BPSC 71st CCE Prelims General Studies · Question 2 of 150

Correct — A, OMEFPA. Set the two words one above the other and replace every letter by its position in the alphabet, A = 1 up to Z = 26. SPRING is 19-16-18-9-14-7 and UNUFRC is 21-14-21-6-18-3. Subtract position by position and the code stops being letters and becomes arithmetic: S to U is +2, P to N is −2, R to U is +3, I to F is −3, N to R is +4, G to C is −4. The step therefore grows 2, 3, 4 while the direction alternates forward, backward, forward, backward — a progressive alternating shift, not the constant shift most candidates are drilled on. Apply exactly the same six operations, in the same order, to MOBILE, which is 13-15-2-9-12-5. M plus 2 is 15, which is O. O minus 2 is 13, which is M. B plus 3 is 5, which is E. I minus 3 is 6, which is F. L plus 4 is 16, which is P. E minus 4 is 1, which is A. The code word is O-M-E-F-P-A, option (a). Now decode it back, because a coding answer you have not reversed is an answer you have not verified: O minus 2 gives M, M plus 2 gives O, E minus 3 gives B, F plus 3 gives I, P minus 4 gives L, A plus 4 gives E — MOBILE, exactly. No wrap-around is needed anywhere here, because the smallest result is E minus 4, which lands on position 1 and stops precisely at A rather than running past the start of the alphabet.

• (b)OMPGNC — The dangerous one, and it is engineered to be. It opens with the same two letters as the answer, O and M, so a candidate who checks the rule on the first pair of letters only, concludes the shift is a constant plus-two/minus-two and stops testing has no way at all to tell it from option (a). Its last three letters give the design away: G, N and C are exactly what a constant plus-two/minus-two rule produces from I, L and E. A candidate reasoning that way actually derives OMDGNC, cannot find it among the options, and settles for the string that differs from it by a single letter instead of going back and re-deriving the rule. The third letter is the entire question — it must be E, because the step has grown to 3 by then, and this option prints P.
• (c)MPQSUL — This option leaves the first letter untouched, M in the plain word and M in the code, and that alone rules it out: all six letters of SPRING moved when it became UNUFRC, so no letter of MOBILE may stay where it is. Past the first letter the shifts it would require are +1, +15, +10, +9 and +7 — no constant step, no growth, no alternation. Its job on the page is to catch the candidate who never derives a rule at all and picks whichever string most resembles the original word.
• (d)SEGRFT — Turning MOBILE into SEGRFT would need shifts of +6, −10, +5, +9, −6 and +15: no constant step, no growing step, no alternation and no mirror. It is deliberate noise, put there so that the four options are not all near-misses of one another, and it is chosen only by candidates guessing on the look and rhythm of the letters. It also fails the cheapest structural check available — not one of its letters sits where any simple forward or backward shift of MOBILE would put it.

Every letter-shift coding question in Indian competitive exams belongs to one small family, and that family is best handled as arithmetic on positions rather than as manipulation of letters. Convert each letter to its ordinal, A = 1 through Z = 26, and a code turns into a sequence of signed differences. Those differences then follow one of a handful of patterns: a constant shift, which is the Caesar cipher that Suetonius records Julius Caesar using with a step of three; an alternating shift, forward then backward; a progressive shift, in which the size of the step grows along the word; a mirrored substitution, where each letter is replaced by its opposite in the alphabet; or a straight letter-to-number substitution. BPSC's SPRING to UNUFRC combines two of these at once — the direction alternates and the magnitude progresses — which is precisely why a candidate trained only on the constant shift finds it slippery. The one working habit that covers the whole family is to write the difference sequence out as numbers before you look at the options at all. Written as letters the pattern is invisible; written as +2, −2, +3, −3, +4, −4 it cannot be missed. A standing hazard applies to all of them: a shift running past Z wraps round to A and one running below A wraps back to Z, so the arithmetic is done modulo 26.

The reasoning is four steps and none may be skipped. First, write the alphabet positions of both words: SPRING is 19-16-18-9-14-7 and UNUFRC is 21-14-21-6-18-3. Second, subtract to get the difference sequence, +2, −2, +3, −3, +4, −4. Third — and this is the step that decides the question — check the differences across all six letters and not merely the first two, because the first pair on its own suggests a constant step of 2 and that misreading is exactly what the paper is testing you against. Fourth, apply the six operations to MOBILE (13-15-2-9-12-5) to get 15-13-5-6-16-1, which reads OMEFPA, and then decode it back to confirm. The single fact that discriminates is the size of the third shift. Options (a) and (b) both begin O-M, so the first two letters settle nothing; the third letter is E if the step has grown to 3 and would have been D if the step had stayed at 2. Option (b) prints neither, so the moment you compute the third letter honestly you are left with one answer. If time is short there is a faster elimination. Every letter of SPRING moved, so option (c), which leaves M as M, is impossible on structure alone; option (d) needs shifts of +6, −10, +5, +9, −6, +15, which is no pattern of any kind. That leaves two options sharing their opening pair, and one letter of arithmetic separates them. A memorised aid makes the whole method fast enough for the exam hall: EJOTY — E is 5, J is 10, O is 15, T is 20, Y is 25 — puts any letter within two counts of an anchor.

• The rule here is a progressive alternating shift: S+2=U, P−2=N, R+3=U, I−3=F, N+4=R, G−4=C, so the step size grows 2, 3, 4 while the direction alternates forward and backward along the word.
• Applied to MOBILE the same six operations give M+2=O, O−2=M, B+3=E, I−3=F, L+4=P, E−4=A, that is OMEFPA — and reversing them returns MOBILE, which is the verification step most candidates skip.
• EJOTY is the standard mnemonic for alphabet positions: E=5, J=10, O=15, T=20, Y=25, with A=1 and Z=26. Five anchors let you convert any letter to its ordinal in under a second, which is what makes the difference-sequence method fast enough to use under time pressure.
• The opposite-letter rule is the second most-tested coding pattern after the constant shift: a letter and its mirror image in the alphabet have positions that always add to 27, so A pairs with Z, B with Y, and M with N.
• Shifts are computed modulo 26 — a step past Z wraps to A and a step below A wraps to Z. No wrap arises in this question because the smallest result, E minus 4, lands exactly on position 1, the letter A. The constant-shift special case is the Caesar cipher, recorded by Suetonius as a shift of three places.

• Verifying the rule on the first one or two letters only, concluding the shift is constant, and never testing whether the step grows — the exact error option (b) is built to reward
• Getting the direction of the alternation the wrong way round, which yields a string with the right letters in the wrong places
• Forgetting the wrap at the ends of the alphabet, so that a shift past Z or below A is computed as an impossible position instead of coming round to A or Z

BPSC sets one letter-coding item in its opening reasoning block in every cycle and keeps it mechanical — one worked example, one word to encode, no conditional clauses and no sentences to decode. UPSC set exactly this family in GS Paper-I until 2010, most cleanly in 1998 with SAND written as VDQG, a plain constant shift of three, and then moved all reasoning to the CSAT Paper-II. The practical difference is the shape of the shift: UPSC's older items favoured a constant step, while BPSC's step grows along the word, so checking the rule against every letter of the example rather than the first pair is what separates a correct answer from a near-miss here.

• practice — not a real PYQ

  In a certain code, CANDLE is written as DCQHQK. How will the word FLOWER be written in the same code?

  • (a)GNRAJX
  • (b)GNRZJW
  • (c)EJLSZL
  • (d)GNSAKX

  Answer(a) GNRAJX — CANDLE to DCQHQK uses the shift row +1, +2, +3, +4, +5, +6; applying the same row to FLOWER gives G, N, R, A, J, X. Note the fourth letter: W plus 4 runs past Z and wraps round to A.

• practice — not a real PYQ

  If in a code language RAIN is written as OXFK, then how will the word CLOUD be written in that language?

  • (a)ZILRA
  • (b)ZIMRA
  • (c)FORXG
  • (d)ZILSA

  Answer(a) ZILRA — every letter moves three places back, wrapping past A to the end of the alphabet, so C, L, O, U, D become Z, I, L, R, A.