In a certain code system, words are transformed according to the following rules based on … — State PSC Exam

In a certain code system, words are transformed according to the following rules based on letter positions indexed $1, 2, 3, \dots, n$ from left to right:
1. Letters at odd position indices ( $1, 3, 5, \dots$ ) are replaced by their reverse alphabetical counterparts (where $A \leftrightarrow Z$ , $B \leftrightarrow Y$ , $C \leftrightarrow X$ , etc.).
2. Letters at even position indices ( $2, 4, 6, \dots$ ) are replaced by the letter shifted forward in the English alphabet by a number of positions equal to their 1-based index (wrapping around from $Z$ to $A$ if necessary).

Based on these rules, what is the correct code for the word 'JOURNEY'?

QQFVMKBAnswer
B
KQXVSKF
C
PQEVLKA
D
KLXISVF

Answer

QQFVMKB

The correct answer 'QQFVMKB' is obtained by applying the two positional rules step-by-step: at odd positions (1, 3, 5, 7), the reverse alphabetical positions (

27 - \text{rank}

) yield Q, F, M, and B respectively. At even positions (2, 4, 6), adding the letter's 1-based index to its rank gives Q (15 + 2 = 17), V (18 + 4 = 22), and K (5 + 6 = 11). Combining these yields 'QQFVMKB'.

Step-by-Step Solution

Identify the 1-based position index and standard alphabetical rank of each letter in 'JOURNEY'

J (pos 1, rank 10), O (pos 2, rank 15), U (pos 3, rank 21), R (pos 4, rank 18), N (pos 5, rank 14), E (pos 6, rank 5), Y (pos 7, rank 25).

Establishing letter positions and numerical ranks is required before applying positional transformation rules.

Apply reverse alphabetical substitution (

27 - \text{rank}

) to letters at odd positions (1, 3, 5, 7)

Position 1:

27 - 10 = 17 \rightarrow \text{Q}

; Position 3:

27 - 21 = 6 \rightarrow \text{F}

; Position 5:

27 - 14 = 13 \rightarrow \text{M}

; Position 7:

27 - 25 = 2 \rightarrow \text{B}

Letters at odd position indices must be replaced by their reverse counterparts.

Apply forward shift by position index

(\text{rank} + \text{index})

to letters at even positions (2, 4, 6)

Position 2:

15 + 2 = 17 \rightarrow \text{Q}

; Position 4:

18 + 4 = 22 \rightarrow \text{V}

; Position 6:

5 + 6 = 11 \rightarrow \text{K}

Letters at even position indices are shifted forward by their respective 1-based index.

Combine the transformed letters in order from position 1 to 7

The encoded string is QQFVMKB.

Assembling the results from all letter positions yields the complete final code.

Key Concept

Positional Hybrid Coding (Reverse Indexing combined with Index-Based Forward Shift)

Estimated Time:1m 30s

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