BPSC 71st CCE Prelims General Studies · Question 10 of 150

Correct — D, 8@7$%6. Build the letter-to-symbol table from the two words you are given, reading each sample position by position because the code preserves letter order. RAIN = 8$%6 fixes R = 8, A = $, I = % and N = 6. MORE = 7#8@ fixes M = 7, O = #, R = 8 and E = @. R appears in both samples and is 8 in both, which is the built-in proof that a single consistent code is at work rather than two unrelated ones. Between them the samples supply seven letters — R, A, I, N, M, O and E — against seven distinct symbols, a clean one-to-one mapping with nothing left to guess. Now check coverage before writing anything: REMAIN is spelt R-E-M-A-I-N and every one of those six letters is in the table, so the word can be encoded exactly. Reading them off in order gives 8, @, 7, $, %, 6, that is 8@7$%6. Two glance-level filters get you there without decoding a single option. First, REMAIN begins with R, so the code must begin with 8 — that alone eliminates options (a) and (b), which begin with @ and % respectively. Second, the only letter the samples supply that REMAIN does not contain is O, coded #, so any option carrying a # is impossible — that eliminates (b) again and (c). A third filter finishes the job for anyone who prefers it: REMAIN's six letters are all different, so its code must show six different symbols, and option (b) repeats %. Every wrong option is the right alphabet in the wrong arrangement or with a foreign symbol smuggled in, which is exactly what this question type is built to punish.

• (a)@$86%7 — Decoded symbol by symbol this reads E, A, R, N, I, M. It draws on the right table and uses six different symbols, but it arranges them in the wrong order, so it spells an anagram of REMAIN rather than REMAIN itself. It is the option built for the candidate who verifies that all the required symbols are present and stops there. The one-second test that kills it: the word starts with R, so the code must start with 8, and this one starts with @.
• (b)%7@%6# — Decodes to I, M, E, I, N, O. It fails three independent checks at once — it does not begin with 8, it repeats % for I although REMAIN has no repeated letter, and it brings in # for O, which is the one letter the samples supply that REMAIN does not contain. Any single one of those three observations rules it out before the full decoding is finished, which makes it the weakest distractor of the set.
• (c)86@$#7 — Decodes to R, N, E, A, O, M. It is the most tempting wrong answer because it opens with 8, so it survives the first-letter screen that removes options (a) and (b). But it introduces # for O — a letter absent from REMAIN — and drops % for I altogether, so both the set of letters used and their order are wrong. Checking which symbols may legitimately appear, rather than only which order they come in, is what separates it from the answer.

This is a one-to-one substitution code: every letter is represented by a fixed symbol, the same symbol always stands for the same letter, and no two letters share a symbol. The two sample words are the key. Because each sample lists its symbols in the same order as its letters, the mapping can be read off position by position, and any letter that appears in both samples doubles as a built-in consistency check — if R had come out as 8 in one sample and something else in the other, the samples would not belong to a single code and the question would be unanswerable. Coding items in these papers come in two families and it is worth knowing which one you are looking at before you start. The substitution family, as here, has no internal logic to discover: the symbols are arbitrary, nothing can be computed, and the only work is to tabulate and look up. The rule-based family — where SAND becomes VDQG because every letter has moved three places forward — has no table at all, only an operation on alphabet positions, and there the samples exist to reveal the shift rather than to be copied. A quick tell: if the code characters are non-letters, as they are here, it must be a substitution table; if they are letters, test the alphabet distance between each plain letter and its code letter before assuming a table. The third thing to check in either family is coverage. A code can only encode letters the samples have already fixed, so a target word containing a letter that appears in neither sample cannot be encoded at all — a variant BPSC and other commissions set deliberately.

Two steps and three screens decide this mark. Step one, write a small table of letter against symbol and confirm that every letter of the target word appears in the samples — here R, E, M, A, I and N are all covered, so nothing has to be inferred. Step two, spell REMAIN out letter by letter and pick up its symbol from the table in that order, resisting the urge to work from the shape of the options. The wrong options are all built from the same symbol pool, so a scrupulous ordering is the only real defence, and decoding each option back into letters is the fastest way to expose them. Before that, though, three screens will usually settle the question on their own. The first letter screen: REMAIN starts with R, R is 8, so the code starts with 8, which leaves only options (c) and (d) standing. The foreign symbol screen: the samples supply exactly one letter that REMAIN lacks, O, coded #, so no # may appear — and that removes (c). The repetition screen: REMAIN has six distinct letters, so six distinct symbols must appear. Applied in that order the three screens isolate option (d) in a few seconds, which matters on a paper that allows about 48 seconds a question and deducts one-third of a mark for a wrong tick. The single discriminating fact is the treatment of #: it is the one symbol in the table with no business in this answer, and two of the three distractors contain it.

• From RAIN = 8$%6, read position by position: R = 8, A = $, I = % and N = 6.
• From MORE = 7#8@: M = 7, O = #, R = 8 and E = @ — seven letters and seven distinct symbols across the two samples, a clean one-to-one mapping.
• R occurs in both samples and is 8 in both, which confirms that the two examples belong to one consistent code and that the table may be merged.
• REMAIN spelt out is R, E, M, A, I, N, which maps in that order to 8, @, 7, $, %, 6 — option (d), 8@7$%6.
• The symbol # stands for O, the only letter the samples supply that REMAIN does not contain, so any option containing # can be rejected at sight; that single screen removes options (b) and (c).
• REMAIN has six distinct letters, so its code must contain six distinct symbols and must begin with 8 — screens that between them eliminate every wrong option without decoding any of them in full.

• Matching symbols to letters by position across two different words instead of merging both samples into one table
• Skipping the consistency check on the letter that appears in both samples, which is the only evidence that the two examples share a code
• Getting the table right and then assembling the answer in the wrong order — the defect option (a) is built on
• Ignoring which symbols are permitted at all: # codes for O, a letter REMAIN does not contain, yet two options include it
• Failing to check coverage first, which matters most in the harder variant where a letter of the target appears in neither sample

BPSC usually pairs a symbol-substitution item with a shift-code item inside the opening reasoning block, keeps the target word short, and builds every distractor from the correct symbol pool so that the mark turns on ordering and on which symbols are admissible rather than on the decoding itself. A harder BPSC variant leaves one letter of the target uncovered by the samples, in which case the correct response is that the code cannot be determined, so coverage should always be checked before any symbol is written. UPSC set exactly this family in the General Studies paper until mental ability moved to Prelims Paper-II in 2011, and that paper has been merely qualifying at 33 per cent since 2015; its versions leaned towards the rule-based side — a fixed alphabet shift to be inferred from two sample words, or a positional transformation to be applied to a fresh string — so the samples had to be analysed rather than merely tabulated.

• practice — not a real PYQ

  In a certain code PLANT is written as 5@3#7 and GRAPE is written as 2$35&. How will the word PEARL be written in that code?

  • (a)5&3$@
  • (b)5@3$&
  • (c)&53$@
  • (d)5&3@$

  Answer(a) 5&3$@ — the samples give P = 5, L = @, A = 3, N = #, T = 7, G = 2, R = $ and E = &, with A = 3 and P = 5 confirmed in both, and PEARL follows in that order.

• practice — not a real PYQ

  In a code language CHAIR is written as %2@8# and TABLE is written as 6@*&4. What is the code for BEACH?

  • (a)*@4%2
  • (b)4*@%2
  • (c)*4@%2
  • (d)*4@2%

  Answer(c) *4@%2 — B = *, E = 4, A = @, C = % and H = 2, and A is @ in both samples, which confirms the table. Option (d) has C and H the wrong way round.