Infosys Placement Paper Whole Testpaper Chitkara Institute, Patiala-27 Jul 2006

Posted by FreshersWorld

7 Jan, 2012

INFOSYS PAPER ON 27th JULY 2006 AT CHITKARA INSTITUTE, PATIALA

I ve appeared for Infosys test at Chitkara institute of Engg. & Technology, Patiala on 27/07/06. I cleared written & appeared for interview next day in chandigarh.

Result will be declared in 4 weeks time.

Here r questions:

1. a,b,c are eating pizza at buffet. A makes 2.4 times as many visits as b, & c makes 6 more visits than b. find total number of visits made by 3 of them?

Ans. As no. of visits should b whole no., so our equation is b+2.4b+b+6

To make this eqn whole no. , we ve to put b=5. This gives no. of visits as 28.

2. 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

This was 8marks ques.

3.Then there was cube problem.

Coloured Cube was cut into 27 cubes by 6 cuts. How many faces will have :

No side colured: 1(centre)

1 side coloured: 6 (one at each side)

3 sides coloured: 8 (all corners)

2 sides coloured: 12

4. Repeated one

A person gets his old book binded. but he finds that the page numbers are cut off.so he starts numbering the pages. During the process he find that 3 is encountered61 times. Can u tell, how many pages were there in book?

ans:-3 encountered once= 3,13,23,30...32,34..39,43,53,63,73,83,93....300=55 times

3 encountered twice=33,133,233= 6 times

Hence answer is 300 pages.

5. If I *I = ME. ME*ME = SHE. Find the value of SHE. All alphebets represent different numbers.

Ans. E is common in both. So I is 4, ME is 16 & hence SHE is 256.

6. . There was a person who smoked a lot .One day he decided to quit his habit, but he had 27 cigarettes with him. So he started smoking them one by one ,to finish them. He had the habit of smoking only 2/3rd of it and leaving the rest butt. later he found out that by joining 3 butts he can form 1 cigarette. So ,tell haw many cigarettes in all he smoked. Sol: from 3 butts he get 1 cigarette. Hence from 27 butts he get 9 (27/3) cigarettes. 3 butts give way 1 cigarette 27 27/3=9 9/3=3 3/3=1 total: 27+9+3+1=40 Ans: 40 (sure)

7. Jack, Daug and Ann, 3 Chledren hadrunning race while returning from school. Mom asked who won the race. Then Jack replied, "I want tell u. I will give u clue. * When Ann takes 28 steps. Daug take 24 steps, meantime I take 21 steps. Jack explained that his 6 steps equals Daug's 7 steps & Ann's 8 steps. Who won the Race?

Answer: DAUG.

8. Then there was magician problem. We ve to calculate age of magician. Simple. "How old are you,Alchemerion?" asked one of the wizards apperences.the wizards answer with a riddle,"I am still Very young as wizards go.I am only three times my son's age. My father is 40 year more than twice of my age.Together the three of us are a 1240 year old".how old is Alchemerion?.

Ans. 360

9. ) There are 3 tribes - sororreans,who always speak truthfully,Narrorean-always false and Midoreans-truth and false alternatively. A: 1) I am a sororean. 2) B is Narorean.

B: 1) I am Midorean. 2) C is Narorean.

C: 1) I am Sarorean. 2) A is Narorean.

Find who is who?

Ans: C is sorean.B is Narorean, & A is midororean.

10. There are 6 people- W,H,M,C,G,F.they are Murderer,Victim,Judge,Police,Witness,Hangman.The Murderer Was later Hanged. 1)M knew both Murderer and the Victim. 2)Judge asked C to Discribe the Murder incident. 3) W was Last to see F alive. 4) police Found G at the Murder site. 5) H and W have never met.

Find who is who? they asked only Who is Murderer Victim Judge Witness

( I am not sure about this one)

W: hangman F: murderer M: Judge C: policemen G: witness Fill the form carefully, as ques in interview are asked 4rm form. English test was easy. Study old papers(very important).

Interview comprised of general question from Resume & form, and 4 puzzles.Pray for my positive result.

Regards,

Vijay Makhija

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INFOSYS PAPER ON 27th JULY 2006 AT CHITKARA INSTITUTE, PATIALA

I ve appeared for Infosys test at Chitkara institute of Engg. & Technology, Patiala on 27/07/06. I cleared written & appeared for interview next day in chandigarh.

Result will be declared in 4 weeks time.

Here r questions:

1. a,b,c are eating pizza at buffet. A makes 2.4 times as many visits as b, & c makes 6 more visits than b. find total number of visits made by 3 of them?

Ans. As no. of visits should b whole no., so our equation is b+2.4b+b+6

To make this eqn whole no. , we ve to put b=5. This gives no. of visits as 28.

2. 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

This was 8marks ques.

3.Then there was cube problem.

Coloured Cube was cut into 27 cubes by 6 cuts. How many faces will have :

No side colured: 1(centre)

1 side coloured: 6 (one at each side)

3 sides coloured: 8 (all corners)

2 sides coloured: 12

4. Repeated one

A person gets his old book binded. but he finds that the page numbers are cut off.so he starts numbering the pages. During the process he find that 3 is encountered 61 times. Can u tell, how many pages were there in book?

ans:- 3 encountered once= 3,13,23,30...32,34..39,43,53,63,73,83,93....300=55 times

3 encountered twice=33,133,233= 6 times

Hence answer is 300 pages.

5. If I *I = ME. ME*ME = SHE. Find the value of SHE. All alphebets represent different numbers.

Ans. E is common in both. So I is 4, ME is 16 & hence SHE is 256.

6. . There was a person who smoked a lot .One day he decided to quit his habit, but he had 27 cigarettes with him. So he started smoking them one by one ,to finish them. He had the habit of smoking only 2/3rd of it and leaving the rest butt. later he found out that by joining 3 butts he can form 1 cigarette. So ,tell haw many cigarettes in all he smoked.

Sol: from 3 butts he get 1 cigarette. Hence from 27 butts he get 9 (27/3) cigarettes. 3 butts give way 1 cigarette27

27/3=9

9/3=3

3/3=1

total: 27+9+3+1=40

Ans: 40(sure)7. Jack, Daug and Ann, 3 Chledren hadrunning race while returning from school. Mom asked who won the race. Then Jack replied, "I want tell u. I will give u clue.

* When Ann takes 28 steps. Daug take 24 steps, meantime I take 21 steps. Jack explained that his 6 steps equals Daug's 7 steps & Ann's 8 steps. Who won the Race?

Answer: DAUG.

8. Then there was magician problem. We ve to calculate age of magician. Simple. "How old are you,Alchemerion?" asked one of the wizards apperences.the wizards answer with a riddle,"I am still Very young as wizards go.I am only three times my son's age. My father is 40 year more than twice of my age.Together the three of us are a 1240 year old".how old is Alchemerion?.

Ans. 360

9. ) There are 3 tribes - sororreans,who always speak truthfully,Narrorean-always false and Midoreans-truth and false alternatively.

A: 1) I am a sororean.

2) B is Narorean.

B: 1) I am Midorean.

2) C is Narorean.

C: 1) I am Sarorean.

2) A is Narorean.

Find who is who?

Ans: C is sorean.B is Narorean, & A is midororean.

10. There are 6 people- W,H,M,C,G,F.they are Murderer,Victim,Judge,Police,Witness,Hangman.The Murderer Was later Hanged.

1)M knew both Murderer and the Victim.

2)Judge asked C to Discribe the Murder incident.

3) W was Last to see F alive.

4) police Found G at the Murder site.

5) H and W have never met.

Find who is who? they asked only Who is Murderer

Victim

Judge

Witness

( I am not sure about this one)

W: hangman F: murderer M: Judge C: policemen G: witness

Fill the form carefully, as ques in interview are asked 4rm form. English test was easy. Study old papers(very important).

Interview comprised of general question from Resume & form, and 4 puzzles.Pray for my positive result.

Regards,

Vijay Makhija