A circle has 29 points arranged in clockwise
manner from 0 to 28. A bug moves clockwise on the
circle according to the following rule. If it is at a point
i on the circle, it moves clockwise in1 sec by (1+r)
places, where r is the remainder (possibly 0) when i
is divided by 17. If starts from position 0, at what
position will it be after 2012 sec?
a) 1 b) 10
c) 15 d)20

• 15 is the answer
  
• 9 years agoHelpfull: Yes(7) No(1)
• After 1st second, it moves 1 + (23/11)r = 1 + 1 = 2, So 25th position
  After 2nd second, it moves 1 + 25/11 = 1 + 3 = 4, So 29th position = 0
  After 3rd second, it moves 1 + 0/11 = 1 + 0 = 1, So 1st position
  After 4th second, it moves 1 + 1 = 3rd position
  after 5th, 1 + 3/11 = 4 So 7th
  After 6th, 1 + 7/11 = 8 so 15th
  After 7th, 1 + 15/11 = 5 so 20th
  After 8th, 1 + 20/11 = 10th, So 30th = 1st
  So it is on 1st after every 3 + 5n seconds. So it is on 1st position after 2008 seconds (3 + 5 x 401) So on 20th after 2012 position.
• 9 years agoHelpfull: Yes(6) No(7)
• Yes NILAGHA DAS i just missed it after 17 sec literally means at the 18th sec it will be at 1 position each time and on dividing 2012 by 18 gives 14 as reaminder . so it will be at pos 1 after 2012-14= 1198 . Now:
  1198- Pos 1
  1199- Pos 2
  .
  .
  .
  2012- Pos 15 [Ans]
  
• 9 years agoHelpfull: Yes(5) No(0)
• @RANOPRIYO NEOGY : U have done everything right except a silly mistake . after every 18sec (Not 17) it will be at 1 position. now 2012/18 gives remainder 14. so if initial pos is 1 then after 14 sec it will be in 15th Position . So C is the correct answer
  
  
• 9 years agoHelpfull: Yes(3) No(0)
• initially i=0
  after 1 sec pos= (1+0/17)=1st
  after next sec pos= (1+1/17)=2=> 3rd pos
  after next sec pos= (1+3/17)=4=> 7th pos
  after next sec pos= (1+7/17)=8=> 15th pos
  after next sec pos= (1+15/17)=16=> 2nd pos
  after next sec pos= (1+2/17)=3=> 5th pos
  after next sec pos= (1+5/17)=6=> 11th pos
  after next sec pos= (1+11/17)=12=> 23rd pos
  after next sec pos= (1+23/17)=7=> 1st pos
  so after 2009 sec it will be at 1st pos
  nd after 2012 sec it will be at 15th pos as bug comes at 1 after every 8 sec
  
• 9 years agoHelpfull: Yes(3) No(0)
• c will be the answer
  
• 9 years agoHelpfull: Yes(2) No(6)
• 20 th after 2012 sec
• 9 years agoHelpfull: Yes(2) No(5)
• i=0 initially
  so r =0/17 i.e remainder=0
  acc to given formula=1+r= 1+0 =1
  So After 1 sec it will b at pos 1
  Case 2:
  Now i=1
  So i/17 gives remainder 1
  So new pos after 2 sec is 1+r i.e 1+1=2
  Similarly in Case 3 . new pos =3 after 3 sec
  But after 17 sec it will again be at pos 1 coz 17/17 gives remainder 0
  and 1+0 =1
  now after every 17 sec it will be at 1 position ...
  now on dividing 2012 with 17 we get 6 as remainder so nearest number divisible by 17 is
  2012-6=2006 , so after 2006 it will be at 1 pos again and like wise after 2012 it will be at 7 th pos ..
• 9 years agoHelpfull: Yes(1) No(1)
• answer is 15
• 9 years agoHelpfull: Yes(1) No(0)
• c)15 is the correct answer..
• 9 years agoHelpfull: Yes(1) No(0)
• plzz xplain the soln...
• 9 years agoHelpfull: Yes(0) No(0)
• posibly ans will be 10
• 9 years agoHelpfull: Yes(0) No(3)
• on 7th position it will be after 2012 sec
• 9 years agoHelpfull: Yes(0) No(1)
• i=0 (initially)
  then 0/17=0; it moves by 1+0=1
  then i=1
  1/17=1; it moves by 1+1=2
  i=2
  2/17=2; moves 1+2=3
  continue like this and it gives again 0 at 17th pos
  divide 2012 by 17 we get remainder=6 then minus 6 to 2012
  2012-6=2006 which is divisible by 17
  so at 2006 it is at 1 pos
  therefore at 2012 it is at 7 pos
• 9 years agoHelpfull: Yes(0) No(0)