If ZERO is encoded as ADSN, then how do you encode STOP?

In the context of the UPSC Civil Services Aptitude Test (CSAT)—officially General Studies Paper-II, introduced in 2011 to evaluate analytical and reasoning skills—coding-decoding questions test a candidate's 'General Mental Ability'.

Why Option B is correct: To solve this cryptographic puzzle, we must first decipher the alphanumeric sequence applied to the reference word, 'ZERO'. By assigning numerical values to the standard English alphabet (A=1 through Z=26), we map the transformation:

• Z (26) + 1 = A (1) (wrapping around the alphabet)
• E (5) - 1 = D (4)
• R (18) + 1 = S (19)
• O (15) - 1 = N (14)

The underlying pattern is an alternating '+1, -1, +1, -1' letter shift. Applying this identical operational sequence to the target word 'STOP':

• S (19) + 1 = T (20)
• T (20) - 1 = S (19)
• O (15) + 1 = P (16)
• P (16) - 1 = O (15)

The generated cipher is TSPO, affirming Option B as the logically correct answer.

Why the other options are incorrect:

• Option A (SPOT): This is incorrect because it represents a straightforward anagrammatic rearrangement of the letters in 'STOP', completely ignoring the numerical shifting rule derived from the reference word.
• Option C (TSOP): This is incorrect because, while it correctly shifts the first three letters, it fails to apply the final '-1' shift to the letter 'P' (which must become 'O').
• Option D (POST): This is incorrect as it merely scrambles the letters to form another valid English word rather than methodically applying the deduced '+1, -1' cipher.

Concluding Takeaway: For rapid decoding in the CSAT examination, utilize the EJOTY mnemonic (E=5, J=10, O=15, T=20, Y=25) to quickly anchor alphabet positional values. Always verify the shifting pattern across all letters of the reference word before calculating the final answer.