(may infi 28th may-kindly solve) 1)one day I was going somewhere.I found charity 1,charity 2,charity 3. for charity 1,I gave half of the money what I have plus Rs 1. than for charity 2 I gave half of the money that was remaining with me plus Rs 2. than for charity 3 I gave half of the money that was remaining with me plus Rs 3. finally I was left with Rs 1 only. what amount I had initially.

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arp001

  • May 25th, 2006
 

start with Rs. 41 give half + 1 so remaining is Rs.20. Now give half of 20 + 2 so remainign is 8 then give half of it+ 3 so remaining is Rs.1

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bindu

  • Jun 9th, 2006
 

may b it is 70

70/2=35+1=36

36/2=18+2=20

20/2=10+3=13

70-(36+20+13)=1

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arp001

  • Jun 9th, 2006
 

You have to donate half and besides half give 1 dollar on first donation then 2 dollar for 2nd donation and 3 dollars for the third donation. If you start with 70 then 35 so you have 35 - 1 =34. then half of 34 i.e.17 you have and  2 more dollars so 15 then half of 15 is 7 or 8 you have and give 3 dollars so ultimately you have either 4 dollars or 5 dollars but you need 1 dollar only at the end.

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jetendra

  • Jul 5th, 2006
 

at first u assume the initial be Xfor the 1charity : u have spend is (X/2)+1 remaining is (X/2)-1for the 2charity : u have spend is (((X/2)-1)/2)+2 remaining is (((X/2)-1)/2)-2for the 2charity : u have spend is (((((X/2)-1)/2)+2)/2)+3 remaining is (((((X/2)-1)/2)-2)/2)-3the last half of the remaining equals rs.1so using the equation (((((X/2)-1)/2)-2)/2)-3 = 1if we solve we get x=42 ans is 42

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sany

  • Jul 24th, 2006
 

ya its 42.....same one came for me tdy in infy paper........

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