A un Financial Software Developer le preguntaron...15 de febrero de 2010

↳

Seems to me the correct answer is five. Assuming that each horse's performance is timed, by running five races with five horses each, you'll know the speed of all 25 horses. The three with the shortest times are the three fastest horses. Most responses assume you need multiple rounds, but these responses seem to assume that the five horses that finish first in the first round are the fastest five overall. That may not be the case. Just because a horse beat its four competitors doesn't mean it's one of the five fastest overall ... just that it was faster than the four it competed against. Menos

↳

7 races. We run five races with five horses. The five winners race in a sixth race while the 4th and fith place finishers are eliminated from further consideration. The sixth race show horse is faster than all the horses that participated in the preliminary races where the 4th and 5th place horses participated and they are eliminated from further consideration. The other horses in the preliminary race where the 6th race show horse participated are also eliminated. The show horse in the preliminary race where the 6th race place horse participated is eliminated since there are at least three remaining horses that are faster. We are left with 6 horses. We know the winner of the 6th race is fastest overall, so that leaves five horses to enter a 7th race for the overall place and show. Menos

↳

12 RACES ARE REQUIRED -------------------------------------- In the worst case scenario the 'best three' can be from a single team of 5 horses. So 5 races in round one. Chose all the first three of each of the 5 races. 3 races of all the 15 horses which were the 3 winners of the first round. Chose the 9 horses which were winners in round two and have 2 races for the 9 winners[ 5 & 4 horses] you get 6 winners. 1 race of 5 horses, out of the 6 round three winners, keeping one standby 1 race of 4 horses of round four winners with the standby. This will give you the best 3 horses out of 25 So you need to have 5+3+2+1+1 = 12 races in order to get the best three horses. Answer 12 races required. Sometimes the 2nd and/or 3rd best athlete do not get selected if they are teamed in a race along with the best athlete. Menos

A un Software Engineer II le preguntaron...19 de junio de 2014

↳

Actually the answer is zero. Plans fly, people don't.

↳

Roughly as many who flew into Chicago.

↳

oops - typo - Planes fly, people don't

A un Financial Software Developer le preguntaron...17 de mayo de 2010

↳

Never....the frog would be dead by day 10 since nothing to eat or drink.

↳

Assuming it doesn't die of starvation, the answer is 28 days.* start of day 1 (0 days elapsed): 0m --> 3m (then falls back 2m by start of day 2) start of day 2 (1 day elapsed): 1m --> 4m start of day 3 (2 days elapsed): 2m --> 5m ... start of day 28 (27 days elapsed): 27m --> 30m start of day 29 (28 days elapsed): 28m --> 31m In other words, 28 days will have elapsed before the frog can jump to a height exceeding 30m.* * This answer assumes the frog is not able to walk away after it hits 30m. I would assume it has no energy left to climb out based on the problem description. If the questioner disagrees with this assumption, then the answer is 27 days. Menos

↳

I agree -- it's 28...because on that morning, he'll be at 27 metres and he can jump to the top in one bound. Menos

A un Software Developer le preguntaron...20 de febrero de 2012

↳

30+25=55 angle between them...

↳

You know that there is 30 degrees between each number on the clock because 360/12 = 30. You know that the minute hand is at exactly 10, while you know that the hour hand is 5/6th of the way to the 12 (from the 11) because 50/60 = 5/6. Therefore to get the total angle, you must add the 30 degrees from the 10 to 11 plus the 5/6 distance from 11 to 12, and you get 30 degrees + (5/6 * 30) = 30 + 25 = 55 degrees. So the final answer is 55 degrees. Menos

↳

Anonymous is on the right track with the hour hand going 0.5 degree/minute, but at 11:50 the hour hand is no longer near the 11, it's near the 12. 10 minutes before 12, it is 5 degrees from the 12. The minute hand is 60 degrees from the 12 in the same direction (360 deg=60 minutes, so 60 deg=10 minutes). Angle between the hands is the difference between these two: 60-5 = 55. Menos

A un Software QA Engineer le preguntaron...7 de octubre de 2010

↳

Swaz answer is almost correct however it does not work in all scenarios. lets assume: box 1 is labelled Oranges (O) box 2 is labelled Apples (A) box 3 is labelled Apples and Oranges (A+O) and that ALL THREE BOXES ARE LABELLED INCORRECTLY" Pick a fruit from box 1, 1) if you pick an Orange: - box 1's real label can only be O or A+O - box 1's current label is O - since ALL LABELS ARE INCORRECT then box 1's real label can not be O - box 1's new label should then be A+O by elimination - since ALL LABELS ARE INCORRECT - box 2's label is changed to O - box 3's label is changed to A - SOLVED 2) if you pick an Apple: - box 1's real label can only be A or A+O - box 1's current label is O - since ALL LABELS ARE INCORRECT then box 1's real label can not be O - this still leaves us with the choice between label A and label A+O - which would both be correct - FAILURE Solution: The trick is to actually pick a fruit from the A+O labeled box Pick a fruit from box 3: 1) if you pick an Orange: - box 3's real label can only be O or A - box 3's current label is A+O - since ALL LABELS ARE INCORRECT then box 3's real label can not be A+O - box 3's new label should then be O by elimination - since ALL LABELS ARE INCORRECT - box 1's label is changed to A - box 2's label is changed to A+O - SOLVED 2) if you pick an Apple: - box 3's real label can only be O or A - box 3's current label is A+O - since ALL LABELS ARE INCORRECT then box 3's real label can not be A+O - box 3's new label should then be A by elimination (not O) - since ALL LABELS ARE INCORRECT - box 1's label is changed to A+O - box 2's label is changed to O - SOLVED Menos

↳

It's easier to draw it out. There are only 2 possible combinations when all labels are tagged incorrectly. All you need to do is pick one fruit from the one marked "Apples + Oranges". If it's Apple, then change "Apple + Orange" to "Apple" The "Apple" one change to "Orange" The "Orange one change to "Apple + Orange" If it's Orange, then change "Apple + Orange" to "Orange" The "Apple" one change to "Apple + Orange" The "Orange" one change to ""Apple" Menos

↳

All the three boxes are names incorrectly. SO the bax lebeled Apples+Oranges contains only Oranges or Only Apples. Pick one fruit from it. If it is Orange then lebel the box as Orange. So the box lebeled Oranges contains Apples and the remaining contains both. Menos

A un Software Engineer Intern le preguntaron...28 de julio de 2009

↳

Two. Split into three groups of three, three, and two. weigh the two groups of three against each other. If equal, weigh the group of two to find the heavier. If one group of three is heavier pick two of the three and compare them to find the heaviest. Menos

↳

2 3a+3b+2 = 8 if wt(3a)==wt(3b) then compare the remaining 2 to find the heaviest if wt(3a) !== wt(3b) then ignore group of 2 discard lighter group of 3 divide the remaining group of 3 into 2+1 weigh those 2 If == the remaing 1 is the heaviest if !== the heaviest will be on the scale Menos

↳

2 weighings to find the slightly heavier ball. Step 1. compare 2 groups of three balls. Case 1. if they are both equal in weight, compare the last 2 balls - one will be heavier. case 2. If either group of 3 balls is heavier, take 2 balls from the heavier side. compare 1 ball against the 2nd from the heavy group result 1. if one ball is heavier than the other, you have found the slightly heavier ball. result 2. if both balls are equal weight, the 3rd ball is the slightly heavier ball. Easy Shmeezi Menos

A un Software Engineer le preguntaron...23 de abril de 2010

↳

Not quite. The volume of the water is not the same as the weight of the elephant. You'd have to estimate the density of an elephant and multiply that by the volume of the water to get the mass, then multiply that by the acceleration due to gravity in water system (SI, English Customary, etc.) you're using. Luckily, mammals are mostly water (humans are around 70% water on average), so about 2/3 of the weight of the elephant would be equivalent to the weight of the water displaced. So you would have to estimate how dense the rest of the elephant is (since it'd be minerals and such, I'd say it's more dense than water) and follow the steps described above. Menos

↳

As it is not specified that the International System of Units must be used, define the Elephant unit (E) as the weight of your elephant. Your elephant then weights exactly 1E. Menos

↳

Simple answer : Use a beam balance . Put elephant on one side and start throwing weights on the other side . When the beam is balanced you got the weight of the elephant equal to the sum of weights on the other ! Menos

A un Software Engineer le preguntaron...6 de junio de 2016

↳

For some questions answers are posted in glassdoor!

↳

Hi . Did they inform about results till now?

↳

Hi ..Yes i got the offer!

A un Junior Java Developer le preguntaron...10 de enero de 2018

↳

please let me know if anyone got Job

↳

36. If you get call then follow this interview experience

↳

How many of you waiting after 3rd round

A un Software Engineer le preguntaron...18 de marzo de 2009

↳

Actually, 16 is not the optimal, nor is 15; you can do it in 14. Here is one solution (there are at least 5 other equivalents): * Drop first bulb at 14th, 27th, 39th, 50th, 60th, 69th, 78th, 85th, 91st, 96th, and (optionally) 100th until it breaks. * Go to the highest floor where the first bulb didn't break. * Repeatedly go up one floor and drop the second bulb. When it breaks, you have your answer. Why is 14 optimal? Since you are decrementing each time, you want (n) such that sum(1..n) barely reaches 100. The answer is n=14. Generally, if the number of floors is (f), the lowest number of drops is ceil((sqrt(8*f+1)-1)/2). This is the best worst-case scenario. An interesting related question is what gives the lowest expected number of drops. And no, I could not have gotten this in an interview. Menos

↳

19 drops is not the best worst case scenario... imagine trying floor 16, if it breaks, you try 1 - 15 and thats 16 tries. if it doesn't break, then try floor 31 and if it breaks, then try 17 - 30 (so 16 tries, including the try on floor 16). And on and on (45, 58, 70, 81, 91, 100). If you reach 91, you'll have tried 7 floors so far and if it doesn't break, then there's 9 more tries to get to 100 (thus 16 in the worst case) Menos

↳

Go to the 100th floor and drop the first bulb. It WILL break. Done, 1 drop. It doesnt say whats the lowest floor it will break at, just at what floor will it break with least drops. Thus floor 100. Menos

desarrollador de flexdesarrollador de flash seniordesarrollador de aplicaciones movilesdesarrollador creativodesarrollador iosdesarrollador interactivodesarrollador multimediadesarrollador webdesarrollador de aplicaciones web uidesarrollador de aplicaciones webdesarrollador de html5desarrollador de juegosdesarrollador web diseño de interfacesdesarrollador de interfacesdiseñador de flashdesarrollador de aplicaciones móvilesdesarrollador uxdesarrollador de aplicaciones web netasdesarrollador html