Odd One Out (SSC CGL 2022 Tier – 1)

(1^st letter - 2 = 2^nd letter) & (2^nd letter - 1 = 3^rd letter)

LJI → (L - 2 = J) & (J - 1 = I)
NLK → (N - 2 = L) & (L - 1 = K)
TRQ → (T - 2 = R) & (R - 1 = Q)

But,

GDA → (G - 2 ≠ D) & (D - 1 ≠ A)
Hence, GDA is the correct answer.

The pattern followed here is :

(1^st number ÷ 5 = 2^nd number) & (2^nd number ÷ 5 = 3^rd number)

(600 - 120 - 24) → 600 ÷ 5 = 120 & 120 ÷ 5 = 24
(300 - 60 - 12) → 300 ÷ 5 = 60 & 60 ÷ 5 = 12
(500 - 100 - 20) → 500 ÷ 5 = 100 & 100 ÷ 5 = 20

But,

(400 - 80 - 14) → 400 ÷ 5 = 80 & 80 ÷ 5 = 16 ≠ 14
Hence, (400 - 80 - 14) is the correct answer.

The pattern followed here is :

{1^st letter + 1 = alphabetical opposite of second letter} and {2^nd letter - 1 = 3^rd letter}
EUT ⇒ E + 1 = F  ↔ U (opposite) U - 1 = T
RHG ⇒ R + 1 = S ↔ H (opposite) H - 1 = G
TFE ⇒ T + 1 = U ↔ F (opposite) F - 1 = E

But,

TRQ ⇒ T + 1 = U ↔ F ≠ R (opposite) R - 1 = Q

Hence, TRQ is the correct answer.

The pattern followed here is :

(1^st number - 1) × 4 = 2^nd number)

(15 - 56) → {(15 - 1) × 4} = 56
(13 - 48) → {(13 - 1) × 4} = 48

(17 - 64) → {(17 - 1) × 4} = 64

But,
(11 - 39) → {(11 - 1) × 4} = 40 ≠ 39

Hence, (11 - 39) is the correct answer.