NIMCET 2023 — Reasoning PYQ
NIMCET | Reasoning | 2023If DENMARK in coded on FCPKCPM Then code SINGAPORE of which option.
Choose the correct answer:
- A.
UGPECNQPG
(Correct Answer) - B.
UGPFCNRPG
- C.
UGPEDNQTC
- D.
UGPECNQTC
UGPECNQPG
Explanation
Let
B=−100amp;−1amp;−1amp;0amp;2amp;−1amp;−1=−I+N,N=000amp;−1amp;0amp;0amp;2amp;−1amp;0,N3=0.
So
B19=(−I+N)19=−I+19N−171N2,
Now
N2=000amp;0amp;0amp;0amp;1amp;0amp;0.
Sum of entries is linear, hence
Sum(B19)=Sum(−I)+19Sum(N)−171Sum(N2)=−3+19⋅0−171⋅1=−174.
Pattern in coding DENMARK – FCPKCPM is alternating:
+2, -2, +2, -2, +2, -2, +2
Apply same to SINGAPORE with 9 letters → pattern becomes:
+2, -2, +2, -2, +2, -2, +2, -2, +2
Apply shifts:
S(+2) = U
I(-2) = G
N(+2) = P
G(-2) = E
A(+2) = C
P(-2) = N
O(+2) = Q
R(-2) = P
E(+2) = G
Final code: UGPECNQPG
Answer: (a) UGPECNQPG
Explanation
Let
B=−100amp;−1amp;−1amp;0amp;2amp;−1amp;−1=−I+N,N=000amp;−1amp;0amp;0amp;2amp;−1amp;0,N3=0.
So
B19=(−I+N)19=−I+19N−171N2,
Now
N2=000amp;0amp;0amp;0amp;1amp;0amp;0.
Sum of entries is linear, hence
Sum(B19)=Sum(−I)+19Sum(N)−171Sum(N2)=−3+19⋅0−171⋅1=−174.
Pattern in coding DENMARK – FCPKCPM is alternating:
+2, -2, +2, -2, +2, -2, +2
Apply same to SINGAPORE with 9 letters → pattern becomes:
+2, -2, +2, -2, +2, -2, +2, -2, +2
Apply shifts:
S(+2) = U
I(-2) = G
N(+2) = P
G(-2) = E
A(+2) = C
P(-2) = N
O(+2) = Q
R(-2) = P
E(+2) = G
Final code: UGPECNQPG
Answer: (a) UGPECNQPG

