In a certain code language, ‘GAME’ is written as ‘$ ÷ * %’ and ‘BEAD’ is written as ‘# % ÷ ×’. How will the word ‘MADE’ be written in that code language?
- (a)$ ÷ × %
- (b)* ÷ $ %
- (c)* ÷ × %
- (d)# ÷ × %
Correct — C, * ÷ × %. This is a positional substitution code and both clues agree, so it takes under a minute once you line them up. Step 1 — read each clue position by position. GAME is '$ ÷ * %', so G = $, A = ÷, M = *, E = %. BEAD is '# % ÷ ×', so B = #, E = %, A = ÷, D = ×. Step 2 — check the clues against each other. The two words share exactly two letters, A and E, and the two codes agree on both of them: A = ÷ in the second position of GAME and the third of BEAD, E = % in the fourth position of GAME and the second of BEAD. Nothing is jumbled and nothing is ambiguous — one consistent map covers both clues, which is what tells you the code is positional rather than a set of symbols to be rearranged. Step 3 — build MADE letter by letter. M = *, A = ÷, D = ×, E = %. Writing them in the word's own order gives * ÷ × %, which is option (C). The efficient route skips even that. G and M are the only letters unique to GAME, and $ and * the only symbols unique to its code, so {G, M} = {$, *}; position fixes which is which. Every wrong option here differs from the right one in exactly one place, and each substitutes a letter that appears in a clue but not in MADE — so the fastest check is not to decode the options but to ask, of each, which letter it has smuggled in.
- (a)$ ÷ × % — $ ÷ × % decodes as G-A-D-E. Only the first symbol is wrong: it carries $, which is G, where MADE needs M's symbol *. This is the closest wrong answer on the paper and the one a hurried candidate takes, because the last three symbols are already correct and the eye stops checking once a pattern looks right. G and M are precisely the pair that positional reading separates.
- (b)* ÷ $ % — * ÷ $ % decodes as M-A-G-E. It gets M and A right, then puts $ — again G — in the third position where D's symbol × belongs. It is the mirror of option (a): the same wrong letter smuggled in at a different place, which is why checking symbol-by-symbol against the word beats matching shapes.
- (d)# ÷ × % — # ÷ × % decodes as B-A-D-E. The last three symbols are right and the first is #, which is B — a letter that appears in the BEAD clue but not in MADE at all. The option is tempting because BEAD and MADE share three of four letters, so a candidate who anchors on the second clue and forgets to swap its opening letter lands here.
A substitution code assigns one symbol to one letter and keeps the order of the word. That second half is what makes an item like this solvable: if the symbols could be rearranged, two clues sharing two letters would not pin anything down. The standard method is to intersect the clues — find the letters two code-words share, find the symbols their codes share, and the two sets must correspond. Here GAME and BEAD share A and E, and their codes share ÷ and %, so {A, E} = {÷, %}; position then says A = ÷ and E = %, consistently in both. Stripping the shared pair out leaves {G, M} = {$, *} from the first clue and {B, D} = {#, ×} from the second, and the whole alphabet of the puzzle is fixed: G = $, A = ÷, M = *, E = %, B = #, D = ×. Six letters, six symbols, no ambiguity. An examiner who wanted the item to be genuinely hard would supply clues that share only one letter, which leaves two maps alive and forces a case split.
Do not decode the options first — decode the clues, then build the target word and read the answer off. Candidates lose time here by taking each option in turn and testing it, which is four times the work and invites the error the paper is fishing for. The discriminating move is positional reading. Every one of the three wrong options has three of its four symbols right, so any method that checks 'does this option use the right symbols' passes all four. What separates them is *where* each symbol sits and which letter it stands for: (a) puts G's symbol where M belongs, (b) puts G's symbol where D belongs, (d) puts B's symbol where M belongs. Build M-A-D-E in order — * ÷ × % — and the question answers itself before you have looked at a single option.
- Rule: this is a POSITIONAL substitution code — symbol n of the code stands for letter n of the word. Check it by taking the letters two clue words share and confirming they carry the same symbol in each; if they do, read straight off, and only if they do not do you fall back on intersecting the sets.
- The full map this item fixes: G = $, A = ÷, M = *, E = %, B = #, D = × — six letters, six symbols, no letter left ambiguous.
- GAME and BEAD share exactly two letters (A, E) and their codes share exactly two symbols (÷, %); that correspondence is the engine of every intersection-method coding question.
- Option (b) uses both $ and * and gives D no symbol at all. Option (d) is not BEAD's code — BEAD is # % ÷ × and (d) is # ÷ × %, the same four symbols in a different order — which is the trap in miniature on a question decided by order.
- Each wrong option differs from * ÷ × % in exactly one position, and each substitutes a letter drawn from the clues: (a) and (b) smuggle in G, (d) smuggles in B. Spotting the intruding letter is faster than decoding four options.
- The same reading is confirmed by BPSC's own habit — in the 71st CCE paper, RAIN = 8$%6 and MORE = 7#8@ are strictly positional, with R = 8 in both, giving REMAIN = 8@7$%6
The highlighted first row is the engine: two shared letters force two shared symbols. The last row is the tie-breaker that separates option (c) from option (a), both of which pass the set test.
- Stopping at the set test: more than one option can carry the right family of symbols, and then only the order decides
- Using both symbols of a pair, which silently codes a letter the target word does not contain
- Picking the string that repeats one of the clue codes verbatim, which always leaves one letter of the target uncoded
BPSC puts reasoning inside the General Studies paper itself rather than in a separate aptitude paper, and its coding items run in two families — the alphabet-shift type and this symbol-substitution type built on two overlapping clue words. The 71st CCE paper carried one of each, at Q2 and Q10. UPSC used to do exactly the same in General Studies Paper I before the CSAT was introduced in 2011, which is why its 1990s papers carry code questions and its current prelims does not; reasoning of this kind now lives in the qualifying Paper II.
If in a certain code SAND is VDQG and BIRD is ELUG, then what is the code for LOVE?
- (a) PRYG
- (b) ORTG
- (c) NPUH
- (d) ORYH
Answer(d) ORYH
The same task — two word-and-code pairs are given, you must recover the rule and apply it to a third word — from the years when UPSC still put reasoning inside General Studies Paper I. There the rule is mechanical, a uniform shift of three places forward, so both clues agree and no tie-break is needed; here the shift is not uniform — six letters take six unrelated symbols — so the map has to be built letter by letter from the two clues rather than inferred from one rule, which is what makes the BPSC item the longer of the two.
- practice — not a real PYQ
In a certain code language, 'FISH' is written as '2 5 8 3' and 'HERB' is written as '3 7 1 9'. Which digit stands for H?
- (a)2
- (b)3
- (c)8
- (d)9
Answer(b) 3 — H is the only letter common to FISH and HERB, and 3 is the only digit common to the two codes, so the two must correspond.
- practice — not a real PYQ
In a certain code language, 'ROSE' is written as '# % @ &' and 'ROAD' is written as '# % + ='. How is 'SEED' written in that code?
- (a)@ & & =
- (b)& @ @ =
- (c)@ & & +
- (d)# % & =
Answer(a) @ & & = — ROSE and ROAD share R and O, and the codes share # and %, so R = #, O = %; reading each clue in order gives S = @, E = &, A = +, D = =, and SEED becomes @ & & =.