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### Reasoning Questions-Person Sitting in Circle-Bank PO and Clerical Exams -

Reasoning Questions on Person Sitting in Circle

Directions – Nine friends L,M,N,O,P,Q,R,S,T are sitting in a circle not necessarily in the same pattern.
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 ·       T sits 5th to the right of R. ·       M sits between P & S. ·       S does not sit immediate to T. ·       N does not sit immediate to either T or R. ·       sits second to right of Q.
·       N sits 4th to the left of P.

Q1- Who is second to the right of M?

a.     R

b.    T
 c.     L
d.    Q

e.     Cannot be determined

Q2 – if all the nine friends sit alphabetically from the L position in anticlock manner, how many are still going to get the same seat.

a.     None

b.    Two

c.     Three

d.    Four

e.     One
Q3 – Four of the five are alike in a certain way based on their sitting pattern. Which one does not belong to the group ?

a.     LP

b.    PN

c.     SP

d.    TS

e.     OQ

Q4 – What is O’s position with respect to P in clockwise manner ?

a.     Firs to the right

b.    Second to the right

c.     Third to the left

d.    First to the left

e.     Second to the left

Q5 – If S:P, then in same pattern  which will hold for M: ___ ?

a.     T

b.    Q

c.     R

d.    S

e.     O

Q6- Four of these five groups are alike in certain way, find the odd one out.

a.     TN

b.    NS

c.     QM

d.    OR

e.      PL
_________________________

HOW TO GET THE ARRANGEMENT –

First you can easily mark T & R as per given condition.

Next you cannot mark N in the positions 9,2,5,7 because they are neighbour of R & T & also 1,6 are already occupied.

So N can go in 3,4,8

SMP go together because M is between S & P.

And S cannot go in position 2 & 9.

N is fourth to the left of P.

Suppose N goes to position 3 – Then since N is fourth to left of P, so P should occupy position 8. Since SMP go together. There is no position for them to go together in this arrangement

so N is not in position 3.

Suppose Now N goes to 8 position, in similar way P will occupy 4, M will occupy 3 and S will go to 2. But S cannot be in 2 because of neighbourhood to T.

So N will also not go in position 8.

Left is position 4. N will go to position 4. As depicted in the figure.

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