Analogy varieties Question TCS (Part – II)

First and last letter are interchanged, third letter is written in reverse alphabetical order and 5 is added and subtracted from the second and fourth letter respectively.

P - R - I - N - T → T - (R + 5 = W) - R - (N - 5 = I) - P

C - L - O - S - E → E - (L + 5 = Q) - L - (S - 5 = N) - C

Hence, EQLNC is the correct answer.

Each number is in the form of [(n³ - 3) : {(n + 2)³ - 3} : {(n + 4)³ - 3}]

[(5³ - 3) : {(5 + 2)³ - 3} : {(5 + 4)³ - 3}] → (122 : 340 : 726)

Similarly,

[(13³ - 3) : {(13 + 2)³ - 3} : {(13 + 4)³ - 3}] → (2194 : 3372 : 4910)

Hence, (2194 : 3372 : 4910) is the correct answer.

{(a² + 4), (b² + 4), (c² + 4)}, Where a, b and c are consecutive odd numbers in reverse order.
So, (173 : 125 : 85) → {(13² + 4), (11² + 4), (9² + 4)}
Similarly,

(365 : 293 : 229) → {(19² + 4), (17² + 4), (15² + 4)}

Hence, (365 : 293 : 229) is the correct answer.

The difference between the sum of digit of the number is same.
(93 : 87) → (9 + 3) - (8 + 7) = - 3
Similarly,

(58 : 55) → (5 + 8) - (5 + 5)  = - 3

Hence, (58 : 55) is the correct answer.

Increase 1 and then reverse the letter.

PORCH → QPSDI and now on reversing we get IDSPQ.
Similarly,

MOIST → NPJTU and now on reversing we get UTJPN.

Hence, UTJPN is the correct answer.

Hence, IH is the correct answer.

23 ↔ 59 ↔ 18
(2 + 3) → 59 ← (1 + 8)

Similarly, Option (b) follows the pattern.

71 ↔ 83 ↔ 21
(7 + 1) →83 ← (2 + 1)

Hence, (71, 83, 21) is the correct answer.

Hence, TTRP is the correct answer.

Each number is in the form of [(a³ - 1) : {(b³ - 1} : {(c³ - 1}], Where a, b and c are consecutive natural numbers

(26, 63, 124) → {(3)³ - 1}, {(4)³ - 1}, {(5)³ - 1} ⇒ (27 - 1), (64 - 1), (125 - 1) ⇒ (26, 63, 124)
Similarly,
(63, 124, 215) → {(4)³ - 1}, {(5)³ - 1}, {(6)³ - 1} ⇒ (64 - 1), (125 - 1), (216 - 1) ⇒ (63, 124, 215)

Hence, (63, 124, 215) is the correct answer.