Hello friends I need answer with working steps for the following questions as soon as possible as I have only one day to prepare for Infosys test.1)There was a person who smoked a lot .One day he decided to quit his habbit,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. latter he found out that by joining 3 butts he can form 1 cigarette. So, tell haw many cigarettes in all he smoked.2)There are 100 men in town. Out of which 85% were married, 70% have a phone, 75% own a car, 80% own a house. What is the maximum number of people who are married, own a phone, own a car and own a house ? 3)One day Harry and I set our watches together. None of us was aware that my watch was getting faster by 2 min per hour and Harry�s watch was getting slower by 1 min per hour. After sometime, we discovered that my watch was 1 hr ahead of Harry�s watch. Can you find out after how long we noticed this ?

Questions by uspriya

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aartee

  • Mar 31st, 2006
 

hello usha...i'm sending ur solutions.....Q1) 27 +(27/3)+(27/9)+(27/27) = 40 cigarettes Q3) difference in time = 3 mins in 1 hr... therefore ur watch will gain 1 hr in (60/3)= 20 hrs..and usha i have my infy xam too plz post the question paper that u get....plzzz

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vidhya

  • Apr 5th, 2006
 

Answer to Question 127/3 = 99/3 = 33/3 =1Totally 13 ciggrates

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bijoy's

  • Apr 10th, 2006
 

hai, firstly 27 cigretes leaving 27butts hence 27/3=9 more cigerettes.so now a total of 36 cigerettes already. now again 9/3 leaving 3 more which again3/3 leaves one more .hence a total=27+9+3+1=40 cigerettes.

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yusufqtv

  • Jun 3rd, 2006
 

According to me the solution is:


Every above friends are forgetting smoker has the habit of smoking 2/3rd of what he has at one time.

so aproach is this way

(2/3)*27+(18/3)+(2/3)*9+(6/3)+(2/3)*3+(2/3)=104/3 cigs.

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