Infosys placement Paper - 26 Nov 2006 - Pune



Infosys placement Paper - 26 Nov 2006 - Pune..

Answer / o prettysuni

Hi guys This infosys paper for freshers.

The test paper consists of:
Section I - Aptitude test of 30 Questions in 45 minutes.
Section 2 - English - 40Questions in 30 minutes.

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Aptitude section:

First five questions are cube based problems.

There is a 10 cm cube,it is painted with a layer of blk colour of 1 cm
alonf its edges and rst of part is paintd wid blue color. Then it is cut
into 1000 pieces.
Q1 How many small cubes have 2 sides painted black?
Q2 How many sides have no sides painted black?
Q3 How many pieces have at least 2 sides painted black?
Q4 How many sides have adjascent sides blue and black resp.
Q5 How many pieces have at least 1 side painted?

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6-10 questions are on Non-verbal reasoning.


11-15 questions on Data Sufficiency problem.


ex: (1)7% of C.P = 9% of S.P


16-20 questions on Data Interpretation.

Two tables were there

A certain factor(product rate factor, prf) was compared among some
countries.

Factors like research,technology, training, production ), were given of
countries like US,UK, germany,india & japan.

All factors had some weightage given e.g. research =0.3,training=.4, etc.
5 questions asked on this. 4 are easy and 1 is tough.
Be fast to analyse as you know the time is very precious.

21-25 questions analysing type.

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Comments from gavaskar about 4 players, sachin,kambli,shastri & dravid.

All have a specific batting style among technical, offensive, stylish, & 1
more type was there.

4 typical strokes played among stret drive, sqr cut, flick & covr drive.

The offensive batsman played flick .
Tendulkar loves street drive.
Kambli & shastri not technically sound.
One more case was there.
Its very easy to do.

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26-30 questions are of logical type.

Logical deduction
medium level
we have to choose any three logical sequences, 4 choices given.

prepare from Time or Ims Cat study material. verbal prepare from RS
agarwal.

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SNAPDEAL QUESTIONS 1. A train is going at a speed of 60kmph towards Delhi and returned back at a speed of 30kmph. What is its average speed? ANS : (2*30*60)/(30+60) = 40kmph 2. How many different 4 letter words can be framed that have at least one vowel? ANS : 264 - 214 (total no of 4 digits words – no of words with no vowels) 3. Write an algorithm to find out a number from an array of numbers where only one number occurs once and rest all occurs twice. ANS : XOR all the numbers ,you will get the number with single occurrences . 4. Which among the following have the product of the distance between opposite sides of a regular polygon and it side equals one fourth of the area. A. hexagon B. octagon C. n=16 D. n=18 ANS : n=16.(area of regular polygon = apothem*perimeter/2 Apothem = distance between opposite sides/2 Area = (opp_side_dist * n * a )/4 Product of opp_side_dist and side of reg. polygon = opp_side_dist * a For n= 16 the ration becomes 1:4 5. Which of the following cannot be a relation between two variables? ANS = 4th diagram. 6.what will be the output of this program Void print (int n) { If (n>0) { printf(“hello”); print(n-1); } printf(“world”); } ANS : N times hello followed by N+1 times world. 7. Which among the following cannot be used for future prediction? ANS : 4th Diagram. 8. There are 25 horses. We have to find out the fastest 3 horses In one race maximum 5 horses can run. How many such races are required in minimum to get the result. ANS : 7 races (A. first run all horses = 5 races, eliminate 4th 5th of all races. B. Run horses who came 1st in those 5 races = 1 race , the horse coming first is the fastest Run horses a. 2nd and 3rd with the fastest horse (in first time race A) b. 2nd and 3rd coming horse in B. c. The horse who came 2nd with the horse(who came 2nd in race B) in race A You will have the fastest 3 horses.) 9. In a game of rolling dice you are given 2 dice and you have to roll them. Whatever is the outcome the player will win that many dollars. What should the game owner charge each player (optimum) so that he doesn’t have to bear any loss? ANS : $7 10. We have a function REV(“string”,m,n).This function is capable of reversing the caharacters in the string from mth location to nth location. e.g. REV(“abcd”,2,3)  the output will be acbd We need to swap a string from a position,e.g. SWAP(“abcdefg”,4)  output needs to be efgabcd. How can the REV function used do this. ANS : L = string length,N= position given in SWAP function. SWAP(“abcdefg”,4) = REV(REV(REV(“abcdefg”,N+1,L),1,N),1,L).

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