separate from each other so the computer can pick out the right ones easy. Also they do math by themselves with the data they hold. Your uncle uses them a lot to make his websites
6) 11?
7) Get as far as I can… and then buy more gas or whatever
Organize them by color or put the most frequently worn ones on one side.
9) Nothing? Wouldn’t the wives know that 99 husbands had been unfaithful so it wouldn’t cause them any concern?
10) I’m not sure how to solve this one but it would be probably in the 75% female range?
11) Not sure about this one either… I’m glad I’m not applying at google (yet)
12) 7.5 degrees (the hour hand is 1/4th of the way between 3 and 4, the angle measure of that is 360/12 = 30 degrees between hours / 4 = 7.5 degrees)
13) 1 and 2 cross, taking 2 minutes, 1 goes back carrying the flashlight = 3 minutes. 5 and 10 cross, taking 10 minutes = 13 minutes, 2 goes back = 15 minutes. 1 and 2 cross again, taking 2 minutes, = 17 minutes
14) No.
15) Ahh a classic. I remember reading there are ~500 piano tuners in new york which has a total population of ~13 million? so 500/1.3E7 = x/6E9 = 230769… of course about 50% of the world is too poor to own a piano or need a piano tuner so the answer is probably lower, my guess is around 125,000.
16) Weigh 3 vs 3. If one side is heavier, weigh one vs one of those three. If that is equal the third is the heavier one (if not, the heavier is heavier). If the 3vs3 is equal, weigh the two remaining balls and there you go.
17) Never heard this one before, perhaps by giving one gold coin to two others who will then agree with the pirate’s plan? Number 5 keeps 98% of the gold this way leaving 2% to be split among two other pirates.
The answer to all th questions is simply.. “I don’t know, Google it” =]
Number 16 is easy:
Take two sets of three balls and weigh them. If the scale stays level weigh the two remaining balls and the heavier one is the heaviest. If the scales do not stay level take the heavier set and put two balls on either side of the scale and keep one off, if the scales stay even then the ball that is not on the scale is heavier, if one goes to one side, than it is heavier.
ya, whatever. about 30% of these are bullshit questions … the type that make you flinch, if you’re the flinching type. the rest are worth thinking about, but your answers to those just mean you know how to read books. ironically, it’s the bs questions, and the way you approach them, is what makes the difference.
hehe, this from a fool that’s never worked for a company larger than 20 other fools.
my favorite is #13 and i dont know the answer, the reason i like it so much is because its hard! at first you say, oh easy you just make the fast one acompany all of them back and forth, but it doesnt add up. if the fastest teleports back magicaly to the begining, then it would add up to 17minutes. the one mintute man(laughs) and the two minute man cross the bridge, making two minutes, the one minute one, returns to the other side, making another minute, so we got three, the one minute one crosses with the five minute one, making 8 minutes, then returns 9 minutes, then the 10 minute. while writing i thought of another way, but now i get less! like 15 minutes or 12 minutes. its so perfectly made to be so tricky, i likes!
For number 17, how exactly can one pirate end up with 98% of the 100 gold coins (98 gold coins) with at least three other people being happy with the deal? If Pirate one has 98, then pirate 2 and pirate 3 both have one how can you say they would be satisfied with that?
I personally think these questions are bull, no offense.
The number of golf balls that can fit into a school bus, is most certainly and defiantly. A Googol. LOL….
Number 11 is easy.
It says “assuming constant default probability”.
So the answer is 0.95
I calculated #10, and it had a surprising result. Half the children are boys and half girls. Here’s the spreadsheet I used:
http://samsclass.info/fun/family_puzzle.htm
Here is how I explain it. Half the families have one boy and no girls, so they create excess boys.
Families with two children have one girl and one boy, so they keep the balance even.
Families with three or more children have more girls than boys, so they create excess girls. Their quantity is exactly sufficient to cancel out the excess boys.
The question with the bridge is easy: Have the fast camper carry the slowpoke.
the #10 i think it’s more simple…”In a country in which people only want boys” means that people in that country is only masculine, so the families are homosexual families and 2 man cannot still have a children, so for me the answer is 100% boys and 0% girls.
What a load of horseshit. Google is doomed.
#10 is interesting.
Assuming each family has 1 boy, then the average number of
girls per family is 1/2 1/4 …. which sums to:
$ echo “import math; print sum(1.0/(2
My previous note on #1 didn’t account for
spaces between the golf balls:
http://mathworld.wolfram.com/SpherePacking.html
Therefore I should multiply by 0.64:
$ units ‘40ft*10ft*7ft*0.64′ ’spherevol(.84in)’
* 1247255.1
/ 8.0176063e-07
The pirate answer has to be worked backwards.
* If there is 1 pirate left, he will get all of the gold.
* If there are 2 pirates left, #2 will keep all of the gold and #1 will get none.
* If there are 3 left, #3 will give #1 a coin and keep 99 for himself. This is the best deal that #1 can get - if he votes against it then #2 will get all the money.
* If there are 4 left, #4 will give #2 1 coin. Since we know what will happen if there are 3 left, this is the best that #2 can get.
* If all 5 are left, #5 will give 1 coin to #1 and #2. #1 will vote for it because he will get nothing otherwise. #2 will get the same thing anyway, so he’ll vote for it as well. #5 keeps 98 coins.
Would it work in real life? Probably not as both #1 & 2 would find the deal unfair and would rather get nothing than a mere pittance…
These questions are insane, i would consider some of them pointless to ask in an interview, i like the database to 8 year old one, thats quite good, specially if it was for a development job
Wouldn’t the answer to number 1 be “all of them”? Unless they’re including novelty huge balls..
I was once interviewed to google over the phone - i got to give’em thumbs up because they where checking on leads from people who have submitted projects to freshmeat - but now i got alittle sidetracks .. So, it was phone interview and i was presented 10 questions and first thing the woman says. Dont Use Google! I scored like 1 point in the questions but just hearing that request made my smile for few days =)
The pirate question is bullshit.
The answer is that the pirates will split the gold evenly. Because no matter how they divide it up, if the other pirates aren’t getting an even share, then they will kill the pirate dividing the gold up, and then divide it up again.
This will continue until they all have an even share and find the deal fair, or pirate number 5 (or 1, whichever is last) is the only one still alive, allowing him to keep all the gold himself.
No pirate would ever allow himself to be bought off for just one coin when they could have a larger share. Pirates are greedy. Unless of course the other pirates didn’t know exactly how many coins were to be divided up, but it’s implied that they do.
I would solve No 11 the following way:
Let x be the probability in 10min. Then you have a chance of x in 10min to observe a car and a chance of 1-x that you don’t observe a car. In the latter case, the chance to see a car in the next 10min is again x. But this only happens with a prob. 1-x, so the total prob. to see a car after 20min is x (1-x)x. Now there is again a chance of 1-(x (1-x)x) that you haven’t seen a car after 20min. Again the chance is x to see a car in the next 10min, so the total probability to see a car after 30min is x (1-x)x (1-(x (1-x)x)x = 0.95. Now just solve this eqn. for x…
you people are all idiots. They dont want you to calculate any of this out. If you think you had to do that then your are not the type of person that google is looking for. They want people with imaginative answers, not those who create spreadsheets.
They want answers like
Gary who said “The number of golf balls that can fit into a school bus, is most certainly and defiantly. A Googol. LOL.”
and
“The answer to all th questions is simply.. I dont know, Google it =]”
how do these questions qualify anyone…all it does is reinforce how much of a nerd you are. What about questions about people skills, accomplishments…real like scenarios?
8. Imagine you have a closet full of shirts. Its very hard to find a shirt. So what can you do to organize your shirts for easy retrieval?
You hang them, duh!
pirate question is simple… he walks to the edge of the plank and offers 1 gold each to two of the pirates.
#10,
Surprisingly enough there should be more boys
prob boy first time =50%
girl/boy 25%
girl/girl/boy - 12.5%
girl/girl/girl/boy - 7.25%
…
50% chance just a boy
50% boy and some girls
because as the number of girls per family go up, the chance drops, the math turns out to being more boys.
A simple way to think: its definite you’ll have a boy, but not definite you’ll have a girl.
the easiest way to think of number ten is this:
you have a 50/50 chance of having either a boy or girl, ergo, the ratio of boys to girls will be 1:1
And I actually thought of applying for Google when they set up post here…
Clearly nobody seems to understand - Microsoft does something similar, and they fully expect you to provide an answer on the spot. As crazy as it might sound, questions like these demonstrate your ability to think through a complex problem with little or no information. They expect you to take an educated guess. Most of the time you can ask them questions like - how many buildings are there in Seattle, or what is the size of the blender.
To Number 10:
——————
“In a country in which people only want boys, every family continues to have children until they have a boy. if they have a girl, they have another child. if they have a boy, they stop. what is the proportion of boys to girls in the country?”
——————
Surprisingly the relation will be 50% boy and 50% girls (if there are enough families for a fair statistic quote).
Think like this for 100 families: There might be 50 families with girls and 50 with boys in the first run. In the second run, the 50 girl families will have another baby, 25 of them male and 25 of them female. So we have now 50f/50m from the first run 25f/25m from the second run …. and so on. It always tends to half/half.
Interesting also, that the total number of children by applying this rule will be 2 per family on average, as by definition only 1 boy can be in a given family, and the above says that for each boy on average there will be also one girl.
For Number 6:
“How many times a day does a clocks hands overlap?”
the answer is 23.
I write the time here when the hands are (approx.) overlaping:
am: 0:00 / 1:06 / 2:11 / 3:17 / 4:22 / 5:27 / 6:32 / 7:38 / 8:43 / 9:48 / 10:53 / 12:00
pm: 1:06 / 2:11 / 3:17 / 4:22 / 5:27 / 6:32 / 7:38 / 8:43 / 9:48 / 10:53 /
To #11
——–
If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?
——-
The correct answer is approx. 63.1%
If the chance to see the car is 10 percent per minute, the first minute you have 10% chance, the second minute you have 10% of 90% = 9% (so total 19%), the third minute 10% of 81% (= 8,1%, total 27,1 %) ….
As the chance for 30 minutes is 95 percent, the chance for 1 minute is 9.5% and for 10 minute 63.1 %
Minute Single chance Cum. chance
1 9.5% 9.5%
2 8.6% 18.1%
3 7.8% 25.9%
4 7.0% 32.9%
5 6.4% 39.3%
6 5.8% 45.1%
7 5.2% 50.3%
8 4.7% 55.0%
9 4.3% 59.3%
10 3.9% 63.1%
11 3.5% 66.6%
12 3.2% 69.8%
13 2.9% 72.7%
14 2.6% 75.3%
15 2.3% 77.6%
16 2.1% 79.8%
17 1.9% 81.7%
18 1.7% 83.4%
19 1.6% 85.0%
20 1.4% 86.4%
21 1.3% 87.7%
22 1.2% 88.9%
23 1.1% 89.9%
24 1.0% 90.9%
25 0.9% 91.8%
26 0.8% 92.5%
27 0.7% 93.2%
28 0.6% 93.9%
29 0.6% 94.5%
30 0.5% 95.0%
Ooops… writing mistake:
——————-
For Number 6:
How many times a day does a clocks hands overlap?
the answer is 23.
———————-
Of course, I mean 22, as you can see in the table below. Sorry…
I do not get the pirate question. One assumptions is needed, and I will go through my logic for each version. The assumption is: How does a pirate vote if he knows his vote does not matter? If he get 1 gold piece either way - does he vote for or against? (#5 is the top pirate, #1 is the bottom one)
——————
Assuming pirates vote for a solution if he gets the same either way:
With two pirates, #2 must give #1 everything, if not #1 votes against, and gets it all anyway.
With three pirates, #3 takes all for himself. #2 votes for as he will get nothing anyway.
With four pirates, #4 takes all for himself. Everyone votes for, as they will get nothing anyway.
With five pirates, #5 takes all for himself. Everyone votes for, as they will get nothing anyway.
———————-
Assuming pirates vote against solutions where he would get the same either way:
With two pirates, #2 must give #1 everything, if not #1 votes against, and gets it all anyway.
With three pirates, #3 takes 99 for himself and gives 1 to #2. #2 votes for as he will get nothing if he votes against.
With four pirates, #4 takes 97 for himself and gives 2 to #2 and 1 to #1. #1 and #2 will vote for - had they gotten anything less then what they got, they would vote against. #3 votes against.
With five pirates, #5 takes 97 for himself. #1 and #3 are the cheapest to buy off - #1 gets 2 (must be more than 1) and #3 gets 1. #4 and #2 get nothing, and votes against, but #1 and #3 votes for, and #5 is safe.
——————-
Either way, I get 100 gold for #5 or 97 - not 98. Where is my flawed logic?
I can say for certain that 1 golf ball _can_ fit in a school bus.
CR - You’re forgetting that less than half of the pirates must agree for the top pirate to be killed. Since there are 2 pirates, the only agreements that can occur is both agree, one or the other agrees, or none agree. Since #2 will always agree (because he is smart), the agreements must be AT LEAST half. Therefore he will allocate himself all of the coins.
#s 9 and 17 are classic game theory scenarios.
#17 has a very important note omitted: that the more senior pirate gets to break ties in a vote. The answer is:
#5 gets 98 coins
#4 gets 0
#3 gets 1
#2 gets 0
#1 gets 1
This assumes that the pirates are actually capable of figuring out the logic. It is easiest to work through backwards:
If only two pirates remain, #2 and #1, #2 will get 100 coins and #1 will get 0 because #2 is more senior and has the deciding vote in a tie. Knowing this, #3, who doesn’t want to be thrown overboard, offers #1 a single coin (which is better than nothing). In that case, #3 would get 99, #2 would get 0 and #1 would get 1. Knowing this, #4 offers #2 a single coin with the distribution being #4 getting 99 coins, #3 getting 0, #2 getting 1 and #1 getting 0. Both #1 and #3 know that their best chance is to vote for #5’s allocation, which is #5 getting 98 coins, #4 getting 0, #3 getting 1, #2 getting 0 and #1 getting 0.
#9) Nothing will happen for 99 days. On the 100th day, every woman in the village will kill her husband. Every woman knows that every other man in the village has cheated, but naively believes her own husband to be faithful. If there were only two couples, Mr and Mrs Jones and Mr and Mrs Smith, Mr Jones has cheated with Mrs Smith and Mr Smith with Mrs Jones. Mrs Jones would know that Mr Smith has cheated and expect Mrs Smith to kill Mr Smith on the first day. Likewise, Mrs Smith knows Mr Jones has cheated and expects Mrs Jones to kill him the next day. When that does not occur, Mrs Jones and Mrs Smith will realize on the second day that their own husbands have cheated. For each couple added, another day is added for everyone to fully grasp the situation.
The questions about the number of golf balls one can fit in a bus and how much one would charge to clean all the windows in Seattle are really about the process. The interviewee is supposed to work through the problem while the interviewer observes their problem solving skills. The actual solution is inconsequential.
What crap. But it’s Google, so it does not surprise me. All of these stupid internet companies try so hard to be edgy and quirky. It’s embarrassing.
What an awful place to work that must be.
Those question were never asked in Google interviews. Are you out of things to write?
4. How would you find out if a machines stack grows up or down in memory?
What’s “a machine”? More to the point, shall we assume it has a user-accessible stack pointer? And does “up” mean that pushing something onto the stack increments the stack pointer?
Answer “A”:
Look at the “machine’s” designer’s guide. That manual, or alternatively its programming manual, will tell you.
Answer “B”:
Assuming “a machine” means a CPU with a program-accessible stack pointer, all CPU’s that employ stacks have an instruction for pushing the contents of the stack pointer onto the stack. So do two such pushes, followed by two pops into separate registers, then subtract the contents of the second register from the contents of the first.
If the result is negative the stack grows up (second stack pointer address greater than first). A positive number means the stack grows “down” (second address less than the first).
Kind of a dumb question in my own opinion…
you guys are awful, these are good questions but unfortunately none of you has a clue. just solidifies the truth about reddit patronage goodbye
For a previous response to the window washing anwsers, who charges $1 to wash a window, come on.. I think these type of questions are very innovative, as they do measure the creativity and then “smarts” of a person. For that is what they are looking for in an employee, not some number cruncher/ programmer. Innovation is the key to success these days. And for all the people that view these questions and say that they are “dumb/stupid” need not to be ignorant, but rather think about it from Google’s perspective.
9. The wives kill the king.
1. All golf balls can fit in a school bus. Not at the same time mind you, but that wasn’t the question
Ok people, Here are your answers
1. About 24, 18 If I want to fix an entire 12 pack in may bag.
2. Pray hard that the owner loves margaritas.
3. A million and 12 green cards.
4. Hire someone who knows.
5. Like the internet, but provides just the information I need.
6. As many as they wish
6.a. You have a clock with hands??
7. Convince someone else they want to get there.
8. Wear the first shirt I find. Be happy.
9. I’d get the hell out of dodge.
10. My kind of odds.
11. Better with a cell phone and taxi company’s number.
12. 360
13. Leave the slow guy, he is a liability anyway
14. Depends if he already knows my birthday
15. More than mandolin tuners, less than guitar tuners.
16. Juggle them, the one you keep dropping is it
17. Accuse pirate 1 of betrayal. When they look at him, flintlock pirate 2 and 3. That just leaves pirates 1 and 4. That seems fair. take 100% of gold.
Well, apparently they are fans of my site since all those questions are up there (no, I didn’t make them up, but they’ve been up there for five years).
http://www.techinterview.org
“4. How would you find out if a machines stack grows up or down in memory?”
If you take the wording of the fourth question into consideration - the answer would be: It is impossible to “grow down”.
Pirate Solution -
Lead pirate hides 98 gold; tells the others the first 2 to agree to the division will get 1 gold each. If 2 don’t agree then they kill him and get nothing (since they won’t know where it is). 2 of the 4 will likely take something (1 gold) instead of nothing.
Pirate question: 98 for number 5, 1 for number 4, 1 for number 3. That leaves pirates number 2 and 1 very unhappy. So only two votes against.
The rule is that if fewer than half of the other pirates (fewer than half of 4) agree with him, he will be killed. This rule does not kick in, because the condition is not met. For the condition to be met, there would have to be only one pirate of the four others agreeing with his plan. But there are two of the four others agreeing, so he is fine.
Now why those two would agree to getting only one coin is beyond me.
“[I]f fewer than half agree” means that half the vote is enough.
17) With only 2 pirates left, pirate 2 allocates 100 coins to himself and votes for himself to seal the deal, so pirate 1 has no interest in letting it get that far.
With only 3 pirates left, pirate 3 need only offer pirate 1 a single coin to buy his vote. Otherwise, he gets nothing from pirate 2. This sucks for pirate 2.
With only 4 pirates left, pirate 4 can offer pirate 2 one coin and pirate 2 should be happy to vote for him, cause pirate 3 won’t be dumb enough to send coins his way. Pirate 1 and pirate 3 aren’t too happy about this idea.
So, with 5 pirates who are smart and cynical, pirate 5 need only offer a single gold coin each to pirate 3 and pirate 1. If they vote against, they get nothing when pirate 2 sells his vote to pirate 4 for one coin on the next round.
I think the answer for #9 cheating husbands one is that for the first 100 nights all is well then after 101th night every husband is killed.
Think about it.
Nobody seems to have answered question 4, which to me seems the easiest, off the top of my head:
test2(int a) {
int i[256];
if ((int)(&i[255]) > a) {
printf(”Stack grows up\n”);
} else {
printf(”Stack grows down\n”);
}
test1() {
int i[256];
test2((int)(&i[255]);
}
The pirate question needs an instruction that all the pirates are perfectly rational individuals. A rational individual will choose something over nothing.
Plus, we have to be sure that it couldn’t be repeated or the pirates would not agree because, clearly, the next pirate would give more of his share.
1. None, if the school bus is filled with bananas.
2. I’ll try to un-hinge the blades off from the rotor with my super-microhuman strength.
3. two venti mocha frap for each window. The caffeine and sugar will keep me wake from that boring pointless job.
4. I’ll pray and ask God to provide me with a vision for the answer. Either that or I could try to generate a stack trace and examine it.
5. a database is like a file cabinet. The files, or data, is stored in it and can be arranged in categories. But unlike an actual file cabinet, you can do a lot more cool stuff with a database like being able to make it accessible thru the internetz.
6. none, if the clock is digital.
7. First, I would try to know “how” to get there. That should help me answer this question.
8. I would put RFID tags on the shirts that represents keywords then build a smart mechanical system in the closet that will automatically retrieve clothes for me by querying keywords.
9. if the wives were communicating with each other beforehand, then nothing will happen because there will be no husbands left in the 1st place. if not, the queen’s announcement will still not cause anything to happen since every wife is already aware that 99 men cheated. At the very least, the only news here is that there probably was only one man who dared to sleep with the queen.
10. it depends on the probability of giving birth to a boy and how many men were present in the beginning. If we assume that the chance of giving birth to a boy is 50% and if the number of men is equal to the number of women in the beginning, then the proportion should be close to 1:1.
11. 31.7%?
14. Yes, only if I knew beforehand that there are at least 6 people that has the same birthday as me.
15. Hmm… wait a sec, I think I remember learning this from music class… Can you fix me a cup of coffee? It helps me refresh my memory. Thank you!
Mike: You’re right about the pirates. In fact, while the brain-teaser-esque answer given by most of the comments is what neoclassical economics would suggest is the answer, studies have shown it’s patently false - people don’t treat one-shot games as one-shot games even when told they’re one-shot games. Humans just don’t think like that, and ego won’t let us think like that. If someone says “Pirate 5 gets 98 gold!”, put them in the math department, but keep them the hell away from UI, PR, marketing, or just about anything having to do with people.
1) Google it.
Google it.
2) Google it.
3) Google it.
4) Google it.
5) Google it.
6) Google it.
7) Google it.
9) Google it.
10) Baidu it.
11) Google it.
12) Google it.
13) Google it.
14) Google it.
15) Google it.
16) Google it.
17) Google it.
#8 The coloured shirts one is easy. Organize shirts alphabetically by the most prominent colour on the shirt. You then do a binary search on the shirts. You probably need to do some more detailed divisions if you have a large amount of shirts on one colour.
#9 This one is easy if you think about a smaller problem and use induction type principles.
2 couples: Cheaters die on the 3rd day. If 1 husband has cheated, he is killed because wife A knows she hasnt slept with anyone therefore if wife B hasnt killed her husband yet wife A’s husband must of cheated. If 2 husbands die there is similiar reasoning.
3 couples: Cheaters die on the 4th day. Similiar reasoning as above
Nth day: Cheaters die on the 5th day.
I dont feel like using proper induction or explaining each step cause I am lazy
For #4 to wash all the windows in Seattle, you can charge as much as you want, as you have a monopoly.
The four questions I always ask when interviewing people for my air conditioning business are:
1. If I offered you the job on the condition that you slept with me, what would you say? (this can be put to both male and female candidates and is designed to see how they deal with moral dilemmas on the spot)
2. We have a tradition at this company (not true but I make it sound plausible) that new employees have to work naked every Friday for the first four weeks. Would you have a problem with this? (This tests their openness to new ideas and new thinking)
3. We have been going through big financial difficulties over the past year (not true, we are actually growing fast) which may mean some months we cannot pay our junior staff. Very occasionally we may have to borrow money from our junior staff. How would you feel about this? (This is designed to test company loyalty)
4. Very occasionally our product malfunctions and can kill people (this rarely happens). For legal reasons we never take responsibility for these deaths. What would you say to placate a man who has just lost his wife because of our air conditioning system, so he doesn’t try to sue us. (This is designed to test negotiating skills)
I have the feeling that if you’re able to anwer any of these question, you are probably still a virgin
Verne,
You’re asking way too loaded questions. It’s only an air conditioning business, man. I doubt you’re sincere since 1 and 2 should almost immediately qualify for sexual harrassment cases against you.
If someone actually did show up naked, would you be open enough to accept them working in the buff for the day? Are you amenable to new thinking yourself?
3 is stupid because I’ve looked at your books before I went to the interview and I know you’re bullshitting me. Let me give you a brief course in HR: nobody gives a damn about the company. People care about the people they work with. It is very rare that they will actually spend mental energy about the company doing well. You’re in an airco business, which means it’s only about money to you. It’s only about money to the people who work for you too, Verne. It’s about paying the bills and feeding the kids. Company loyalty, that’s a joke, right?
4 would be your responsibility. Unless you hired me to be your spokesperson, I’m not making your excuses for you. Build a better product.
The sad part is, these are the questions for the janitor interview. They get harder as you move up.
8. Imagine you have a closet full of shirts. Its very hard to find a shirt. So what can you do to organize your shirts for easy retrieval?
If it’s hard to find a shirt when you have a closet full of shirts, do as follows:
1. Open your eyes.
2. Put the lights on.
3. Grab a shirt (with your hand).
You’ve just found a shirt. Hallelujah!
someone else may have said this already, but these questions are designed to see HOW YOU THINK. The correct answer is not required, you just need to “show your work” by thinking out loud.
What role was your friend interviewing for? I’m pretty sure they don’t ask such questions in engineering interviews, and I’m speaking from my own experience here.
It seems that i would be the next employer of Google…
xd
“You are shrunk to the height of a nickel and your mass is proportionally reduced so as to maintain your original density. You are then thrown into an empty glass blender. The blades will start moving in 60 seconds. What do you do?”
Didn’t see this one answered..
You Jump Out!
Any small rodent or large insect can easily jump several times his own height. Since you still have the same proportions, that means you have more muscle than the average small mammal, hence, you can easily jump it.
So many of these (and more) are in or from “How Would You Move Mt. Fuji?”
If anyone asked any damnfool questions like that at an interview I would strike them with my fists. I swear to God I’d beat the bejabers out of them.
Since I’m a security guard and that’s the only job I’d be applying for this would show them how tough I could be defending their trashy company. I’d write my resume all down the side of their fat face and let’s see how they like that.
I don’t take that kind of garbage from no one. I come to work to do a job and then go home at 4:30pm. I don’t need to know how to cross a bridge with a flashlight or how to escape from a blender. When was the last time you got shrunk to the size of a nickel?
These people should be locked up asking bloody crackhead questions like these.
The idea for #13 is to pair slower and faster people together.
First trip:
1 & 2 minute cross, 1 comes back.
3 minutes
Second trip:
5 & 10 cross, pass the light to 2 who comes back.
15 minutes
Third trip:
1 & 2 cross.
17 minutes.
On the blender one: I’ve been shrunk and someone is trying to kill me by blending to death. My chances of survival are nil.
On the pirate one: The fifth pirate effectively gets all the coins. You can “split” them by offering a raffle to the next pirate: we’ll pick a random number out of 1000 and if it’s 8 then you get the coin. It’s still better than nothing.
doesn’t it seem like the window washing in seattle question is actually a reference to microsoft?
On 11: it is indeed 50%. A mathematical argument goes like:
1 boy in each family.
In 1/2 of the families, 0 girls.
In 1/4 of the families, 1 girl.
In 1/8 of the families, 2 girls…
On average, 1/4 2/8 3/16 … = \sum_{n=0}^{\infty} n/2^(n 1) = 1 girl per family.
I’m ashamed to not be entirely sure how to do the series, but Maple told me the sum is 1, so…
new york has around 8 million people, not 13 million.
For number 10: The odds will be 50-50
The chance for a family to have a boy the first time is 1/2; the second time - 1/4; the third time - 1/8… That is, 1/2^n. The expected return, “measured in girls”, would then be (1/2)*0 (1/2^2)*1 (1/2^3)*2…. So, then the expect number of girls in a given family would be the limit of the series (1/2^n)*(n-1), where n goes to infinity. I think the limit of that series is 1.
So, the average number of girls per family is 1. The average number of boys, by definition, is also 1. Therefore, the odds of boys to girls are 50-50.
What do you say?
I also had a gut feeling that the odds should be something like 1/e.
The answer to #9 is, if there are n couples, all the wives kill their husbands after n days. The logic leading to this is left as an exercise to the reader.
Some of these are from “How Would You Move Mount Fuji” which are questions for Microsoft.
I think there are no correct answers but instead they just try to understand how you would face/answer a problem/question. I think they can tell a lot about the person depending on how they answer not the answer itself.
I got asked some of these questions at a rather large employer out here in Portland Oregon… (hint: they’re located in Hillsboro) they never told me which ones I got right or wrong but I could tell & I did end up getting the contract
The trick is to not think too hard about the answer and ask off-key questions. They’ll give you some hints so long as they’re not designed to give away the answer and they (the answers) are more obvious than one would expect.
for number one you can always fit golfball in a bus. it doesnt say all at once.
2.
if i was as small as a nickle i would duck underneath the blades.
3.seatle has alot of windows, considering alot peoples windows will be dirty becuase its the city. i would guess 1 dollar a window. becuase with so many windows. thats alot of money.
4.what machine. and you cant grow down
5. i dont have an 8 year old nephew
6. uhm would if its digital?
7.to get to point b you should google it to figure out how
8. if you have a whole closet of shirts. why would it be hard to find a shirt. pick one up. and there you go.
9.well if everyone cheats there would be no men.
10. i think there would be more girls if people always stop at the first boy. unless they kill the girls. then there would be more boys
11. why would i care about observing a car?
12. 180? this one i have no clue o.o
13. how did they get to that side of the bridge in the first place? if they got over there they should be able to get back.
14..no
15. how is this somthing i need to know to work at an internet company. i guess it would make more sence if the company was for oneils music or somthing
16. weigh 3 :3 then wich evers side is heaviest weigh 1:1 with one still remaining if the two weigh the same the one you didnt way must be heaviest.
17.prates are greedy. and if they killed the first guy who poposed a way to split it. they will probably kill the second and third. which then there would remain only one more pirate who would get the 98%
( can somone explain where the two percent went?)
so many fucking noobs in this thread
For some reason, the plus signs in my post did not come through
(1/2)*0 (1/2^2)*1 (1/2^3)*2 … should be
(1/2)*0 plus (1/2^2)*1 plus (1/2^3)*2 plus…
For the blender one, you can’t go under the blades, when they start you’ll get sucked into them. You have to get above them; when the blades start, they should blow you out the top (as long as it’s not enclosed)
It appears to me that what we have here are a bunch of overthinkers. In the company I work for (a software company), around 5 of our developers have relocated to the bay area to work for Google. These questions aren’t designed to find out how mathematically ingenious you are, or for a correct answer. They are designed to find out how you think. For instance:
1. How many golf balls would fit in a school bus. Yes, many of you have figured out really great mathematical equations. However, you’re forgetting a couple things, such as - are the seats in the bus? Is it just a shell? Are there any passengers in the bus? What size school bus? All of these variables would give a different answer. There is no one correct answer to this question.
Also, #13. If the flashlight only has enough juice for 17 minutes - fast could carry slow, then hold the flashlight in a way to illuminate the bridge for the other 2 to come across. However, again, this one doesn’t have any definite answer. It’s a question designed to see how well you would work in a team situation.
I would have to say, though, that my favorite is #7. You have to get from point A to point B. You dont know if you can get there. What would you do?… I’d look up the directions on Google Maps! lol
Re. Blender: I don’t see why you’d fall between the blades and the plastic–a blender can easily take out something the size of a nickel.
I’d probably move to the center before the blades were started. Most blenders have blades that slope outwards a short distance from the center, so material in the center is safe. I might get dizzy as I rode the blades, but it wouldn’t be that bad–like sitting in the center of a merry-go-round.
9. Nothing. Tho it depends, if wives ask 99 other women each if they slept with her husband and thus get a proof. Then they kill’em all
10. I agree on 50%
11. It’s easy 0.95/3 = 31.7. Probabilities just add up over time. Simple question just ppl expect something huge so they complicate in their answers.
Boys/Girls is 50-50. Imagine you try to simulate it with coin flips. Write your results on a piece of paper. Heads means boy. Start flipping. When you land on heads, group all results since last ‘boy’ flip as a ‘family’. This is probabilistically equivalent to the problem. Do you imagine simply grouping the results will change the random distribution?
#10 I seems like people are having a hard time reasoning about #10. This is the easiest way to think about it.
The series you have to sum is as follows:
1/4 2/8 3/16 4/32 5/64 6/128 7/256 …
that is equivalent to (1/4 1/8 1/16 …) (1/8 1/16 1/32 …) …
which ends up being 1/2 1/4 1/8 1/16 1/32 …. which equals 1
The answer, then, is that 1/2 of the children are boys and 1/2 are girls.
You can’t do an integral (the integral ends up being slightly over 1) because it is a series and is discontinuous.
Alternatively, after one child, half the children are boys, and half the couples have another child. After those couples have a child, half the children are boys, and half of those couples have another child. After those couples have a child, half the children are boys, and half those couples have another child. After those couples have a child…
At each iteration, you only ever add the same proportion of boys or girls to the population; you only really need to resort to series if you’re going to work out the average size of a family.
Pointing out which couples happen to continue to spawn tends to lead people into something like the gambler’s fallacy, which is why I hate such questions in interviews. I’m an engineer, so I know the answer I first think of is usually wrong.
For the blender, if you maintined the same mass you’d weigh the same and probably break the blender, and if not, you can just easily push it over by throwing your weight, but try to be careful not to get cut.
ah wait, read it wrong, mass is reduced too. then yeah, probably just jump out.
interesting questions
> 1. How many golf balls can fit in a school bus?
Assuming the balls are stacked like: oooooo and assuming the bus is
50′x8×81 and assuming golf balls are 1.5″ in diameter:
50*8^2*12^3 / 1.5^3
You don’t use the spherical volume of the balls if they are stacked oooo. You
assume they are cubes.
> 2. You are shrunk to the height of a nickel and your mass is proportionally reduced so as to maintain your original density. You are then thrown into an empty glass blender. The blades will start moving in 60 seconds. What do you do?
Go as close to bottom center of the blender as possible. The air will be moving at the slowest speed the nearer the axis of blades, so there is less chance a small particle will be swept up on the wind current the spinning blades make.
> 3. How much should you charge to wash all the windows in Seattle?
Say there are one million people, and on average each person lives in a home with 4 windows per person, drives a car with 6 windows, or rides a bus or train with an average 0.5 windows per rider, and works / attends school somewhere with 2 windows per worker/student). Say 50% drive, and 50% ride. Say it takes 5 minutes to clean a window, both sides, and I want $10/hour.
10^6 * (4 2 (6 0.5) / 2 ) * 5/60 * 10.
> 4. How would you find out if a machines stack grows up or down in memory?
static unsigned intptr_t a1; a2;
sub()
{
int v2;
a2 = &v2;
}
main()
{
int v1;
a1 = & v1;
sub();
if (a1 > a2) {
printf (”stack grows down in memory\n”);
} else {
printf (”stack grows up in memory\n”);
}
}
> 5. Explain a database in three sentences to your eight-year-old nephew.
A database stores tables of information in a computer like the name of each kid in your school, and the names of their parents, and their phone number. The database is set to make it easy and fast to answer questions like which kids have a certain phone number, or whether a kid has at least one brother or sister. For the second question, the principal could search for your name in the database, and find your parents names and then search for the parents names to find all kids with those parents.
> 6. How many times a day does a clocks hands overlap?
A clock has 12 hours, but there are 24 hours in a day. I’ll analyze the A.M.,
first. The AM starts at 12:00. The hour hand
then has to go past an hour 11 more times. The minute hand goes fast,er so will
pass the hour hand 11 more times, thus overlaps 11 more times. So 12 times
in the A.M. The hour hand then hits NOON, starting the P.M. That is 13 times
so far. Similarly in the PM, the minute hand passes and overlaps the hour
hand 11 more times, so 24 times.
> 7. You have to get from point A to point B. You dont know if you can get there. What would you do?
I assume this is asking if there is a route from A to B. It’s basically a maze
problem. In a maze problem, you are place somewhere in the maze and told to find the exit, if there is one, or if there isn’t one, decide there isn’t one. In the maze problem, you keep track of which squares you visited, and move to an adjacent square that you haven’t visited. You keep doing this until you either find the exist, or find there are no more unvisited adjacent squares. If there are no more unvisited adjacent squares, you back track your steps to till you find an adjacent unvisited square, marking each square you back track as both visited and a dead end.
In the real world, you could treat each road segment between two intersections (or between an intersection and a dead end, or between two dead ends) as a maze square and each intersection as a gateway to 3 other maze squares (assuming 4 way intersections, and the roads in the intersection aren’t one way streets; adjust the treatment of gateways accordingly). The problem is that the real world has many more roads than a typical maze has squares. We want to try to optimize. So if point A is somewhere on a road that is not at an intersection, move in the direction that is closest to the direction of point B. When you find an intersection, select the road in the intersection that is closest in the direction of point B.
> 8. Imagine you have a closet full of shirts. Its very hard to find a shirt. So what can you do to organize your shirts for easy retrieval?
Gee why do you have to find a particular shirt? Let’s suppose there are
a set of attributes of each shirt you are interested in: e.g. sleeve length, color, buttons (no buttons, fully button, partially buttoned from collar to chest level).
Let’s say the closet is a simple wall closet with a single closet rod running the entire length of closet. On the left you put all the short sleeve shirts, and on the right the long sleeve shorts. You separate then long and short sleeve sides with a specially marked coat hanger. Then you separate each group into no buttonoed, partially buttoned, and fully button, using more specially marked hangers. Then each sub group is separated into colored and monochrome sub-sub-groups (specially marked hangers aren’t needed for separators unless you are color blind) Then each colored group is sorted left to right according to the color spectrum: ROYGBIV: red, orange, yellow, green, blue, indigo, violet. Each monochrome ggroup is sorted left to right: white on the left, black on the right, and shades of grey in the middle, the darker greys on the right, the lighter on the left.
So now I can instantly and visually find say a orange short sleeved T-shirt.
> 9. Every man in a village of 100 married couples has cheated on his wife. Every wife in the village instantly knows when a man other than her husband has cheated, but does not know when her own husband has. The village has a law that does not allow for adultery. Any wife who can prove that her husband is unfaithful must kill him that very day. The women of the village would never disobey this law. One day, the queen of the village visits and announces that at least one husband has been unfaithful. What happens?
The problem is poorly stated I think.
Before the queen arrived, every man had cheated, so all 50 wives knew that
at least one husband had cheated. The law also says that if a wife can prove her hubby cheated she has to kill him. The proof would be easy: if the husband cheated with another man’s wife of the village, that another man’s wife would come forward, and so every man would have been killed. This didn’t happen. The queen apparently is not one of the 100 married couples and she says at least one husband has been unfaithful. The
conclusion is that the man ( men ) the queen says cheated cheated with a woman (women) outside the village, perhaps the queen herself but she apparently isn’t telling.
So, nothing happens. The queen’s assertion that there has been cheating doesn’t add information the women didn’t already know.
> 10. In a country in which people only want boys, every family continues to have children until they have a boy. if they have a girl, they have another child. if they have a boy, they stop. what is the proportion of boys to girls in the country?
Let’s say the country has 1024 moms. Let’s say the odds of a conception producing a boy or girl are equal. Let’s also assume moms birth just one child per pregnancy.
512 moms first child was a boy. And they stop conceiving.
We now have 512 boys and 512 girls .
512 moms now conceive again. 256 get a boy for the second child. They stop conceiving. We now have 256 more boys. We also have 256 more girls.
256 moms now conceive again. 128 get a boy for the third child, and stop conceiving. We now have 128 more boys and also 128 more girls
128 moms conceive again. 64 get a boy for 4th child, and stop. We now have
64 more boys, and 64 more girls.
Each iteration produces the exact same number of girls and boys, even though
all families of 3 or more child have more girls than boys.
The ratio of boys to girls is 1 to 1.
> 11. If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?
The probability of not observing a car in 30 minutes is 1 - 0.95 = 0.05.
Just for illustration, the probability of not observing a car in 90 minutes is
therefore 0.05 * 0.05 * 0.05, and so the probability of observing a car
in in 90 minutes is 1 - 0.05^3 or 0.999875.
Just as 30 minutes is 1/3 of 90 minutes, 10 minutes is 1/3 of 30 minutes.
Leaving it for an exercise for the reader from google won’t win the job. OK.
x is the probability a car won’t be seen in 10 minutes. So x^3 is the
probability a car won’t be seen in 30 minutes. And 1 - x^3 is the probability
a care *will* be seen in 30 minutes, which we know to be 0.95.
And of course 1-x is the probability a car *will* be seen in 10 minutes.
So:
0.95 = 1 - x^3.
0.95 - 1 = - x^3
-0.05 = -x^3
0.05 = x^3
x^3 = 0.05
x = cubic root of 0.05
x is the probability that a car won’t be seen in 10 minutes.
1 - x = 1 - cube root of 0.05 = 0.63159685013596133942201771664202
> 12. If you look at a clock and the time is 3:15, what is the angle between the hour and the minute hands? (The answer to this is not zero!)
The 3 position on the clock is the same
position as 3:15 where minute hand is. As for the hour hand,
The :15 in 3:15 represents one quarter hour. Therefore the hour hand
is 1/4 of the way from 3 to 4. A clock has 360 degrees and 12 hours.
360 / 12 = 30 degrees per hour. 1/3 of thje way from 3 to 4 is thus 30/4
or an angle of 7.5 degrees between the two hands.
> 13. Four people need to cross a rickety rope bridge to get back to their camp at night. Unfortunately, they only have one flashlight and it only has enough light left for seventeen minutes. The bridge is too dangerous to cross without a flashlight, and its only strong enough to support two people at any given time. Each of the campers walks at a different speed. One can cross the bridge in 1 minute, another in 2 minutes, the third in 5 minutes, and the slow poke takes 10 minutes to cross. How do the campers make it across in 17 minutes?
Speedy carries the slow poke across the bridge, while
pokey holds the light. That uses 16 minutes.
When speedy crosses back there are 15 minutes left.
The 2 and 5 minute walker then cross together, no
need for anyone to carry each other this time, and there
are 10 minutes left. The two minute walker walks back
leaving 8 minutes. The one and two minuter walk cross
leaving 6 minutes to spare.
Even if speedy’s speed is cut down due to carrying slow poke,
there is still plenty of time to spare.
>14. You are at a party with a friend and 10 people are present including you and the friend. your friend makes you a wager that for every person you find that has the same birthday as you, you get $1; for every person he finds that does not have the same birthday as you, he gets $2. would you accept the wager?
I’m assuming birthday means month and day of month, and I’ll ignore
leap years (assume no one is born on Feb 29).
The odds that one person shares my birthday are equal to
1 - the probability no one shares my birth day.
The probability no one shares my birthday is (364/365)^9.
That is 0.9756109649554942106741643672081 and
the probability one person shares my birthday is thus
0.024389035044505789325835632791899.
My friend gets $2 for every person who does not have
my birthday. The odds are over 97% he will get 9 * 2 = $18
from me. Whereas I get $1 for every person who has my
birth days. The odds or slightly over 2% I will get $1.
1 * 2% = $0.02 and $18 * 97% is over $17.
No I don’t accept his wager.
> 16. You have eight balls all of the same size. 7 of them weigh the same, and one of them weighs slightly more. How can you find the ball that is heavier by using a balance and only two weighings?
split the balls into one group of 2, and two groups of 3.
Weight the two groups of 3. If the balance says the groups weight
the same, the heaviest ball is one of the other two
balls. Weigh the remaining two balls. The heavier balls is heaviest.
If the balance says one group of three is heavier, then take
two balls from the heavier gorup of three and weight them. If
one ball is heavier that is the heaviest ball. If the balanbce says
both balls weigh the same, the remaining ball of the heavier group of
three is the heaviest ball.
> 17. You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How should he allocate the gold in order to maximize his share but live to enjoy it? (Hint: One pirate ends up with 98 percent of the gold.)
Each pirate values life more than gold.
The 5th ranked pirate will never have to put forth a plan for
others to vote on. He is guaranteed to live. His focus is thus on
plans to get gold.
The 4th ranked pirate is seemingly in the most precarious situation
because if we get down to two pirates, the 5th ranked pirate can
vote down the 4th ranked pirate’s plan, and thus have the joy of
killing and end up with 100 coins. On the other hand,
if the situation gets down to three pirates, the 4th ranked pirate
can ensure his survival by voing for the 3rd ranked pirate’s plan.
The 4th ranked pirate will vote for any plan that gives him coins
and/or ensures survival.
The 3rd ranked pirate is in an interesting position. If the
plans of the 1st and second ranked pirates are defeated, then the
3rd ranked pirate’s plan needs just one vote. He could allocate 100 coins to
himself and have a nearly certain chance that the 4th ranked pirate
supports the plan, because otherwise the the 4th ranked pirate becomes
top pirate and ends up in the precarious situation noted above.
If the second ranked pirate ends up as top pirate,
then his plan has to win 2 of 3 votes. The 5th and 4th ranked pirates are
motivated to accept plans that offer them coins, because they know
if the second ranked pirate’s plan is defeated, they will get
zero coins. The 3rd ranked pirate will be happy with any plan that
gives him coins, but he is certain to vote no to anything that gives
him less than 100 coins, and even then he might vote no just for the
pleasure of killing the second ranked pirate. The second ranked pirate
knows if he proposes 50 coins each to the 4th and 5th pirates, they are
almost certain to accept since both no they will get zero if the 3rd
pirate ends up top pirate. Similarly, the second pirate knows he is certain
to get no vote from 4th or 5th rank pirate if he proposes a zero share.
In terms of maximizing his own share and enjoy it there is probably
some combination of a split of non-zero coins for each of the
5th, 4th, and 2nd ranked pirates, but it is unclear what that is,
and it depends on the rationality of the 5th and 4th pirates. The safest
is to assume total irrationality, and give each of the 4th and 5th
pirates 50 coins. It is also possible he could survive by proposing
say 1 coin for the 5th ranked pirate and 99 for the 4th ranked pirate,
but who is to say thee 5th ranked pirate won’t think it is worth one
coind to see the 2nd ranked pirate dead.
The first ranked pirate is aware of the analyses the other 4 pirates
will have to go through. He needs just 2 of 4 votes for
his plan to win. He proposes 1 coin for himself, 1 for the
second pirate, and 98 for the 4th pirate. The second pirate balances the
certainty of life and gold versus probable life and probably no gold,
and supports the plan. The 4th pirate knows that most coins he can expect
based on the above analysis is 99, and the chances of being offered that
many in second round are improbable. This is a sure thing, and one less
coin makes little difference. Giving the 98 coins to the 5th pirate
is nearly as good, but the 1st pirate has to consider the slight
possibility that the 5th pirate might irrationally reason that he has one more possible voting round than the 4th pirate so maybe a better deal is coming.
Giving 99 coins to either the 4th pirate makes little difference
to the 1st pirates chances of survivability and deprives him of a coin,
giving him nothing to enjoy life with. Similarly giving one coin to the
5th pirate, and zero to himself isn’t likely to change the outcome and just
deprives the first pirate of a coin.
#1 Depends on a size of the BUS (maybe none, maybe all-even at the same time!)
#2 Nothing, I would be shocked by the fact that I’m the size of a nickel (probably freek out)
#3 25$ an hour, weekend and hollidays off
#4 Why do I care, and if I needed to know I would ask the person who built the “MACHINE”
#5 Why would my 8 year old nephew want to know that! Besides, I have a 5 month niece - she’s tooooo cute :))
#6 I have a digital watch :)) Probably 22?!?
#7 Find out
#8 Buy hangers!!!
#9 Nothing, they all allready know that at least one husband has been unfaithfull!!! But knowing women sooner or later they would tell each other everithing and the men would all die! :))
#10 It depends, how many girls and boys are there? (Eventually all boys, and the nation dies out)
#11 0.95
#12 7.5 degrees
#13 first 1,2; 1 back; then 5,10; 2 back; then 1,2. 2 1 10 2 2=17
#14 Do they mean the party or anywhere? Since my english is not that good, do they mean BIRTH DATE? Are birthday and birth date the same?
#15 As many as there are people, offcourse, not all of them can tune a piano well :))
#16 Weigh 3 vs 3. If one side is heavier, weigh one vs one of those three. If that is equal the third is the heavier one (if not, the heavier is heavier). If the 3vs3 is equal, weigh the two remaining balls and there you go. Good answer, easier than typing :))
#17 This is really a good one. No1 says :”No5 and No4 will get nothing, and No2 and No3 will get the money, and the same amount!”. No2 and No3 agree to that and he gives them 1 coin, No4 and No5 votes don’t count becouse he got majority! To tell you the truth I’m not sure this is the right answer (if there is one) but it’s the best I could think of in such a short notice!
My assumptions in answering the pirate riddle.
1) The top pirate does not get a vote in the plan. The riddle says “the others” will vote on the plan.
2) If exactly half of the non-top pirates do not agree with the plan, then the plan is agreed to. The riddle says “fewer than half.”
This basically amounts to saying the plan is agreed to iff and only give 51% or more of the total number of living pirates, including the top, agree to it. That’s how it works in regular democracy, so it’s fair to assume it works this way as well.
3) All the pirates will act in their best interests.
4) All pirates will vote against a plan if he lacks a good reason to vote for it.
5) All pirates are rational.
So here’s my solution.
If only 1 is left, he gets 100gp. At no time during this process will his own life be in danger. Therefore, he has no reason to vote for any plan that results in his getting less than 100gp.
1 votes yes IFF 1 gets 100gp.
If 1 and 2 are left, then 2 must give 1 all 100 gp in order to stay alive. If 2 says 99 for you and one for me, 1 will kill him to get that last gold piece. So if 1 and 2 are left, 2 gets zero, and any gold he gets must be given by another. He has no reason to vote for a plan that gives him zero, as that’s what he’s getting anyway. However, if a plan is proposed where he gets 1 gp, then he’ll take it, even if 1 gets 99, as that’s better than what he’ll get if the plan is not agreed to and the pirate above him is killed.
2 votes yes IFF 2 gets 1gp or more.
If 1, 2 and 3 are left, then 3 only needs one other vote for his plan to go through. He can give 1 all 100gp, or give 2 all 100 gp, which would give him his life but no gold. He can divide it 33/33/34gp, and regardless of who gets the 34gp, 2 will agree to the plan. To maximize his profit, however, 3 can give himself 99, and give 1gp to 2. 2 will agree, because otherwise he gets nothing. If 3 gives himself 100gp, then 1 obviously won’t agree, and 2 has no reason to so he’ll reject.
3 votes yes IFF 3 gets 99gp or more.
If 1, 2, 3 and 4 are left, then 4 needs the votes of two other pirates to stay alive. 1 won’t vote yes unless he gets all of it, and if that happens, 2 and 3 will vote now. The only way for 4 to get two votes is to give 1gp to 2, and 99gp to 3. So this places 4 in the same boat as 2; if it’s down to him being the top pirate, he can escape with his life but nothing else, so whatever he gets must come from someone else, and he’ll agree to receive anything.
4 votes yes IFF 4 gets 1gp or more.
If all five pirates remain, then it should be fairly obvious what 5 proposes: he proposes 1gp for 2, 1gp for 4, and 98gp for himself. 1 and 3 obviously won’t agree, but 2 will agree because he’ll get no better plan from anyone else, and 4 will agree because withholding his vote will give him nothing. So the answer is:
1 - 0gp
2 - 1gp
3 - 0gp
4 - 1gp
5 - 98gp
Microsoft asked these questions long before GOOG was ever conceived.
Lorenzo on September 6th, 2007
the #10 i think its more simpleIn a country in which people only want boys means that people in that country is only masculine, so the families are homosexual families and 2 man cannot still have a children, so for me the answer is 100% boys and 0% girls.
I just love political correctness. Even morons jump on the bandwagon. If there’s 100 gay male couples, how exactly are they going to produce ANY boys? The culture is doomed. Unless they have cloning. And there was no mention of cloning in the question. Nor gays for that matter.
C’mon guys, the purpose of this is to USE YOUR BRAIN :), not to show off with more or less whacky answer in style “google it”.
2. The blender one:
Noone actually considered, how much a power of human muscles would INCREASE (per unit of mass) if human body would be shrinked that much. Power of muscless depends on their square section which goes with the second power of dimension, and mass of human body goes with third power of dimension (that’s why it is not as impressive as we are told when a certain flea can jump own 100 lengths, or when a cockroach can lift a mass 10 times his own. If you enlarged it to the size of 1 or 2 meters, his own legs would probably break under own weight).
So, the answer is: you would most likely be able to JUMP out of the blender
10. Problem with kids
It is quite simple once you do fully understand it. Ilian, your solution is good, but still wrong. The infininte sum you have to calculate is:
1/2 plus (1/2)*1/2^2 plus (1/3)*1/2^3 etc… Why? Because:
50% of children will be boys at the beginning. 50% (1/2) of them will be girls, and their follower can be either a boy (25%=1/2^2) or a girl (25%) and so on. So, probability of having a row of n-1 gilrs and 1 boy is 1/2^n, so number of boys will be 1/n-th of it (only the last kid). That brings us to upper formula, which solution is Ln(2)=0.69.
More than two boys per girl :). China, anyone?
17. p1 gets 98, p4 gets 1, p5 gets 1
For p1 survival, two other pirates agreement needed.
So, p4 and p5 expected no coins to get.
If they can get 1 coin, they will feel happy and support p1.
P2 and p3 expected they can get 1 or more coins.
If p1 wants their support, they request more coins than p4 and p5.
It’s fantastic… Google must be full of quiz players … Any programmer inside?
When I was eighthen I went to an interview to get a job in a factory of cleaning products. After a normal interview of about fifteen minutes the last queston of the company director was “Are you capable of killing a chicken?”
My answer was, after five or ten seconds of thinking, “Yes, if I need it to feed myself.”
I showed to him that I can understand a problem, think about it and find a good solution. (the solution is not about killing the chicken, but the logical answer I gave to him! That’s what this strange questons are all about, finding solutions to unexpected problems.)
P.S.: I got the job, a friend of mine went to the same interview after me in the same hour, He stayed so puzzled with this question that He was unable to answer it with some logic in useful time.