17 February 2007 Infosys Written Test Paper In Pondichery
Answers were Sorted based on User's Feedback
Answer / poonam
I attended the Infosys test on 17th Feb. Of course it was a bit tough
one. Don't beleive in old question papers. They don help much.
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collect the following materials and start preperation.
1. basic cat prep. materials given in TIME .
2. or any cat book.........
3. R.s.agarwal for verbal non verbal reasoning.
4. wren and martin high school English grammar book with key.
Some models will come from these books.
Aptitude Section:
They have given the patterns unlike other question papers.
Data interpretation was easier than any other part
Given a puzzle test on a t.v. channel and its shares and revenues which
was somewhat tough.
series completetion problem was given .
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Andn R.s. agarwal verb and non verb .
Another puzzle on tennis tournament which is a moderate one.
syllogism.
English Section:
Well Prepared with basic grammar .
2 reading comprehensions were given .
one regarding romeo and juliet which was very easy .
other Comprehension regarding dramas which was somewhat tough.
correction of sentences.
Tenses based questions.
choose a suitable alternate which best fits in the given sentence. check
both tenses and meaning before filling.
conclusions.
synonyms.
500 attended the test and only 26 were selected for the interview.
Luckily i was 1 among them.. I performed well in the interview also .
They told they will inform the results through mail.
Regards
Poonam Gupta
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| Is This Answer Correct ? | 4 Yes | 3 No |
Answer / soubhagya ranjan
please send me question paper of written test.
| Is This Answer Correct ? | 2 Yes | 1 No |
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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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