Puzzles, arrangements & general mental ability

Answer: (a) 5

The clues chain into the block P-Q-R-S-T, leaving three seats for U-V-W with V in the middle. W must be opposite S, which fixes the full circle as P, Q, R, S, T, U, V, W in clockwise order, so Q passes R, S, T, U and V before W.

1. P and R sit on both sides of Q, and R and T on both sides of S: this gives the block P-Q-R-S-T (5 seats).
2. The remaining 3 seats go to U, V, W with V in the middle (U and W both next to V).
3. Number the seats 1 to 8: P1, Q2, R3, S4, T5. The seat opposite S (4) is seat 8, so W8, V7, U6.
4. From P, going the way P → Q → R → S → T meets R before T, so this direction is clockwise.
5. Clockwise from Q: R, S, T, U, V, then W. Q crosses 5 persons.

Remember · In circular seating, first join 'adjacent' clues into blocks; the 'opposite' clue then fixes the rest.

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

Answer: (d) bb

The string has 15 places, so the repeating block has length 3 or 5. Length 3 clashes (place 2 is b but place 11 is a); length 5 fits every known letter and gives the block 'abbab', so places 7 and 8 are b and b.

1. Number the 15 places: 1 ×, 2 b, 3 ×, 4 a, 5 ×, 6 a, 7 ×, 8 ×, 9 a, 10 ×, 11 a, 12 ×, 13 b, 14 a, 15 b.
2. Block length 3 fails: places 2 and 11 should match but are b and a.
3. Block length 5: places 1, 6, 11 → a; 2, 7, 12 → b; 3, 8, 13 → b; 4, 9, 14 → a; 5, 10, 15 → b.
4. Block = abbab, repeated three times: abbab abbab abbab.
5. Places 7 and 8 = 2nd and 3rd letters of the block = b, b.

Remember · For a repeated pattern, try block lengths that divide the string length; letters one block apart must match.

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

Answer: (a) 16

With pieces left unmoved, four plane cuts can give at most 15 pieces (1 + 4 + 6 + 4), so 16 is impossible. The other values arise easily from identical pieces after three cuts: 5 and 8 from four equal slabs, 12 from eight small cubes.

1. Identical pieces after three cuts: 4 equal slabs (three parallel cuts), 6 equal blocks (two parallel + one perpendicular), or 8 small cubes (three mutually perpendicular cuts through the centre).
2. Four plane cuts on an unmoved solid give at most 1 + 4 + 6 + 4 = 15 pieces; equivalently, the fourth plane can cross at most 7 of the 8 small cubes, giving 8 + 7 = 15.
3. So x₄ = 16 is not possible.
4. Check the rest: 4 slabs + a fourth parallel cut through one slab = 5; 4 slabs + a perpendicular cut through all four = 8; 8 cubes + a cut parallel to a face through four of them = 12.

• ✗ (a) Not possible: four plane cuts on unmoved pieces give at most 15 pieces.
• ✓ (b) Possible: after three central perpendicular cuts (8 cubes), a fourth cut parallel to one face through one layer splits 4 cubes → 12.
• ✓ (c) Possible: after three parallel cuts (4 equal slabs), a fourth cut perpendicular to them splits all 4 → 8.
• ✓ (d) Possible: after 4 equal slabs, a fourth cut parallel to them inside one slab → 5.

Remember · Unmoved pieces: plane cuts give at most 2, 4, 8 and 15 pieces for one to four cuts.

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

Answer: (d) न तो I और न ही II

With P = 4 and R = 1, the ranks 2, 3 and 5 remain for Q, S and T, and only two orders fit the conditions: (Q, S, T) = (2, 3, 5) or (3, 5, 2). In each case the 'then' part of the statement is the opposite of what actually happens.

1. P = 4, R = 1, so Q, S, T share ranks 2, 3 and 5.
2. Q ≠ 5 → Q is 2 or 3; S ≠ 2 → S is 3 or 5; T ≠ 3 → T is 2 or 5.
3. Q = 2: then T ≠ 3 forces T = 5 and S = 3.
4. Q = 3: then S ≠ 2 forces S = 5 and T = 2.
5. S = 3 happens only with T = 5, so I is wrong; Q = 3 happens only with T = 2, so II is wrong.

• ✗ I S = 3 occurs only in the order Q = 2, S = 3, T = 5, so T is 5, not 2.
• ✗ II Q = 3 occurs only in the order Q = 3, S = 5, T = 2, so T is 2, not 5.

Remember · List all arrangements that satisfy the restrictions; there are usually very few, then test each 'if…then' against them.

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

Answer: (c) I और II दोनों

For 'possible', one valid layout is enough. The layout Q4, Q5, Q3, Q1, Q2 beside P3, P5, P1, P4, P2 meets every condition, and it has Q3 in the middle and P2 at an end, so both are possible.

1. Number the portions 1 to 5 along the rods. Try Q3 (and so P1) at portion 3, with Q5 next to it at portion 2.
2. Put Q4 at portion 1 with P3 beside it (Q4 must match P3 or P5).
3. Put Q1 at portion 4 with P4 beside it (P4 must match Q1 or Q2), and Q2 at portion 5.
4. The remaining P5 goes to portion 2 and P2 to portion 5.
5. Layout: Q-rod Q4, Q5, Q3, Q1, Q2; P-rod P3, P5, P1, P4, P2. All conditions hold.
6. Here Q3 is in the middle (I possible) and P2 is at an end (II possible).

• ✓ I Possible: in the layout Q4, Q5, Q3, Q1, Q2, Q3 occupies the middle portion and every condition holds.
• ✓ II Possible: in the same layout, P2 sits at portion 5, an extreme end.

Remember · 'Is possible' needs just one example that satisfies every condition; 'must be' needs proof for all cases.

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

Answer: (b) केवल I और III

B cannot be in Y or with A in X, so B joins C in Z. E and G ride alone together, which leaves them car Y; D and F then fill X with A. So cars are X: A, D, F; Y: E, G; Z: B, C — D is not the only one with F, since A is there too.

1. B is not in Y and not in X (A is in X and A, B are apart), so B is in Z with C.
2. E and G travel together with nobody else; X has A and Z has B, C, so E and G take Y.
3. X has A and two others: the only ones left are D and F.
4. Cars: X = A, D, F; Y = E, G; Z = B, C.
5. I true (no car has one person); II false (A also travels with F); III true (Z has only B and C).

• ✓ I Each car carries two or three persons.
• ✗ II F shares car X with both A and D, so D is not the only one.
• ✓ III Car Z carries only B and C.

Remember · Place the person with the most 'not' conditions first; the rest usually falls into place.

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

Answer: (a) 2

Each weighing has three outcomes (left lighter, right lighter, balance), so split the cubes into three groups: 3, 3 and 1. One weighing narrows the light cube to a group of at most three, and a second weighing of one against one finds it.

1. Weigh 3 cubes against 3. If they balance, the 7th cube is the light one.
2. If one side rises, the light cube is among those 3.
3. Weigh 1 against 1 from that group: the lighter pan shows it; if balanced, it is the third cube.
4. So 2 weighings always suffice; 1 cannot, since one weighing separates at most 3 possibilities, not 7.

Remember · A balance gives three outcomes: with k weighings you can find one odd (known-lighter) item among up to 3ᵏ items.

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

Answer: (d) 47

Build the sequence by adding the last two terms each time: 4, 7, 11, 18, 29, 47. The sixth term is 47.

1. Term 3 = 4 + 7 = 11.
2. Term 4 = 7 + 11 = 18.
3. Term 5 = 11 + 18 = 29.
4. Term 6 = 18 + 29 = 47.

Remember · Short sequences: just write the terms out; count positions carefully (option 29 is the fifth term).

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