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(#M40031973) ELITMUS Numerical Ability question Keep an EYE

Q. The method of selecting a member is to make stand all member in a straight line and are alotted roll no from right so person at right most is 1 and last is 513 if ther are 513 members so now one will start from right and will remove all alternate like 2,4 will be eliminated than when it reaches otherside will do same to right ie 512 514 will remove okk alternately so if person have to b elected where will person should stand

Asked In Elitmus siddharth (10 years ago)

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rohitashwa  (10 years ago)

1 is the answer consider numbers as 5in place of 513 and do d same. 1,2,3,4,5. 1st =4
2nd=2
remaining(1,3,5)
3rd=3 we removed 3 from right so its now lefts turn
remaining nos 1,5(5 at even pos so removed )answer at 1st position was it in option???

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vikash kumar  (10 years ago)

513 will be the ans , as in each round will eliminate the no's which are at even position and each round we get odd number of no's and in last round we will be left with 1 and 513 so in last 1 will be eliminated and person at 513th position will win.

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Wasim Akram Khan  (10 years ago)

it will be left most number ie 513 eg
9 8 7 6 5 4 3 2 1
after removing from right
9 7 5 3 1
after removing from left
9 5 1
again from right
9 1
again from left
9 left most number....

Is this solution Helpfull? Yes (4) | No (1)  

parul  (10 years ago)

513...is the answer

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K Sangamma  (10 years ago)

The member has to stand in odd positions

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DHAKAD NEETENDRA SINGH  (10 years ago)

left most no is the correct answer = 513

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prerna  (10 years ago)

no. starting from 1st and at the difference of 4 and so on will be the answer. for ex: 1,5,9,13,17....513.

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somia jain  (10 years ago)

answer should be at first position..

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Avigyan  (10 years ago)

If the number at even places is only deleted then in first round..all even no will be deleted left with..
513 511 509 ........... 5 3 1
in 2nd round.. again the no at even places if deleted then.. no left will be..
513 509 505..........5 1
in 3rd round ..no at even places will be deleted then left with...
513 505 ....9 1 .. and so on.. finally the 2 no's which are left is..
513 1 ... no if no at even places are deleted then .. obviously 513 will be deleted and only the 1 will re remaining.. so 1 would be the ans.. i think so.

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