Puzzles, arrangements & general mental ability

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