A circle has 11 points arranged in a clockwise manner from 0 to 10. A bug moves clockwise on the circle according to following rule. If it is at a point i on the circle, it moves clockwise in 1 sec by (1 + r) places, where r is the remainder (possibly 0) when i is divided by 2. If it starts at 4th position, at what position will it be after 2012 seconds?

(a) 7 (b)9 (c)5 (d)1

• after 4 second it comes to 0 . And then after every 6 second it comes back to 0 again.
  2012-4=2008
  2008=334*6+4 (it moves 334 times the clock cycle and remaining 4 seconds it will move 0-1-3-5-7)
  thus answer is 7.
• 9 years agoHelpfull: Yes(12) No(3)
• (D) 1
  cycle= 1 +7n after this cycle always is 5th position.
  now after 2008 sec point is 5th position.
  now next 4 sec 1st position.
  or
  Trick:- starting position - 3 = present position. maximum cases.
• 9 years agoHelpfull: Yes(10) No(4)
• (C) At starting it is at position 4,and remainder r is 0 so after four it moves to 5 , after 5 it moves to 7 then 9,1 ,3 thus the pattern is 4 and repeating 5 7,9,1,3,.....
  so 2012-1=2011 and 2011/5 remainder is 1 so answer is 5.
• 9 years agoHelpfull: Yes(2) No(6)
• starts at i=4>>4%2=0...1+0=1 goes to 5
  i=5>>5%2=1...1+1=2 goes to 7
  i=7>>7%2=3...1+3=4 goes to 0
  i=0>>0%2=0...1+0=1 goes to 1
  i=1>>1%2=1...1+1=2 goes to 3
  i=3>>3%2=1...1+1=2 goes to 5
  i=5>>5%2=1...1+1=2 goes to 7
  i=7>>7%2=1...1+1=2 goes to 9
  ......further cycle repeats 0-1-3-5-7-9-0-1-3-5....
  2012-1=2011(4 is not in cycle)
  2011%6=1...therefore after 0 1st iteration i.e 1
  answer d
• 4 years agoHelpfull: Yes(1) No(0)