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.A circle has 29 pts arranged in a clock wise manner numbered from 0 to 28 as shown in figure.
A bug moves clockwise around the circle,according to the following rule (i.e) if it is in the position i then it moves i+1+r position for next one second… so that in 2012th sec what position the bug will be
Read Solution (Total 12)
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- there is no figure here.
pls give link for figure. - 12 years agoHelpfull: Yes(7) No(0)
- i dn't know
- 12 years agoHelpfull: Yes(2) No(1)
- 2012/29 = 69.37931 --> round off =70
than i chose my starting point 28th ..
1 cycle is completed in 28th position..we want to complete 70 cycle in circle for 2012
than 28+28 = 56
2 cycle completed
now than for 70 we have 14 pts more.. (70-56 = 14)
but in question there are 29 point and we start with 28th so for 29 = 14+1 = 15 ans. - 12 years agoHelpfull: Yes(2) No(1)
- answer is 15
- 12 years agoHelpfull: Yes(1) No(3)
- 2012/29 = 69.37931 round off =70
now we take 28th position as our initial position...
i cycle is completed in 28th position
than 28+28 = 56
2 cycle completed
now for complete 70 cycles ..we have only 14 more..(70-56 = 14)
than we start with 28th position and count with zero ...so final 13 is ans . - 12 years agoHelpfull: Yes(1) 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.
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