• (cs) in reply to hmmmm...
    hmmmm...:
    (I'm posting this for nostalgic purposes, I also suspect that some of my ex-colleagues read this site, - might see you Friday ;-)

    If I have 3 pieces of bread and I want to toast them on an old-fashioned grill, what's the quickest (most efficient) way of doing so ? The grill heats from above only (I want both sides toasted) and you can only fit 2 pieces of bread on it at a time.

    Unless you remove the top cover from the grill: Six sides in total means minimum three batches, so:

    1. 1a, 2a
    2. 1b, 3a
    3. 2b, 3b
  • Shinobu (unregistered) in reply to hmmmm...

    In three heating sessions, but I won't tell you folks how. ;-)

  • Shinobu (unregistered)

    Oh, and for you bitcounters out there, I've got a bit of a disappointment for you: it turns out the most prozaic methods are fastest. snif

  • s (unregistered) in reply to akatherder
    akatherder:
    Take the problem of weighing the airplane. There is no correct and practical answer. But how would YOU approach it?

    The correct technical and practical answer would be reading out the weight sensors inside the airplane. A plane of this class MUST have a device to read out its weight before takeoff, and pressure in hydraulics of the wheel gear is easy enough to convert to weight.

    being a smartass, I prefer googling for mass of boeing 747. The cargo is not a part of the plane, and they asked me to weight the plane.

  • s (unregistered)

    The table of transitions of values in a binary encoder is, for rotation left: channel: A B 0 0 0 1 1 1 1 0 for rotation right: A B 0 0 1 0 1 1 0 1

    What bit operation (one mnemonic!) would enable you to determine the direction of rotation of the encoder?

    How to easily implement error checking (illegal transitions)?

  • B (unregistered) in reply to nini
    nini:
    Fill the 5 litre jug. Pour from the 5L jug into the 3L jug till it's full. Empty the 3L jug. Pour the remaining 2L from the 5L jug into the 3L jug. Refill the 5L jug. Fill up the 3L jug (i.e. 2 already in there, 1 from the 5L). There is 4L left in the 5L jug. QED.

    You can do this the other way too. Fill up the 3l jug. Pour from it to the 5l jug. Refill and repeat. After two fills, you will have 1L in the 3l jug and a full 5l jug. Empty the 5l jug and pour the 1L from 3l jug into the 5l jug. Fill the 3l again and pour into the 5l jug - you now have 1L + 3l = 4l in the 5l jug.

    The mathematical treatment would be to find a,b such that 5a - 3b = 4 or 3a - 5b = 4. If you observe that multiples of 5 necessarily end with 5 or 0, you can get the smallest integer pair satisfying given condition as 2,2 in case 1 (your solution) or 3,1 in case 2 (my solution).

  • Flo (unregistered) in reply to I can so relate...
    I can so relate...:
    Actual questions I got at an interview (not with MS):
    • Why is the sky blue?
    • How would you estimate the weight of Earth?
    • How many people are there in the world?

    I was told the point of these questions was to see how comfortable I am to admit that I don't know the answer.

    • http://en.wikipedia.org/wiki/Rayleigh_scattering
    • Measure the gravitational constant, find your own mass and the radius of the earth, use Newton's law of universal gravitation. Simple.
    • ~6,500,000,000 ?

    There are people who don't know the answers? oO

  • Nicolai Sanders (unregistered) in reply to Grandpa

    the strategy is the same for all people. if the hats of the other two people are different colours, they pass. if the other hats are the same colour, they guess that their own hat is the opposite colour. this way they win 75% of the time.

  • Davey (unregistered)

    This whole 8 coins thing - i always remember it being 12 coins

    12 indentical looking coins 1 is heavier or lighter, but you dont know which You get 3 weighings, no more. Find the odd coin out, and determine whether it is heavier or lighter than the others.

  • Dhericean (unregistered) in reply to s
    s:
    being a smartass, I prefer googling for mass of boeing 747. The cargo is not a part of the plane, and they asked me to weight the plane.
    Of course what you are showing here is that you assume details that you are not given (in order to simplify your job), and do not clarify unclear specifications.

    This would let you fit right in as a programmer in certain environments in which I have worked.

    <cliche>Just remember that when you assume you make an ASS out of U and ME.</cliche>

  • (cs) in reply to Flo
    Flo:
    - Measure the gravitational constant, find your own mass and the radius of the earth, use Newton's law of universal gravitation. Simple.
    Simple and quite wrong, since the Earth's mass is not concentrated in a point and thus not all at the same distance from you.

    Addendum (2007-05-16 07:02): A better approach would be to compute the Earth's volume and multiply it by a guesstimated average density. IIRC most kinds of rock are about 6 or 7.

  • s (unregistered) in reply to Dhericean
    Dhericean:
    Of course what you are showing here is that you assume details that you are not given (in order to simplify your job), and do not clarify unclear specifications.

    More or less, yes. I usually clarify "If that is what you wanted, here's your trivial, express answer, in 2% of the time allocated for the project. If this is not what you wanted, I want a precise, detailed specification of the project on my desk. I'm paid for writing programs to specification, you're paid for writing the specification, don't make me do work of us both.

    This is a kind of tactics of educating the business units and project offices what is their role - not just get an idea out of blue sky and tell programmers "do it", but analyse, prepare details, find caveats and so on. Otherwise you'll keep getting orders more and more vague and get all the blame for the result not working as the manager imagined it (but never bothered to tell).

    If the question was "devise a way of finding the mass of ladden airplanes including ones of size of boeing 747, within an airport, precision no worse than 2% of the mass of the plane, it should be implemented as an airport facility" I could start thinking further, because I see the project makes sense. If they want "finding weight of an 747", I suggest finding, aka googling it.

  • s (unregistered) in reply to brazzy
    brazzy:
    Flo:
    - Measure the gravitational constant, find your own mass and the radius of the earth, use Newton's law of universal gravitation. Simple.
    Simple and quite wrong, since the Earth's mass is not concentrated in a point and thus not all at the same distance from you.

    Radius of earth about 6000km, diameter 12000, round down to 10000km that is 10^7m, approximate sphere with a cube, 10^21m volume, about 1 ton per cubic meter of soil, earth mass is around 10^21 tons, plus minus an order of magnitude. What was the allowed tollerance again?

    alternatively: http://www.google.pl/search?q=mass+of+earth=

  • (cs) in reply to Bouk
    Bouk:
    Grandpa:
    Three players enter a room and a red or blue hat is placed on each person's head. The color of each hat is determined by a coin toss, with the outcome of one coin toss having no effect on the others. Each person can see the other players' hats but not his own.

    No communication of any sort is allowed, except for an initial strategy session before the game begins. Once they have had a chance to look at the other hats, the players must simultaneously guess the color of their own hats or pass. The group shares a hypothetical $3 million prize if at least one player guesses correctly and no players guess incorrectly.

    What strategy would you use?

    If a player sees two hats of the same color, guess the other color. Otherwise, pass.

    In 6 out of 8 possible hattings this will give one correct answer and 2 passes, hence a win. I believe this is the best option, but to prove it is best is of course a whole lot harder.

    Problem is, you have no way of knowing if the proportion blue vs red is evenly matched. It's guided purely by chance.

    It could also be that the coin is fixed. :)

  • Mike (unregistered) in reply to nini

    Exactly, haven't y'all played Knights of the Old Republic? Or you just all played the dark side?

  • PBalis (unregistered)

    Therre is a simple truth that often is overlooked. Not evrything in life is worth doing well. Silly riddles are at best only entetainment,

  • htg (unregistered) in reply to Grandpa
    Grandpa:
    Three players enter a room and a red or blue hat is placed on each person's head. The color of each hat is determined by a coin toss, with the outcome of one coin toss having no effect on the others. Each person can see the other players' hats but not his own.

    No communication of any sort is allowed, except for an initial strategy session before the game begins. Once they have had a chance to look at the other hats, the players must simultaneously guess the color of their own hats or pass. The group shares a hypothetical $3 million prize if at least one player guesses correctly and no players guess incorrectly.

    What strategy would you use?

    Initial strategy session: Person A is the guesser, B and C will pass. In the room, if B stands to C's left, A's hat is red, otherwise it is blue.

  • Gareth Pitt-Nash (unregistered)

    Weight of a 747, hmmm ...

    A South African or European 747?

  • (cs) in reply to MR
    MR:
    poochner:
    And it's pretty obvious you don't understand what you're saying. Proof by example: strategy of "no one guess." Fails 100% of the time.

    You can't prove by example. What you did is disprove by counterexample ;)

    Not so! :-) I proved my assertion that he didn't understand understand his own statement, by giving an example that didn't meet his criteria! :-)

    There. Ran rings around you logically.

  • (cs) in reply to Fuz
    Fuz:
    (can't take credit for it... search for HACKMEM 169)

    Alternatively, search for HAKMEM 169 - might give you better results. :)

  • Anonymous (unregistered)

    Ok, this has many WTFs in its own, but let's extend the boat ride riddle:

    In one side of the river there are:

    • a father with his son
    • a mother with her daughter
    • a security guard with a prisoner

    The boat has a capacity of two, but only the father, mother and security guard can operate it.

    The father can't stand the girl if her mother is on the other side of the river The mother can't stand the boy if his father is on the other side of the river The prisoner will kill any one if the guard is on the other side of the river

    How can they all get to the other side of the river?

    CAPTCHA: tacos (the riddle reward)

  • Anonymous (unregistered) in reply to kimos
    kimos:
    Given an opaque box with three light bulbs inside and three switches outside, how would you determine which switch corresponded to which bulb if the box could be opened only once and only after all the switches were permanently set?

    That's the one that I've heard of. Took me a long time to get the answer and would never have thought of it in the span of an interview.

    Turn on switch A for a couple minutes. Turn A off then turn on switch B. Quickly open the box. The bulb that's on is switch B, the bulb that is off but is still warm to the touch from being lit is switch A, and the remaining bulb is switch C.

    Mind you, with LED and compact fluorescent bulbs becoming more popular because of the regulation on power usage for incandescents, this solution will quickly become invalid.

    It's a clever answer, but when I first heard the riddle I threw my hands up in the air because I didn't assume that each switch controlled one and only one light bulb. It's a black box! Who knows how it's wired?

  • gpg (unregistered) in reply to Harvey

    Chicken -> Grain -> Chicken <- Fox -> Chicken ->

    Done ...

  • Me! (unregistered)

    Weigh a 747? Use load cells or strain gauges. Elegant, accurate and repeatable. If you didn't know that such things exist, you would be using a sub-optimal solution. So maybe the best answer would be to ask someone who knows how to arrive at the correct answer so you could do it properly the first time rather than reinventing the (irregularly shaped) wheel.

  • javascript jan (unregistered)

    As far as 0.9 recurring = 1: presumably those who don't accept that numbers may have several numeral representations would also claim that 1 is not equal to 1.0000 recurring?

    Did some FE analysis for a chap who, amongst other things, made manhole covers about 10 years ago. His professional opinion: nobody gives a shit if some idiot drops a manhole cover into a manhole, because they're the one who has to fetch it out. Manhole covers are round (in the cases where they are) because round covers distribute the stress better (you don't have to reinforce corners) and are therefore slightly cheaper to make.

  • JM (unregistered) in reply to Matthew

    "NB: this is definitely against the manufacturer's recommended guidelines."

    Not only that, it isn't safe sex, and against the premise of the question. While you may be ok, woman 2 has an exposure risk.

    I'm not sure I'd work for someone who proposes riddles that come complete with inbuilt misogyny and irresponsible behaviour.

  • (cs) in reply to s
    s:
    akatherder:
    Take the problem of weighing the airplane. There is no correct and practical answer. But how would YOU approach it?

    The correct technical and practical answer would be reading out the weight sensors inside the airplane. A plane of this class MUST have a device to read out its weight before takeoff, and pressure in hydraulics of the wheel gear is easy enough to convert to weight.

    being a smartass, I prefer googling for mass of boeing 747. The cargo is not a part of the plane, and they asked me to weight the plane.

    The interviewer would tell you that the weight sensor in the plane is broken. Now how would you figure it out? Or the interviewer could be a dick and just let you fail miserably on that question and continue on. It's like the old riddle... how do you get an elephant in a fridge? Open the door and put him in! The interviewer wants to see you apply some analytical thinking. Maybe if you prefaced your solution with that trivia, you would be in good shape.

  • Troy Mclure (unregistered) in reply to Rodyland
    Rodyland:
    - You have a 3 litre jug and a 5 litre jug. How do you get exactly 4 litres into the 5 litre jug?

    Hire Bruce Willis and Samuel L Jackson to do this for me. Ala Die Hard 3.

    Sorry if someone already said this, but I'm not reading 325 replies.

  • JM (unregistered) in reply to brazzy

    Actually, it's perfectly correct, the mass of the Earth is concentrated at it's center. While the side of the Earth you're standing on is 1 radii closer to you than the center and therefore attracting stronger, the side directly opposite is 1 radii further away and is attracting weaker.

    So if you sum all the attractions from every grain of sand and piece of rock over the entire volume you'll find that it all balances out, and the result is identical to the graviational attraction of a single massive pellet at the center with you suspended in space 1 radii above it.

    Physics 101. Comes in handy sometimes.

  • Someone (unregistered) in reply to JM
    JM:
    "NB: this is definitely against the manufacturer's recommended guidelines."

    Not only that, it isn't safe sex, and against the premise of the question. While you may be ok, woman 2 has an exposure risk.

    I'm not sure I'd work for someone who proposes riddles that come complete with inbuilt misogyny and irresponsible behaviour.

    It is only misogyny and irresponsible behavior if you use an incorrect solution.

  • Plasmab (unregistered) in reply to Anonymous Coward

    I disagree. Brainteasers are what managers and HR people think as being lateral thinking. I get brainteasers, or exams im likely to either walk out of the interview or turn down the job.

    My favourite question to ask is "Why do you work here", especially in group interviews. Very few can answer it!!!

  • Lorad (unregistered) in reply to vt_mruhlin

    that is the longest way, but it works. If you want the shortest (weighing 3 times at the most)

    if 1&2 == 3&4 {
  if 1&2 == 5&6   {
    if 1 == 7     {
      8 is different
    }
    else     {
      7 is different
    }
  }
  else if 1 == 5   {
    6 is different
  }
  else   {
    5 is different
  }
}
else if 1 == 3 {
  if 1 == 4   {
    2 is different
  }
  else   {
    4 is different
  }
}
else if 1 == 4 {
  3 is different
}
else {
  1 is different
}
  • PBalis (unregistered) in reply to PBalis

    The hat problem is easy. No one can know what color hat they have so forget about your own hat. This is simple misdicrection. You need to follow the rules. Only person has to be correct and they don't have to prove how they know what color they have and they are in no way onbligated to be serious about trying to guess.

    The first player agrees to say his hat is the color of someone elses hat that he knows to be true because he can see it. To make the game efficent agree to say the most common color is their hat. In turn all the rest of the people agree to say they have the same color as what the first player said what his hat color. Sooner or later the person with the hat that the first player saw will say they have the correct color. They don't have to provve how they knew.

  • (cs) in reply to PBalis
    PBalis:
    The hat problem is easy. No one can know what color hat they have so forget about your own hat. This is simple misdicrection. You need to follow the rules. Only person has to be correct and they don't have to prove how they know what color they have and they are in no way onbligated to be serious about trying to guess.

    The first player agrees to say his hat is the color of someone elses hat that he knows to be true because he can see it. To make the game efficent agree to say the most common color is their hat. In turn all the rest of the people agree to say they have the same color as what the first player said what his hat color. Sooner or later the person with the hat that the first player saw will say they have the correct color. They don't have to provve how they knew.

    This one I have to comment since you ignored not just one but two of the premises. Everyone will lose if anyone gives a wrong answer, and all the guesses are made simultaneously.

    What's more, it should be obvious to anyone that without either of those two premises, the riddle isn't much of a riddle at all.

  • s (unregistered) in reply to akatherder

    The interviewer would tell you that the weight sensor in the plane is broken.

    I'd get it fixed. Having the sensor in working order can't be less important than knowing the mass of the plane and fixing it would automatically solve the problem. Besides, measuring the pressure in the hydraulics would be about the least hassle-free, cheapest method of weighting the plane, without ever moving it from a place.

    No pressure in the hydraulics? Seal it, and repressurize it. Sorry, moving the plane anywhere in current condition (in order to weight it) would break its gear beyond repair, so it needs to be done anyway. And so on :]

    Measuring the goddamned time the barometer takes to fall from the top of the tower. "No, sir, you won't ass me into putting the goddamned plane on a goddamned barque, no matter how hard you try."

  • Richard@Home (unregistered)
    Put on A and B, have sex with 1. Remove B, have sex with 2. Turn B inside out, have sex with 3.

    Um, nope - that doesn't work because you will infect yourself with whatever 1 was carrying. (The outside of condom B would become the inside when you were having sex with 3)

    The real WTF here is that I've actually spent time thinking about this O.o

  • Andreezer (unregistered)

    How about this to weigh the Boeing 747:

    • measure the pressure in a tire
    • measure the area of contact with the ground
    • pressure * area == applied force (convert to mass in your unit of choice)
    • multiply by number of tires in that landing gear
    • repeat for the 4 other landing gears
    • add up the results
    • ???
    • profit!
  • (cs) in reply to Richard@Home
    Richard@Home:
    Um, nope - that doesn't work because you will infect yourself with whatever 1 was carrying. (The outside of condom B would become the inside when you were having sex with 3)

    Incorrect. Condom A never comes off. The outside of B comes in contact with woman #1. The outside of A with #2, and the inside of B with #3. At this point, the outsides of A and B are in contact with each other (B is inside-out), and you're still wearing A correctly, having never removed it.

  • JM (unregistered) in reply to JM

    Sorry, this comment: Actually, it's perfectly correct, the mass of the Earth is concentrated at it's center. While the side of the Earth you're standing on is 1 radii closer to you than the center and therefore attracting stronger, the side directly opposite is 1 radii further away and is attracting weaker.

    So if you sum all the attractions from every grain of sand and piece of rock over the entire volume you'll find that it all balances out, and the result is identical to the graviational attraction of a single massive pellet at the center with you suspended in space 1 radii above it.

    Physics 101. Comes in handy sometimes.

    Was in response to this:

     brazzy:

    Flo:
    - Measure the gravitational constant, find your own mass and the radius of the earth, use Newton's law of universal gravitation. Simple.


Simple and quite wrong, since the Earth's mass is not concentrated in a point and thus not all at the same distance from you.
  • Dealing with cheaters (unregistered) in reply to Zlodo
    Zlodo:
    Good luck proving that the "no communication at all" rule have been broken then.

    Um, you put three strangers together, and you monitor their strategy session. Good luck in breaking the rules THEN.

  • (cs) in reply to XML Hater
    XML Hater:
    Did M$ start doing this after Google had their "IQ Test" job applications? Or was M$ actually an innovator? ;-)

    Microsoft was asking brain teasers before Google was so much as a thought.

    Of course, in the past few years, I haven't been asked a brain teaser, so I think it's no longer something that they (or at least most groups) do. I'm assuming that this is because the answer to most of these brain teasers are online and their usefulness is limited to people who don't search the internet to prepare for an interview.

    The purpose behind asking the brain teaser is to evaluate how you approach problems. Most interviewers could care less if you manage to get the right answer. In fact, I don't think that I've ever correctly answered a brain teaser while interviewing.

    One of my favorites:

    There are three vending machines. Each dispenses one item per quarter. One machine dispenses apples, one oranges, and one has a mixture of apples and oranges. There are signs on each machine indicating what is dispensed, however the signs are wrong. How many quarters would it take to determine which sign goes on which machine?

    Oh, and I love how you substituted the $ for an S. You're just sooooo cool.

  • Richard@Home (unregistered)
    Incorrect. Condom A never comes off. The outside of B comes in contact with woman #1. The outside of A with #2, and the inside of B with #3. At this point, the outsides of A and B are in contact with each other (B is inside-out), and you're still wearing A correctly, having never removed it.

    Bravo! :-D

    captcha: muhahaha

  • JM (unregistered) in reply to Anonymous

    First of all, the whole hat thing has evolved to be quite a WTF itself... It's funny to see how many intelligent(?) people just can't accept what has been said to them like million times in this thread :)

    Anonymous :
    Ok, this has many WTFs in its own, but let's extend the boat ride riddle:

    In one side of the river there are:

    • a father with his son
    • a mother with her daughter
    • a security guard with a prisoner

    The boat has a capacity of two, but only the father, mother and security guard can operate it.

    The father can't stand the girl if her mother is on the other side of the river The mother can't stand the boy if his father is on the other side of the river The prisoner will kill any one if the guard is on the other side of the river

    How can they all get to the other side of the river?

    Let's try (f=father, s=son, m=mother, d=daughter, g=guard and p=prisoner):

    m, d, g, p ---(f, s)-->
m, d, g, p <--(f)------ (s)
d, g, p    ---(f, m)--> (s)
d, g, p    <--(m)------ (f, s)
m, d       ---(g, p)--> (f, s)
m, d       <--(f)------ (s, g, p)
d          ---(f, m)--> (s, g, p)
d          <--(g, p)--- (f, s, m)
p          ---(g, d)--> (f, s, m)
p          <--(g)------ (f, s, m, d)
           ---(g, p)--> (f, s, m, d)
  • Anonymous Coward (unregistered)

    Assuming you mean that all signs are guaranteed to be wrong on the vending machines, you only need 2 quarters.

  • Ken (unregistered) in reply to unknown
    unknown:
    Yes,

    whenever you choose to answer, the prob = 0.5 that you are correct.

    If all 3 answer then the prob is 0.875 that someone guesses wrong and you all lose.

    If 2 answer (doesn't matter which) prob is 0.75 that you lose.

    If 1 answers (dones't matter which one) prob is 0.5

    If 0 answer , prob is 1.0 that you lose

    Wrong.

    Given the agreed-upon rules, you only answer if you see two hats of the same color.

    If all 3 answer, there is a 100% chance you lose. You cannot have 2 answer. If 1 answers, there is a 100% chance you win. You cannot have 0 answer.

    There is a 25% chance that all 3 will answer, and a 75% chance that 1 will answer.

    Net result: there is a 75% chance you win.

  • Ken (unregistered) in reply to Matthew
    Matthew:
    I didn't see a serious answer to the two condoms, three women question. Here's my attempt:

    Put on condom A, then condom B over the top of it. Have sex with woman 1. Remove condom B, turn it inside out and replace. Have sex with woman 2. Remove condom B and have sex with woman 3.

    NB: this is definitely against the manufacturer's recommended guidelines.

    Woman 3 is exposed to woman 1's "fluids" which were placed on condom A by the inside-out condom B.

    Here's my attempt:

    Put on A and B. Have sex with 1. Remove B. Have sex with 2. Place B inside-out over A. Have sex with 3.

    By delaying the use of the inside-out B, you have not "contaminated" A until after it has been used.

  • Ken (unregistered) in reply to Chaz
    Chaz:
    For those of you saying that you can look at the other hats and use statistics to find your own hat's color... the line "with the outcome of one coin toss having no effect on the others." means that the events are independent. Thus, despite how it may seem, there is no greater or lesser chance of your hat being a color based on the color of their hats. As far as I can tell, the feedback of the other hats is useless and you're just as well off with one person guessing randomly. There may be tricky wording somewhere in the question, however.
    The only "tricky wording" is the "at least one correct answer" part. That means that you do not have to answer!

    Yes, the color(s) of the other hats has no effect on the probability of your hat being a particular color. However, the fact that you will not give an answer 50% of the time is relevent, and is what changes the odds from 50-50 to 75-25 in your favor.

  • Flo (unregistered) in reply to brazzy
    brazzy:
    A better approach would be to compute the Earth's volume and multiply it by a guesstimated average density. IIRC most kinds of rock are about 6 or 7.
    I think my approximation will me more accurate. The earth core consists of liquid metals which have a higher desitiy than rocks.
    brazzy:
    Simple and quite wrong, since the Earth's mass is not concentrated in a point and thus not all at the same distance from you.
    It is not concentrated, but it can be threated as if it is. From outside everything acts like its mass is concentated on a single point. -> Physics
  • Jno (unregistered) in reply to kimos
    kimos:
    Given an opaque box with three light bulbs inside and three switches outside, how would you determine which switch corresponded to which bulb if the box could be opened only once and only after all the switches were permanently set?

    That's the one that I've heard of. Took me a long time to get the answer and would never have thought of it in the span of an interview.

    Turn on switch A for a couple minutes. Turn A off then turn on switch B. Quickly open the box. The bulb that's on is switch B, the bulb that is off but is still warm to the touch from being lit is switch A, and the remaining bulb is switch C.

    Mind you, with LED and compact fluorescent bulbs becoming more popular because of the regulation on power usage for incandescents, this solution will quickly become invalid.

    That's fine if the switches are wired up one to one with bulbs in three discrete circuits. We don't know that. How about two circuits, one with two switches in parallel and one bulb, and the other with two bulbs and one switch? And do we assume there is an electrical power source in every circuit? :)

  • Jno (unregistered) in reply to Flo
    Flo:
    I can so relate...:
    Actual questions I got at an interview (not with MS):
    • Why is the sky blue?
    • How would you estimate the weight of Earth?
    • How many people are there in the world?

    I was told the point of these questions was to see how comfortable I am to admit that I don't know the answer.

    • http://en.wikipedia.org/wiki/Rayleigh_scattering
    • Measure the gravitational constant, find your own mass and the radius of the earth, use Newton's law of universal gravitation. Simple.
    • ~6,500,000,000 ?

    There are people who don't know the answers? oO

    2/3. You were asked to estimate the weight of the Earth, and you have outlined a method for measuring its mass. Nice try, would you consider the intern position? :))

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