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Analytical Interview Questions

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Analytical Interview Questions

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16 Oct 2013

Technical Support Engineer at Oracle was asked...

18 Jul 2012

17 Jun 2010
 How many rotations does earth make on its axis while going around the sun for one year.12 AnswersDepends on the Leap Year or non Leap Year 365 for Non Leap Year 366 for Leap YearThe question is flawed. It should either ask how many days earth takes in one revolution around sun, or how many days are in a year. The leap year is something imaginary humans have concocted to make the calendars look good. You can't say once is 4 years earth takes an extra day to go around the sun.It's to get you thinking about casting... the calender is a double casted to an int!Show more responses365.25 / yr I wouldn't say 365/ 366 leap year. Jain has a point.^It's either 364.25 or 366.25 depending on whether the earth spins in the same direction as it goes around the sun or not (I don't remember which it is and can't deduce it from the information that's available to me without looking it up). Imagine if the Earth did not spin at all on its axis so it would seem as if a year was equivalent to one day. It will take the Earth rotating either 264.24 or 366.25 times depending on the direction to seem like the 365.25 days that it does take.Every year it takes 365.25 rotations but while making calendar we ignore that 0.25 for 3 years and add an extra day in 4th year.365.25 is WRONG!!! Actually Anonymous is right! Just think about this (lay it out on your desk):
• How many days would there be if the earth would not revolve around its axis?
• How many days would there be if it turned once around it's axis?
• Now do again one, change the direction the earth revolves it's axis, but keep going the same direction around the sun ...
The concrete right answer is 366.25 Here you find details on the exact number (with other proof): :)This question makes no sense.It's other way around, only when earth complete one round, we call it a year. To make human calculations easy, we introduced leap yrs.366 rotations per year with respect to a distinct star.366 rotations in a no leap year with respect to distinct star. One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

19 Aug 2012
 How'll you find the weight of an object attached to another object without detaching them and without using any weighing machines. Both the weights are unknown8 AnswersIf the objects are linked by a thread, I'd just put each object one side of my arm and let the heavier one go down while the lighter one goes upTheMigthyBat, no they're attached as in bolted together. You don't know the weight of either of them. And it's not about finding the heavier of them, it's about finding the exact weight of one of them without disturbing them.Cant we just see the amount of water displaced by the object and then calculate the weight? I think its easy that way, as we can control the volume immersed in water. (let it be attached in any way)Show more responsesI think we can take a perfect model of both objects and weigh them separately...by immersing them in a liquid....one by one....and checking the amount of volume displaced.......and applying principles on the liquid collectedimmersing in the liquiud is the ans to thi s ques ... bt still then they are connected together so the total volume obtained would the sum of both objects .. then how could we find the volume of each one ..Water displacement theory seems to be nice. Since there is no information regarding how they are attached we can do the following way. catch one weighted and dip the other weight in water. Some water will be displaced. Now repaeat the same in revesrese order. i.e catch the second side and dip the other weight. Now calculate the amount of water displaced in both the cases. greater displacement - lesser water displacement. Now immerse completely both the objects and find the displacement of water. by this you can arrive at the answerits simply and easy. Let us use the simple way to determine which is heavier. Let us assume the weights are connected with the thread. And use the water displacement. And we can use the six standard mathematics (A+b) whole square formulae. We can calculate the weight of the object

Senior Software Engineer at Virtusa was asked...

19 Sep 2010
 if there are 6 people in a team, how many handshakes will be there8 Answers15There will be 30 hand shakes. In total we have 6 people, so it will be 6 * (6-1) (i.e 1 person will shake hands with 5 people)15 A | B | C| D | E | F | A will perform 5 B will perform 4 C will perform 3 D will perform 2 E will perform 1 F ultimately hand shakes with everyoneShow more responses6c2==6!/(2!*4!)=15lets consider A,B,c,d,e,f so a shakes hand with the other 5 ..b with oter 4 and so on so at last 5+4+3+2+1=1515 people handshakes at a time you have given 6 person first of all the peoples is arranged in row 6 number people have an 5 option for handshakes after that 6 number person out then total number of people 5 again....again....same procedure apply.......Each one will shake hand 5 times, 6x5 -> 30 times6c2= 15

Applications Developer at Oracle was asked...

10 Sep 2011
 Tell me a palindrome date before 10-02-2001 (mm-dd-yyyy)7 Answersi got it to be in the 14 th century ...12-31-132109-31-1390Show more responsesSorry it should be 08-31-138010-01-1001how can the month no. will be more than 12 i think the answer should be 29-11-1192 :) correct me if i am wrong.....U r wrong. Mmddyyyy

10 Jun 2012

Computer Operator at Reliance Industries was asked...

23 Mar 2013
 this is moet important that the company must take some decesion to gave chance to talented student.7 Answerscompony can gave chanceFirst of all it's most not Moet if you are looking for a talented people there are about 1.3 billion people living in India who are smart and better to think and give different ideas and no mention hard workingHardworkShow more responsesInternship is alow company??I am electric engineering studentYesYes I operate in computer since 2011.

BAA at ZS Associates was asked...

15 May 2013
 How can you find 3 fastest horses among of 25 horses with 5 race tracks given to you to find out by race them up(in minimum races)?7 AnswersI answered at 8 races. Divide 5 groups of five horses. Race one by one group and will get a fastest horse from every group.(5 races) Race between all 5 fastest horses of all group. (5+1 races). Race between all 5 second fastest horses of all group. (5+1+1 races). Race between all 5 third fastest horses of all group. (5+1+1+1 races).You missed out the probability that all 3 horses might be of the same groupDivide 5 groups of 5 horses. Get top 3 in each group(5). You are left with 15. Then 3 more races in that group of 15 (5+3). You are left with 9. Then 2 more races in that group of 9(5+3+2). You are left with 6. Then 1 more race in that group of 6(5+3+2+1). You are left with 4. Then a last and race to decide best 3 of 4.(5+3+2+1+1). So, total comes to 12.Show more responses9 is the lowest I can figure out. Run 5 races with five random groups of horses, call the groups A B C D and E. Take the winners from each group and put them in their own group (Winner's group). Also put the second place winners in the Second Place group, and the third place winners in the Third Place Group. At this point there have been five races. The key to the problem is to recognize that while the Winner's Group may or may not contain the first, second, AND third fastest horse (or the first and second, or just the first), the Second Place Group cannot contain the fastest horse (since every horse in the Second Place Group has been beaten by at least one horse). The Second Place Group MIGHT, however, contain the second and third fastest horse. Similarly the Third Place group cannot contain the first or second fastest horse, since each horse in the TPG (third place group) has been beaten by 2 horses. So, now at this point, we can easily find the fastest horse by simply racing the Winner's Group, to bring us up to a total of 6 races. Now race the SPG (second place group) and TPG, for 2 more races, bringing our total to 8. Take the second and third place horse from that race, the first and second place horse from the SPG, and the fastest horse from the TPG, and race them (9 races). The winner of that race is the second fastest horse, and the runner up the third fastest.1 race! Make all 25 horses run on one race track. The top 3 are your fastest horses.5+1+1 = 7 Races. 5 races: 5 random groups( A,B,C,D,E). After the race assign them numbers as per finished position ( A1, A2, A3 ,A4, A5). 6th race: Take 1's of all 5 groups ( A1, B1, C1 , D1, E1) and race them. Lets assume A1 comes 1st , B1 comes 2nd and likewise E1 comes fifth. A1 occupies 1st place and need not race again. It is the fastest horse Now the possibilities for 2nd place are A2 (or) B1 (2nd position of 6th race) ( A2 & B1 - 2 horses selected to race again) Possibilities for 3rd place : A3 (or) B2 (or) C1 ( 3 horses are selected to race again ) 5 horses are selected 7th Race : Race all these 5 horses ( A2, B1, A3,B2,C1) and the winner would be 2nd fastest while runner up would be 3rd fastest6 races

9 Jul 2012