How do you determine the risk profile of a company?1 Answer
financial statements, cash flows, financial ratios, historic events and company developments
What markets do you think Cisco should get into and why?1 Answer
Cisco is already touching the SMB and consumer segment, esp. with its managed services. Public TelePresence at a low enough price would be extremely attractive to consumers in less developed areas. With Cisco's experience in using games to promote its products, offering game hosting as a managed service is not such a big leap.
How would you handle an upset patient coming up to the front desk and complaining?3 Answers
I certainly understand why you would be upset. Let me see if I can get a nurse to speak with you more about this.
I would empathize with the customer and diffuse the situation by addressing his/her concern.
I let them get their complaints and anger out that way they know that I am concern about their problem and so I can better understand them why they are upset. I give them a sincere apology then help them by resolving the problem.
Tell me about a time when you were working in a team and your opinion was challenged.1 Answer
I asked the team member to explain their opinion. No one is perfect, so keeping an open mind when listening to a team member’s response is key. There were times a team member could convince me they were right and other times when I could explain why my opinion may be the better option.
Suppose you had eight identical balls. One of them is slightly heavier and you are given a balance scale . What's the fewest number of times you have to use the scale to find the heavier ball?48 Answers
3 times. (2^3 = 8)
Two. Split into three groups of three, three, and two. weigh the two groups of three against each other. If equal, weigh the group of two to find the heavier. If one group of three is heavier pick two of the three and compare them to find the heaviest.
Brian - this would be correct if you in fact were using a weighing scale, and not a balance scale. The ability to weigh one group against another with a balance scale allows Marty's answer to be a correct answer. Although - the question as worded provides a loophole. If it had been worded as "What's the fewest number of times you have to use the scale to CONSISTENTLY find the heavier ball", then Marty's answer would be the only correct answer. However, it is possible that you could get lucky and find the heavier ball in the first comparison. Therefore, the answer to the question as stated, is ONE.
This question is from the book "How to move Mt Fuji".... Marty has already got the right answer.
Actually Bill, by your interpretation of the question the answer is zero, because you could just pick a ball at random. If you get lucky, then you've found the heaviest ball without using the scale at all, thus the least possible amount of times using the scale would be zero.
The answer is 2, as @Marty mentioned. cuz its the worst case scenario which u have to consider, otherwise as @woctaog mentioned it can be zero, u just got lucky picking the first ball....
None- weigh them in your hands.
Assuming that the balls cannot be discerned by physical touch, the answer is 3. You first divide the balls in two groups of 4, weigh, and discard the lighter pile. You do the same with the 4 remaining, dividing into two groups of 2, weighing, and discarding the lighter pile. Then you weigh the two remaining balls, and the heavier one is evident.
2 3a+3b+2 = 8 if wt(3a)==wt(3b) then compare the remaining 2 to find the heaviest if wt(3a) !== wt(3b) then ignore group of 2 discard lighter group of 3 divide the remaining group of 3 into 2+1 weigh those 2 If == the remaing 1 is the heaviest if !== the heaviest will be on the scale
With the systematic approach, the answer is 3. But, if you randomly choose 2 balls and weigh them, and by coincidence one of these two is the heavier ball, then the fewest number of times you'd have to use the scale is 1. Although the real question is: are the balls truly identical if one is heavier than the rest?
just once. Say you are lucky and pick the heavy ball. One use of the scale will reveal your lucky choice
so once, or the creative answer zero if you allow for weighing by hand
Without judging by hand: Put 4 balls on one side, and 4 on the other. Take the heavier group and divide again, put 2 balls on one side, and 2 on the other. Take the 2 that were heavier, and put one on each side. You've now found the heaviest ball. This is using the scale 3 times, and will always find the right ball.
None. They are identical. None is heavier.
2 weighings to find the slightly heavier ball. Step 1. compare 2 groups of three balls. Case 1. if they are both equal in weight, compare the last 2 balls - one will be heavier. case 2. If either group of 3 balls is heavier, take 2 balls from the heavier side. compare 1 ball against the 2nd from the heavy group result 1. if one ball is heavier than the other, you have found the slightly heavier ball. result 2. if both balls are equal weight, the 3rd ball is the slightly heavier ball. Easy Shmeezi
Fewest - get lucky and pick the heaviest one. But wait! How would you know it is the heaviest one by just weighing one ball? Your logic is flawed. Two groups of four. Split heavier one, weigh. Split heavier one, weigh. 3 times.
i think its 3. i would take it like this OOOO OOOO then OO OO then OO problem solved. i do this everyday. bye. praise be to allah. thats it.
It's 2. Period. If you can't figure it out look it up online or in "How Would You Move Mount Fuji" (like somebody else said). This is one of the most basic brainteasers you could be asked in an interview.
The answer is 2. 1) Divide the balls into 3 groups. 2 piles with 3 balls each, 1 pile with 2 balls. 2) Weigh the 2 piles of 3 balls. If both piles are the same weight, discard all 6 and weigh the last 2 to find the heavier one. 3) If 1 pile of 3 is heavier than the other, discard the lighter pile and the pile of 2 balls. Weigh 2 of the remaining 3 balls from the heavier pile. If both of the weighed balls are equal, the last ball is the heavier one.
2=if all the balls are identical and you pick up the first...weigh it and the second one is lighter or heavier then you've found the heavier ball in the least amount of attempts.
1=if all the balls are identical and you pick up the first...balance it and the second one is lighter or heavier then you've found the heavier ball in the least amount of attempts.
Amy is 100% correct for the following reason: everyone (except Amy) is solving the theoretical problem. The practical side of the problem - notwithstanding jimwilliams57's brilliant observation that if one weighs more than the others IT IS NOT IDENTICAL (would have loved to see the interviewer's face on that one) - in order to 'weigh' them on a scale, one has to pick them up, therefore, you will immediately detect the heavier one without weighing: pick-up three and three ... no difference, no need to weight. Pick-up the remaining two to determine the heavier one. Steve
First off, take yourself through the process visually and forget square roots, that doesnt apply here and here is why: The question is the Minimum, not the MAXIMUM. BTW, the max would be 7 ( 8-1); you are comparing 2 objects, so 1 ball is eliminated automatically in the first step. Anyway, you have a fulcrom of which you are placing 2 of 8 objects on each end. If by chance you pick the slightly heavier object as one of the two balls, you have in fact, found the slightly heavier one in the first round... btw dont be a smartass with your interviewer, he is looking for smarts not smarmy;)
Respectfully, the folks who are answering "3" are mathematically modeling the nature of the balance incorrectly. Performing a measurement on a balance scale is not binary. It is trinary. Each measurement gives you one of three responses: The left is heavier, the right is heavier, or they are equal. So while you do need three binary bits to specify a number from one to eight, you need only two TRINARY-DIGITS Formally, you want the smallest value of n such that 3^n >= 8. The answer is 2. Note that you could add a ninth ball, and still, you'd only need to make two measurements. Of course, the smarty pants answer would be one. Just pick two balls at random and be lucky to have chosen the heavy one. But you're not guaranteed to be able to do it in just one measurement.
English isn't my mother tongue... What is a balance scale? If looking up on Google, I find some devices with two bowls on a bar bearing in the center. Hence, the answer is once (if I'm luck enough to select the heavier ball in teh first measurement) If a balance scale allows to measure only one ball at a time, then it would take two measurements, unless you'd have more information on the weight, which is not listed here, hence doesn't exist in the context of the question^.
3 times. Not having looked at the other comments, hopefully, I am the 26th to get this right. Put the balls 4 and 4 on the scale, Take the heavier side and place those balls 2 and 2 on the scale. Take the heavier side and place them 1 and 1 giving the heaviest ball.
OK, now I read the comments and see that the people, like the question are divided into to groups, systematic approach people that say 3 (like I did) and analytic people that say 2. It takes a systematic person (me) a minute to get the answer. I'm guessing it took the analytic 5 minutes just to interpret all the ramifications of the question, i.e. they aren't idenitical if..., do it by hand..., get lucky.
minimum is 1 (if lucky - 25% chance by picking 2 balls at random) & max is 2 (using most efficientl process to absolutely determine without luck - 3/3/2 scenario)
While Symantec was busy weighing my balls I took a job with NetApp.... They need to focus on hiring good, capable security engineers, not weighing their balls.
The point of these interview questions is to both check your logical brain function and to hear how you think. Most of you are just posting jerk off answers trying to be funny, or you are really dumb. These answer get you nowhere with me in an interview. Think out loud, go down the wrong path back track try another logic path, find the answer. None of this "0 if you use your hands". That is fine if you are interviewing for a job in advertising where creativity is desired, nobody wants you writing code like an 8 year old.
You have 12 balls, equally big, equally heavy - except for one, which is a little heavier. How would you identify the heavier ball if you could use a pair of balance scales only twice?
The problem is based on Binary Search. Split the balls into groups of 4 each. Choose the heavier group. Continue till you get the heavier ball. This can be done in log(8) (base 2) operations, that is, 3.
Since there is only one scale available to weigh. You first divide the balls in half. Weigh each group, take the heaviest group. This is using the scale twice so far. Now, divide the previous heaviest group into half, weigh both groups. Take the heaviest. Divide this last group and take the heaviest. This is the heaviest ball. We have used the scale 5 times.
Would it be wrong to say for a sample size as small as 8, we might as well not waste time thinking about an optimal solution and just use the scale 7 times, as this will be more efficient than coming up with an ideal solution prior to using the scale?
I stumbled across this while looking for something else on Google but I had to answer. It is 2, split balls into 2,3 and 3. weigh the 2 groups of 3 against each other. If equal weigh the group of 2 and the heaviest is obvious. If they are not equal keep heavy group of 3 and weigh 2 of the balls. if equal heaviest ball is one you didn't weigh. If not equal the heavy ball is obvious.
2 times. 8 balls. 1st step:    2nd step: [  ] [  ] [ ]
The fewest number of times to use the scale to find the heavier would be Eight to One times ?
It will actually be 1 because the question asks what's the fewest amount of times which is one because you could just get lucky you can use any method you want it would still be one because that is the fewest amount of turns you can have
It's one. The fewest number of tries on using a balance scale would be one. If you put one ball on each side and one is heavier, you have the found the heavier ball.
Use an equilateral triangular lamina which is of uniform mass throughout. It is balanced on a pole or a similar structure. Steps: Place 2 balls at each corner (total 6 balls) i. if the odd ball is one of those, one side will either go up or go down. Now repeat the process with one ball at each corner including the 2 unbalanced ones. ii. if balance is perfect, repeat the process with the remaining two balls and one of the already weighed balls.
test answer 2016-01-12 00:34:07 +0000
You would not be able to find a ball heavier than the others. All eight balls are identical; therefore, they must all be the same weight.
Correct answer has already been posted. I just want to contribute some theoretical analysis. Given N balls, one of them is heavier. Finding out the ball requires log3(N) trit of information. (trit is the 3-base version of bit). Each weighing may give you one of the three outcomes: equal, left-heavier, right-heavier. So the amount of information given by each weighing is upper-bounded at 1 trit. Therefore, theoretical lower-bound for number of weighings in the worst case is log3(N), which is actually attainable. So 27 such balls need only 3 weighings and 243 balls need only 5 weighings, etc.
2 as many have indicated above. The 3 is the kneejerk reaction but 2 is the correct answer.
Marty's answer is correct, but he does not explain why. The logic of the balance scale is three-valued: . Its most efficient use is the recursive application of the three-valued logic until there is only one item left. The integral ceiling of ln(x)/ln(3) thus gives the fewest number of times you have to use the balance scale to find the uniquely heaviest ball of x balls. Ceiling(ln(8)/ln(3)) = 2.
How many square feet of pizza is eaten in the US each year?27 Answers
800 billion Sq.ft
A pizza is roughly 1 sqft. If the average american eats 1/3 of a pizza and eats pizza 3 times a month, that would mean 12sqft a year. Times 200M americans that gives you 2.4 billion sqft.
Sorry, i must be stupid. I didn´t know pizza were made of feet.
These market estimation questions are pretty common in IB and consulting; so this one is not an oddball. The critical part is the logical reasoning used to estimate rather than how accurate you are. Try these: how many diapers used everyday in Canada how many commercial planes are currently flying in US air space
All of them.
Pizza is not served in square feet in the U.S.
If i know & see it right, Pizza's are not square in shape so sq feet is not the correct UOM (Unit of measure). Pl correct ur question in terms of circle/diameter/radius & I shall revert with the answer. Trust your question is answered.
Unless i am going for a pizza delivery or pizza marketting job which Goldmann Sachs is not, i would blatantly say would either of us care?
How the hell would I know ? Let me do some research and get back to you on it.
MOST PIZZA IS SEVED IN THE SHAPE OF TRIANGLE (FOR A SLICE) OR IN THE SHAPE OF A CIRCLE - NEED TO DO SOME RESERACH AND THEN SOME MATH. LET ME GET BACK TO YOU ABOUT IT.
Way to much lousy square feet - that's for sure. Hungry ? What's your favorite pizza there Bossman ? You buy - I'll fly. Whadya say ?
pie r squared
Assume average pizza is 16", or 8" radius. That is 0.66 feet. Square it and mulitply by 3.1 to about (0.4)*3.1 = 1.2 ft^2. Assume 100 million families in US eating one pizza 10 times a year (this is the weakest assumption) = 100,000,000*10*1.2 = 1.2 billion square feet of pizza, +-0.3 billion.
$38 billion Annual pizza sales in America, according to Pizza Today 3 billion Number of pizzas sold in the U.S. each year, according to NAPO 350 Slices of pizza sold every second, according to NAPO 46 Slices of pizza the average American eats each year, according to Packaged Facts 23 Pounds of pizza the average American eats each year 93 Percent of Americans who eat pizza at least once a month 70 Percent of Super Bowl viewers who eat at least one slice during the game 251.7 million Pounds of pepperoni Americans consume each year 36 Percent of pizza orders that specify pepperoni as a topping 70,000 Number of pizzerias in the U.S., according to Pizza Today 24 Percent of those pizzerias owned by Pizza Hut, Domino's, or Papa John's 9,000 Number of pizzerias in New York 17 Percent of restaurants in America that are pizzerias, according to Food Industry News 40,010 Number of subscribers to Pizza Today, the leading pizza industry magazine
150 million square feet - Math is not the answer to your question - Google is. http://bit.ly/MoP5xR
Well, I consume about 400 square ft by myself so...
There are none, because I was live at Indonesia
If I wanted the job, I'd say I would research the answer directly. If not found, I would use whatever statistics I can find. An example would be: (# of pizza's sold) times (average sq feet per pizza) times (average ratio of pizza eaten vs thrown away) If I didn't want the job, I'd say "I'm going to research this, then leave for the day, score some good acid, return 24 hours later and make up a number".
Around 30 Million in India
The answer is none. Pizza is sold by volume. The correct unit of measure for volume is cubic feet, not square feet. You cannot eat something that has no volume.
Pizza eaters population * size of pizza in square feet
Answer is zero, or at least it can't be answered. Question said "US" - not U.S. US is not a measurable subject (whose US?) Question did not ask United States (U.S.) because there was no punctuation marks in their subject US.
There's about 320m people in the US who live to about age 80. That means there are 4 million people at each age. I would say prime pizza years are about age 10 through age 60, which is a range of 50 years, meaning about 200 million people eat pizza regularly. I assume that people, on average, eat pizza twice a month, meaning that there are 200m * 24 = 4.8 billion instance of pizza eating every year. The service size for pizza is about a quarter of a pie, a pie has a diameter of about 2 feet, meaning pi/4 or about 0.75 square feet of pizza are eaten per sitting. The area of pizza eaten in the US every year is about 4.8b * 0.75 = about 3.75 billion square feet. According to this website, 3 billion pizzas are sold every year in the US. http://www.statisticbrain.com/pizza-statistics/ I might be off by a factor of 3 or so in my answer depending on the average size of a pizza, but I am pretty close all things considered. That is how you answer these questions, people.
How many tennis balls are in this room and why?14 Answers
There is no correct answer to this question. It's all based on your problem solving abilities.
The method for "solving" this involves making assumptions and then calculating an answer. The assumptions could be wrong, which isn't as important, and the interviewer will likely help guide your assumptions if they felt it was necessary. The assumptions here would be about the approximate size of a tennis ball, the approximate volume of the room, etc. Getting fancy, you might begin discussing how the weight of the tennis balls would begin compressing the lower levels, giving more potential room at the top, as well as how the second level of tennis balls would rest in a lattice structure on the first level, sitting in the gaps, and thus not being as high as simply multiplying the diameter of the tennis ball by 2 to get the height of two layers.
I agree with both prior posts. An interviewer asking this question is trying to determine your problem solving skils. The worst thing you could do is blurt out an actual number without explaining your thought process. I'm not sure the interviewer is as interested in the quantity as much as how you arrive at it. The worst answer "Wow, I have no idea".
Everybody is reading this question wrong. It doesn't say how many tennis balls CAN FILL this room. It says how many tennis balls are IN THIS ROOM and why. The question is probably meant to see if the candidate is actually fully paying attention to the questions (unlike everybody who has answered it here). It is a trick. I would gather to say that there are NO tennis balls in the room...and why? Because it is an office, not a tennis court.
Hi wildfiresg, Maybe you should ask the interviewer to show the insides of his pockets? You never know, maybe there is a tennis ball in there!
I think wildiresg is the only one above who's really paying attention (which is CRITICAL in an interview) - give him the job!! And Igor makes a very good but funny point....the candidate should also ask to check any drawers, boxes or closets in the room!!!
In the original post the candidate did mention "how many tennis balls can fill this room?" as part of the interview. The first answers were tricked into reading this question as the same in the story. Good job wildiresg!
"If the room is already filled with sand, none. Do you want to know if I can actually write code?"
If the question really is how many "are in this room", then make an observation about how many you can see. I'm going to assume "zero" is a safe bet for most interviews. If, however, the question was suppose to be how many *can fit* in the room, then please ask (type, in this forum) the right question, or the people who are smart enough to actually listen to the question you're asking will never give you a response you'll be happy with.
Assume: 1) this is a theoretical determination of how many the room would hold & 2) not an exercise of observation & 3) there is no compression ... then this is a simple mathematical problem (as long as you know the formulas) and you have the relevant metrics to hand. A tennis ball is a sphere & its volume is calculated at 4/3 Pi r(cubed). Packing ratios are 66 & 1/2% (there will be space between the balls when packed). With the dimensions of the room you can then calculate how many tennis balls would fit into it. This can be used to work out the fairground game of how many marbles in a jar or even m&m's in a jar (though that is a different formula & is for an oblate spheroid & not a sphere)
This is a metaphorical question. There is one tennis ball in the room. The interviewer is serving and the candidate is returning. The game will continue until the ball is missed.
As any schoolboy could tell you: count the number of chairs in the room and multiply by 4. The tennis balls are on the bottom of the chair legs so they don't scuff the floor!
sound weird but they i want to be sure that u r focus
metaphorical question..because there aren't balls in the room. but the balls are keeps on rolling, because it has its determination to keeping that ball rolling..
Suppose we hire you, and you and the rest of the new interns decide to go buy a cup of coffee. Each intern purchases one cup of coffee. One of the interns suggests everyone play a game. Everyone will flip a fair coin, dividing the group of interns into two subgroups: those that got heads and those that got tails. The game is this: whichever group is smaller evenly splits the cost of everyone's cup of coffee (i.e. if there are 5 interns, 3 get H, 2 get T, then the two interns that got tails each buy 2.5 cups of coffee). However, nothing says you need to play this game. You can choose to buy your own cup of coffee and not play the game at all. The question: Should you play this game? (Note: You may assume that there is an odd number of interns, so there are no ties, and that if everyone gets H or everyone gets T, then everyone loses and just buys their own cup of coffee).18 Answers
Hint: Despite its look, this is not a math question.
I dont get it, Would you please provide more hints?
Assume each coffee costs $1, for simplicity. So this is effectively a choice between two outcomes: paying $1 with probability 100%, or paying $0 with some probability and paying more than $0 with some probability. So you ask yourself: what is your expected cost in the second case? Give that a try and see if you can figure it out. However, I want to remind you that the question is "should you play this game?" The answer to this question isn't just a math question. If you only work out expected values, you've missed the point. For example, a separate question (with the same kind of flavor as the direction I'm trying to lead you) is this: suppose I give you a choice of two outcomes. Either you get $1 with 100% probability or I give you $500,000,000 with probability 1/100,000,000 and 0 otherwise. Which would you pick? Now what if it was the same first choice, but the second choice was $50 with prob 1/10 and 0 otherwise? Now which would you pick? These are the kinds of things you want to think about while answering this kind of question. Let me know if you have anymore questions. And if you want me to post the answer, just let me know.
I think I get it. It's about investor's risk appetite. Investors is likely to take guaranteed gains, here is $1.
Well yes and no. This is indeed a risk aversion question. If you work it out, you'll find that the EV in each case is exactly the same (your EV is -1 cup of coffee in both scenarios) but that's not the end of it. It's also really a question for them to test your risk aversion. You can really support either answer, and *should* comment on the validity of either answer. My answer was to go with buying my own cup of coffee, and followed it up with a story where a friend of mine had tried to get us to play credit card roulette (which is similar in spirit to this game) and I told him that I did, in fact, say no in that instance and why I said no. However, traditionally people are risk-averse when faced with gains and risk-seeking when faced with losses, so many would probably choose to play the game. But this is as much a question about your psychology as it is about your math skills. And that's why this is such an awesome question, and is probably a question that kills most people they ask it to.
Would you please explain to me how do you get -1 for the second scenario? There are 50% H and 50% T. Therefore players have 50% opportunity in the winning group. Given 5 interns, there are two combination of lossing group (4 vs. 1 and 3 vs. 2). EV=0.5*0 + 0.5*(0.5*-5 + 0.5*-2.5) = -1.875.
The probability of winning isn't 50%. It's actually slightly above 50%, but that's not the way to look at it. The total number of coffees that need to be bought is n, where n is the number of interns. Going into this game, every intern is the same so they each have the same expected value, and the sum of the expected values must equal -n. So everyone has an EV of -1 as claimed.
Thank you very much. I finally got it.
I'm still not sure why you said no if the EV is the same?
It's a matter of risk preference. See the example I gave in my Mar 19 posting: "Suppose I give you a choice of two outcomes. Either you get $1 with 100% probability or I give you $500,000,000 with probability 1/100,000,000 and 0 otherwise. Which would you pick? Now what if it was the same first choice, but the second choice was $50 with prob 1/10 and 0 otherwise? Now which would you pick? These are the kinds of things you want to think about while answering this kind of question." You would probably not play in the first instance and consider possibly playing in the second. (And those instances even have the risky game with a HIGHER EV). This coffee game is the same kind of game, and even if the EV is the same in each case, the volatility is not the same. It is usually a good rule of thumb to take the lower volatility outcome if the EV is the same (think Sharpe Ratio here! Or think efficient frontier here! Both get the point across I think.). Does that help?
I say yes. To play the game. Only because I'm prepared if I lost. The fact that I can afford to loose, makes me want to try my chance at wining
I would play the game. Consider the expected price with n total interns splitting a total cost of P. That is P/2*E[1/(N)|you're paying]. This is simply equal to P/2*E[1/(N+1)] where n is a binomial now corresponding to n-1 interns. Notice that 1/n+1 is a concave function. That means that P/2*E[1/(N+1)] <= P/2*(1/E[N]+1)=P/2 * 1/( (n-1)/2 + 1)) = P / (n +1). So it pays to play on average
Apologies, but both of the above answers are incorrect. The EV of playing is the same as the EV of not playing.
No way would I play! The most money I save is a dollar the most money I can lose ( in the case of five interns) is 4 dollars... That's a 400 percent downside verse a 100 percent upside ( kind of you cannot technically compute your return on zero dollars invested). Plus, I do not even drink coffee!
Stupid...I don't have time to play games at work.
The EV in both scenarios are not the same: it's clear when think about the case where there are n = 3 interns.
I've already explains why they are the same. In either scenario there will be n cups of coffee bought (3 in your case) so total EV is -n (-3 in your case). In the game each person is the same as any other so their EV must all be the same. That is why in the game the EV is -1 (same as if they buy their own cup of coffee). To argue otherwise is to argue that either the total EV is not the number of coffees purchased or that someone has an unfair advantage in the game. Incidentally, if you are going to claim that someone who has given you the answer is wrong, you should provide more of a response than "go think about it."
if v = pay by game -1 . Then all have five intern have same distributed V by symmetry 5EV=0 It's zero sum.
Why is a man hole cover round?9 Answers
Weight displacement. So the cover does not fall into the hole no matter where it is or what is on it.
There are a few answers that would qualify as "correct", but I suspect this question won't be making the rounds much longer as this is becoming the stereotypical case example question which is meant to "see how you think". A circular manhole cover can never fall into the hole, as a square cover could by turning it on a diagonal, for example. Also, a worker can easily roll the cover without every having to lift it.
In addition to Adam's answers, they are also round because they will fit on the manhole regardless of which way you put them on.
The real reason manhole covers are round is because the manhole is round. It certainly wouldn't do to put a square cover on a round hole. (However, that's likely not the answer they are looking for.)
Because the manhole pipe is round.
Because if it were star-shaped it would have pointy bits that would be ouchy and tear your clothing.
In addition to Adam's observations (which contain the very best answers to this question), a round manhole cover has no sharp corners, so if for any reason it did not sit right in the manhole, and a car or truck ran over it, it would be less likely to puncture the vehicle's tire.
This was a Microsoft interview question at one point. It is round so that the cover can't fall into the hole the way a square or rectangular cover could if inserted diagnally. It's also easier to move (by rolling it).
It's also the maximum coverage for the minimal amount of material - making it the cheapest shape in terms of material cost.
how many times the hour and minute hands of the clock form right angle during one day?10 Answers
each hour creates 2 right angles...2 x 24 = 48 times a right angle is formed in one day
Wrong. Think what's happening around 3 and 9 o'clock
Let me show you a mathematical approach. Common sense dictates that the minute hand moves at a faster rate of 5.5 degrees a minute (because the hour hand moves 0.5 degrees a min and the minute hand moves 6 degrees a minute). We start at 12 midnight. The hands are together. For subsequent 90 degree angles to occur, the minute hand must "overtake" the hour hand by 90 degrees, then 270 degrees, then 360 + 90 degrees, then 360 + 270 degrees, then 360 + 360 +90 degrees.. and so on. This can be re-expressed as: (1)90, 3(90), 5(90), 7(90), 9(90), 11(90)... n(90). The number of minutes this takes to happen can be expressed as (1)90/5.5, 3(90)/5.5, 5(90)/5.5, 7(90)/5.5, 9(90)/5.5, 11(90)/5.5... n(90)/5.5. In one day, there are 24 hr * 60 mins = 1440mins To find the maximum value of n, n(90)/5.5 = 1440 n = 88 but as seen from above, n must be an odd number (by pattern recognition and logic) hence n must be the next smallest odd number (87) counting 1,3,5,7,9,11......87, we see that the number of terms = (87-1)/2 +1 = 44. In other words, the minute hand "overtakes" the hour hand on 44 occasions in 24 hours in order to give a 90 degree angle. Therefore the answer to your question is 44.
i agree with right_ans. although i got lost in his explanation, though i'm sure it is correct. i found another answer here: http://brainteaserbible.com/
Each hour has 2 occurrences of 90 degrees. In 12 hrs, it overtakes 24 times. BUT hrs 2 to 3 has only 1 NOT 2. Also hr 8 to 9 has only 1. So subtract 2 from 24. You get 22. In half a day (12 hrs) you get 22 times. Therefore in 1 day ie 24 hrs, it cross 22 * 2 = 44 times.
Relative speed is 5.5 degree/min. Time is 24*60 mins. Total distance is 5.5*24*60 degrees. How many full circles it is? 5.5*24*60/360 = 5.5*4 = 22. Each full circle contains 2 right angles (90 and 270). So answer is 22*2 = 44.
Calm down, we first must convert time to angle, two different units. The hour hand completes one full revolution each 12 hours (considering a 12 clock). So, theta_h = 360 x h/12, where theta_h is the angle that the hour hand makes with 12 hs mark and h is the number of hours since 0hs. So at 0 hs, the angle is 0, and at 12, the angle is 360 = 0. Since h = m/60, where m is the number of minutes since 0h, we have: theta_h = 360 x m/60 / 12 = 360 m / 12 x 60 = 0.5m Now, given the number of minutes since 0h, we can tell the angle of hour hand using theta_h. The minute hand angle, theta_m is: theta_m = 360 x m / 60 = 6m So the difference between theta_h and theta_m is |theta_h - theta_m| = 5.5m. Now given the minutes since 0h, we can tell the angle between hour and minute hand. Now, how many minutes we need to make |theta_h - theta_m| = 5.5m = 90? About 16.36. Since we have 12 x 60 minutes in a day, we have 12 x 60 / 16.36 gaps that satisfy 90 degrees, which is 44.