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) Neither I nor 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.

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Answer: (c) Both I and II

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) I and III only

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).

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Answer: (b) July, April

Counting forward, the gap between successive months shrinks by one each time: 12, 11, 10, 9, 8, 7, 6, 5 months. Nine months after October is July, and six months after October is April.

1. Write the months as numbers: 1, 1, 12, 10, X, 3, 10, Y, 9.
2. Count forward from each month to the next: January → January is 12 months, January → December is 11, December → October is 10.
3. The gap falls by one each time: October + 9 months = July (X); July + 8 = March; March + 7 = October.
4. October + 6 months = April (Y); April + 5 = September, which matches the last term.
5. So X = July and Y = April.

Remember · For month or weekday series, convert to numbers and study forward gaps, wrapping around after 12 (or 7).

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) Both I and II

The fewest cuts come from dividing the cube into 3 × 4 × 5 slices, which takes 2 + 3 + 4 = 9 cuts. The inner block untouched by paint is (3 − 2) × (4 − 2) × (5 − 2) = 6 pieces, so both statements are correct.

1. If the cube is cut into a × b × c pieces, it needs (a − 1) + (b − 1) + (c − 1) cuts. With a × b × c = 60, make a + b + c as small as possible.
2. Factors closest to one another give the smallest sum: 3 × 4 × 5 = 60 with sum 12 (2 × 5 × 6 gives 13, 2 × 2 × 15 gives 19). Minimum cuts = 2 + 3 + 4 = 9.
3. Unpainted pieces are the ones strictly inside: (3 − 2) × (4 − 2) × (5 − 2) = 1 × 2 × 3 = 6.
4. Both statements are correct.

• ✓ I The 3 × 4 × 5 split needs 2 + 3 + 4 = 9 cuts; every other way of making 60 pieces needs more.
• ✓ II Removing the painted outer layer leaves an inner block of 1 × 2 × 3 = 6 pieces with no paint.

Remember · a × b × c pieces need (a + b + c − 3) cuts; unpainted pieces = (a − 2)(b − 2)(c − 2).

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 65

Both examples are Pythagorean triples: 7, 24, 25 and 12, 16, 20. So a b means √(a² + b²), and 16 63 = √(256 + 3969) = √4225 = 65.

1. 7² + 24² = 49 + 576 = 625 = 25², and 12² + 16² = 144 + 256 = 400 = 20². So a * b = √(a² + b²).
2. 16 * 63 = √(16² + 63²) = √(256 + 3969) = √4225 = 65.
3. Check: 65² = 4225.

Remember · When results look like hypotenuses, test √(a² + b²); know common triples such as 7-24-25 and 16-63-65.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) B, C, B, A, C

The 18-letter series is the block ABBCCC written three times. Filling the five blanks to fit this block gives B, C, B, A, C.

1. The series has 18 places. Split it into three blocks of six: AB_CC_ | A_BCCC | _BBC_C.
2. The middle block A_BCCC suggests the repeating block ABBCCC.
3. Fill every block as ABBCCC: first block needs B and C, second needs B, third needs A and C.
4. In order the blanks are B, C, B, A, C.
5. Check: ABBCCC ABBCCC ABBCCC.

Remember · For letter series with blanks, count the length, try equal blocks, and start from the most complete block.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 12

Each term is the previous one multiplied by 0.5, 1, 1.5, 2 and 2.5 in turn. So X = 24 × 0.5 = 12, and 12 × 1 gives the next term, 12.

1. From 12 onwards: 12 × 1.5 = 18, 18 × 2 = 36, 36 × 2.5 = 90. The multiplier rises by 0.5 each time.
2. Going backwards, the earlier multipliers are 1 and 0.5.
3. So X = 24 × 0.5 = 12, and X × 1 = 12 matches the third term.
4. Check: 24 × 0.5 = 12, 12 × 1 = 12, 12 × 1.5 = 18, 18 × 2 = 36, 36 × 2.5 = 90.

Remember · If differences do not work, try ratios; ratios rising by a fixed step (0.5, 1, 1.5 …) are a common UPSC pattern.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) QRPS

Replacing the letters as in (b) gives 60 − 15 × 3 + 20 ÷ 4 = 60 − 45 + 5 = 20. The other orders give 60, 61.75 and −73.

1. Test each option, doing ÷ and × before + and − (BODMAS).
2. (a) SPRQ: 60 ÷ 15 + 3 × 20 − 4 = 4 + 60 − 4 = 60.
3. (b) QRPS: 60 − 15 × 3 + 20 ÷ 4 = 60 − 45 + 5 = 20, which matches.
4. (c) QRSP: 60 − 15 × 3 ÷ 20 + 4 = 60 − 2.25 + 4 = 61.75.
5. (d) SPQR: 60 ÷ 15 + 3 − 20 × 4 = 4 + 3 − 80 = −73.

Remember · In symbol-substitution items, test each option with BODMAS; a rough size check quickly rules out most options.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) Both I and II

The points force P to have one win and one draw and R one draw and one loss; the goal condition rules out Q winning a match, so Q drew both. P's only goal beat R 1–0, leaving P–Q at 0–0, and the Q–R draw must be 1–1 — so both statements hold.

1. Three matches at 2 points each give 6 points (3 + 2 + 1 checks). Each team scored exactly one goal, so there were 3 goals in all.
2. P's 3 points = one win + one draw. R's 1 point = one draw + one loss. Q's 2 points = one win + one loss, or two draws.
3. If Q had a win and a loss: P's draw and R's draw must be the same match, P vs R, so P beat Q and Q beat R. P's only goal came against Q, so P–R was 0–0; Q's only goal came against R, so R scored in neither match — but R must score one goal. Impossible.
4. So Q drew both of its matches, and P beat R.
5. P's only goal was needed to beat R (1–0), so P–Q was 0–0. Q's only goal came in the drawn Q–R match, so it ended 1–1 and R's goal came against Q.
6. Both statements are correct.

• ✓ I P scored only once and needed that goal to beat R, so P–Q was a 0–0 draw.
• ✓ II Q's single goal came in the Q–R match, which was a draw, so it ended 1–1 and R's one goal was against Q.

Remember · In tournament puzzles, list the win/draw/loss patterns from the points, then use goal totals to eliminate cases.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 29

The gaps between terms are consecutive prime numbers: 2, 3, 5, 7, 11, 13. So X = 18 + 11 = 29, and 29 + 13 = 42 confirms it.

1. Differences: 3 − 1 = 2, 6 − 3 = 3, 11 − 6 = 5, 18 − 11 = 7 — consecutive primes.
2. The next primes are 11 and 13, so X = 18 + 11 = 29.
3. Check: 29 + 13 = 42, the last term.

Remember · Try differences first; gaps of 2, 3, 5, 7 signal a prime-number pattern.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 9

64 identical pieces means a 4 × 4 × 4 arrangement. Each direction needs 3 cuts to make 4 slices, so the least number is 3 + 3 + 3 = 9.

1. With a, b and c straight cuts in the three directions, the number of pieces is (a + 1)(b + 1)(c + 1).
2. We need (a + 1)(b + 1)(c + 1) = 64 with a + b + c as small as possible; the total is least when the three factors are equal: 4 × 4 × 4.
3. So a = b = c = 3, and the number of cuts = 3 + 3 + 3 = 9.
4. Check: other splits cost more, e.g. 8 × 4 × 2 needs 7 + 3 + 1 = 11 cuts.

Remember · Pieces = (a + 1)(b + 1)(c + 1) for a, b, c cuts in three directions; cuts are fewest when the factors are equal.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 0

Zero itself can be reached: 5 − 4 − 3 + 2 × 1 = 0, using minus twice and plus and times once each. No non-negative value can be smaller than 0.

1. The smallest possible non-negative value is 0, so first try to make 0.
2. 5 − 4 − 3 = −2, so the last part must give +2; and 2 × 1 = 2.
3. 5 − 4 − 3 + 2 × 1 = 5 − 4 − 3 + 2 = 0 (multiplication is done first).
4. Check the rule: − is used twice, + once and × once, so no symbol is used more than two times.

Remember · For a ‘smallest value’ question, test the smallest option first; one valid construction settles it.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) A, A, C, D

Split the sequence into blocks of five letters. Each block is ABCD with one letter doubled, and the doubled letter moves along: AABCD, ABBCD, ABCCD, ABCDD. The blanks are A, A, C, D.

1. Write it in blocks of five: A_BCD | _BBCD | ABC_D | ABC_D.
2. The second block has B doubled in ABCD, so each block is ABCD with one letter doubled.
3. The doubled letter moves A → B → C → D: AABCD, ABBCD, ABCCD, ABCDD.
4. So the blanks are A (block 1), A (block 2), C (block 3) and D (block 4).
5. Check: the completed series is AABCD ABBCD ABCCD ABCDD, a clean pattern.

Remember · In letter series with blanks, split the string into equal blocks (try the length of the repeating unit plus one).

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 3 times

The hands coincide once every 12/11 hours, and in this window that happens at about 10:54:33, 12:00 and 1:05:27. There is no separate meeting in the 11 o’clock hour, because the meeting after 11 falls exactly at 12:00.

1. The hands coincide every 12/11 hours = 65 5/11 minutes, counting from 12:00.
2. Working back and forward from 12:00: 10:54 6/11, 12:00 and 1:05 5/11; the next one, 2:10 10/11, is after 2:00 p.m.
3. So the hands coincide 3 times between 10:00 a.m. and 2:00 p.m.
4. Check: the hands meet only 11 times in 12 hours, so one hour-slot (11 to 12) has no meeting of its own.

Remember · Hands coincide 11 times in 12 hours; the meetings of the 11 o’clock and 12 o’clock hours are the same one, at 12:00.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 2031

2025 is an ordinary year. Adding odd days year by year (2025: 1, 2026: 1, 2027: 1, 2028: 2, 2029: 1, 2030: 1) reaches 7 at the end of 2030, so 2031 begins on the same weekday. It is also an ordinary year, so its calendar matches 2025.

1. An ordinary year has 1 odd day; a leap year has 2.
2. Odd days from the start of 2025: 1 + 1 + 1 + 2 (2028 is a leap year) + 1 + 1 = 7, a whole number of weeks, by the end of 2030.
3. So 1st January 2031 falls on the same weekday as 1st January 2025 (a Wednesday).
4. 2031, like 2025, is not a leap year, so the whole calendar repeats.
5. Check: by the start of 2029 only 5 odd days have gathered, and by the start of 2033 there are 10, which leaves 3; neither gives a match.

Remember · For an ordinary year, add odd days until the total is a multiple of 7, then confirm the target year is also ordinary.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 17.5°

At 4:25 the minute hand is at 25 × 6° = 150° from 12, and the hour hand at 4 × 30° + 25 × 0.5° = 132.5°. The angle between them is 17.5°.

1. Minute hand moves 6° per minute: 25 × 6 = 150° from the 12.
2. Hour hand moves 30° per hour plus 0.5° per minute: 4 × 30 + 25 × 0.5 = 132.5°.
3. Angle = 150 − 132.5 = 17.5°.
4. Check with the formula |30H − 5.5M| = |120 − 137.5| = 17.5°.

Remember · Angle between the hands = |30H − 5.5M|; if the result is more than 180°, subtract it from 360°.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) −25

Replace every sign at once: 10 + 30 − 100 × 50 ÷ 25 becomes 10 − 30 × 100 ÷ 50 + 25. By BODMAS, 30 × 100 ÷ 50 = 60, so the value is 10 − 60 + 25 = −25.

1. Swap the signs: + becomes −, − becomes ×, × becomes ÷, ÷ becomes +.
2. 10 + 30 − 100 × 50 ÷ 25 becomes 10 − 30 × 100 ÷ 50 + 25.
3. 30 × 100 ÷ 50 = 3000 ÷ 50 = 60.
4. 10 − 60 + 25 = −25.

Remember · Rewrite the whole expression with the new signs first, then apply BODMAS; never substitute and calculate at the same time.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 2

Q − S = 2S means Q = 3S, so (Q, S) is (15, 5) or (12, 4). Only (15, 5) lets T = R + S + 3 work, with R = 4 and T = 12, which leaves P = 10. So P + R − T = 10 + 4 − 12 = 2.

1. Q − S = 2S gives Q = 3S. From 4, 5, 10, 12, 15, the options are (Q, S) = (15, 5) or (12, 4).
2. Case Q = 12, S = 4: T = R + 7 with R and T from 5, 10, 15; no pair works.
3. Case Q = 15, S = 5: T = R + 8 with R and T from 4, 10, 12; R = 4, T = 12 works, leaving P = 10.
4. P + R − T = 10 + 4 − 12 = 2.

Remember · Use the tightest equation first to list the cases, then eliminate cases with the second equation.

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Answer: (a) 27

125 cubes make a 5 × 5 × 5 block. A cube surrounded on every side cannot lie on any outer face, so remove one layer from each side: the hidden core is 3 × 3 × 3 = 27 cubes.

1. 125 = 5 × 5 × 5, so the block is 5 cubes long, wide and high.
2. Every cube in the outer layer has at least one face exposed.
3. Removing one layer from each side leaves (5 − 2) × (5 − 2) × (5 − 2) = 3 × 3 × 3 = 27 cubes, each touched by cubes on all sides.

Remember · Fully hidden cubes in an n × n × n block = (n − 2)³.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 6

Add the three numbers and then add the digits of the total: 26 → 8, 50 → 5, 49 → 13. For 23, 4 and 15 the total is 42, and 4 + 2 = 6.

1. 7 ⊕ 9 ⊕ 10: 7 + 9 + 10 = 26, and 2 + 6 = 8 (fits).
2. 9 ⊕ 11 ⊕ 30: 9 + 11 + 30 = 50, and 5 + 0 = 5 (fits).
3. 11 ⊕ 17 ⊕ 21: 11 + 17 + 21 = 49, and 4 + 9 = 13 (fits).
4. 23 ⊕ 4 ⊕ 15: 23 + 4 + 15 = 42, and 4 + 2 = 6.

Remember · For invented operators, test simple rules — sums, products, digit sums — on every given case and accept only one that fits all.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) All five

Chaining the statements gives C = D > F > A > B, with E the youngest, so C and D are the oldest. Every statement is needed: drop any one and some person could be older than D, or C could not be placed.

1. From 4, 5 and 1: D > F > A > B. From 2: C = D. From 3: E is the youngest.
2. So C and D are the oldest.
3. Without 1, B is not linked to anyone and might be older than D; without 3, E might be older than D.
4. Without 5, A and B are cut off from D; without 4, F, A and B are cut off from D; without 2, C's age is unknown.
5. So all five statements are required.

Remember · To fix 'the oldest', every person must be linked below the candidate; test by dropping one statement at a time.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) Wednesday

Counting today (Sunday) as day 1, the 10^10th day comes 10^10 − 1 days later. Since 10 leaves remainder 3 on division by 7, 10^10 leaves the same remainder as 3^10, which is 4; so 10^10 − 1 leaves 3, and three days after Sunday is Wednesday.

1. Take today, Sunday, as day 1. Then day N falls N − 1 days after Sunday.
2. 10 leaves remainder 3 on division by 7, so 10^10 leaves the same remainder as 3^10.
3. Remainders of powers of 3 by 7: 3, 2, 6, 4, 5, 1, then repeat (cycle of 6). 3^10 = 3^(6+4) leaves the same as 3^4, i.e. 4.
4. So 10^10 − 1 leaves remainder 3: the day is 3 days after Sunday, i.e. Wednesday.
5. Note: counting 10^10 days after today (not counting today) would give Sunday + 4 = Thursday; UPSC's key uses the reading in which today is the 1st day.

Remember · Day questions: reduce the count modulo 7 (powers via short cycles), and be clear whether today counts as day 1.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) I

Z appears 2 times, Y 3 times and so on, up to A 27 times, so there are 377 terms and the middle one is the 189th. The first k letters fill k(k + 3)/2 places; for k = 18 this is exactly 189, so the middle term is the 18th letter counted from Z, which is I.

1. The kth letter from Z appears k + 1 times: Z (2), Y (3), …, A (27).
2. Total terms = 2 + 3 + … + 27 = 377, so the middle term is the (377 + 1)/2 = 189th.
3. The first k letters take up 2 + 3 + … + (k + 1) = k(k + 3)/2 places.
4. k = 17 gives 170 and k = 18 gives 189, so the 189th term is the last copy of the 18th letter from Z.
5. Counting back from Z: Z, Y, X, W, V, U, T, S, R, Q, P, O, N, M, L, K, J, I — the 18th is I.

Remember · For blocks of growing length, find the total, locate the middle position, then use running totals to see which block holds it.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) DDCA

With the blanks filled as D, D, C, A the series splits into groups of five: ABCDD, ABCCD, ABBCD, AABCD. Each group is ABCD with one letter doubled, and the doubled letter moves back from D to C to B to A.

1. There are 16 printed letters and 4 blanks: 20 letters, which suggests four groups of five.
2. The unbroken stretch DABBCD contains ABBCD — the block ABCD with B doubled.
3. So the groups should be ABCD with one letter doubled: ABCDD, ABCCD, ABBCD, AABCD (D, C, B, A doubled in turn).
4. Fill the blanks to match: ABC(D)(D) ABC(C) DABBCD(A) ABCD, i.e. D, D, C, A.
5. Check: ABCDDABCCDABBCDAABCD splits exactly into ABCDD | ABCCD | ABBCD | AABCD.

Remember · In letter series, count the total letters and try equal groups (4, 5, 6); the stretch without blanks usually shows the rule.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error