INFOSYS PAPER ON 7th AUGUST 2006 AT  NMAM INSTITUTE OF TECHNOLOGY, NITTE

 

Hello my friends.. I am Ashwinee.K ,  from NMAMIT College of Engineering Nitte studying in Information Science Branch

I got into Infosys,my dream company,through the campus interview held in our college on 7th Aug...

There were 10 puzzles asked..I attempted all 10 but one was wrong in that(3 marks)...All the 10 puzzles were from prevous question papers..

So PLEASE GO THROUGH ATLEAST 30 to 40  QUESTION PAPERS from THIS SITE...It helped me a lot...If I am placed today that is bcoz of God's grace and this site..I am very greatful to this...and I am sure u will make it..

Ok now coming to PUZZLES...

 

1.. There is a die..position the no.s on it such that pairs 1 & 6, 2 & 4 and 3 & 5 are placed on opposite faces...So in how many ways u can   

   arrange the no.s?                      (3 marks)

 

Ans: BEWARE in many papers the answer is given to be 24...ITS WRONG....the answer is 48.

          Approach : this problem deals with permutation combination..consider 3 pairs fixed as three separate no.s.. now u can arrange these 3

                               no.s in 3! ways i.e 6 ways..now come to each pair of the no.s.. in each pair u can arrange the no.s in 2! ways..and for each of

                               these 2 ways u can have the previously said 6 ways..and so on for other pairs,therefore the answer is 2!*2!*2!* 3! i.e 

                               8*6=48...

 

2.. A donkey was going with a speed of  12miles per hour without any load on it.. but after some distance it was made to carry a load and 

     hence its speed was reduced to 4miles per hour.. What is the average speed of the donkey?   (3 marks)

 

Ans: 6 miles per hour..

          Approach : just see Aggarwal book.. there is a direct formula for this kind of question..the formula is (2*x*y)/(x+y)..where x & y are 2 

                               speeds..here substitute x as 12 and y as 4. u will get the answer...

 

3.. There are 2 systems..A & B.. If system A reads some 34 degree sys. B reads some say 86 degrees..if A reads 84 then B reads 136 degrees.

     (SORRY not sure about numbers...)..At what degree both the systems read same value...?      (may be 6 marks...cant remember)

 

Ans: 52.5 degrees...this question u can find in most of the previous papers..

          Approach :   if the question is.... A          B

                                                                   34         86

                                                                   84         136     then the formula to be used is

                                x + 34y = 86, x + 84y = 136...solving these two equations u get the value of x & y.....(say i've got x = 3, y = 4..for example) 

                                now substitute the value of x & y in any of the above equation say  x + 34y = 86  .. so u get  the equation as

                                3 + A4 = A....(this is bcoz at this temparature both systems measure same value...hope u understood..i can explain only

                                this much)....solving this equation u get A = 52.5..........

 

4.. Some problem regarding PILOT... u can find this in previous papers... There were 3 persons...Kim Jones...Jan Robinson and Pith Smithare..

    There were 3 occupations..one of them was pilot, one engineer and one more biochemist....   

    they gave some conditions...sorry i cant remember the conditions...Finf who is PILOT....    (8 marks)

 

Ans : Jan Robinson was the pilot....

           Approach : one of the imp condition was that Jan Robinson defeated the Engineer in  chess...this says that he is not engineer

                                hence from one more condition u can clearly make Jan is the pilot...Other conditions are not required to b considered.

                                DONT FORGET to draw TABLE as in George Summer..even if you dont know just draw table and explain something..

                                this is enough....even i dint know how to solve this..but i had seen previous paper n hence knew the answer...and drew

                                table according to the answer....

 

5.. Some Kibi Problem....U can find this question in almost all the previous papers.....   (8 marks)

 

Ans : Kibi is a girl...and Mother Spoke first....

           Approach : I am not misguiding you people... but still i will say u what trick i used... i dint know how exactly to solve...but i knew

                                the answer...DRAW TRUTH TABLE................

 

6.. There is a person who walks 4kmph for some distance along the road and 3kmph uphill...and returns in the same way with 6kmph

      downhill...he starts at 3p.m from his house and returns back at 9 p.m....what is the ONE WAY distance?  (6 marks ...may be)

 

Ans : 12km

           Approach : just observe the question... its not total distance travelled by him...but its ONE WAY distance...

                                total time taken by him was 6 hours..(9-3)...let x be his distance with 4kmph..and y the distance with 3 and 6kmph..

                                hence the formula is...6 = ( x/4 + y/3 + y/6 + x/4 ) ... u get x+y=12.....

 

7..  Same old gems problem.

     A jeweller makes a window display. he has 7 gems of which he needs to display 6. three on the left side of the pane on three on the right.     with following conditions. let the gems be Armanet, diamond, emerald , sapphire , garnet, opal and ruby. armanet should be always displayed on left side diamond always on the right ruby should not be kept with diamond or garnet. emerald and sapphire should always be on the same side.

1]what are the possible combinations on the possible for left pane.

a] armanet,garnet , opal

b] armanet, ruby, garnet

c] diamond, emerald , sapphire

d] armanet , emerald, garnet

 

2]what are the possible combinations on the possible for right pane.

a]diamond, garnet , opal

b]diamond, ruby, opal

c]armanet,emerald, sapphire

d]diamond,garnet,ruby

 

3]if garnet, diamond, opal are on the right side what is possible

option for right side.

a]ruby,armanet,sapphire

b]armanet,emerald,ruby

c]armanet,emerald,sapphire

d]ruby,emerald,sapphire

 

4]if armanet, ruby, opal are on the left side what is possible

option for left side.

a]diamond, garnet, emerald

b]diamond, sapphire, garnet

c]garnet, sapphire, emerald

d]diamond, emerald, sapphire

 

Ans 1]a 2]a] 3]c 4]d

         Approach : PLEASE do TABLE and show them the answer....u gety 2 cases or 2 ways in which u can arrange gems..and can find answers  from them...               

 

8..Two watches were set together...one was faster by 1 min per hour..other was slower by 2 mins per hour..after some time two watches had

     one hour difference...after how many hours it showed this one hour difference?                (3 marks)

 

Ans : 20 hours..

           Approach : the total  time error for one hour was (2+1)=3 mins per hour...

                                hence for 60mins it took how many hours?...cross multiply... answer is 20..(60*1/3)..

         

9...series problem