Coding-Decoding for Punjab Police SI

Answer: 1 — CLUE

Compare the pairs letter by letter. LUTE → MUTE: only the first letter changes, L → M, one step forward. FATE → GATE: again only the first letter changes, F → G, one step forward. The rule is therefore 'move the first letter one place ahead in the alphabet and leave the rest alone'. Applying it to BLUE, B → C, giving CLUE. Options 2, 3 and 4 change the first letter by more than one step (B → G, B → F, B → S).

Answer: 2 — HPSSLFW

Line the code up under the word and take the shift at each position: A→Y (−2), C→D (+1), Q→O (−2), U→V (+1), I→G (−2), R→S (+1), E→C (−2). The rule is simply −2 on the odd positions and +1 on the even positions. Apply it to JOURNEY: J−2 = H, O+1 = P, U−2 = S, R+1 = S, N−2 = L, E+1 = F, Y−2 = W → HPSSLFW. Option 1 (HPSTLFW) mishandles position 4, giving T instead of S; option 3 (HPSSFLW) swaps positions 5 and 6; option 4 (HPSSLFX) moves the last letter by −1 instead of −2.

Answer: 2 — MSP

Every letter moves two places back in the alphabet. Z−2=X, I−2=G, P−2=N → XGN. P−2=N, E−2=C, T−2=R → NCR. Apply to OUR: O−2=M, U−2=S, R−2=P → MSP. Option 1 (NSP) shifts only the first letter by one; option 3 (NTQ) shifts every letter by one.

Answer: 4 — I9K7

The rule alternates by position. Letters in odd positions move two places forward; letters in even positions are replaced by the digit sum of their alphabet number. Check JUTE: J(1st) → L (+2); U(2nd) = 21 → 2+1 = 3; T(3rd) → V (+2); E(4th) = 5 → 5. Gives L3V5. ✓ Check SILVER: S → U; I = 9 → 9; L → N; V = 22 → 4; E → G; R = 18 → 1+8 = 9. Gives U9N4G9. ✓ Now GRIP: G(1st) → I; R(2nd) = 18 → 9; I(3rd) → K; P(4th) = 16 → 1+6 = 7. So GRIP = I9K7. Option 1 (I8K7) mis-sums R, and option 3 (I9K8) mis-sums P as 8.

Answer: 4 — CQB

The three positions are shifted forward by 1, 2 and 3. CAT: C + 1 = D, A + 2 = C, T + 3 = W, giving DCW. DOG: D + 1 = E, O + 2 = Q, G + 3 = J, giving EQJ. For BOY: B + 1 = C, O + 2 = Q, and Y + 3 runs past Z and wraps round to B, so the term is CQB. Option 3 (CQZ) is the trap — it stops the wrap one letter short at Z (Y + 1) — while option 1 wraps too far to A and option 2 mis-shifts the first letter.

Answer: 2 — ODWT

Compare the letters position by position. G(7) → I(9) is +2; M(13) → J(10) is −3; T(20) → X(24) is +4; R(18) → M(13) is −5. The second pair obeys the same rule: F(6) → H(8) +2, L(12) → I(9) −3, U(21) → Y(25) +4, T(20) → O(15) −5. Apply +2, −3, +4, −5 to MGSY: M(13)+2 = O, G(7)−3 = D, S(19)+4 = W, Y(25)−5 = T. That gives ODWT. OIVT keeps only the first letter right; RIVS and RDWS start from a wrong shift on M.

Answer: 1 — ECLAN

Two things happen at once. The first and last letters move +3 in the alphabet; the three middle letters are rotated one place to the right. TWINE: T+3 = W, E+3 = H; middle W, I, N → N, W, I. Code = W NWI H = WNWIH. WIRED: W+3 = Z, D+3 = G; middle I, R, E → E, I, R. Code = Z EIR G = ZEIRG. BLACK: B+3 = E, K+3 = N; middle L, A, C → C, L, A. Code = E CLA N = ECLAN. DDKZM is what you get by shifting every letter and reversing — neither example supports that.

Answer: 1 — GRP

The three letter positions get different shifts: +2, +3, +2. AGE: A+2 = C, G+3 = J, E+2 = G → CJG. CAB: C+2 = E, A+3 = D, B+2 = D → EDD. EON: E+2 = G, O+3 = R, N+2 = P → GRP. GQP (option 4) shifts every letter by 2 and GSP shifts the middle by 4 — both ignore the uneven pattern.

Answer: 1 — BESRD

Split the five letters into the two outer ones and the middle three. Outer letters move 4 places back: FACED → F − 4 = B, D − 4 = Z. CHIME → C − 4 = Y (wrapping past A), E − 4 = A. Both match the first and last letters of the codes. Middle three rotate one place to the left: A, C, E → C, E, A ✓ and H, I, M → I, M, H ✓. Apply to FRESH: F − 4 = B, H − 4 = D; middle R, E, S rotates to E, S, R. Code = B + E S R + D = BESRD.

Answer: 4 — NYWXMJC

Each letter moves 4 places forward in the alphabet. F+4=J, O+4=S, R+4=V, E+4=I, I+4=M, G+4=K, N+4=R gives JSVIMKR; the same check works for HISTORY → LMWXSVC. For JUSTIFY: J+4=N, U+4=Y, S+4=W, T+4=X, I+4=M, F+4=J, Y+4=C, so the code is NYWXMJC. Options 1 and 2 use the same seven letters but with the middle ones swapped out of order, which is the usual trap in this pattern.

Answer: 3 — MGWXJX

Rule: each letter moves forward by its own position number in the word — 1st letter +1, 2nd +2, 3rd +3, and so on. Check: P+1=Q, A+2=C, P+3=S, E+4=I, R+5=W gives QCSIW; C+1=D, L+2=N, O+3=R, T+4=X, H+5=M gives DNRXM. LETTER has six letters, so L+1=M, E+2=G, T+3=W, T+4=X, E+5=J, R+6=X, i.e. MGWXJX. The tempting option 1 (MHWXJX) puts +3 on the second letter and +4 on the third, which breaks the progression.

Answer: 1 — ZXGRMT

Each letter is replaced by its opposite letter in the alphabet, i.e. position 27 − n (A↔Z, B↔Y, C↔X, ...). Check FACE: F(6) → 27 − 6 = 21 = U; A(1) → Z; C(3) → 24 = X; E(5) → 22 = V. That gives UZXV ✓. Check LAMB: L(12) → 15 = O; A → Z; M(13) → 14 = N; B(2) → 25 = Y. That gives OZNY ✓. Now apply it to ACTING: A(1) → Z, C(3) → X, T(20) → 27 − 20 = 7 = G, I(9) → 18 = R, N(14) → 13 = M, G(7) → 20 = T. So ACTING becomes ZXGRMT. All four options begin ZXG, so the item is decided by the last three letters; option 4 (ZXHRMT) slips on T → H instead of G, and options 2 and 3 slip on I and G respectively.

Answer: 1 — NRXGRE

Replace every letter by its opposite letter (A↔Z, B↔Y, …, i.e. position 27 − n), then interchange the first and the last letter of the result. UNABLE → opposites F, M, Z, Y, O, V = FMZYOV → swap the ends → VMZYOF. ✔ WEALTH → opposites D, V, Z, O, G, S = DVZOGS → swap the ends → SVZOGD. ✔ VICTIM → V→E, I→R, C→X, T→G, I→R, M→N gives ERXGRN → swap the ends → NRXGRE. Option 2 (NXRGRE) has the second and third letters interchanged, so it fails the opposite-letter step.

Answer: 4 — Y1O3

Two different treatments alternate by position in the word. Odd positions (1st, 3rd) → the opposite letter (A↔Z, i.e. 27 − position). Even positions (2nd, 4th) → the digit sum of the letter's own alphabet position. Check against ARC → Z9X: Letter  ·  Place  ·  Treatment  ·  Code A  ·  1  ·  opposite of A  ·  Z R  ·  2  ·  R = 18 → 1+8  ·  9 C  ·  3  ·  opposite of C  ·  X Check against RAIN → I1R5: R(1st) → I; A(2nd) = 1 → 1; I(3rd) → R; N(4th) = 14 → 1+4 = 5. ✓ Now BALL: B(1st) → Y; A(2nd) = 1 → 1; L(3rd) → O; L(4th) = 12 → 1+2 = 3. So BALL → Y1O3. Option 1 (Y1O4) is the trap — it writes the last L as the opposite-letter count or a miscounted digit sum.

Answer: 2 — VUVU

Rule: every letter goes to its opposite letter (A↔Z, B↔Y, C↔X, D↔W, E↔V, F↔U), and every digit is first read as that position in the alphabet and then turned into its opposite too (1 → A → Z, 2 → B → Y, 3 → C → X, 4 → D → W). Check: AB12 → Z, Y, Z, Y ✓ and CD34 → X, W, X, W ✓. Now EF56: E → V, F → U, 5 → E → V, 6 → F → U, giving VUVU. UVUV reverses the order of the pair and VVVV/UUUU ignore the second letter, so all three fail the worked examples.

Answer: 4 — REGNAMUI

Split each eight-letter word into blocks of 3, 2 and 3 and reverse each block on the spot. MAR|IG|OLD becomes RAM|GI|DLO = RAMGIDLO ✓, and HIB|IS|CUS becomes BIH|SI|SUC = BIHSISUC ✓. Now GER|AN|IUM becomes REG|NA|MUI = REGNAMUI. Option 1 (REGINAMU) is the trap of reversing 3-3-2 instead of 3-2-3.

Answer: 2 — BURP - URBP – WPDN

Two steps are at work. Step 1 — the first three letters rotate left, the fourth stays: BRUT → RUBT (BRU → RUB), DINS → INDS (DIN → IND). Step 2 — apply +2, −2, +2, −2 to the four letters: RUBT → (R+2)(U−2)(B+2)(T−2) = TSDR; INDS → (I+2)(N−2)(D+2)(S−2) = KLFQ. Test the options: 1. TORN → ORTN is correct, but ORTN should give QPVL, not QORP. 2. BURP → URBP is correct, and URBP gives (U+2)(R−2)(B+2)(P−2) = WPDN. Both steps hold. 3. FILE should rotate to ILFE, not ILEF — it fails at step 1. 4. LENS → ENLS is correct, but ENLS should give GLNQ, not CPJU.

Answer: 1 — DGSAJ

The rule has two parts: the first and last letters of the word swap places, and the three middle letters move back by 2, 3 and 4 places. FETCH → swap F and H → H E T C F → E−2 = C, T−3 = Q, C−4 = Y → HCQYF. ✔ GRAPE → swap G and E → E R A P G → R−2 = P, A−3 = X, P−4 = L → EPXLG. ✔ JIVED → swap J and D → D I V E J → I−2 = G, V−3 = S, E−4 = A → DGSAJ. Option 4 (DHTAJ) keeps the correct end swap but shifts the middle letters by −1, −3, −4, so it fails on the first middle letter.

Answer: 1 — TRPEQI

Reverse the word first, then shift the letters by 1, 2, 3, 4, 5, 6 in order. BANGLE reversed is ELGNAB; E+1 = F, L+2 = N, G+3 = J, N+4 = R, A+5 = F, B+6 = H, giving FNJRFH. Check on the second example: NUMBER reversed is REBMUN; R+1 = S, E+2 = G, B+3 = E, M+4 = Q, U+5 = Z, N+6 = T, giving SGEQZT. Apply it to CLAMPS: reversed it is SPMALC; S+1 = T, P+2 = R, M+3 = P, A+4 = E, L+5 = Q, C+6 = I, so the code is TRPEQI. Shifting without reversing produces the distractor spellings in options 2 to 4.

Answer: 2 — IPUQZ

The first four letters are swapped in pairs and the fifth is moved 5 places back in the alphabet. CLAMP: C-L becomes L-C, A-M becomes M-A, and P − 5 = K, giving LCMAK. BORED: B-O becomes O-B, R-E becomes E-R, and D − 5 counts back past A to Y, giving OBERY. For PIQUE: P-I becomes I-P, Q-U becomes U-Q, and E − 5 wraps to Z, so the code is IPUQZ. Options 1 and 3 keep J instead of I, i.e. they shift the first letter as well; option 4 fails the pair swap.

Answer: 2 — XELMER

Break each six-letter word into three pairs. GLAZED = GL, AZ, ED and its code LGWDDE = LG, WD, DE. The first pair is simply reversed (GL to LG) and so is the third (ED to DE). The middle pair is not reversed: A − 4 = W (counting back past A to W) and Z + 4 = D (counting past Z to D). Verify on QUILTS = QU, IL, TS to UQEPST = UQ, EP, ST: QU reverses to UQ, TS reverses to ST, and in the middle I − 4 = E, L + 4 = P. Now code EXPIRE = EX, PI, RE. Reverse the first pair: XE. Middle pair: P − 4 = L and I + 4 = M, giving LM. Reverse the last pair: ER. The code is XELMER. Option 3 (XMLRED) mixes up which middle letter moves back and which moves forward.

Answer: 1 — DDALW

The rule has two parts: the first and last letters of the word simply change places, and the three middle letters move forward by +3, +5 and +7 respectively. Check GRAPH: outer letters G and H swap, giving H at the front and G at the end. Middle letters R + 3 = U, A + 5 = F, P + 7 = W. Result H U F W G = HUFWG. Correct. Check FLECK: outer letters F and K swap, giving K at the front and F at the end. Middle letters L + 3 = O, E + 5 = J, C + 7 = J. Result K O J J F = KOJJF. Correct. Now WAVED: outer letters W and D swap, so D goes in front and W at the end. Middle letters A + 3 = D, V + 5 = A (V, W, X, Y, Z, A — the alphabet wraps round), E + 7 = L. Result: D D A L W. Option 3 (EDALV) shifts the outer letters instead of swapping them, and option 2 (DEBLX) uses the wrong middle shifts (+4 and +6) and moves the last letter forward to X instead of ending in W.

Answer: 4 — NEVOY

Write the word backwards, and move the letter that was originally first two places forward in the alphabet. Check PROBE: reversed it is E B O R P, and the original first letter P becomes P + 2 = R, so the code is E B O R R = EBORR. Correct. Check PLUMB: reversed it is B M U L P, and P + 2 = R, giving B M U L R = BMULR. Correct. Now WOVEN: reversed it is N E V O W, and the original first letter W becomes W + 2 = Y, so the code is N E V O Y. Option 1 (NEVOW) is a plain reversal with the +2 step left out — that is the trap, since both worked examples happen to start with P and only the last letter of the code reveals the shift.

Answer: 2 — RUST - TSUR – VVXV

Two rules chain together. Rule 1: the second group is the first group reversed. CRAB → BARC, SINK → KNIS. Rule 2: the third group comes from the second by shifting its successive letters +2, +3, +4, +5. Check: BARC → B+2 = D, A+3 = D, R+4 = V, C+5 = H = DDVH ✓. KNIS → K+2 = M, N+3 = Q, I+4 = M, S+5 = X = MQMX ✓. Now test the options: GRIP → PIRG ✓; PIRG → P+2 = R, I+3 = L, R+4 = V, G+5 = L = RLVL ✓. BART → TRAB ✓; TRAB → T+2 = V, R+3 = U, A+4 = E, B+5 = G = VUEG ✓. MONK → KNOM ✓; KNOM → K+2 = M, N+3 = Q, O+4 = S, M+5 = R = MQSR ✓. RUST → TSUR ✓, but TSUR → T+2 = V, S+3 = V, U+4 = Y, R+5 = W = VVYW, while the option prints VVXV. The reversal is correct in all four, so only the shift step separates them.

Answer: 4 — ELGAC

Reverse the word, then move the middle (third) letter 5 places forward. PLACE reversed is ECALP; the middle letter A + 5 = F, giving ECFLP. FINER reversed is RENIF; the middle letter N + 5 = S, giving RESIF. For CABLE: reversed it is ELBAC; the middle letter B + 5 = G, giving ELGAC. The trap is to reverse only and stop at ELBAC, or to pick GLADE, which is merely an anagram of the letters.

Answer: 3 — THLKRF

The rule has two halves. The outer letters simply swap in pairs: 1st with 6th and 2nd with 5th. The middle two letters stay in place but move forward in the alphabet — the 3rd letter by +3 and the 4th letter by +4. Check on WALKED: 1st and 6th give D and W, 2nd and 5th give E and A, L+3 = O, K+4 = O, so DEOOAW. Check on ENJOYS: S and E, Y and N, J+3 = M, O+4 = S, so SYMSNE. Apply to FRIGHT (F R I G H T): 1st and 6th give T and F, 2nd and 5th give H and R, I+3 = L, G+4 = K, giving T H L K R F. Option 2 (TKHFLR) is the trap for anyone who only reverses the word and then shuffles at random; the plain reversal THGIRF is not an option at all.

Answer: 4 — LMJ

First reverse the word, then shift the three letters by +1, -1, +1. AID reversed is DIA: D+1 = E, I-1 = H, A+1 = B, giving EHB. URN reversed is NRU: N+1 = O, R-1 = Q, U+1 = V, giving OQV. INK reversed is KNI: K+1 = L, N-1 = M, I+1 = J, giving LMJ. Option 1 (JMJ) is what you get if you forget to reverse the word first.

Answer: 4 — UQTOK

Rule: take the five letters in pairs — swap letters 1 and 2, swap letters 3 and 4, and shift the fifth letter forward by 3. Check: TRUNK → TR becomes RT, UN becomes NU, K+3=N, giving RTNUN; BLAZE → LB, ZA, E+3=H, giving LBZAH. For QUOTH: QU becomes UQ, OT becomes TO, H+3=K, giving UQTOK. Option 2 (UPQOH) leaves the last letter unshifted and scrambles the middle pair.

Answer: 2 — 86

Add the alphabet positions of the letters and double the total. CUT = 3 + 21 + 20 = 44, and 44 × 2 = 88. NUT = 14 + 21 + 20 = 55, and 55 × 2 = 110. Both given codes fit, so the rule is confirmed. BUT = 2 + 21 + 20 = 43, and 43 × 2 = 86. Option 3 (43) is exactly the trap: it is the plain sum, with the doubling step forgotten.

Answer: 4 — 435939713927

Each letter becomes the digit sum of its position in the alphabet. Check on PLAYER: P is 16 and 1 + 6 = 7; L is 12 and 1 + 2 = 3; A is 1; Y is 25 and 2 + 5 = 7; E is 5; R is 18 and 1 + 8 = 9. That spells 731759, the code given. Now do MUNICIPALITY letter by letter: M 13 → 4, U 21 → 3, N 14 → 5, I 9 → 9, C 3 → 3, I 9 → 9, P 16 → 7, A 1 → 1, L 12 → 3, I 9 → 9, T 20 → 2, Y 25 → 7. The code is 435939713927. All four options share the first six digits, so the work is in the tail. Option 3 turns the P into 1, option 2 turns the L into 8 and option 1 ends the Y as 0.

Answer: 4 — 26-24-18-23

Find the rule from PLAN. P is the 16th letter and is coded 11; L is 12th and is coded 15; A is 1st and is coded 26; N is 14th and is coded 13. In every case code = 27 − (position of the letter), i.e. each letter is replaced by the position of its opposite letter counted from Z (A↔Z = 26, B↔Y = 25, and so on). Apply it to ACID: A = 1 → 27 − 1 = 26; C = 3 → 27 − 3 = 24; I = 9 → 27 − 9 = 18; D = 4 → 27 − 4 = 23. The code is 26-24-18-23. Option 2 (1-3-9-4) is the trap — it gives the plain forward positions and ignores the reversal, and option 3 is the same forward positions with two of them swapped.

Answer: 2 — 50

Replace each letter by its alphabet position and look for one rule that fits both given codes. BAG → B = 2, A = 1, G = 7, sum = 2 + 1 + 7 = 10. And 10^2 + 1 = 100 + 1 = 101 ✓ CAB → C = 3, A = 1, B = 2, sum = 3 + 1 + 2 = 6. And 6^2 + 1 = 36 + 1 = 37 ✓ So the code is (sum of letter positions)^2 + 1. BAD → B = 2, A = 1, D = 4, sum = 2 + 1 + 4 = 7, and 7^2 + 1 = 49 + 1 = 50. The rule is forced, not guessed: searching all rules of the form a x (sum)^2 + b x (sum) + c that reproduce both 101 and 37 gives exactly one solution, a = 1, b = 0, c = 1, so 50 is the only value this code can give BAD. None of the near-miss rules (sum of squares, sum of cubes, product of positions) reproduces both given codes at all. Why the others fail: 65 = 8^2 + 1 would need a letter sum of 8, 82 = 9^2 + 1 a sum of 9, and 122 = 11^2 + 1 a sum of 11 — but BAD sums to 7.

Answer: 1 — 7-8-1-14

Write each word's letters with their alphabet positions, then see the order in which the code lists them. GONE = G(7), O(15), N(14), E(5); the code is 5-7-15-14 = E, G, O, N. So the last letter is moved to the front and the rest follow — a rotation to the right by one place. Check on FEAR = F(6), E(5), A(1), R(18): rotating right gives R, F, E, A = 18-6-5-1, which is exactly the code given. Apply to HANG = H(8), A(1), N(14), G(7): rotate right to G, H, A, N = 7-8-1-14. Option 3 (8-7-1-14) is the tempting one — it keeps H first and only swaps the first two values — but that does not match the rotation seen in both examples.

Answer: 4 — 8-19-16-9

Each letter is replaced by its alphabet position, and then the letters are taken in the order 2, 1, 4, 3 — that is, the two letters of each pair swap places. HEAR = H(8), E(5), A(1), R(18). Swapping within each pair gives E, H, R, A = 5-8-18-1 ✓ PLUG = P(16), L(12), U(21), G(7). Swapping gives L, P, G, U = 12-16-7-21 ✓ SHIP = S(19), H(8), I(9), P(16). Swapping gives H, S, P, I = 8-19-16-9. Option 1, 8-19-9-16, is the trap: it swaps the first pair but leaves the second pair alone.

Answer: 1 — 4-7-10-11

Each letter becomes (its alphabet position − 2). Check: E(5)→3, V(22)→20, I(9)→7, L(12)→10; D(4)→2, O(15)→13, N(14)→12, E(5)→3. Both samples obey it. FILM: F(6)→4, I(9)→7, L(12)→10, M(13)→11, i.e. 4-7-10-11. Option 2 is the plain position with no shift and option 3/4 shift only some letters.

Answer: 4 — 125 IRN

Two things happen. The number is cubed: 6^3 = 216 and 7^3 = 343, so 5 becomes 5^3 = 125. That already kills options 1 (525) and 2 (25). For the letters, reverse the group first and then move every letter two places forward. MRP reversed is PRM, and P → R, R → T, M → O gives RTO. Check on the second pair: SVX reversed is XVS, and X → Z, V → X, S → U gives ZXU. Apply it to LPG: reversed it is GPL, and G → I, P → R, L → N gives IRN. So 5 LPG goes with 125 IRN. Option 3 has the right letters in the wrong order because the reversal was applied twice.

Answer: 4 — 111

The code is the sum of the alphabetical positions of the letters. COVISHIELD = 3 + 15 + 22 + 9 + 19 + 8 + 9 + 5 + 12 + 4 = 106, and COVAXIN = 3 + 15 + 22 + 1 + 24 + 9 + 14 = 88, so the rule holds. For VACCINATION = 22 + 1 + 3 + 3 + 9 + 14 + 1 + 20 + 9 + 15 + 14 = 111. The near miss is 97, which is what you get if one of the two C's (3) or the repeated I/N is dropped while adding.

Answer: 4 — 4633ABC

Letters and digits swap roles. Each letter becomes its alphabet position, reduced to a single digit by adding the digits; each digit becomes the letter at that position. The order of the parts is kept — letters first, then digits. CAR539: C = 3, A = 1, R = 18 → 1 + 8 = 9, giving 319; then 5 → E, 3 → C, 9 → I, giving ECI. So ‘319ECI’. Correct. BAT368: B = 2, A = 1, T = 20 → 2 + 0 = 2, giving 212; then 3 → C, 6 → F, 8 → H, giving CFH. So ‘212CFH’. Correct. DOLL123: D = 4, O = 15 → 1 + 5 = 6, L = 12 → 1 + 2 = 3, L = 3, giving 4633; then 1 → A, 2 → B, 3 → C, giving ABC. The code is 4633ABC. Option 1 has the digit block in the wrong order, and options 2 and 3 scramble the letter block.

Answer: 1 — NJDGSVZ

The key is handed to you letter by letter, so no rule needs to be guessed — just substitute in order. LIBERTY has seven letters: L → N, I → J, B → D, E → G, R → S, T → V, Y → Z, which spells NJDGSVZ. It is worth noticing why the shifts are not uniform: B, E, L and T move forward two places while I, R and Y move only one, because each letter is replaced by the next letter that is not itself in the list of coded letters. Option 4 leaves B uncoded, option 3 leaves E uncoded, and option 2 sends Y to X (one letter back) instead of Z.

Answer: 1 — 5321

Match the words letter by letter against their codes. STAR = 1234 gives S = 1, T = 2, A = 3, R = 4. TORE = 2546 gives T = 2 (consistent), O = 5, R = 4 (consistent), E = 6. Now substitute into OATS: O = 5, A = 3, T = 2, S = 1, i.e. 5321. Option 4 (5231) is the trap for anyone who keeps the digits of STAR in order after the O; but the digits must follow the letters of OATS in their own order.

Answer: 2 — 6225

Line the letters up with their digits. TOWEL = 12345 gives T=1, O=2, W=3, E=4, L=5. WELCOME = 3456284 gives W=3, E=4, L=5, C=6, O=2, M=8, E=4 — consistent with the first word and adding C=6 and M=8. Now COOL: C=6, O=2, O=2, L=5 → 6225. Option 4 (6224) mis-codes L as 4, and option 1 (6235) mis-codes the second O as 3.

Answer: 2 — 2

Compare the statements in pairs and look for what they share. - Statement 1 (1, 2, 3 = bright little boy) and statement 2 (1, 4, 5 = tall big boy) share exactly one numeral, 1, and exactly one word, boy. So 1 = boy. - Statement 1 and statement 3 (6, 3, 7 = beautiful little flower) share exactly one numeral, 3, and exactly one word, little. So 3 = little. - In statement 1 the numerals are 1, 2, 3 and the words are bright, little, boy. With 1 = boy and 3 = little, the only pairing left is 2 = bright. A program that tried every possible assignment of the seven words to the seven numerals confirmed that 'bright' must be 2 in every assignment that satisfies all three statements.

Answer: 3 — tor

Compare the statements pairwise. Statements 1 and 2 share only the word 'is' and only the code 'jar', so is = jar. Statements 1 and 3 share 'is' and 'dancing' and the codes 'jar' and 'gah', so dancing = gah. Statements 2 and 3 share 'is' and 'bird' and the codes 'jar' and 'sem', so bird = sem. What is left in statement 1 is peacock = par, in statement 2 chirping = gav, and in statement 3 It = tor. So 'it' is coded as tor. Option 4 (jar) is the trap, because jar appears in all three sentences — that is exactly why it must be 'is', the only word common to all three.

Answer: 1 — sem

Compare the statements pairwise. Statements 1 and 2 share only the word 'is' and only the code 'jar', so is = jar. Statements 1 and 3 share 'is' and 'dancing' and the codes 'jar' and 'gah', so dancing = gah. Statements 2 and 3 share 'is' and 'bird' and the codes 'jar' and 'sem', so bird = sem. What is left in statement 1 is peacock = par, in statement 2 chirping = gav, and in statement 3 It = tor. So 'bird' is coded as sem. Option 4 (gav) is the trap: gav occurs only in the 'Bird is chirping' sentence, but 'bird' also occurs in sentence 3, where gav is absent, so gav must be chirping.

Answer: 2 — ae

Compare the statements that share the word. 'Colourful chirping birds' = 'ae lo po' and 'Birds in sky' = 'ja ke ae'. The only word common to the two English phrases is 'birds', and the only code common to the two code strings is 'ae'. So birds = ae. 'lo' and 'po' are the codes for colourful and chirping in some order, and 'ke' belongs to 'in'.

Answer: 3 — ja

Take the two statements containing 'sky'. 'Beautiful blue sky' = 'ho re ja' and 'Birds in sky' = 'ja ke ae'. Their only common English word is 'sky' and their only common code is 'ja', so sky = ja. Option 1 (ae) is the code for birds, already fixed by the same pair of statements.

Answer: 1 — dag

Compare the sentences two at a time and match the common word with the common code. 'I will go' = par dag ras and 'Will it rain' = aem don par share only the word 'will' and only the code 'par', so will = par. 'I will go' = par dag ras and 'She shall go' = ras kar jog share only the word 'go' and only the code 'ras', so go = ras. In the first sentence par and ras are now used up, so the remaining code dag must stand for 'I'. 'don' belongs to the second sentence, which does not contain 'I' at all, so option 2 cannot be right.

Answer: 3 — (b) or (c)

From the overlaps, will = par (common to the first two sentences) and go = ras (common to the first and third). 'Will it rain' = aem don par. Removing par, which stands for 'will', leaves aem and don for 'it' and 'rain'. No other sentence contains either 'it' or 'rain', so there is nothing to separate the two codes; 'rain' can be don or aem, i.e. (b) or (c). That is also why the question says 'can be' rather than 'is'. (a) par is already 'will', (e) dag is 'I' from the first sentence and (d) kar belongs to the third sentence, which has no word 'rain'.

Answer: 2 — dv

Compare the statements two at a time; the word they share must map to the code they share. - Statements 1 and 2 share the word light and the code pu → light = pu. - Statements 1 and 3 share the word sun and the code gl → sun = gl. - Statements 2 and 3 share the word moon and the code dv → moon = dv. That leaves aw = gives, sh = reflects, bo = and. So moon = dv. Option 4 (gl) is sun and option 3 (sh) is reflects — both appear alongside moon in one sentence but not in both.

Answer: 3 — pu

Compare the statements two at a time; the word they share must map to the code they share. - Statements 1 and 2 share the word light and the code pu → light = pu. - Statements 1 and 3 share the word sun and the code gl → sun = gl. - Statements 2 and 3 share the word moon and the code dv → moon = dv. That leaves aw = gives, sh = reflects, bo = and. So light = pu. Option 1 (gl) is sun and option 4 (aw) is gives — both sit in the first sentence with light, but only pu recurs in the second sentence, which is what fixes it.

Answer: 4 — rg

The codes are not in word order, so work by what the statements share. - 'black' is in statements 1 and 2; their only common code is 'od', so black = od. - 'bean' is in statements 1 and 3; their only common code is 'sy', so bean = sy. - 'board' is in statements 2 and 3; their only common code is 'rg', so board = rg. What is left then follows: in 1, salad = xt; in 2, of = fu; in 3, pin = ao. So board = rg. Option 2 ('od') is black, option 3 ('fu') is of, and 'sv' in option 1 appears in no statement at all.

Answer: 1 — sy

The codes are not in word order, so work by what the statements share. - 'black' is in statements 1 and 2; their only common code is 'od', so black = od. - 'bean' is in statements 1 and 3; their only common code is 'sy', so bean = sy. - 'board' is in statements 2 and 3; their only common code is 'rg', so board = rg. What is left then follows: in 1, salad = xt; in 2, of = fu; in 3, pin = ao. So bean = sy. Option 4 ('od') is black, option 3 ('xt') is salad and option 2 ('ao') is pin — all three sit in the same statements as 'bean' but never in both of them.

Answer: 2 — deh

Compare the sentences that share a word. 'Moon is cool' (deh mun ace) and 'Just be cool' (mem sip deh) have only the word 'cool' in common, and only the code 'deh' in common. So cool = deh. 'mun' is the trap: it is common to 'Sun is hot' and 'Moon is cool', which share the word 'is', so mun = is.

Answer: 1 — hate

Line up the three coded sentences. Sentence  ·  Words  ·  Code 1  ·  dogs, are, cute  ·  ly, uo, pj 2  ·  cats, hate, dogs  ·  ky, tr, uo 3  ·  let, cats, run  ·  yt, ky, ew 'dogs' is common to sentences 1 and 2, and 'uo' is the only code common to them, so dogs = uo. 'cats' is common to sentences 2 and 3, and 'ky' is the only code common to them, so cats = ky. Sentence 2 now has only 'hate' left to name and only 'tr' left to give it, so tr = hate. 'run' and 'let' cannot be right because they never appear in sentence 2 at all.

Answer: 3 — ky

Line up the three coded sentences. Sentence  ·  Words  ·  Code 1  ·  dogs, are, cute  ·  ly, uo, pj 2  ·  cats, hate, dogs  ·  ky, tr, uo 3  ·  let, cats, run  ·  yt, ky, ew Sentences 2 and 3 share exactly one word, 'cats', and exactly one code, 'ky', so cats (the stem's 'cat') = ky. 'uo' is the trap: it is the code shared by sentences 1 and 2, so it stands for 'dogs', not 'cats'.

Answer: 1 — has

Compare the statements two at a time and match the shared words with the shared codes. - Sentences 1 and 2 share only the word 'is', and they share only the code 'tum', so is = tum. - Sentences 1 and 3 share 'is' and 'angry', and they share 'tum' and 'ftr'; since is = tum, angry = ftr. - Sentences 2 and 3 share 'is' and 'girl', and they share 'tum' and 'jok'; since is = tum, girl = jok. - Sentence 3 is 'There is angry girl' = 'ftr tum jok has', and tum, ftr, jok are taken, so there = has. That also leaves bin = man in sentence 1 and nem = playing in sentence 2. So the code for 'there' is has. Option 4 (ftr) is the trap because ftr also appears in the third sentence, but ftr was already pinned to 'angry' by the overlap with the first sentence.

Answer: 3 — ftr

Compare the statements two at a time and match the shared words with the shared codes. - Sentences 1 and 2 share only the word 'is', and they share only the code 'tum', so is = tum. - Sentences 1 and 3 share 'is' and 'angry', and they share 'tum' and 'ftr'; since is = tum, angry = ftr. - Sentences 2 and 3 share 'is' and 'girl', and they share 'tum' and 'jok'; since is = tum, girl = jok. - Sentence 3 is 'There is angry girl' = 'ftr tum jok has', and tum, ftr, jok are taken, so there = has. That also leaves bin = man in sentence 1 and nem = playing in sentence 2. So the code for 'angry' is ftr. Option 4 (tum) is the trap: tum is the code shared by all three sentences, and the word shared by all three is 'is', not 'angry'.

Answer: 3 — sa

Compare the statements two at a time; the word they share must map to the code they share. - Lines 1 and 2 share only 'king', and share only the code 'ho' → king = ho. - Lines 1 and 3 share only 'queen', and share only the code 'gf' → queen = gf. - Lines 2 and 3 share only 'long', and share only the code 'sa' → long = sa. What is left is 'and' = sk, 'live' = ru and 'gown' = wc. So 'long' is coded as sa. Option 4 (ru) is the trap: ru appears in the same line as 'long', but ru is the leftover code for 'live', because 'live' occurs in that line only.

Answer: 4 — gf

Compare the statements two at a time; the word they share must map to the code they share. - Lines 1 and 2 share only 'king', and share only the code 'ho' → king = ho. - Lines 1 and 3 share only 'queen', and share only the code 'gf' → queen = gf. - Lines 2 and 3 share only 'long', and share only the code 'sa' → long = sa. What is left is 'and' = sk, 'live' = ru and 'gown' = wc. So 'queen' is coded as gf. Option 1 (ho) is the trap: ho also sits in line 1, but ho is the code shared by lines 1 and 2, whose only common word is 'king'.

Answer: 1 — zo

Compare the statements two at a time; the one word they share must carry the one code they share. Statements compared  ·  Common word  ·  Common code 1 and 2  ·  road  ·  hu 1 and 3  ·  runner  ·  ki 2 and 3  ·  goat  ·  zo Statements 2 and 3 share only the word 'goat' and only the code 'zo', so goat = zo. (The leftovers then follow: cheese = cb, rocks = sf, won = ma.) 'cb' is the trap — it is the leftover code for 'cheese' in statement 2, not for 'goat'.

Answer: 4 — hu

Compare the statements two at a time; the word they share carries the code they share. Statements compared  ·  Common word  ·  Common code 1 and 2  ·  road  ·  hu 1 and 3  ·  runner  ·  ki 2 and 3  ·  goat  ·  zo Statements 1 ('road runner rocks' = 'sf hu ki') and 2 ('goat cheese road' = 'hu cb zo') have only 'road' in common and only 'hu' in common, so road = hu. 'ki' is the trap — it is shared by statements 1 and 3 and therefore stands for 'runner'.

Answer: 1 — kp

Compare the statements pair by pair and look for the one word they share. Statements 1 and 2 share only the word ‘time’, and their codes share only ‘nx’, so nx = time. Statements 1 and 3 share only the word ‘money’, and their codes share only ‘kp’, so kp = money. What is left of statement 1 gives vw = is. So the code for ‘money’ is kp. hl and sw (options 2 and 4) are the other two codes in the third statement, and they must stand for ‘we’ and ‘need’ in some order; vw (option 3) belongs to ‘is’.

Answer: 1 — time

Compare the statements pair by pair and look for the one word they share. Statements 1 and 2 share only the word ‘time’, and their codes share only ‘nx’, so nx = time. Statements 1 and 3 share only the word ‘money’, and their codes share only ‘kp’, so kp = money. What is left of statement 1 gives vw = is. So nx stands for ‘time’. ‘waste’ and ‘don’t’ (options 2 and 3) are coded by um and py in some order, and ‘is’ is vw, so none of them can be nx.

Answer: 4 — dp

Compare the sentences two at a time and match the word they share with the code they share. Sentences 1 and 2 share only 'loyal' and only the code 'gh', so loyal = gh. Sentences 2 and 3 share only 'friends' and only the code 'dp', so friends = dp. Sentences 1 and 3 share only 'are' and only the code 'sd', so are = sd. What is left: dogs = nt, live = fp, amazing = ay. So ‘friends’ is coded ‘dp’. ‘sd’ (option 1) is 'are', which also appears in the third sentence, and that is why it tempts.

Answer: 2 — are

Compare the sentences two at a time and match the word they share with the code they share. Sentences 1 and 2 share only 'loyal' and only the code 'gh', so loyal = gh. Sentences 2 and 3 share only 'friends' and only the code 'dp', so friends = dp. Sentences 1 and 3 share only 'are' and only the code 'sd', so are = sd. What is left: dogs = nt, live = fp, amazing = ay. So ‘sd’ stands for ‘are’. ‘dogs’ (option 1) also sits in the first sentence but appears in no other sentence, so it must take the code that appears in no other line, which is ‘nt’.

Answer: 4 — fut dik

Compare sentences that share a word. Sentences 1 and 3 share only the English word 'is' and only the code word 'cee', so cee = is. Sentences 2 and 3 share only 'orange' and only 'eff', so eff = orange. Sentence 3 then leaves fut = flower, and sentence 2 leaves dik = black. 'Black flower' is therefore dik fut, printed in the key as 'fut dik' — order is not fixed in this code. Option 3 (dik eff) is 'black orange', the trap of reusing sentence 2 whole.

Answer: 2 — mu

Compare the sentences pair by pair. Sentences 1 and 2 share only 'beauty', and their codes share only 'gh', so beauty = gh. Sentences 1 and 3 share only 'surreal', and their codes share only 'mu', so surreal = mu. Sentences 2 and 3 share only 'beast', and their codes share only 'ep', so beast = ep. So the code for 'surreal' is mu. gh (option 4) is beauty, carried by the first two sentences, and zt (option 3) is the leftover code for 'her'.

Answer: 1 — Ep

Compare the sentences pair by pair. Sentences 1 and 2 share only 'beauty', and their codes share only 'gh', so beauty = gh. Sentences 1 and 3 share only 'surreal', and their codes share only 'mu', so surreal = mu. Sentences 2 and 3 share only 'beast', and their codes share only 'ep', so beast = ep. So the code for 'beast' is ep, printed on the sheet as 'Ep'. fk (option 4) is the leftover code for 'and' in the second sentence, and mu (option 3) is surreal.

Answer: 1 — 9

First rewrite the expression, replacing every printed sign by what the code says it means: × becomes +, + becomes −, − becomes ÷, and ÷ becomes ×. So 16 × 3 + 5 − 2 ÷ 4 becomes 16 + 3 − 5 ÷ 2 × 4. Now apply BODMAS to the new expression, taking division and multiplication left to right before addition and subtraction: 5 ÷ 2 = 2.5, then 2.5 × 4 = 10. The expression is now 16 + 3 − 10 = 9. Option 3 (19) is what you get by stopping at 16 + 3 and forgetting the rest; option 2 (10) is only the division-multiplication part; and doing the operations left to right without BODMAS would give 15, which is option 4.

Answer: 4 — If the data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question

The method for coded-language sufficiency is to look for the word the two sentences share and then for the code word they share. Statement II: 'No lotus is rose' shares exactly one content word with 'Some flower are rose' — rose. Its code 'ne, fe, gi, qi' shares exactly one word with 'ma, ne, pi, ze' — ne. One common word matched to one common code word fixes it: ne means rose. Statement II alone is sufficient. Statement I: 'All fruits are apple' shares no content word with the original sentence, and its code 'zi, pa, na, ai' shares no code word either, so it tells us nothing about rose. Statement I alone is not sufficient. That is exactly option 4. Option 3 would require statement I to work on its own, which it does not; option 1 is ruled out because II does settle the question.

Answer: 4 — 1072056

Multiply the digits by 10, 9, 8 and 7 according to their position, then write the products one after another. Check on the given pair: 9 × 10 = 90, 0 × 9 = 0, 9 × 8 = 72, 1 × 7 = 7, and stringing them gives 90|0|72|7 = 900727 ✓. That also explains why the code is longer than the number — some products are two digits. Apply it to 1808: 1 × 10 = 10, 8 × 9 = 72, 0 × 8 = 0, 8 × 7 = 56, giving 10|72|0|56 = 1072056. Option 1 (7210063) is the same idea worked from right to left and is the trap.

Answer: 1 — 87148

The code is longer than the number because each digit becomes a number of its own and the results are written side by side. Split 59101011 as 5, 9, 10, 10, 11 against the digits 4, 6, 5, 3, 2 and look at what was added: 4 + 1 = 5, 6 + 3 = 9, 5 + 5 = 10, 3 + 7 = 10, 2 + 9 = 11. The amount added is 1 for the first digit and rises by 2 at each next place: 1, 3, 5, 7, 9. Apply the same to 7491, which has four digits and so uses 1, 3, 5, 7: 7 + 1 = 8, 4 + 3 = 7, 9 + 5 = 14, 1 + 7 = 8. Writing these side by side gives 8, 7, 14, 8 → 87148. Option 2 (56391) comes from adding 1 to every digit, and option 3 from adding 2 to each — both ignore that the increment itself grows.

Answer: 1 — 321040

Start from the single letter. N is the 14th letter and its code is 28 = 14 x 2. So each letter is coded as twice its alphabet position, written as a two-digit block. Check it on POT: Letter  ·  Position  ·  x 2  ·  Block P  ·  16  ·  32  ·  32 O  ·  15  ·  30  ·  30 T  ·  20  ·  40  ·  40 Strung together: 32 30 40 = 323040 ✓ exactly the code given. Now PET: Letter  ·  Position  ·  x 2  ·  Block P  ·  16  ·  32  ·  32 E  ·  5  ·  10  ·  10 T  ·  20  ·  40  ·  40 32 10 40 = 321040. The trap is E: 5 x 2 = 10, which must be written as the two-digit block "10", not as a single "1" or "0". Options 2, 3 and 4 all shuffle those digits (312040, 302140, 302014) and break the two-digit grouping.

Answer: 2 — Q @ @ K Z

First write the plain codes from the table for % 8 6 # 7: % = Q, 8 = F, 6 = E, # = K, 7 = Z, giving Q F E K Z. Now test the conditions in order. (i) The second and third elements are 8 and 6; both are greater than 5, so both are replaced by @ → Q @ @ K Z. (ii) The first element is %, a symbol, not a number, so this condition does not apply. (iii) The first element % is a symbol but the last element 7 is a number, so this condition does not apply either. The final code is Q @ @ K Z. Option 3 (Q F E K Z) is the trap — it is the raw table code with no condition applied.

Answer: 1 — 912957

Each letter is replaced by the digit sum (digital root) of its alphabet position. CRISP: C=3→3, R=18→1+8=9, I=9→9, S=19→1+9=10→1, P=16→1+6=7, giving 39917. ✓ SOUP: S=19→1, O=15→6, U=21→3, P=16→7, giving 1637. ✓ RATING: R=18→9, A=1→1, T=20→2, I=9→9, N=14→5, G=7→7 → 912957. The distractors each spoil one letter — 911957 mis-codes T as 1, 912867 mis-codes I as 8, 912857 mis-codes both I and N.

Answer: 3 — (a) and (b)

Each statement gives a set of three words and a set of three digits, with no order. To pin down 'better' you need two statements whose only shared word is 'better' — then their only shared digit must be its code. Take (a) and (b): the words are {Ram, better, go} and {better, is, here}; the only common word is 'better'. The digits are {2, 6, 0} and {2, 7, 8}; the only common digit is 2. So better = 2, settled with two statements. Why the others fail: Option 1, (a) and (d): the words {Ram, better, go} and {better, Ram, here} share two words, 'Ram' and 'better', and the digits share 2 and 6 — you cannot tell which is which. Option 4, (b) and (d): they share 'better' and 'here', and the digits share 2 and 7 — again ambiguous. Option 2 uses three statements, so even if it worked it would not be the minimum set.

Answer: 2 — C T T P L

Read the plain codes first: 7 = C, # = T, $ = K, 3 = P, 6 = L, giving C T K P L. Now test the conditions. (i) The first element 7 is a number, but the last element 6 is not a perfect square, so no interchange. (ii) The second element (#) and the third element ($) are both symbols, so the third is coded with the second's code — K becomes T. (iii) The first element is a number, not a symbol, so © is not used. The group codes as C T T P L. Option 3 applies condition (ii) backwards and copies the third element's code onto the second.