Puzzles, arrangements & general mental ability

Answer: (c) 8

The only cut that works is a 4 × 4 × 4 cube with four 2 × 2 × 2 blocks. To keep the big blocks off both blue faces they must fill the middle two layers, leaving 16 small cubes in the top layer and 16 in the bottom. Only the 4 central cubes of each of these layers have a single painted face.

1. 36 cubes of two sizes fit a 4 × 4 × 4 cube: 4 big cubes of 2 × 2 × 2 (4 × 8 = 32 units) and 32 unit cubes; 32 + 32 = 64.
2. Let the two blue faces be the top and the bottom. A big cube is 2 units tall, so it avoids both only if it sits in the middle two layers.
3. The 4 big cubes fill the middle 4 × 4 × 2 block; the top and bottom 4 × 4 × 1 layers hold 16 + 16 = 32 small cubes.
4. Each big cube stands at a corner of the middle block and shows one yellow and one black face — two painted faces.
5. In the top layer, 4 corner cubes have 3 painted faces, 8 edge cubes have 2, and the 4 central cubes have only the blue top painted. The bottom layer is the same.
6. Cubes with exactly one face painted = 4 + 4 = 8.

Remember · Fix the cut first, place the pieces with the restriction, then count painted faces layer by layer.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) Thursday

Only one start works for both clues: 1 April a Sunday, which makes 1 November a Thursday. Counting forward from there, 20 December is a Thursday.

1. April has 30 days (4 weeks + 2 days), so it has a 5th Monday only if 1 April is a Monday or a Sunday.
2. November also has 30 days, so it has a 5th Thursday only if 1 November is a Thursday or a Wednesday.
3. Days from 1 April to 1 November = 30 + 31 + 30 + 31 + 31 + 30 + 31 = 214 = 30 weeks + 4 days, so 1 November falls 4 weekdays after 1 April.
4. 1 April Monday → 1 November Friday (no 5th Thursday). 1 April Sunday → 1 November Thursday, which fits.
5. 1 December = 1 November + 30 days = Thursday + 2 = Saturday.
6. 20 December = 1 December + 19 days = Saturday + 5 = Thursday.

Remember · In a 30-day month only the weekdays of the 1st and 2nd occur five times; link months by counting odd days.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 5:36 p.m.

The clock shows 24 h 10 min for every 24 real hours. It shows 58 hours gone, which is 58 × 1440/1450 = 57.6 real hours (57 h 36 min), so the correct time is 5:36 p.m. on Wednesday.

1. From 8:00 a.m. Monday to 6:00 p.m. Wednesday the clock shows 48 + 10 = 58 hours.
2. Shown time : real time = 24 h 10 min : 24 h = 1450 : 1440.
3. Real time = 58 × 1440/1450 = 57.6 h = 57 h 36 min.
4. Correct time = Monday 8:00 a.m. + 57 h 36 min = Wednesday 5:36 p.m.
5. Check: gain = 10 min × 57.6/24 = 24 min, and 6:00 − 0:24 = 5:36.

Remember · For a fast or slow clock, convert shown time to real time by the ratio real : shown, not by a flat subtraction.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) cbbb

With c, b, b, b the series reads ccacc | aabaa | bbcbb — three five-letter blocks, each of the form x x y x x. No other option keeps this shape.

1. The series has 15 places; split into blocks of five: c _ a c c | a a _ a a | _ b c _ b.
2. Each block has the shape x x y x x: c c a c c needs c; a a b a a needs b.
3. The third block b b c b b needs b in both gaps.
4. Gaps in order: c, b, b, b.
5. Check: option (a) would make the first block c a a c c, which breaks the pattern.

Remember · For letter-gap series, split the string into equal blocks (4, 5 or 6 letters) and look for a repeating shape.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 2015

A year repeats 2009's calendar when the odd days carried forward total a multiple of 7 and it is also a non-leap year. From 2009 the odd days add up to 1 + 1 + 1 + 2 + 1 + 1 = 7 by the start of 2015, and 2015 is not a leap year.

1. Odd days carried by each year: 2009 → 1, 2010 → 1, 2011 → 1, 2012 (leap) → 2, 2013 → 1, 2014 → 1.
2. Total to the start of 2015 = 7, a full week, so 1 January 2015 falls on the same weekday as 1 January 2009.
3. 2015, like 2009, is not a leap year, so the whole calendar matches.
4. Check: 2016 would carry 8 odd days (≡ 1) and is a leap year, so it cannot match.

Remember · Same calendar = total odd days ≡ 0 (mod 7) and the same leap-year status.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) jUlY

From B onwards every alternate letter — B, D, F … the even-numbered letters — is written small, and the odd-numbered letters are capitals. In JULY, J (10th) and L (12th) are small, U (21st) and Y (25th) are capitals: jUlY.

1. Small letters: B, D, F, H, J, L, N, P, R, T, V, X, Z — the even positions.
2. Capitals: A, C, E, …, U, W, Y — the odd positions.
3. The first month of the second half of the year is July.
4. J = 10th → j; U = 21st → U; L = 12th → l; Y = 25th → Y.
5. Result: jUlY.

Remember · Turn 'every alternate letter from B' into positions — the even-numbered letters — then check each letter's position.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 1

The sequence has three 5s. The first is preceded by 1 and followed by 7; the other two are both preceded by 3. Only one 5 meets both conditions.

1. 5s are at positions 2, 5 and 9: (1, 5, 7), (3, 5, 7), (3, 5, 7).
2. Position 2: preceded by 1, followed by 7 → counts.
3. Positions 5 and 9: preceded by 3 → excluded.
4. Count = 1.

Remember · Mark every target symbol first, then test both neighbours of each; don't count by eye.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) Fifth

If F is second, C must be first, and then E is last. Places 3, 4 and 5 go to D, A and B in that order (D above A, A above B), so B is fifth.

1. Either C or F is first; F is second, so C is first.
2. C first → E is sixth (least).
3. D, A and B fill places 3–5 with D above A and A above B.
4. Order: C, F, D, A, B, E — B is fifth.

Remember · Apply the 'either–or' rule first; it fixes the top and bottom, leaving a short chain to slot in.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) Second or third

If C were first, E would be last; B is last, so F is first. D and A are close competitors — adjacent, with D just above A — and C is above A, so C must be above the D–A pair. With E free to go anywhere in places 2–5, C is second or third.

1. B last means C is not first (C first would force E last), so F is first.
2. Places 2–5 hold C, D, A and E; D is just above A (close competitors) and C is above A.
3. C cannot sit between D and A, so C is above D.
4. Possible orders: F, C, D, A, E, B / F, C, E, D, A, B / F, E, C, D, A, B.
5. C is second or third.

Remember · 'Close competitors' means adjacent ranks; treat such a pair as one block when listing arrangements.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) C gets more marks than E

E third means E is not last, so C cannot be first; F is first and B last. D and A must sit together in places 4 and 5, leaving place 2 for C — above E.

1. E is third, so E is not last; hence C is not first (C first forces E last).
2. So F is first and B is sixth.
3. Places 2, 4 and 5 remain for C, D and A, with D immediately above A.
4. D and A take places 4 and 5; C takes place 2.
5. Order: F, C, E, D, A, B — C scores more than E; A is fifth and D fourth, so (c) and (d) fail.

Remember · Use each 'whenever' rule in reverse too: if E is not last, C cannot be first.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error