Latest TCS Aptitude Question SOLUTION: A circle has 29 points arranged in a clockwise manner numbered from 0 to 28 . A bug moves clockwise around the circle according to the following rule.If it is at a point i on the c

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(#M40102930) TCS Logical Reasoning question Logical Reasoning Keep an EYE

A circle has 29 points arranged in a clockwise manner numbered from 0 to 28 . A bug moves clockwise around the circle according to the following rule.If it is at a point i on the circle, it moves clockwise in 1 second by (1+r) places , where r is the remainder (possibly 0) when i is divided by 17. Thus it is at position 5 , it moves clockwise in one second by (1+5) places to point 11. similarly ,if moves (1+11) or 12 places to point 11 in one second.

If it starts at point 0,at what point will it be after 2012 seconds?

Asked In TCS SHEELA S (9 years ago)

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Arvind gupta  (9 years ago)

Ans is 15th place
0 0
1 (1+0)=1 1
2 (1+1)=2 2+1=3
3 (1+3)=4 4+3=7
4 (1+7)=8 8+7=15
5 (1+15)=16 16+15=31 2
6 (1+2)=3 3+2=5
7 (1+5)=6 6+5=11
8 (1+11)=12 12+11=23
9 (1+23%17=6)=7 7+23=30 | 1

so after every 8 second it reaches to position 1st,
2012/8 gives 4..
2012+1 1
2012+2 3
2013+3 7
2012+4 15

Is this solution Helpfull? Yes (20) | No   

Gaurav Kumar  (9 years ago)

here it starts from position 0
so the table will be as follows:
time present position next position
0 0
1 0 0+(0+1=1
2 1 1+(1+1)=3
3 3 3+(1+3)=7
4 7 7+(1+7)=15
5 15 15+(1+15)=31(2)
6 2 2+(1+2)=5
7 5 5+(1+5)=11
8 11 11+(1+11)=23
9 23 23+(1+6)=1
10 1 1+(1+1)=3

and the position will continue to repeat in the same manner.
Hence after every 9 sec the bug will come to 1.

so 2012/8 gives 4 as reminder.
so after 5 seconds the bug will be at 15. answer!

Is this solution Helpfull? Yes (5) | No   

Rishita  (9 years ago)

Time Position
0 0
1 (1+0)=1 1
2 (1+1)=2 2+1=3
3 (1+3)=4 4+3=7
4 (1+7)=8 8+7=15
5 (1+15)=16 16+15=31 2
6 (1+2)=3 3+2=5
7 (1+5)=6 6+5=11
8 (1+11)=12 12+11=23
9 (1+23)=24 24+23=47 18
10 (1+10)=11 11+18=29 0
After every 10 sec it comes back on 0
Now, (2012/10) gives remainder 2
And after 2 sec it reaches at 3rd position.

Is this solution Helpfull? Yes (4) | No (5)  

lipsita priyadarshini pradhan  (9 years ago)

starts from 0 position,
rule is=1+(previous position/17)|r,r=remainder
1.0 position
2.1+(0/17)|r=1+0=1,position=0+1=1st
3.1+(1/17)|r=1+1=2,pos=1st+2=3rd
4.1+(3/17)|r=1+3=4,pos=3rd+4=7Th
5.1+(7/17)|r=1+7=8,pos=7th+8=15th
6.1+(15/17)|r=1+15=16,pos=15th+16=21st
7.1+(21/17)|r=1+4=5,pos=21st+6=26th
8.1+(26/17)|r=1+9=10,pos=26th+10=36th position=0th position
so after every 8 sec it reaches to its 0th position bcoz it starts from 0
2008 sec again reaches to 0 position(:2012/8|r=4)
if 2008 sec=0th position
2009 sec=1st
2010 sec=3rd
2011 sec=7th
2012 sec=15th position


ans-15th place

Is this solution Helpfull? Yes (3) | No   

VIKASH KUMAR  (9 years ago)

0 1 2 3 4 5 6 7
0 1 3 7 11 16 22 0
aftr every 7 sec it comes back on 0 ... (2012/7) = gives remainder 3
now after 3 sec it reaches at 7 ..

Is this solution Helpfull? Yes | No (2)  

SHEELA S  (9 years ago)

How did u take 7 sec ?..can u elaboratly explain me..

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purvi singh  (9 years ago)

submit the right answer....

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