Q. 3 If ππππ = ππππ;πππ = πππ; and ππ = ππ, then πππ =____ .
(A) JDE
(B) JED
(C) JDC
(D) JCD
Solving IMHO = JNIP Coding Puzzle: IDC Equals JED
IMHO = JNIP, IDK = JEL, and SO = TP form a letter-shift pattern where each letter advances by one position in the alphabet. Applying this to IDC yields JED as the correct code. This reasoning puzzle tests pattern recognition skills common in competitive exams like GATE.β
Pattern Analysis
Each coded group replaces every original letter with the next letter alphabetically: I (9th) to J (10th), M (13th) to N (14th), H (8th) to I (9th), O (15th) to P (16th) in IMHO β JNIP. IDK follows suit with IβJ, D (4th)βE (5th), K (11th)βL (12th), while SO β TP shifts SβT and OβP. This +1 shift rule holds consistently across all examples.β
Correct Answer: JED
For IDC, apply the +1 shift: IβJ, DβE, C (3rd)βD (4th), resulting in JED. This matches the established pattern without exceptions.β
Options Breakdown
| Option | Code | Explanation |
|---|---|---|
| (A) JDE | JDE | Shifts IβJ and DβE correctly but CβE skips one position, violating the +1 rule.β |
| (B) JED | JED | Matches perfectly: IβJ, DβE, CβD as per all given examples.β |
| (C) JDC | JDC | Correct IβJ but fails on DβD (no shift) and CβC (no shift).β |
| (D) JCD | JCD | Shifts only IβJ correctly; DβC and CβD reverse the pattern.β |
Exam Tips
Practice spotting uniform shifts in letter codes to solve similar verbal ability questions quickly. Review GATE ME 2019 papers for more examples, where this appeared in General Aptitude.β
