Analyst Interview Questions in United States
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When a hot dog expands, in which direction does it split and why? 99 AnswersThe stress in the circumferential/tangential direction is always larger than the stress in the axial direction, so it splits in long seams down its length rather than in hoops around its circumference. The equation for hoop stress is pr/t, whereas the equation for axial stress is pr/2t. It is a good exercise to derive these equations yourself. To its weakest part Splits lengthwise because the diameter is less than the length. Show more responses It splits along the long axis, the shear stress is greatest there. Materials always fail in shear, which is a combination of normal and tangential forces. "That way!" Then wag my pointer finger in the air It splits along the slits I put in it before I started to grill. Clearly you need someone that knows how to grill. So for the most part answer one is correct. But it is not the stress its the strain that causes the failure. Plus most dogs have a slit which is a built in "defect" weak spot and it fails at the weakest point. It splits in the path of least resistance. It expands & cooks proficiently where resistance is greater, yet achieves desired results. Much like I will do as a member of your team. And as a comforting side note, although my confidence in my abilities is apparent and justified, my co-workers have never labeled me a hot dog. It explodes on both ends, trust me , I have seen this happen. Otherwise, down the middle. Right down the middle like any good deep fried Texas weiner should. th from inside out Show more responses It splits vertically. I only answer this way because it's what I've seen, although I don't know why. It usually breaks lengthwise because the casing has some elasticity to it but there is more stretch in the length than the diameter because there is more casing in that direction, therefore it breaks lengthwise.. Heat makes it expand in both directions. The weakest point is at the middle that splits alongside due to pressure produced by vaporization of water. Both Ways Oposite your mouth. If it's a quality dog made with authentic ingredients, it'll split along the seem. If it's a cheap dog made with synthetic ingredients, it'll break in half because it doesn't have the integrity of a genuine product. It will split along the length because there is more of the "skin" to stretch than if it were to split the short way. In both deriction It depends on what type of casing the hotdog has. If it is man0made casing then they tend to split length wise. The man-made casings usually have a casing line down one side which seems to be the weak point in the product. 1, I have no idea. 2. When objects are heated, they expand due to molecules moving fast. The outer csing of a hotdog will split open if heated longer than necssary becoause of a seam in the casing. 3.I have no idea. I don't eat hotdogs. Show more responses lengthwise, expansion around smallest diameter of dog is greatest, and small tear forms and continues as expansion continues They split at the weakest seam, just like my apostrophe in my last name on Unix servers. It splits apart, letting the heat escape from the core That's an outstanding question, which I am extremely happy to answer for you, in detail!! But first, let me explain how a flux capacitor works, and more specifically how it applies to 'time travel'.... Does this question have anything to do with the position? If not, thanks for the interview. Don't call me, I'll call you. If yes, lengthwise. Depends in the breed. It splits upward because of the gases that release upward through heat. horses are for ride .....and above all question does not suggest they are attacking so no question for fighting The Hotdog splits lengthwise, but I don't know why. Most likely at one end horizontally Show more responses Lengthwise Why is it expanding? Is somebody cooking it or has it just been lying around getting rotten? Did somebody toss it outside of a spacecraft or are they filling it with air? Not enough info. The last time I over cooked a hot dog it split at the middle along the length of the dog A frankfurter will split along the longitudinal axis because that's how it's always been done. Actually. I know it has something to do with the skin/casing is stronger latitudinally, but I can't explain why. #1 is correct (hoop stress vs axial stress) and the main reason for the lengthwise split, but you should also mention (for a manufactured casing) that the machine direction (as the casing was extruded) has better crosslinking than the cross direction (and thus can take more stress.) This is the reason a plastic bag tears in a vertical line, not horizontally. (they are extruded as a tube and sealed, so the machine direction is vertical) And as another answer pointed out, surface stress concentrations (such as the "grill" marks) will also cause the casing to fail. Down the middle lengthwise and later at the ends. The casing is cut open to remove the cooked dog leaving a weak spot down the center. The ends are weak where the dog is "pinched" to size in the extruder. Next... Okay, forgot the part about steam expanding. Seriouly, it's a hot dog and a bs question. diam expands / casing is weaker on diam horizontally because of the pressure of the casing around the hotdog, and less pressure horizontally Show more responses It splits in a seamlike fashion across the middle line along the top, lengthwise in horizontal position, due to the fact that steam rises to the top. I splits length wise because how the meat expands inside the skin. vertical depends on the manufacturer of the hot dog, its SKU, the price you paid for it, what day you bought it, from which store you bought it, how long you have had it, how quickly you have heated it, and other factors too many to list. Along the seAmsterdam of the dog. It splits lengthwise, because the circumference is far less than the length, so there is less ability to stretch in that direction. The skin of the hot dog can more easily stretch lengthwise. Hence it splits perpendicular to the circumference. Hot dogs split horizontally down the middle and occurs due to operator error as a result of cooking at too high of a temperature. In both directions if it gets really excited. The middle because there is room to stretch. The ends aren't. Middle Show more responses It splits lengthwise because - as its circumference expands - it gets thicker. The skin casing becomes tight as the water molecules heat up, and eventually splits from one end to the other. The pressure on the inside of the hotdog is evenly distributed from one end to the other. That is why the split occurs lengthwise, as opposed to across the width of the hotdog. outward lengthwise, because it plumps and is weaker on the long side than it is on circumference I eat Ball Park franks. They split first on the ends because that is their weakest point as the meat inside seeks to expand. It splits lengthwise, which is the path of least resistance, similar to a fault line in the earths crust. In my experience I have never seen it split any other way than down the middle. Why? No idea and I don't care. when a weiner gets hot and expands it spits in the direction of the bun Along the seam. What is causing the hot dog to expand? Heat? Decrease in atmospheric pressure? Injection of foreign material? More information is required to adequately answer the question. I'd say it splits lengthwise because that's what I've observed in microwaved hot dogs. I'm guessing that's basically with the grain. In other words, a hot dog is weaker that way. Show more responses axially It splits where it is the hottest and also where the casing is the weakest. Why do you ask? Thinking the hotdog is me, expand outward. I would like to expand all over the company. I would want to know the ups and down, also inputs and outputs of the company. It expands in width not length. At top heat rises Hunh. A badly phrased question. From an Engineering company. Who would have thunk ? As far as hot dogs go, it varies, by brand and chance; My initial flash was that a hot dog splits lengthwise because the two endpoints are weak where the casing is sealed; I recall an image of a hot dog split along the long axis starting from one of the endpoints (Imagine a tea kettle). But this is obviously much more complex. Reading the various previous answers has opened my vision to such an explosion of factors : (a) Direction of the cylinder relative to the heat -- heat rises, steam pressure is upward? -- wow; (b) Lengthwise seam exists in some brands, who knew? -- (c) the casing is stronger depending on direction of extrusion of casing material at manufacture, somewhat like some woven fabrics, okay, that's news -- Hunh. So really we'd have to experiment with the subject hot dog(s) under relevant conditions. There isn't any other reliable method. There are too many factors that could go into the strengths and weaknesses of a hot dog design. Nice question I guess, or dumb question maybe, depending on what the question was designed to elicit. If SpaceX is founding its designs on hot dog dynamics though, that's a disappointment. It splits along the length of the dog. It expands in all directions. Why are you eating hot dogs? Get some proper vittles. It splits outwards if cut down the middle. It depends WHY it expands and HOW you define expansion. Why 'how expansion is defined' matters: [Most] People are answering based on heating/cooking which would increase the volume of the entire hot dog which = expansion. If expansion, however, can include individual cross-sectional expansion rather than 'single unit' expansion, then as you bite into a [presumably cooked] hot dog, the internal structural change within the hot dog from the pressure of your teeth will cause cylindrical cross sections near the bite point to expand. It will split [presumably] horizontally with your teeth. Even if you assume the hot dog would have already split during the heating process, it would expand in this way again when you bite and would split again when you bit it. Also, how do you define 'split'? Won't go there now.... Why 'why it expands' matters: Based on the above, the hot dog can be cooked or uncooked, so to expand it in order to split it, you can cook it, bite it, or smash it with a hammer and all of those will result in different 'directions of split' and also, maybe even multiple splits. So my concise answer: It depends how it expands and how you define expansion, and even how you define split (split the surface vs split into separate pieces). To be [arguably] persnickety, when a hot dog expands it will only split once a critical point is hit, so it could also expand and NOT split. Show more responses Out as i haw seen they can do when boiled and skin pop. The real answer is why would you want to work for someone that ask such stupid questions put it in a microwave... it doesn't split... it explodes. Assuming the insides of the hotdog are more-or-less fluid, then pressure will be exerted equally normal to the casing at every point on the surface. However, the force which pulls the casing apart is related to the gradient of the normal vector and not just the normal itself. As the curvature is much sharper axially, the gradient will be higher for it to split lengthwise. This will cause a seam that grows quickly, resulting in long, length-wise splits. However, we should also take into account any anisotropic structures in the casing itself. If the casing is made out of a bunch of long strips, it will split lengthwise even easier. If it is made out of a bunch of stacked rings, it will have hard time doing this and it probably wont split length-wise quite as easily. This is a question of intestinal physiology, and I suppose it depends on how the muscles work. It splits lengthwise because each hot dog link is made by twisting the casing at the ends during the manufacturing process. This makes the ends more rigid and structurally sound than the rest of the casing. It splits lengthwise at top - it is the casing that splits as someone else mentioned, but I think that what is responsible for the split are the juices that begin to boil inside the hotdog, and which will naturally rise to the top creating the pressure to split it. Once the pressure is released, it is done expanding. The Educated Answer: Most people prepare a hotdog by pricking it, or slicing it lengthwise to reduce the circumferential and axial expansion due to the gases released from the heat source and the ingredients of the hot dog.. Most heat sources are due to: grilling, boiling, or microwaving. The expansion is also affected by the contents of the hot dog: chicken, turkey, beef or combination. I have noticed significantly less expansion of gaseous expansion circumferential and axial in beef kosher products with less fillers , sodium and preservatives. It is most likely to see a lengthwise gaseous expansion in most hot dogs depending on the cooking temperature, method,preparation and ingredients of the hot dog. Lengthwise, because it has to expand in the direction that gives it the most opportunity for growth. This question essentially asks you whether or not you understand stresses in pressure vessels and the most common failure mode involved. The failure will be propagating lengthwise (unless "meat product" is not homogeneous) when far away from ends. Show more responses It would split lengthwise because the ends are scrunched up and thicker than the middle part. From the middle outward & lengthwise, following the seams of the hotdog being that it is the weakest part & mostly in the middle because the ends are compressed. I'm a vegetarian so I'm not interested in hot dogs! Besides to its entirety of expansion and direction, it always depends when it's cooked or uncooked,also it depends on its longevity of cooking and when this was processed as well. How about inviting me to the company barbecue? We can test several different brands and hypothesize in real time. The hotdog splits in a vertical pattern first and then horizontally if too much heat or pressure is applied. The outer layer splits first which is the vertical aspect of the reaction. Then the split extends horizontally on the hot dog resulting in the cracked skin and loss of juices. It splits along the middle, because that's easiest for it, physically. Hoop stress > Longitudinal stress Similar to 99% of question asked during interview. They are not interested in the answer unless specified, they just want to see how is your reaction and response to any difficult questions. So re-phrase the question as how do you response to stress under extreme situation. Diversion & delay in response are one of the best technique as not to commit anything under duress situation that might cost company suffer a huge lost. vertical Show more responses I agree with #1 that [the strain induced by] hoop stress vs. axial stress is the main cause of the longitudinal hot dog failure. As stated by others above, possible additional contributions include localized stress concentrations, weak seams from the hot dog production process, pre-cut slits, and (as stated by "plastic bag manufacturer") machine-direction vs. non-machine-direction effects in the hot dog casing. Given the assumed context of thermal expansion in the process of grilling a hot dog, this is the correct answer. When asked this question in an interview, the actual correct response is to first clarify the assumptions behind the question by asking a critical follow-up question or two: By what means is the hot dog being induced to expand? The default assumption for most people would be thermal expansion due to heating on a grill, but the situation is quite different if the hot dog is expanding due to the Poisson effect from someone squeezing it. What is the surrounding environment relative to the hot dog? The default assumption is that the hot dog is being heated on a grill, but the situation is quite different if the hot dog is expanding in a pressure chamber under controlled conditions. It expands sideways and splits from the center, long ways. This may be the most simple answer to the question but the actual answer will depend on the relevance of this question to the job applied for. Ends first, lengthwise next and if cooked long enough the possibility across the center of the hot dog. As the meat expands due to cooking, tension on the casing will cause the cut ends to split first due to them being the weakest points. This allows the meat inside to easily expand out the ends reducing tension across the width of the casing . However tension lengthwise continues until the casing splits. Cooked long enough the hot dog can split across the hot dog as well. The method in which the hot dog is cooked also plays a factor. Microwave or open flame will cause more tension on the casing due to moisture leaving the casing. Boiling will allow the casing to be more pliable allowing the casing to expand more before a split occurs. equally in all directions upwards, the portion of the hot dog that is opposite of the side that is laying on the heat source, because pressure builds and that is the side of least resistance. First it depends on how it’s presented, horizontally or vertically. If it’s presented horizontally then you could answer the same way and Visa versa. You could simply answer “longways” too. The “and why” part of the question could be answered technically if you know your physics, or simply state, “Because you can’t teach an old dog new tricks...” to hopefully put a smile on their face. One or more comments have been removed. |
Research Analyst at Aksia was asked...
If you had a machine that produced $100 dollars for life what would you be willing to pay for it today? 75 AnswersNo more than 100$ for 100$ for life Depends entirely on how fast the machine produces the bills. Current price will be determined by estimated future yield which isn't defined here. I would pay nothing for it. $100 dollars in a life? I can make that in one day... Show more responses I would not pay anything. Only the Federal Reserve can legally produce $100 dollar bills. I would pay for everyone in my family to get a four year college degree and then their master's degree. I would cover the cost of books, supplies and dorm rooms or apartments near campus. I would buy a new car if they held a GPA over 2.00. I would buy them nice clothes and shoes so that they looked nice while attending class. I would also buy them a new coat to keep warm and an umbrella to keep the rain off of their heads. I would value education more than I can afford to do so now. I already have the machine, why would I pay for it?? Nothing, I already have it. absolutely nothing. $100 dollars for a life time is ridiculous What is "$100 dollars"? That reads as "one hundred dollars dollars." What's a dollar dollar? Is it double value? Say $100 or one hundred dollars; do not say both, in order to avoid being redundant. As much as I could lay my hands on. It could all be repaid instantly I would pay nothing for a machine that produced dollars at a cost of $100 each. What would you offer me to haul the machine away for you? Nothing. Fiat money is either rendered worthless by uncontrollable inflation, or $100 over a lifetime is not worth much. Firstly, if it made $100 for life, it wouldn't be worth anything. Secondly, if it's making money, it's illegal. Show more responses 10 to 20 years is what I would have to pay for a machine like that, life is only for murder with special circumstances. Does such a machine exist? What? No? Then why are you asking this question ... MORON. No. Not worth it. Why is this considered a top 10 odd interview question? It's a basic accounting question that applies to any applicant at a financial institution. Let's assume the proper phrasing of the question is "If you had a machine that produced a free $100 dollars per year for life, what would you be willing to pay for it today?" Given that Aksia is a financial firm, they're basically asking what is the present value of a perpetuity with a $100 annual payment. PV=pmt/r where: PV=PResent value PMT= payment per period r= discount rate Given current US fed reserve discount rate is 0.75%, the Present value of such a device would be $13,333.33 Answer varies obviously if discount rate changes or if proper phrasing was meant to be $100 for a different time period. Nothing. According to the question, I already have it. I am assuming the question is stated incorrectly. It should read: "$100 per year for life". Let's say I expect to live another 50 years. Then in nominal terms, the machine produces $50,000. My time-preference rate is 4 percent per year and let's suppose that the Federal Reserve does a good job of keeping inflation in their target range and averages 2 percent. Then the total discount rate is 6 percent per year. In present-value terms, the machine produces $100 this year, $94.34 next year [e.g. $100/((1+r)^n) ], $89.00 in two years, etc, for a sum of $1,676.19. This is the present value of the machine and I would pay this much or less to own the machine. Now suppose that the machine actually produces real $100 dollars (so increases the amount according to inflation). In that case the discount rate is 4 percent and I would pay a maximum of $2,248.22. Obviously I'm using a calculator (Excel) to make these calculations, but the idea is the same. What are talking here? $100 per day? Week? Month? Only once in a lifetime? Oh, and does the IRS have to know about this? Be specific. Does it produce $100 every minute of your life or just once in your life? Will it produce enough for me to make bail and hire a lawyer after I'm arrested for counterfeiting? Press NPV button on calculator. 1) As many have stated, this is illegal. But putting that aside, 2) The way the question is presently phrased, technically "you" already have the machine so why would you pay anything? And also, there is a lack of clarity regarding the frequency of the machine's production of $100 - daily, weekly, monthly, etc? The output frequency changes the entire question (because if you only receive $100 once well why bother wasting x% of that $100 by paying for the machine itself?), 3) But assuming this machine provides $100 more than once, does it print a single $100 bill or 100 $1 bills (or other denominations)? Because based on that, what are the operating costs of this machine? How much does the paper cost? How much does the ink cost? How much does it cost to maintain this machine? The interviewer may or may not answer those questions depending on the interview style and format so I'd just go with "It's an illegal practice and I'd never do something so horrific and undermine the honest principles this country was founded upon!" :P Show more responses Somebody has been watching too much Twilight Zone. Whoever formulated this question does not speak good English. The question has no real discernible meaning. They should be fired and a qualified person should be given the job. I will pay with a blank cheque; the seller puts whatever he/she wants; I just have to print it & deposit to my bank account. I assume I can print the $100 dollars unlimited number of times a day. I would tell the manufacturer that I can't pay for the machine now but I'm willing to pay them $1,000 per day for the rest of my life. That'll come up to just 10 pcs for them each day and I'll print another 100 pcs for myself each day. That'll be $10,000 for me everyday! Please define what $100 dollars is. Is it $100 or 100 dollars? USD? When does it produce this undefined amount? For my life or its? What are its operating and disposal costs? Is that in $ dollars also? Where do you spend $ dollars? There is nothing to base an answer on. Life is worth more than a hundred dollars! If I already have it, to whom would I pay anything and why? Silliness aside, you would need to know at least: Payment frequency Your life expectancy Some discount yield (subjective or otherwise) Then it's a matter of PV annuity, or PV perpetuity (if you are allowed to pass on payments after death) It produces $100 for each life it takes? Probably evil, would avoid. Show more responses I already have it so I don't need to pay for it. But if I didn't have it and had to buy its details as mentioned above need to be answered such as $100 a day month year number of times u can take it out. But you are basically putting your own money into a machine to give it back to you when you need it, sort of like an atm Nothing, I already have it! The wording makes no sense. Saving money by using a hybrid car, LED bulbs, efficient appliances and HVAC, or solar is easier to calculate and pays way more than, $100 a year. With that 100$, i will buy another such machine. Now i have 2 machines, each of them generates 100$, which is 200$. With these 200$, i will buy another 2 such machines, now i have 4 machines. I will buy another 4 machines, with 400$, so on.. I will multiply so on, until i have enough money to solve everyone's problems of their lifetime !!! I would use it to buy a truckload of English usage manuals and distribute them to people who insist on uninformed mash-ups like "$100 dollars". Are they anything like "ATM machines" or "square acres"? Charity I'd be really worried that if it put out $100 on the first day, I'd be dead the next day. Show more responses Honestly most people here haven't truly read the question. If I already had a machine that would make $100 bills for the rest of my life, why would I have to pay any money at all to get it? The majority of ppl answering this question, haven't even comprehended the question properly. Seriously. I would pay $6,500 for it. I plan to live to about 85 so I'd set up a payment plan and pay it off with the machine 1 hundred every year i will not pay anything. $100 is the least I could get for free $100. That's all it's worth if it pays that for life. In the first place this is a trick question. It wasn't stated that the machine would produce 100 dollar bills. It stated 100 dollars for life. In the second place it would be counterfeiting so possession of this machine could get you a prison sentence, and no amount of money is worth loosing years of your life. I've noticed most of these questions are trick questions like how many people flew people don't fly plane's and birds do. One dollar. Why? Because I know a value when I see one. My deal, my value. Secure that one and you have it made. Show more responses Anything less than 100$ I'd want at least a 500% profit so $20 at most. I'd pay someone to stop coming up with these stupid questions. Makes me not want to work for anyone. No wonder so many jobs go unfilled. Who wants to work with or for people that start off our first meeting with stupid head games. Chances are that all the job involves is pushing buttons, referring to manuals, taking calls, reading emails and forwarding and escalating issues to other employees. If I wanted to solve riddles, I'd apply to the Riddle Factory where such questions would be relevant. $100USD Dollars for life Well, if I had the machine, as the question indicates, then I would not have to pay for it if I already owned the machine, which the beginning of the questions indicates. I'd leave the interview with a company that produces such an incredibly poorly-worded question. "$100 dollars?" So "one hundred dollars dollars?" Ok. Also, it doesn't specify any sort of rate at which it produces money. It doesn't say if it produces $100 per day, nor does it say if it produces X amount of dollars per day until it reaches $100 in a lifetime. As it is written, this question makes no sense, and the writer of it should be ashamed. I wouldn't pay anything for it today, or the day I got it. What are "dollars for life" anyway? And why do they cost $100? I want to earn money from my work not from machine Who is life, and why would I want to pay for his/her machine? I would pay up to $50 for it. Make 60,70,80 or $90 then sell it for $50+ Show more responses Why would I pay for something that I already have? I would ask first the frequency of distribution, if that's daily/weekly/monthly etc. I would say that I have a policy of never buying something without testing it first and then when I was stood in front of the machine I would say how much do you want for it? Money is no object! :) I would first ask how easy is it to reproduce the machine. If it is less than $100 to reproduce the machine, I would just recreate as many machines as possible and earn the difference. Then I will do some math to calculate the optimized price to pay for a desired annual income. And maybe in the future, invest some of the income to reduce production cost, reuse of old machines, etc. I think the question has probably been misquoted . Probably meant "How much would you pay for a machine giving you $100 per year for the rest of your life ". Its basically a PV of future cashflows ; simply if into perpetuity : = coupon / discount rate. The coupon being the $100 . The Discount rate is the rate of the prevailing interest rate - usually take the yield of 30 yr UST here (say 3% for arguments sake) and add in a bit more if you fell there is any risk in holding the machine. (like some fool with a loaded gun trying to steal it of you) ie 100/ 0.03 = $3,333 . This will be the maximum one should pay. I'll pay for another machine which produce money in such an easy way for sure. Nothing. It clearly is a scam Learn to write a coherent question and then get back to me. Show more responses 1) why would I pay for a machine I already own 2) $100 for life is just a one time payment of $100 One or more comments have been removed. |
25 horses, 5 race tracks. How many races you have to run to select top 5 horses. 60 Answers7 Races First race all 25 in groups for 5 to figure out top 3 in each group. (15 left) Call them A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3 Here A1 is faster than A2. A2 is faster than A3. The same applies for the rest of the groups. Now race A1, B1 C1 D1 E1 Lets say the order of the horses according to ranking was A1, B1,C1,D1,E1 So A1 is No 1 Now A1 is faster than B1 and B1 is faster than C1. So we can get rid of the entire D and E groups 9 horses remaining Also A1 is faster than B1 and B1 is faster than C1. So we can get rid of C2 and C3 7 horses remaining A1 is faster than B1. B1 is faster than B2. get rid of B3 6 horses remaining Out of these, We know A1 is the fastest. So now race A2, A3, B1,B2,C1 to figure out No2 and No3 positions One race is the answer Why 1? Show more responses because, the top 5 are the fastest right ? so you run one race and take the best five times. It's a trick question to see if you'll make a mountain out of a mole hill. The first question to ask to clarify is: how many horses can fit in one track? Technically, one track can be used for all 25 horses, but this is too crowded for an efficient race. Let's say track size is not a factor, then, you need 1 race. Without overcomplicating things - the answer is 1 race. All other answers are based on dissecting the problem into, imo, unnecessary details. There is no mention of the capacity of each race track or number of gates available etc. Even if there are less than 25 gates available at a track, I'd say fetch enough number of gates so you can find the outcome with single race. This isn't a trick question. Normally this question has a track limit of 5 horses. I agree with Anuradha's answer. Anuradha's solution still has problems. (Even if we go with Anuradha's assumptions that you can only race one horse per track, and also assuming that we don't have a stopwatch and must compare horses placing positions) What if the fastest five horses are A1, B1, C1, D1, and E1 ? In Anuradha's second step, he elminates two of the fastest horses (D1 and E1) . He's assuming that A2, B2, or some of the other horses from the other heats are faster, but he hasn't actually tested to see if that is true. @anuradha: I think there are 2 problems: 1. How do you know A4 is not faster than B1 .... Suppose A1,A2,A3,A4,A5 are the fastest. 2. Using the logic posted we can get the answer in 6 races, 5 races for finding A1,B1,C1,D1,E1 and 1 race to find the ranking. Assuming 1 track for 1 horse then it need 5 races to select top 5 horses. Each race for 5 horses and count the time for each race. The top 5 fastest horses are the top 5 horses. 2 max: top 4 from first race, if 5th is a tie, then two, or else just one. Show more responses You guys are not doing CS! 10 runs is my answer. 1. randomize 5 groups, each of 5 horses 2. rank them within each group, I will use Anuradha's notation (5 races) 3. pick the best of each group, race to figure the 1st place, call it A1 (1 race) It should be clear, it wins all times, every one lost once. 4. remove it. substitute 2nd best in. repeat 3 (in my eg. A2,B1,C1,D1,E1) now you have second place. keep going, you get the first 5 and ranking! So, 5+5=10 races in total. The answer is 9. Assuming: - There's no time measuring (stopwatch), just relative places. - The horses perform consistently. - A maximum of 5 horses per race. First we need 5 races (A to E) to get relative scores for all 25 horses. Let's take a worst scenario: the list was already ordered (A1 fastest and E5 slowest), so race A contained the top 5. The 6th race would be the winners of the 5 races (A1, B1, C1, D1, E1), and would give A1 as the fastest of all. This would also mean that some horses can be excluded (only 4 more places to fill): B5 C4, C5 D3, D4, D5 E2, E3, E4, E5 For the 7th race, A2 would replace A1, and A2 would be appointed as the runner-up (of all). We also can exclude some more (only 3 more places to fill): B4 C3 D2 E1 For the 8th race, A3 would replace A2, but as E1 has been excluded, we got a vacancy. Let's add C2 for worst case scenario. The winner would be A3, and we can exclude more horses (only 2 more places to fill): B3 C2 D1 At this point there're only 5 horses who have not yet been classified or excluded, so the winner and runner-up of the 9th race would give 4th and 5th overall. To Patrick: I had to slightly modify your almost correct solution and got 10 races. You has an excellent idea to remove subset of candidates after final selection of the i-th horse, and my modifications are not an essential ones. Actually it is necessary to fix two small things: 1. get rid off the suggestion about the ordered list To avoid this suggestion I would rename groups from the fastest A to the slowest E with order by 1st horse after final selection of the i-th horse. So after race #6 we have order: A1, B1, C1, D1, E1, but there is no suggestion that all A are better than all B and so on. And after every race #7, #8, #9 we have to rename groups. With this modification your method works fine until final selection of horse #5 2. Without this suggestion you need race #10 to find the last 5th horse. To test both methods you may use 25 randomly selected integers and select the 5 minimal or maximum ones. I can't prove that 10 is the minimum number of required tests, but it looks very convincing. Thank you. Maybe Ziqi Dai also meant the same method, but it was difficult to understand without a couple of details. To Leo, I checked some random sequences with a spreadsheet, and all the remaining (not placed or excluded) horses run in the 9th race, so there's no need for a 10th race. Sometimes there're only 4 or even 3 horses left in the 9th race. I didn't check every possibility, but there was no indication a 10th was ever necessary. This answer is 1 regardless of track. If it can fit all 25 then you only need one race. If you make the assumption that only 5 can race at a time then you put 5 on each track and start the race on each track at the same time. Either way, the fastest 5 horses win. Using complex math on this is pointless since a pragmatic approach is available. @ Anuradha -- this is a great solution to the wrong problem! The classic problem is to find the THREE fastest, and that's what your solution is. However, the question posed in this thread is (likely incorrectly quoted from the interview) to find the FIVE fastest of the group. why would you make an assumption that makes this more difficult? I mean given this is an IQ test of sorts, making it harder than stated is........ Attention to details, Anuradha! You're answer would've been correct had the question asked for top 3 horses. This is like merge sort. In the first 5 races (when we run 5 horses per race) we get 5 sorted sublists. Each sub-list contains 5 horses sorted by their relative speeds within that sub-list. Now just merge these 5 sorted sublists to get your answer. Every "merge" means 1 race to find the fastest horse from the front of these 5 sorted sublists. To summarize: 1. Creating 5 sorted sublists of 5 horses each = 5 races 2. Getting fastest horse = 1st round of merge = 1 race 3. Getting second horse = 2nd round of merge = 1 more race 4. and so on... Show more responses Why is this harder than having five races with five horses, recording the times of each horse, and sorting the list of observed values to get the top 3? it is seven race For 3 fastest horses no doubt the answer is 7 but for 5 fastest I worked it out to be 9. And pretty sure too 1 only if you can do good flow control and horse will start with different time / space out based on best estimate. It is kind of feeding the horse to the gate. Simply 5 races only, i.e. 5 horses in 5 race tracks per race. Finally, select top 5 horses by best fastest timing at finish line after 5 races. As a BA, the answer I would give is: We don't yet have enough information to provide a solution. There is critical information missing that could affect the solution. And flushing out all the critical factors of the problem are one of the most important responsibilities of the BA. As a BA, it is very important to NOT start building solutions to problems that you don't yet fully understand. Actually, it's one of the biggest mistakes inexperienced BAs make: solving the wrong problem because they rushed to solutions and did not fully investigate and understand the problem! Once they're committed to the wrong problem, they end up with a beautifully done inadequate solution. 5 races, because you can fit 5 horses in 1 race and there are 25 horses. You just note the timings of each horse and compare them after all the races are done. That way, you won't have to arrange more than 5 races and stopwatch is a common tool for every phone!!! First of all guys, you guys are making critical mistakes. You assume we have a timer but that would make it too easy, right? I'm just assuming that 5 horses a track and no timer. So I have two potential solutions: 1. You could always just race them all at the same time on 5 tracks...but that'll be too easy. Then there's my solution: --------------------------------------- First, group all the horses into five groups of five. It doesn't matter which horse goes into which group, it honestly doesn't matter. Then, race all the horse in one groups against each other. It helps to name them A-E(Group name) and the number place they came in (ex: 1st place a group would be A1). So now we have A1 to A5 and B1 to B5 and so on until E5. Then race all the 1s against each other. You should get something like A1, C1, B1, E1, D1 and then you should arrange their groups in collumns from left to right based on that race (1st on left, 5th on right). Here is where it starts getting complicated, so I will just assume it is A1 fastest in that race and E1 slowest and so on just for simplicity. So so far this is 6 races. Get E1 and race them against D2-5. You can ignore E2-5 because no matter what they can never be in the top 5 because A1, B1, C1, and D1 is the fastest of their groups and you know that the fastest five can come from those groups but not E2-5 because E1 is the slowest of the 1's and there's no way those horses can be in the top five. SO just throw them away. Since there are four spots available max you race the slowest four of the D which happens to be D2-5. Get the place numbers and arrange them. EX: D1 D2 E1 D3 D4 --- D5 E1 can be anywhere based on E1's speed but you immediately want to discard the D5. You might be saying, "Hey, why can't we discard anything below D2 since A1 and B1 and C1 are all faster than D1 and E1?" But...you could be unlucky and have A2-5 and B2-5 and C2-5 be all dirt slow. So just keep them for now. So the rule now is to keep all the horses that are in the top 5 and discard the rest until you get to A, when you know, "Oh, A1 has to be the fastest horse." 7 races. So we have our list above just for examples. Next you want to race D4 against C5 to see if our group can EVEN pass the C's, cause we might get super unlucky and see that C1-5 is faster than D1...if D4 loses in 5th place we just throw this all away and our new list becomes: C1 C2 C3 C4 C5 But otherwise we'll just assume that our D4 luckily got to 3rd place. It becomes: C1 - Our group can go anywhere inside the lines so race them again C2 minus D4 since its in 3rd place to see where they go. Let's assume we got SUPER lucky and C2 became last place in that race. - C3 C4 C5 8 races. We raced this group at that time and left out D4 because D4's fate was already decided: D1 D2 E1 D3 C2 It turnes into this: C1 - D1 D2 The single dash indicates where our new group was spliced, ofc the triple E1 dash lines show where the cutoff is, so discard everything under the line. D3 -(---) C2 D4 C5 9 races. Kay, so now we have the five fastest in the E, D, and C groups. Now continue with the B groups and let's say that our D3 passed B4 only... It becomes: B1 B2 - With our top five so far group racing against B4 B3 - D3 B4 B5 10 races. And then with another race you get this with lets say...only our top two making it through with B3 getting ironically first place: B1 B2 B3 C1 D1 --- D2 E1 Yep...E1 got kicked out. Aw well. SO NOW our to group D3 is the group above the triple dash marks. B4 B5 11 races. Now we're up against the A group. Assume that they are all wusses and that our D2 beat all of them except for A1. So we got lucky and the new lineup is this: A1 B1 B2 B3 C1 --- It was inevitable that A1 would become first but now this is how you do it! D1 D2 A2 A3 A4 A5 12 races/13 if our B1 didn't beat all of the other A's on the first try. Conclusion: 5 races (initial races)+1 race(for initial group leader ranking)+1 races(E-->D)+2 races(D-->C)+2 races(C-->B)+2 races(B-->A) =13 races/10 races if you get insanely lucky and have all our previous groups beat all the #2s of the groups. Show more responses Oh nvm ignore that it was all crushed into one place with no spaces...I'll stick a better one later. The answer is 8 races assuming that you can fit only one horse on one track. Divide the group into 5 horses. In five races we can find the 5 fastest horses in the 5 groups. The 6th race is among the top 5 fastest horses. The top three fastest horses in the 6th race are faster than the horses in the group of the 4th and the 5th fastest horse. The horse that came third in the 6th race may or may not be faster that the horses in the group of the fastest and the send fastest horse. To determine that, we hold 2 more races. So 6+2 = 8 races is what I think is the correct answer ****************ANSWER -- 3 RACES.************* Given : 25 horses, 5 tracks Assuming : 5 horses to a track Five groups of 5 horses for each track (5 tracks, tracks ABCDE, each track with horses labeled "X#") First race allows you to find out the top 5 horses (however this isnt the true top 5) [A1,B1,C1,D1,E1] Second race the top horse is discovered (lets say its A1) Now you have to find out if B1 is better than A2, B2 is better than C1, etc. (each predecessor of each winner) We know that B1 is better than [(CDE)1] Third race we have A2,B1,B2,C1,D1 Conclusion : If both of the predecessors are NOT last place, then we have our answer at race 3 by taking the top 4 to be the bottom 4 of the top 5. Otherwise, repeat this process one more time to find the 5th placed horse. Ans is 14 Patrick gets the right number but his logic is not the most efficient. Initially follow the standard procedure and race 5 groups of 5 (races 1-5) and then race the winners (race 6). The winner is the fastest overall. Now name the groups with the horses in the winners group A1,A2,A3,A4,A5, the first four horses in the runner-ups group B2,B3,B4,B5, the first three horses in the third group C3, C4, C5 down to E5 for the winning horse from the slowest group. The numbers indicate the highest possible position a horse could attain (i.e. C4 is definitely slower than C3, B2 and A1 but may be slower than several others). The rest of the horses are eliminated and so don't need to be labeled. Now race (race 7) the horses labeled 2 or 3 (there are 5 - A2,A3, B2, B3, C3) The top two are second and third overall and the loser is eliminated. You can eliminate a total of 5 or 6 horses - those known to be slower than the fourth or fifth place finisher and any two positions slower than the third place finisher in race 7. There aren't that many options here so you can just inventory all the possibilities if you wish. At this point you have a maximum of 7 horses for the remaining two positions You just race any 5 (race 8) to eliminate at least 3 and race the rest (race 9) to get the fourth and fifth fastest overall. If you choose all the potential fourth place finishers in race you may not need race 9. As noted elsewhere Anuradha gets the right answer to a different question than the one asked. I have stepped among a lot of smart folks here. Maybe even several analyst's. So here it goes. I am looking at this as who the question is coming from and for what. A Business Operations Analyst for Google. Everyone is so focused on answering the question based on assumptions. Ie. That every horse and rider are at their best condition and that every track is exactly the same. That you are running all the races in the same day and that horses, riders, and tracks dont change. Horses and riders get tired, not to mention the possibility of injuries, and track conditions can change so forth and so on. So to me. In order to (analyze) to get the best possible answer. I would start by asking questions and not assuming. The more information you have leads to the best possible answer. Guys think as a programmer, (mind my language and grammar) algorithm would be of these steps ->group the horses (5 horses each) -> get the best runner of each group (five races) -> store the horses in winning orders to five stacks (a,b,c,d,e) ->until the top 5 are found, repeat ->race all the 5 on top of the stack ->pop the winner from its stack and store in top five that all, now lets see how it will work. lets have 5 horses per group and group them as a,b,c,d,e. 1=a1 a2 a3 a4 a5 a1 (winner) 2=b1 b2 b3 b4 b5 b1 3=c1 c2 c3 c4 c5 c1 4=d1 d2 d3 d4 d5 d1 5=e1 e2 e3 e4 e5 e1 now we have best of each group, let them race 6=a1 b1 c1 d1 e1 a1 for every winner get the best candidate for the group, and race again, until top 5 are found 7=a2 b1 c1 d1 e1 a2 8=a3 b1 c1 d1 e1 a3 9=a4 b1 c1 d1 e1 a4 10.a5 b1 c1 d1 e1 a5 for any scenario just get the so the ten races. its like finding five highest numbers from 25 numbers, when you can only compare 5 at a time. Answer- 7 Races Give names H1....H25 Divide into 5 groups as 5 can run at a time H1, H6 , H11, H16, H21 are fastest in there groups Now we can remove last two from each groups as we need fastest 3 now we left with 15 horses 6th Race- between all fastest in groups After 6th Race H1> H6 > H11> H16,>H21 Now H21 was fastest in group but is 5th in 6th race so H22 and H23 who already slower than H21 can't be in first 3 Same thing applied for H17 and H18 as H16 was fastest in group race but 4th in 6th race Same applied for H12 and H13 as H11 was fastest in group race but 3rd in 6th race Already three horses are faster than H16 and H21 so they both also can be rules out Now we left with H1, H2, H3,H6, H7, H8 and H11 But H8 is slower than H7 and H7 is slower than H6 in group race and we have H1 on top already so H8 also can't be in fastest 3 Now we left with H1, H2, H3,H6, H7, H11 H1 we know is fastest..so lets 7th race between H2, H3, H6, H7 and H11 and find the other two After 7th race you will get fastest 3 Make a group of 5 horses per track and race them all at once. The winners of each track totals to "the top 5" ! The answer is 7. Assuming: - There's no time measuring (stopwatch), just relative places. - The horses perform consistently. - A maximum of 5 horses per race. Race r1: A1 A2 A3 A4 A5 (By order of winners) Race r2: B1 B2 B3 B4 B5 Race r3: C1 C2 C3 C4 C5 Race r4: D1 D2 D3 D4 D5 Race r5: E1 E2 E3 E4 E5 Race r6: A1 B1 C1 D1 E1 (rename R1 R2 R3 R4 R5) Here comes the tricky race. From race 6, we already know the ranks. As a demonstration, assume R1 = A1, R2 = B1, R3 = C1, R4 = D1, R5 = E1. Then: 1) D1 D2 D3 D4 D5, E1 E2 E3 E4 E5 are out of the contest. Because D1 and E1 are ranked 4th and 5th in a race. 2) C2 C3 C4 C5 are out of the contest. Because two horses already are faster than C1 ==> 3 horses are already faster than C2 C3 C4 C5. 3) B3 B4 B5 are out of the contest. Because one horse already faster than B1 ==> 3 horses already faster than B3 B4 B5. Remaining potential horses are: A1 A2 A3 B1 B2 C1. Here, A1 is redundant because we already know it is fastest. Race 7: A2 A3 B1 B2 C1. The top 2 will be added to A1 to be the fastest three. Show more responses Take a look at previous horse race statistics from horse newspapers. Make your analytic and you have your top 5 horses without any new race :) It will take min 9 race not 7: first five race and find top 3 @ every slot now total race=5; one race between first five, now we find top 3 and take 2nd and 3rd from the race as 1st is declared. race=1; 1 race between 2nd finisher of all slot and take 2 from them , race=1; 1 race between 3rd finisher of all slot and take top finisher, race=1; now we have race between 5 horses(2+2+1) and find 2nd and 3rd; so total no of races= 5+1+1+1+1=9 ; The question is not a good one! Let those horses be marked X1~X25. First divide them into 5 groups X1~X5, X6~X10, .... Assume the first X1, X6, X11, X16, X21 are the fast ones in each group after the first round. Also, X2 > X3, X6>X7, during the first round. We then, during the 2nd round, form the first group with those fastest from each group in the first round. Assuming that X1 > X6 > X11 > X16 > X21 after the first try of the 2nd round, we will then have X2, X3, X6, X7, X11 to form a new group and determine which are fastest 2 in this new group. Nevertheless, during the 2nd try of 2nd round, it can happen that X3 > X2 or X11 > X6. The inference fails all the relation set in the previous try. ( that the X2 > X3 ,after the first round, and X6 > X11, after the first try of 2nd round). It would be better to change all those horses to be boxes carrying different integer value. You want to sort those boxes to find which carrying the largest 3 numbers and, at any time, you can only open 5 boxes to inspect . You cannot mark the content of boxes but you can put down the relation between boxes during the inspection. Then, the number of inspections needed to find boxes with the largest 3 number will be 7# This seems like a simple answer, but I'm pretty sure it's 9. Race 5 groups of 5, then race all the 3rd place, then the winning 3rd races the seconds, then you have either have five firsts and two seconds left or 5 firsts a second and a third left from the original racing groups. Worst case scenario you have one more race of three ones and two twos or two ones and the 2nd and third place to find the fastest there out of that last group. 9 races. So driving down the road I figured out I should have started from the top after the first five races. Then worst case scenario you run three more races to find out the top three. 11111, 21111, 31111. Damn. Six races, six groups. First group will have five horses, the 5 others groups will have only 4 horses. The winner of the first group of 5 horses will pass to the second race, then the winner of the second race will pass to the third race and so on, bay the end of the 6 race you will have top 5 horses. Assuming no limit to how many horses on a track, 1, because the fastest 5 will obviously finish in 1st, second, third, fourth, and fifth place. If there is a limit to the number of horses on a track, then it will take (25/number of horses to a track) rounded up to the next integer races because you can just compare the times of all of the horses and pick out the five fastest times. There are a lot of wrong answers on here or answers for different questions. Answer for top 5 is 8 races. You can find explanations online for this. Show more responses Friends only 5 races are required As mentioned we have 5 tracks and 25 horses. So divide them into set of 5 And winner of each set is fastest so only 5 races are required. If there in no limit on track one race is enough. But if 5 horses can run at a time we need 6 races. 5 races to find top of each group. And additional 1 race among those top 5 horses to find out top 3. answer is 8 suppose we have limit of 5 horses per track and all horses run together on 5 track so *-5 races-* conducting on five track, the top one of each track is selected for further race and now race is conducting between the all top 5 horses which is *-6th race-* and the remaining 4 horses will race again for 2nd and 3rd position hence total no. of race are 5(all five races)+1(6th race)+1(7th race)+1(8th race). So many wrong answers here, first of all the question is about FIVE fastest horses, not three, so all the posts giving 7 as an answer are wrong. All the answers above 9 are also wrong. You need 7 races to find 3 fastest horses, after that by using logic you come down to either 6 or 7 remaining horses as candidates for places 4-5, and you need 2 races to figure out which 2 of them are the fastest. So it's 9 races in total. 6 races: 25 Horses - categorize in 5 races of 5 horses each (if there is a 5 horse limit) merge and sort timings for top 3 of each of the above races. Now race 6--> race 5 horses with top 5 timing --->Winner 5 races in groups of 5 (as not to crowd the track). Placement means nothing only time matters "If your not first your last" One or more comments have been removed. |
You have a birthday cake and have exactly 3 slices to cut it into 8 equal pieces. How do you do it? 39 AnswersCut in half, stack, cut in half, stack, cut in half. All you have to worry about is the 45 degree rotation of one of the 4 pieces after the second cut. Blade can be kept in place, like a paper cutter, as to minimize the margin for error. Slice it horizontally across the middle creating two equal halves top and bottom. Then simple two slice cross from above like normal. Cut each slice into 3 slices. Then eat one of them. Show more responses This is really an easy one. First cut into half across the top, then cut the halves in half also across the top (you now have 4 equal pieces) then cut across the middle = 8. No, Jason and Sharon, you will only wind up with 6 slices. z, too many stacks. You need to cut in half, then make another cut - to get four pieces. NOW you stack these four pieces and make the last third cut - and you get 8 pieces. A rather easy lateral thinking question. Alina's got it. The stacking seems to be the "right" answer. But this is a stupid question. Who stacks cake? The frosting from the bottom slice would meld with the stacked slice, thus making the cake inseparable. I wouldn't get the job because I would swear at the interviewer for asking a dumb question. What good is an answer to this question if it wrecks the cake? Assuming the cake is square: slice 1: cut horizontally to create 2 equal pieces slice 2: cut vertically to create 4 equal pieces slice 3: line up all 4 pieces of cake side by side and cut horizontally to create 8 equal pieces. don't stack, it will ruin the frosting. With a knife Alina would be penalized for not being able to count past 6. But then, she could get a job at another company where they appreciate people who say someone's idea is wrong, then put forth the same idea and take credit for it. A new solution for you: who says slices have to be a straight line and not circular? I would cut a concentric circle in the middle (would have to calculate the radius compared to the whole), and then slice an X with the remaining 2 cuts. It will look like a target. If done correctly the sizes will be the same, it says nothing about having the same shape! I guess it's much easier if you think of two planes: first cut in half, then to quarters. That's the easy part. Now look at the cake from the side, and cut it across... Each quarter is cut into two and all pieces are equal. I would consider the interviewer's emphasis on "equal pieces." While my first thought was to bisect across each of the three dimensions, half of the pieces would end up with less frosting than the other half. Stacking the pieces would result in frosting transfer, which would also screw up some of the pieces. Therefore, I'd go with lining the pieces up and have a large knife on hand for each bisection. Hey, people get crazy about their cake slices. Show more responses I would take a different approach to this. First of all the question asks for the cake to be cut into 8 equal pieces, not 8 identical or 8 of the exact same size. Equal doesn't always have to mean the same, just equivalent to. First I would find out who I am cutting the cake for, if its 4 old ladies and 4 young guys, equal pieces would not mean that they all needed to be the same. If everybody wanted the exact same amount of cake I would figure out a way to give everybody 12.5% of the entire cake volume, but if some wanted larger or smaller pieces I would come up with a way to satisfy each individuals desire. If you know how to cut an arbitrary shape in half, you have the solution. After every round, plan the cuts for each individual piece. Then align them so the proposed cuts are in one straight line. Make a cut. This way you can cut any cake into 2^n equal size pieces with n cuts. An interesting question is, if you start with one connected piece, will you always be able to end up with connected pieces. Think of cutting letter S in the middle, like this: $. You end up with two equal figures that are not connected (or, in other words, with 4 pieces). It's also easy to design a cake that can be split into 8 pieces with one straight cut. I LOVE Andrew's answer! Eat one of the freakin' pieces. That's the corporate way in America anyway. Mike is either a socialist, or works in non-profit, or government. I thought "slice them long ways" but then someone misses out on the freaking frosting, which is the best part. UNLESS, it's a layer cake. First I'd yell at whoever cut the cake incorrectly to start with. He's ruining the party. Then I'd squish the 3 pieces of cake together and re-cut the cake into the required 8 pieces. Boom. This is as easy as pie. Viewing the cake from the top, make 1 cut vertically down the middle of the cake and another horizontally. Viewing the cake from the side, make your third cut horizontally through the middle; QED three slices and 8 pieces of cake with a beamsplitter and prisms it could be done in 1 (with a laser) That's an oddly presented question that is understood 2 ways: - 3 cuts allowed to cut one cake in 8 pieces. Which yields cut in 4 parts, then split those in 1/2 again with the last cut, either by stacking, realigning the slices or making a round cut. - 3 pieces of cake must be re-cut to make 8 equal parts. Which is an impossibility unless one piece is 2/3 smaller than the other 2. That yields different answers like cutting in 3, and eat one to leave 8 pieces. They key to answering any of the brain teaser questions is to ask a few critical questions before even attempting to answer. I would start with: Is the original cake round or square? If square, line up all three pieces and recut to be 4 equal widths of cake. If round, was the original cake cut in 6 or 8 pieces? If 6 pieces you have 1/4 of a cake = easy to redivide into 8 equal slices. If 8 pieces, you have 1/3 of a cake and a little math needs to be applied to create 8 equal slices. First slice a strip off each of the 3 slices to create a fourth slice. then divide each in 1/2 to make 8 equal slices. Cut each slice into 8 small slices. Then give each person 3 small slices. Show more responses There are some posting above who seem to not have a good grasp of numeric's. The answer is not that difficult. First, presume the pieces are not equal size (nothing states they are). Second, presume two pieces are of equal size and the third piece is twice that size. Third, cut vertically (the most usual manner in which to cut cake) the 2 equal pieces (we now have 5 pieces -4 the same size and one larger piece). Fourth, cut the large piece in half, then those two pieces in half again. Fifth, voila, one now has eight equal pieces of cake. Hmmm, I always assumed they meant three knife cuts by the word "slice" Kind of interesting to me that others assumed the cake was given to you in 3 parts as defined by the word "slice" I'd say that you should cut the cake horizontally using the knife as a measuring device to find the exact center of the circle, then cut vertically using the same method, then take each quarter , using the knife as a straight edge, build an alignment diagram that places each set of two quarter pieces point to point along an axis that defines their center lines, and cut all 4 quarter slices with a single cut of the knife ( defined by the word slice ). Put it all in a blender. Pour each of the resultant mixture onto a plate or into a bowl. 1. assemble 3 sliced cakes into a big cake (original shape) 2. cut it half (don't care about the indentation) , you would get 2 piece of cake 3. cut it half again, you would get 4 pieces of cake 4. cut all of them half again, that's finish. To those who think it means you start with 3 pieces ("slices") of cake, READ IT AGAIN. It says (emphasis added): “You have a birthday cake and have exactly 3 slices to cut IT into 8 equal pieces." Get it? "... to cut IT [the cake] into 8 equal pieces."... There is NO WAY it means to cut 3 slices of cake, otherwise it would say "...exactly 3 slices to cut into...". And Mike: "equal doesn't mean same, just equivalent". That's the funniest (and stupidest) thing I've heard all day. Anyway, since 2^3 = 8, you have to stack. Assuming a round cake: Cut (or "slice") 1 creates 2 semi-circles. Stack them. Cut 2 creates 4 quarter-circles. Stack then Cut 3 creates 8 1/8th circles. is the 3 slices equal in size? Great answer and explanation here: http://www.programmerinterview.com/index.php/puzzles/birthday-cake-8-pieces/ Make sure the guest of honor has Blown out candles 1st! (was not specified but hey so were many other things) If the birthday boy/girl is under the age of 10, I am not too sure you want to be messing with their cake!! Cakes come in many different sizes and shapes...ESPECIALLY Birthday Cakes!!! ACK They even come in characters and shapes you can NEVER get into equal pieces but, back to the solution! Will use 2 shapes: Round and Square! Cut 1: Parallel to cutting board and horizontal to create 2 layers of equal depth; Cuts 2 and 3: Perpendicular to cutting board once then rotate 90 degrees and repeat! Now give it to the Kid in the high chair to for quality control/assurance! Like Andrew, I would eat one piece and then cut the 2 in four equal pieces. Remember 1 whole cake, 3 slices with a knife = 8 equal pieces Place cake flat on table. Grab a knife big enough to cut the cake horizontally. 1st cut - Cut the cake horizontally leaving the cake flat on table as if the cake still in one single piece. Now you have 2 cakes instead of one. 2nd and 3rd cut - cut through the cake vertically in the form of a cross. Now you have 8 equal pieces of cake. As if you had cut 2 cakes in 4 pieces each. Show more responses I would state that I only eat cakes in the shape of circle. then 3 equal cuts across the middle. think of it like a pizza... or a pie.... or a pizza pie. Remember these questions are made to have you think outside the box. Not all cakes are square. cut the diagonal portion then centre line of cake Cut each of the 3 slices into 8 equal parts which makes the slice count 3*8 = 24, Divide the 24 between 8 people 3 each. Cut 2 slices into 4 pieces each, cut the third one into 8 pieces. Separate 2 of the 3 slices, dividing 2 slices for 8 people is easy (each slice in 4) Then with the left piece I divide it in 8 and offer one to each person. One more: I could put all the 3 pieces together and re slice into 8 equal pieces. One or more comments have been removed. |
Associate Analyst at Gallup was asked...
What do you think about when you are alone in your car? 35 Answersgetting out I evaluate design work I have done for clients to see if its ideal or needs to be perfected. I try not to be distracted and pay attention to the road. Show more responses I hope I will never have to work at HR and have to create smart unexpected questions. The fact that because no one will hire me I'm so in debt that the only way out for my family would be for me to drive off that overpass so they can cash in on the life insurance since apparently I'm worth more dead than alive. Why the hell is the dipwad in front of me driving 10 miles below the speed limit and not using his turn signal? I don't think about anything - just driving, because I want to come by my car, not by ambulance. My family, the music or programs on the radio, and story ideas. How crazy life would be if I didn't have a driver's license. Driving. "I hope nobody catches me." Show more responses I don't have a car. Which exit do I need to take to get away from Sallie Mae forever? Must thought "would have been better if some one accompanied in car" I just try to concentrate on road while driving. Life is wonderful.live it once by safe driving.The power is in present moment. I wonder what it's like to be a bird I wonder how long it will be before that body in my trunk starts smelling. Sex. Do you want me to go into more detail on that? There's no backseat driver, I am free. about my destiny Show more responses wheres my chauffeur? He ought to be here now.. how i am different frm other and what should i do to make it About the escape plan if someone attacked me when the doors are closed. :P nothing specific nothing I would be thinking about Flo, from Progressive Insurance, and how many ways to get rid of her fat behind. Drop her off at an all you can eat buffet, and call it a night? Make her one of Revlon's makeup models, and watch their stock drop? Her hitting on the plumber, under the cabinet, and wonder who has the bigger plumbers crack? Take her water skiing, and watch her start a Tsunami? Maybe watch her dance to a song like, "I like big butts and I cannot not lie......." by Sir Mix-A-Lot? Gotta Go..............Beep-Beep, Beep-Beep, Beep-Beep (Flo doing Michael Jackson's moon walk)? it depends at present wtz in my mind... and i cant sit more time freely in car. or hearing by the music i wil be in an enjoyable mood it depends... Its depending upon the mood at that moment........ I spend my time driving to plan my day or recap my day depending on the time. I also rehearse conversations I anticipate may be difficult and try to anticipate responses. On really long drives, I'll listen to books on Audible or practice my Portuguese with language tapes. But if the weather's good, and I'm in the right mood, I'll just open the sunroof and blast Hendrix or he Clash and sing along. Ouch! That BUG will never have the GUTS to do that again! Show more responses How would be my next map in Portal 2. I'd be thinking... what next........ what do I need to do next! One or more comments have been removed. |
Business Analyst at Capital One was asked...
You have 3000 bananas at point A, which is 1000 feet removed from point B. You must move as many bananas to point B, but you can only carry 1000 bananas at any time, and traveling 1 feet requires you to eat 1 banana. You can drop off bananas at any point between A and B, and pick them up later. 31 Answers500 bananas 500? How? Could you pls explain? Don't worry it...got it Show more responses Not very clear formulation of problem . I move V (in thousand) bananas on L feets., what I get in point L 1.V-L/V Or 2.(V-L) In first case I spend 1/V moving on 1 feet, but in second energy do not depend of load. Sorry my Error not 1/v but V and not (V-L/V) but (V-V*L) how do you mathematically formulate it? I mean I see how it's 500 from intuition. But, if I were to approach the problem from a purely mathematical way, how would I formulate it? @Shilpa, Could you share what you found? Thanks! It's not 500. Not sure if correct but I calc 666. It should be around 666 bananas. How did you arrive at 666? let us say you move x feet before you drop your bananas and then go back to point A, and let's say you pick 1000 bananas first time. You eat x bananas when you get to point x, and since you need go back to Point A, you will keep another x bananas, then you drop 1000-2x bananas at point X, next time suppose you pick y bananas, when you get to point X, you will eat y-x bananas, and pick former dropped 1000-2x bananas, all these bananas add up to 1000, so we got y-3x+1000=1000, that's y-3x=0. Since we can pick max 1000 bananas, so max y is 1000, x is 333, that is say, we can go as far as 333 feet and then we must go back. since after we get to point b and don't have any more bananas for us to go back to point A, seems 333 is the answer 1000-2x+1000-2x+1000-x=2000 x=200 (first stop point @ 200 feet) 1000-2y+1000-y=1000 y=333 (second stop point @ 533 feet) so you can move 1000-(1000-533)=533 to Point B. Hey ltss587ATgmailDOTcom - I understand how you got to 1000-2x+1000-2x+1000-x (3000-5x) which is effectively number of banas left at Point X. Why did you equate it to 2000? what is the rationale to it? Also how did you get the second equation? 1000-2y+1000-y=1000 Wei - In your equation, though it makes sense, but u didnot go back the third time to pick up the 1000 bananas. At this time, y-3x=0., you still have a 1000 bananas left? Could you please explan. I really appreciate it. Show more responses Hey G It goes like this 1000-2x+1000-2x+1000-x=2000 ( equate it to 2000 because after this u will wont need to go back third time as the maximum u can carry will be over in two time) x=200 (first stop point @ 200 feet) 1000-2y+1000-y=1000 ( equate it to 1000 because after this u will wont need to go back second time as the maximum u can carry will be over once y=333 (second stop point @ 533 feet) so you can move the remaining 1000 distance in (1000-533)=533 to Point B. So 533 bananas left sorry guys its 833 bananas You have to do it a foot at at time to max out You'll have to carry 3 loads until you eat 1000 bananas so 3x = 1000 or x = 333 ft. next you have to carry 2 loads until you eat the next 1000 bananas so 2x = 1000 or x = 500 ft. Now you have exactly 1000 bananas left so only one load fo the remaining 167 feet this gives you 833 bananas left. It is 333 bananas!!! 833 is the maximum left and best explanation. Do it on a number line. Maximize each load and trip. The second step must be 2 trips closest to the goal ending with 1000 bananas because there be one final trip at maximum load the shortest distance. So stops at 333+500+167 gets the most product moved at least cost. Logistics. Technically this isn't really asking anything. Technically you don't have to eat any bananas. If you move them less than a foot each time, you don't have to eat any bananas. 3000, pickup 1000 bananas and walk 1/2 feet, go back get another 1000 and put them @ 1/2 ft location and then get the rest bunch. 833 is correct. Starting at 1000 feet mark. Carry 1000 bananas drop 999 at 999 foot mark. Go back and bring 2*1000 similar way and drop at 999 ft. Now you have 3*999 bananas at 999 feet. Keep doing this until 3*x exceeds 1000. You are now at 666 foot mark with 1998 (3000-334*3) bananas. Now do two trips and drop bananas at each foot traveled. At 177 (833) foot mark you will have 1000 bananas. Now you will loose 177 bananas to end Show more responses Incomplete question but this is making the presumption that the answer required is 'what is the largest number of bananas I get get to point B'? Te answer is 2 You must eat one banana per foot traveled. Whether forward or backwards, a foot is still a foot, so a banana must be eaten. If I take 1000 bananas and drop 334 of them off after 333 feet, I will have just enough to get back to point A. I grab another 1000. I walk 333 feet, arriving with 667 bananas. I pick up 333 of those I left the first time, leaving one, and walk another 333 feet. I reach the 666 feet mark with 667 in my possession. I leave two there. One my way back, I must pick up the remaining banana from the 333 foot mark in order to get back to point A. (665+1) I pick up the remaking thousand. At 666 feet, I have 334 left. I pick up the 2 that I left there, then eat the 334 necessary for me to reach point B. I have 2 left I grab the remaining 1000. I reach the 666 mark with 334 bananas. I take the one I left there. I walk 334 feet, arriving at point B with a single banana Sorry for the fat fingers in my answers! I ate too many bananas and was feeling ill, lol Second apology. Somehow seems as if my response sequence got messed up. I'll try this again! I must eat a banana whether traveling to or from point A. I take 1000 and walk 333 feet I arrive with 667 and leave 334 of them I get back to point A empty handed and grab another 1000. At 333 feet, I take 333 of those I left the first time, resuming my journey with 1000 bananas. I walk to the 666 feet mark arriving with 667. I leave 2 and head back with 665 in my possession. I pick up the one I left at the 333 point mark and make it back to point A empty handed. The last thousand is enough fuel to get me from A to B, so I arrive at point B with the two bananas I had left at 666 feet. The answer is 2 The answer can be googled, they offer a generalized form. For this particular example it is 533 3000. Since the direction of the distance between point A and point B is not specified, we can assume point A is at a higher elevation than point B. We can then either 1) slide the bananas down the slope from point A to point B if we assume the their is a slope with a low enough coefficient of friction, or 2) assume that Point A is directly above point B and simply drop the bananas. Since in both scenarios we do not have to move, there is no need to eat any bananas. (Note: for the second scenario, the problem does not specify the condition the bananas should be in when they arrive at point B.) One or more comments have been removed. |
People Analyst at Google was asked...
how many basketball can you fit in this room 29 AnswersDepends ... should we assume they're inflated? Determine the volume of the room. For example if the room is 10ftx10ftx10ft the volume would be 1000ft cubed. The average mens basketball has a diameter of 25cm. There is approximately 30cm in a foot. Therefor you could fit one inflated basketball in a 1 foot cubed space. Therefor you could fit 1000 inflated basketballs inside a room with a volume of 1000 ft cubed. If we can deflate the basketballs and flatten them down to one inch thick, this would allow us to place 12 flattened basketballs in a 1 foot cubed space. Therefore you could fit 12,000 basketballs in a room with a volume of 1000 ft cubed. It would be more complicated depending on the shape of the room but the process for figuring out the solution would be the same. Actually, what you want to do is figure out the ratio of the diameter of the basketball to a full foot. Basketballs are 10 inches, and 10 is .8333 of 12. Remember that, then figure out the volume of the room by multiplying height by width by length---divide this number by .8333. BAM. So, if its a room of 12 x 10 x 8, you get a volume of 960. Divide by .8333. 1152 is your answer. And this will get it quickly. Show more responses Oops, I didn't tightly pack the basketballs, it would be closer to 1728 inflated basketballs if they were all tightly packed in a 10ftx10ftx10ft room. The correct answer is 1. You only said "basketball" in the question. The actual question is: "How many basketball[s] can you fit in this room” my take on this as follows: 1. Assume the size of the room 10' (ft)x 10' x 10' = 120" (inch) x 120" x 120" = 1728000 cu inch 2. size of the Basketball = 9" (inch) x 9" x 9" = 729 (basketball size ref @ http://www.nba.com/canada/Basketball_U_Game_Court-Canada_Generic_Article-18039.html) 3. 1728000 / 729 = 2370.370 ~ 2370 Basketballs (you cannot add 0.370 balls) Thanks Instead of working it out, anyone ever replied with something along the lines of: I'm sure you don't need to know how many basketballs can fit in here. Why don't we skip this question and save some time? I can put infinite number of basketballs, as long as I can keep enlarging the room. I don't need to do all the math or know the size of the basketball (which can be very small anyways - given the current technology I assume people can make a "basketball" in nano scale? Your are assuming the room is "empty"... What about you and the interviewer and the desk and the chairs and so forth. Those things take up space as well. Think about it. Because a basketball, like most matter in the universe, is comprised of 99.9999999% empty space (really, just a few atoms bound together by a "gluing" force we know little about, separated by empty space), if you manage to make a basketball dense enough, you could fit nearly an infinite number of them into any room. I wouldn't want to be around when that happens, because such things tend to create their own gravitational pulls, and event horizons. Between 1 and 1 million basketballs. You did not specify your error tolerance or accuracy requirement. It is always a mistake to jump to conclusions and provide answers without collecting data and doing the backup analysis. Google it. not too many if we want to adhere to fire code requirements. let's google the local FD and ask for guidance. Show more responses An infinite number if the door is left open. I can fit more basketballs in this room than beach balls, but certainly less than footballs or baseballs. I guess that highest packing efficiency can be achieved if the spheres [or in this case 'basketballs'] are packed in a face centered system. So we can divide the whole room in the number of cubes with each cube's side = diameter of basket ball. Since one cube contains 4 balls so answer should be = 4 * No of cubes. Bingo! I guess Two. Two basketballs will fit in nicely in the room. Or five. Or 10. Or three. All are valid responses. The question doesn't ask what is the maximum number of basketballs that could fit in the room. Think back to high school chemistry. The most efficient packing schemes - hcp and fcc - where spheres nestle in the valley created by the spheres of an adjacent layer, fill almost 3/4 of the available volume - the actual figure is pi/(3*sqrt(2)), a smidgeon over 0.74. That only holds in regions of uninterrupted packing of course; a lack of fractional balls around the edges of the room will introduce complications we can only ignore if the length of the room's sides are all much larger than the ball diameter. So if the room volume is h*w*l and the volume of the ball is (4/3)*pi*r^3, and {h,w,l} >> r, and the packing coefficient is pi/(3*sqrt(2)), then the number of balls is: h*w*l * pi/(3*sqrt(2)) ----------------- (4/3)*pi*r^3 after cancelling stuff out this reduces to: h*w*l ----------------- 4*sqrt(2)*r^3 That's a bugger to do in your head, but if you're happy with an approximate packing density of 3/4 (close enough since the aforementioned edge effects will make the true value arbitrarily lower anyway), you can arrive at this simplified and much more memorable formula for the number of basketballs:: volume of room * sqrt(2) / volume of basketball-sized cube I can fit in just one basketball in the middle of the room and enjoy watching it! Or, I can maybe order one large basketball the size of the room, get it inside the room and inflate it till it fills the room...like MM said, it's a question of how many and not what's the max no of basketballs one can fit in that room With a gleam in my eye and a wry smile, I'd ask, "When is your next vacation?" I'd give out any random number lets say 100... If he say I counted and thats what is the answer (confidently). If he denies then ask him to prove you wrong by getting 100 basketballs :D say a big number based on the room if they ask how or y tell them to check it out....... I would ask myself how many basketballs I have first. Show more responses I think the answer is ZERO.firstly if u use standard basketball then by calculation you can get the answer, but its an interview not an exam. n if u try to fit a basketball of the size of room then too some space will be left on the corners. so i think its ZERO floor[room_length/basket_ball_diameter] * floor[room_width/basket_ball_diameter] * floor[room_height/basket_ball_diameter] 425 and prove me I’m wrong no help at all no help at all no help at all |
Programmer Analyst at Goldman Sachs was asked...
How many square feet of pizza is eaten in the US each year? 27 Answers800 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. Show more responses 2.4 billion 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. Too many 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? Need Input 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. Show more responses 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 Not enough. 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 Show more responses 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. One or more comments have been removed. |
Sell me your pen. Tell me a time you did something for someone and you didn't expect anything in return. Why American? If you were CEO of American how would you improve American Airlines day 1? 23 AnswersWhen did you get the offer? I went to the same interview, but I have not heard anything from AA yet. I am very nervous that I didn't get it, I believe that I did well at the interview. I interviewed last Thursday, and they called me Wednesday to offer me the position. They are hiring multiple people, so hopefully you all get one too! If not they said they hire for RM in October and March. Anything else I can help with just let me know here! Show more responses after the phone interview, how long did it take to call you for the f2f interview? It was a pretty standard time, about a week-op I have the excel test tonight, any pointers?? I can't really give you any help, but I can tell you every test is different. Make sure to relax and read the directions first. If you didn't get everything right, that is okay there not looking for a 100%. Good Luck! I did it last night, I feel confident about it. It was pretty mucy exactly what I thought it would be, just different format. Hoping for a call about the interview. Nice!!! Please keep me updated on here and let me know how it goes! I'm rooting for you! Thanks! I will. My friend who works there said we would not start working until January 2018 though? I graduate in 2 months though? That start date does not seem right to me, unless they believe you are a Dec grad. If you get a phone interview I would ask at the end "what is the projected start date for this position." If offered they might be able to work with you on a start date if the have availability to start earlier. Can you provide us more questions you have been interviewed during the on-site interview? I would but the point of the onsite interview if to see if you can think on your feet and problem solve. If you already know everything they are going ask you, they don't get to see these things. I can tell you more about it though. 1 interviewer pushed me until i didn't know how to solve a problem and then watched to see how I worked though it or if I shut down. I was asked many basic statistic questions along with pattern recognition. They also asked estimation questions to see my thought process. I was not asked many behavioral questions though. The final type was standard RM questions like EMSR, Demand Forecasting, buying patterns of different types of customers etc........ Show more responses Thanks a lot!!! When will you start to work? No problem, I'm happy I can help out! March 27th yall are such a help, Thanks for your conversations, I got my 2nd interview coming up Having my phone screening tomorrow. Thank you all for the rich info and nice discussion! All the best to all! To the person on your second interview is that in person???? To the Phone screening I hope you did great!!!!!! To answer the question on the second interview, It was a phone interview which went pretty well, BUT I haven't heard anything from AA 2 weeks and counting. Not sure what's going but I did follow and still waiting. After the on site interview, my application status said "in progress" on the application site. A week later it is saying "disqualified". I have not received an email saying I didn't get the position. Anyone else have this issue? I have an excel based case study on the 16th. Any pointer ? How will it be ? This is the first time I’m going through some thing like this I have an Excel-based case study in a couple of days. Any suggestion how should I prepare? What kind of questions can I expect in this case study? What is the selection criterion? How do they evaluate all 3 interviews? |
Analyst at Petco was asked...
How would you direct someone else on how to cook an omelet. 21 AnswersI would confirm their capabilities, confirm what type of omelet they wanted, and confirm the risks. Then design the solution based on their needs, the desired result, and the risks involved. First I would hand them documents on how to cook an omelet. Then I would show them how it is done and after that coach them as they try it for themselves, going over mistakes as they happen or praise during moments of enlightenment. [Listen to the question} I would direct them to Google.com, advising to search for "Best omelet recipe." [As a manager, never do the work if someone else has already done it.] Show more responses Ask my wife. :-) I make terrible omelets, I'm like the patient 0 of bad omelets, don't make me spread bad omelets to anyone else. Go and ask your mom or your Aunt... I would direct them to an internet search engine and suggest they figure it out for themselves, just like I did! First, if I had the chance to talk to the person wanting to make the omelet, I'd ask, "Have you made an omelet before?". That person's answer to determine what I'd do next. If I couldn't meet the person, I'd try to find out if they had a computer. If they did, I'd tell them to use their favorite search engine plus YouTube to find instructions and a video demonstration. If I couldn't have any contact with the person, I'd suggest that the person either use the internet tools I mentioned, go to a local library and look for a relevant cookbook, or find someone they know who's made an omelet to help him/her. My first question would be if they had a computer. If so, I'd talk about the wonders of Google. If it was my job to answer that question, then I'd ask some probing questions to find out how much cooking experience they had before, what supplies they have on hand, and what they like in their omelette. Then based on their own abilities, I'd give simple clear directions. I excel in simplifying, and have experience communicating with children and the disabled. If I didn't know how to make an omelette, I'd look it up myself, while asking those questions, so that I could assist them that way. omelet is already cooked.. Verbally ;-) Try using eggs First I would congratulate them on an excellent choice, ask them what ingredients they proposed, then confirmed their need for my assistance. If they had the ingredients and were certain, I would let them cook it. If they had never done it before, I would take them through the ingredients and the steps until they were comfortable to procced, and then congratulate them on their first perfect omelette when completed Show more responses For the quickest way to achieve this task, I would ask the coworkers if any of them made omelets for breakfast. Next, based on coworker confidence, quick response, and amount of ingredients that they used for the filling, then i would make my choice for teacher. I would assign that employee the task of teaching the individual on how to make an omelet. Assigning the task shows leadership skills, and that is what they are looking for in the first place. My job is to sample the omelet, and to see how well the teacher and the individual did in instructing and following the directions. Or, just give that person $50 to go to the local diner. Knock on the back door, and talk to the cook. Have him offer the cook the money, for him to teach the individual how to make an omelet. When the individual is hired, I would deduct the $50 from his/her paycheck, saying that it was just a loan for their skills training. I personally would not do this to anyone, just so I can have a title with the word "ANAL" in it. Firstly i will give him the step to cook. after all they don't get it i will go to him and cook with him. Obviously if I want to eat an omelette and am asking some one else to make it for me. I would ask him/her to beat well three eggs in a bowl; then finely mince onion, tomato and chilly- one of each, two florets of cauliflower, and two mushrooms. Heat a tsp oil in a non stick pan, saute the ingredients in high flame, add two pinches of salt; spread the mixture evenly, add the beaten eggs, turn down the flame and cover the pan for three minutes. An yummy omelette is ready for me! Of course I will tell him/her that egg, oil, pan and fire are primary , rest including salt is optional and to taste and open to experiment. Refer them to my blog www.mommysrecipe.com refer them to my blog http://www.mommysrecipe.com @ dileep....Looking at your blog site, have you ever had an egg omelet in your lifetime? The one dish, at the beginning of the blog, looks like a scene from the movie "Alien". I am a vegetarian person so I would direct him some food made of green vegetables Very Carefully |