RAIN is written in a code as 8$%6 and MORE is mentioned as 7#8@. How will REMAIN be mentioned in code language?

Question

UnlockIAS · Accepted Answer

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.

Why the other options are wrong — (a) 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) 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) 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.

Concept: 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.

State PCS 2025 Q10 — Science & Technology PYQ with Answer

State PCS 2025 Q10 — Science & Technology PYQ with Answer