Logical & analytical reasoning

Answer: (a) I only

Flip each rule into its contrapositive: Z correct → X correct → W incorrect → Y correct. So X correct forces Y correct (I holds), and Z correct also forces Y correct, which makes II wrong.

1. Rule 1: X incorrect → Z incorrect; contrapositive: Z correct → X correct.
2. Rule 3: W correct → X incorrect; contrapositive: X correct → W incorrect.
3. Rule 2: Y incorrect → W correct; contrapositive: W incorrect → Y correct.
4. Chain: Z correct → X correct → W incorrect → Y correct.
5. I: X correct → Y correct — true. II: Z correct forces Y correct, so 'not necessary' is false.

• ✓ I X correct → W incorrect (rule 3) → Y correct (rule 2).
• ✗ II Z correct → X correct (rule 1) → Y correct; so Y is necessarily correct.

Remember · 'If P then Q' equals 'if not Q then not P'. Rewrite each rule this way and chain them.

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Answer: (d) Q is a sister of the husband of S

S is a mother, so S is female and her spouse R is male — P's son. P's other child Q must then be the daughter, making Q the sister of S's husband R.

1. S is the mother of T, so S is female.
2. R is S's spouse, so R is male.
3. Q and R are P's two children, and P has one son and one daughter. R is the son, so Q is the daughter.
4. Q is R's sister, and R is S's husband: Q is a sister of the husband of S.

Remember · In blood relations, fix genders first from words like mother, son, spouse; then place each person on a family tree.

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Answer: (d) P, Q and R

The statements form a chain of sets: red inside blue, blue inside green, green inside yellow. Anything inside a smaller set is inside every larger set, so all three conclusions follow.

1. Red ⊂ Blue ⊂ Green ⊂ Yellow.
2. P: Blue ⊂ Green ⊂ Yellow — follows.
3. Q: Red ⊂ Blue ⊂ Green — follows.
4. R: Red ⊂ … ⊂ Yellow — follows.

• ✓ P Every blue is green and every green is yellow.
• ✓ Q Every red is blue and every blue is green.
• ✓ R Following the whole chain, every red is yellow.

Remember · 'Every A is B, every B is C' gives 'every A is C'; draw nested circles to see it at once.

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Answer: (d) 6 km, North

Track each turn: East 4 km, right to South 3 km, left to East 2 km, left to North 3 km. The south and north legs cancel, leaving the person 6 km east of the start, facing North.

1. Start facing East: 4 km East.
2. Turn right → South: 3 km South.
3. Turn left → East: 2 km East.
4. Turn left → North: 3 km North.
5. Net: 4 + 2 = 6 km East; 3 South − 3 North = 0. Distance 6 km, facing North.

Remember · Direction items: write the facing direction after every turn, then net out opposite legs.

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Answer: (b) TSPO

Letters shift alternately forward and back by one: Z → A (+1), E → D (−1), R → S (+1), O → N (−1). Applying +1, −1, +1, −1 to S, T, O, P gives T, S, P, O.

1. Z + 1 = A, E − 1 = D, R + 1 = S, O − 1 = N.
2. S + 1 = T, T − 1 = S, O + 1 = P, P − 1 = O.
3. STOP → TSPO.

Remember · Coding items: write the shift for each letter position; alternating patterns are common.

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Answer: (d) 3 only

All three statements belong to the same climate debate, so relationship 2 is false — and every option except (d) contains it. S3 does not argue against S1; it only notes that extreme voices within activism give deniers a target. The open question in S2 about how much humans contribute can explain why both alarmism and denial thrive.

• ✗ 1. S3 does not deny that protecting the environment is vital; it describes extreme views within activism that deniers seize on. That is a comment on the debate, not a counter-argument to S1.
• ✗ 2. S3 is about environmental activism and climate change deniers — the same subject as S1 and S2.
• ✓ 3. The lack of scientific consensus on the human contribution (S2) leaves room for alarmist and extremist positions and for deniers to target them (S3).

Remember · When one relationship is clearly false, strike every option containing it first; often a single option survives.

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Answer: (d) Neither I nor II

A comparison is certain only when every sign between the two letters points the same way. B and E are separated by a change of direction at C, and S = T ≤ U = V allows S to equal V, so neither conclusion is always true.

1. Statement I: between B and E the chain reads B > C < D > E. The direction changes at C, so B and E cannot be compared.
2. Example: B = 5, C = 1, D = 10, E = 8 fits the chain, and here B is less than E. So I is not always true.
3. Statement II: between S and V the chain reads S = T ≤ U = V, which gives only S ≤ V.
4. If T = U, then S = V, so S is not always less than V. II is also not always true.
5. Neither statement is correct.

• ✗ I B > C and D > E, but C < D breaks the chain; B can be smaller than, equal to or larger than E.
• ✗ II S = T ≤ U = V gives S ≤ V, so S may equal V; 'always less than' does not follow.

Remember · Trace the signs between the two letters: a change of direction means no conclusion; any ≤ or ≥ rules out strict < or >.

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Answer: (d) Cannot be concluded

The item gives no rule about who tells the truth. Even if we assume every innocent person speaks the truth, both 'P stole' and 'Q stole' fit all the statements, so the thief cannot be identified.

1. The item says only one person is guilty but gives no rule on who tells the truth or lies.
2. Test the most helpful assumption: innocent people speak the truth. If R stole, innocent P's claim 'Q stole it' would be false — so R is ruled out.
3. If P stole: innocent Q says R and Q did not steal (true), and innocent R says he did not steal (true). Consistent.
4. If Q stole: innocent P says he did not steal and Q stole it (true), and innocent R's statements are true. Consistent.
5. Both P and Q remain possible, so the thief cannot be concluded.

Remember · In truth-and-lie puzzles, test each suspect as culprit for consistency; if more than one survives, the answer is 'cannot be concluded'.

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Answer: (c) Only four

As siblings, P, Q and R share S as mother and T as father, so S and T are a couple. Only P's gender is known, so the statements that need Q to be male or R to be female are not definite; the other four are.

1. P, Q and R are siblings, so R's mother S and P's father T are the parents of all three — and a couple.
2. P is male (a brother). The genders of Q and R are not given.
3. Definitely true: I (S and T are a couple), III (T is Q's father), IV (S is P's mother), VI (P is S's son).
4. Not definite: II (Q may be a daughter) and V (R may be a son).
5. So four statements are definitely true.

• ✓ I S is the mother and T the father of the same siblings, so they are a couple.
• ✗ II Q's gender is not given; Q could be T's daughter.
• ✓ III T is P's father and Q is P's sibling, so T is Q's father.
• ✓ IV S is R's mother and R is P's sibling, so S is P's mother.
• ✗ V R's gender is not given; R could be T's son.
• ✓ VI P is male and S is his mother, so P is S's son.

Remember · Note every gender actually given; a statement that needs an unknown gender is never 'definitely true'.

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Answer: (a) 399000

The code multiplies the alphabet positions of the letters: 14 × 15 = 210, then × 20 = 4200, then × 5 = 21000. Adding S (19) gives 21000 × 19 = 399000.

1. Alphabet positions: N = 14, O = 15, T = 20, E = 5, S = 19.
2. NO = 14 × 15 = 210. NOT = 210 × 20 = 4200. NOTE = 4200 × 5 = 21000. The code is the product of the positions.
3. NOTES = 21000 × 19 = 399000.
4. Check: 4200 ÷ 210 = 20 (T) and 21000 ÷ 4200 = 5 (E), confirming the rule.

Remember · When a code grows as letters are added, divide consecutive codes: the ratio gives each new letter's value (here its alphabet position).

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Answer: (d) 71813

Each letter keeps a fixed digit across both words: Y = 7, E = 1, M = 8 and N = 3. So YEMEN is 7-1-8-1-3, i.e. 71813.

1. From FRANCE = 654321: F = 6, R = 5, A = 4, N = 3, C = 2, E = 1.
2. From GERMANY = 9158437: G = 9, E = 1, R = 5, M = 8, A = 4, N = 3, Y = 7 (E, R, A, N agree with the first word).
3. YEMEN = Y, E, M, E, N = 7, 1, 8, 1, 3 = 71813.

Remember · For letter-to-digit codes, build a lookup table from the given words and read off the new word; check shared letters agree.

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Answer: (c) 31 km

Plotting the places on a grid shows two road routes from D to F: via C, A and B (6 + 10 + 6 + 9 km) and via E and G (5 + 14 + 12 km). Both total 31 km.

1. Put A at (0, 0), with east = +x and north = +y.
2. B = (0, 6); C = (10, 0); D = 6 km south of C = (10, −6); E = 5 km west of D = (5, −6); F = 9 km west of B = (−9, 6); G = 12 km south of F = (−9, −6).
3. Route D → C → A → B → F: 6 + 10 + 6 + 9 = 31 km.
4. Route D → E → G → F: 5 + 14 + 12 = 31 km (E to G runs from x = 5 to x = −9).
5. Either way the distance is 31 km.

Remember · For direction-and-road items, fix one place as the origin and write coordinates; route lengths then follow by simple addition.

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Answer: (b) 1331

Both examples turn a cube n³ into (n + 3)³: 4³ becomes 7³ and 6³ becomes 9³. So 512 = 8³ becomes 11³ = 1331.

1. 64 = 4³ and 343 = 7³; 216 = 6³ and 729 = 9³.
2. The rule: n³ is written as (n + 3)³.
3. 512 = 8³, so it is written as 11³ = 1331.

Remember · Recognise squares and cubes at sight (up to 20² and 12³); number codes often shift the base.

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Answer: (c) South

Assume he starts facing North and trace the walk: the office ends up 200 m West and 200 m South, i.e. South-West. For it to be North-East the whole path must be turned through 180°, so he actually started walking South.

1. Try starting North: 100 m North; left, so 100 m West; left again, so 300 m South; right, so 100 m West.
2. Net: 300 − 100 = 200 m South and 100 + 100 = 200 m West, so the office would be South-West.
3. The office is actually North-East, exactly opposite, so rotate everything by 180°: the first walk was South.
4. Check: 100 m South; left, 100 m East; left, 300 m North; right, 100 m East. Net 200 m North and 200 m East, i.e. North-East.

Remember · Assume North, solve, then rotate the whole figure to match the given final direction.

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Answer: (d) North-East

Walking West and turning left, he faces South. A 225° clockwise turn from South (bearing 180°) brings him to bearing 405° − 360° = 45°, which is North-East.

1. Walking West, a left turn makes him face South.
2. Clockwise from South: 90° brings West, 180° brings North, and 225° is 45° beyond North towards East.
3. So he now faces North-East.
4. Check with bearings: 180° + 225° = 405°, i.e. 45°, which is North-East.

Remember · Convert to bearings (N 0°, E 90°, S 180°, W 270°), add clockwise turns, and subtract 360° if needed.

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Answer: (d) Neither Conclusion–I nor Conclusion–II follows

Being the largest producer says nothing about exports or imports: a country may consume most of what it produces, and may still import some. Neither conclusion follows from the statement.

• ✗ I Large production does not mean large exports; home consumption may take most of it. The statement gives no trade data.
• ✗ II Producing the most does not rule out importing, for example particular milk products or to meet demand in some years.

Remember · A conclusion must follow from the statement alone; production figures never settle questions of trade.

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Answer: (c) Both SP and RQ

Under ‘either a Director or a Producer’, ruling out one role forces the other. So ‘not a Producer’ implies ‘a Director’ (SP), and ‘not a Director’ implies ‘a Producer’ (RQ).

• ✓ SP If Pradeep is not a Producer (S), the Main Statement leaves only Director (P). S implies P.
• ✓ RQ If Pradeep is not a Director (R), he must be a Producer (Q). R implies Q.

Remember · For ‘either A or B’, the valid pairs are ‘not A, therefore B’ and ‘not B, therefore A’.

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Answer: (d) Neither 1 nor 2

Statement 1 gives 2x ≥ 3y and 3x = 4z, which lead to 9y ≤ 8z, so ‘9y > 8z’ fails. Statement 2 gives x < 2y and y ≤ z, which still allow x > z (for example x = 1.5, y = z = 1). Neither is correct.

1. Decode: P is >, Q is <, R is ≤ (not greater than), S is ≥ (not less than), T is =.
2. Statement 1: 2x ≥ 3y gives 9y ≤ 6x; 3x = 4z gives 8z = 6x. So 9y ≤ 8z, which rules out 9y > 8z. Incorrect.
3. Statement 2: x < 2y and y ≤ z. Take x = 1.5, y = 1, z = 1: both conditions hold, but x ≤ z fails. Incorrect.
4. So neither statement is correct.

• ✗ 1. From 2x ≥ 3y and 3x = 4z, 9y ≤ 6x = 8z, so 9y can never be greater than 8z.
• ✗ 2. x < 2y and y ≤ z only give x < 2z; x = 1.5, y = z = 1 satisfies both conditions but has x greater than z.

Remember · Decode the symbols, bring both sides to a common term (here 6x), and use one counter-example to kill an ‘always’ claim.

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Answer: (a) 11880

Each letter stands for its position in the alphabet and the code is the product: 1 × 2 × 3 × 4 = 24 and 5 × 6 × 7 × 8 = 1680. So IJKL = 9 × 10 × 11 × 12 = 11880.

1. ABCD: 1 × 2 × 3 × 4 = 24, which matches.
2. EFGH: 5 × 6 × 7 × 8 = 1680, which matches.
3. IJKL: 9 × 10 × 11 × 12 = 90 × 132 = 11880.

Remember · When letters become numbers, test alphabet positions with sum and product first.

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Answer: (d) AANERA

Each word is reversed and an A is put in front: POT → TOP → ATOP, and TRAP → PART → APART. ARENA reversed is ANERA, so the code is AANERA.

1. POT reversed is TOP; with A in front it is ATOP, which matches.
2. TRAP reversed is PART; with A in front it is APART, which matches.
3. ARENA reversed is ANERA; with A in front it is AANERA.

Remember · When a code looks like a real word, check for reversal plus a fixed added letter.

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Answer: (d) SODBHU

Each letter moves 3 places forward in the alphabet: Z → C, E → H, R → U, O → R. Moving each letter of PLAYER 3 places forward gives SODBHU.

1. Z + 3 = C (wrapping round after Z), E + 3 = H, R + 3 = U, O + 3 = R, so ZERO → CHUR.
2. PLAYER: P + 3 = S, L + 3 = O, A + 3 = D, Y + 3 = B, E + 3 = H, R + 3 = U.
3. So PLAYER is written as SODBHU.

Remember · In letter coding, find the shift for each letter (remember the wrap-around after Z); a constant shift applies letter by letter.

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Answer: (c) All four statements together are sufficient to answer the Question.

A and C are sisters and D is C's father, so D is A's father too, and E, D's son, is A's brother. A, a woman, is married to B, so B is A's husband and E is B's brother-in-law. Every one of the four statements is used in this chain.

1. Statement-4: A and C are sisters, so both are female. Statement-2: D is C's father, so D is A's father as well.
2. Statement-3: E is D's son, so E is the brother of A and C.
3. Statement-1: A and B are married; A is female, so B is A's husband.
4. E is the brother of B's wife, i.e. E is B's brother-in-law.
5. Without Statement-2 or Statement-4, E cannot be linked to A; without Statement-1, B is unconnected. So all four are needed.

• ✗ (a) Without Statement-4, nothing links A (and so B) to C, D or E.
• ✗ (b) Without Statement-2, D (and so E) is not linked to A and C.
• ✓ (c) Together they give: E is A's brother and B is A's husband, so E is B's brother-in-law.
• ✗ (d) The four statements together do answer the Question.

Remember · In blood-relation sufficiency questions, trace one chain from the first person to the second; every link used is a required statement.

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Answer: (b) Only Conclusion-2 follows from the Statement.

Decode the symbols: + is >, − is ≤, × is ≥, ÷ is < and ± is =. The Statement gives Q ≤ P ≤ T < R = S, so S > Q follows, but Q = T does not, since Q may be smaller than T.

1. Meanings: A + B → A > B; A − B → A ≤ B; A × B → A ≥ B; A ÷ B → A < B; A ± B → A = B.
2. Statement: P × Q → P ≥ Q; P − T → P ≤ T; T ÷ R → T < R; R ± S → R = S.
3. Chain: Q ≤ P ≤ T < R = S.
4. Conclusion-1 (Q ± T, i.e. Q = T): we only know Q ≤ T — does not follow.
5. Conclusion-2 (S + Q, i.e. S > Q): S = R > T ≥ P ≥ Q — follows.

Remember · Translate coded symbols into >, ≥, <, ≤, = first, build one chain, then test each conclusion against it.

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Answer: (d) Neither 1 nor 2

Since 'at least one of R and S was selected' is false, neither R nor S was selected. That leaves only P, Q and T as possible selections — at most three people — so neither conclusion can be drawn.

1. The false statement means neither R nor S was selected.
2. So only P, Q and T could have been selected: at most three candidates.
3. Conclusion 1 (at least four selected) is impossible.
4. Conclusion 2 (S selected) contradicts the false statement.
5. Check: the third statement (at most two of R, S, T) holds, since only T can be among them.

• ✗ 1. With R and S both out, at most P, Q and T can be selected — never four.
• ✗ 2. The false statement tells us S was not selected.

Remember · The negation of 'at least one of X and Y' is 'neither X nor Y'. Negate the false statement first; it usually settles the case.

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Answer: (a) 1 only

If P and R were both true, P would make S true, and R with S would make T false. So whenever T is true, P or R must be false. 'If Q is true, then P is true' merely reverses statement I, which never follows.

1. Test Conclusion 1: suppose T is true while P and R are both true.
2. By I, P true makes Q and S true; so R and S are both true.
3. By II, R and S true makes T false — contradicting T true. So Conclusion 1 follows.
4. Test Conclusion 2: take Q true, P false (S, R, T anything consistent). Statement I is not violated, since it says nothing when P is false.
5. So Q can be true without P; Conclusion 2 does not follow.

• ✓ 1. P and R both true would force S true and then T false; so with T true, P or R must be false.
• ✗ 2. It is the converse of statement I; Q can be true while P is false.

Remember · 'If A then B' gives 'if not B then not A', never 'if B then A'. Test a conclusion by assuming it fails.

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Answer: (d) OFXPMLD

Reversing the alphabet swaps each letter with its mirror, whose position adds up to 27 with its own: L→O, U→F, C→X, K→P, N→M, O→L, W→D, giving OFXPMLD.

1. In the reversed alphabet, the letter at position p becomes the letter at position 27 − p (A ↔ Z, B ↔ Y, …).
2. L (12) → O (15); U (21) → F (6); C (3) → X (24); K (11) → P (16).
3. N (14) → M (13); O (15) → L (12); W (23) → D (4).
4. LUCKNOW → OFXPMLD.
5. Check: the second letter separates the options — U (21) → F (6), which rules out (a) and (b); the last letter W (23) → D (4) rules out (c).

Remember · Opposite letters in the alphabet have positions adding to 27 (A–Z, B–Y, …).

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Answer: (c) I

The series is built in blocks A | AB | ABC | ABCD | …, the nth block having n letters. The first 13 blocks use 91 places, so the 100th place is the 9th letter of the 14th block, which is I.

1. Split the series into blocks: A | AB | ABC | ABCD | ABCDE | … The nth block runs from A to the nth letter and has n letters.
2. Places used by the first n blocks = 1 + 2 + … + n = n(n + 1)/2. For n = 13 this is 13 × 14 ÷ 2 = 91.
3. So the 100th place is the 100 − 91 = 9th letter of the 14th block (A to N).
4. The 9th letter of the alphabet is I.
5. Check: the 14th block ends at place 91 + 14 = 105, so place 100 does fall inside it.

Remember · Growing-block series: find the largest n with n(n + 1)/2 below the position; the remainder names the letter.

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UPSC dropped this question from evaluation in its final answer key.

UPSC dropped this item from evaluation. The idea is still worth learning: Q and R are P’s parents and P and U are T’s parents, so T is Q’s granddaughter, while Q’s and R’s genders are never stated.

1. R’s son is P and R is Q’s spouse, so Q and R are P’s parents — the first couple.
2. P’s daughter is T and U is T’s mother, so P (male, being R’s son) and U are T’s parents — the second couple. S is U’s other daughter.
3. The genders of Q and R are never given, so (a), (b) and (c) cannot be decided.
4. T is the daughter of P, who is Q’s son, so T is Q’s granddaughter — as in option (d).

Remember · Blood relations: draw the tree with gender marks; any option that needs an unstated gender cannot be definitely correct.

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Answer: (d) None of the Conclusions follows

“All pens are books” and “No chair is a pen” say nothing definite about chairs and books. Conclusions I and IV are only possibilities, II contradicts Statement–2, and III is impossible, so none follows.

1. Draw the circle ‘pens’ inside ‘books’ (All pens are books) and ‘chairs’ outside ‘pens’ (No chair is a pen).
2. ‘Chairs’ may lie wholly inside ‘books’ (outside the pens), partly inside, or wholly outside — the statements do not decide.
3. So Conclusion–I and Conclusion–IV are possible but not certain; Conclusion–II contradicts Statement–2; Conclusion–III would make every pen a chair, which Statement–2 forbids.
4. None of the conclusions follows.

• ✗ I Possible (chairs may lie inside books, away from the pens) but not certain.
• ✗ II Directly contradicts Statement–2: no chair is a pen.
• ✗ III If all books were chairs, the pens (which are books) would be chairs — impossible under Statement–2.
• ✗ IV Possible (chairs may lie wholly outside books) but not certain.

Remember · Syllogisms: a conclusion follows only if it holds in every possible diagram, not just in one.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) Both Conclusion–I and Conclusion–III

The doctors who are teachers are, by the chain of ‘all’ statements, also engineers and scientists. So “some engineers are doctors” and “some scientists are doctors” both follow; “all engineers are doctors” does not.

1. Some doctors are teachers, and all teachers are engineers, so those doctors are engineers: some engineers are doctors (III follows).
2. All engineers are scientists, so the same doctors are scientists: some scientists are doctors (I follows).
3. Nothing says every engineer is a doctor, so II does not follow.

• ✓ I The doctor-teachers are engineers and hence scientists, so some scientists are doctors.
• ✗ II Only some engineers (those who are teacher-doctors) are known to be doctors; ‘all’ goes beyond the statements.
• ✓ III The doctors who are teachers are engineers, so some engineers are doctors.

Remember · A ‘some’ link passes down a chain of ‘all’ links: some A are B and all B are C, so some A are C.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error