Logical & analytical reasoning

Answer: (c) a, c, c, e

With a, c, c, e filled in, the string splits into four blocks of five letters — ababa, bcbcb, cdcdc, deded — each alternating two neighbouring letters. No other option gives a regular pattern.

1. The completed string has 20 letters; try blocks of 5.
2. Fill a, c, c, e: a b a b a | b c b c b | c d c d c | d e d e d.
3. Each block alternates two consecutive letters (ab, bc, cd, de), starting where the last block ended.
4. Check option (d) b, c, d, e: the first block becomes ababb, which breaks the alternation.

Remember · For letter-gap series, try splitting the full string into equal blocks; the blank count and length often reveal the block size.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) R is S’s wife.

Q, a mother-in-law, belongs to the older couple, so P (U's parent) is her husband. Their children are U, T and one married child whose spouse is R. With three males already P, T and S, R must be female and married to S.

1. Q is R's mother-in-law, so Q is a woman of the older generation and R is married to Q's child.
2. Two couples: the older couple and R with Q's child. U is P's daughter, so P is the older couple's other member — P is Q's husband (male).
3. Males: P, T and S = three, so Q, U and R are the three females.
4. R, a female, is married to a son of Q. T is unmarried, so that son is S — R is S's wife.
5. The remaining two, U and T, are the unmarried siblings.
6. Check (a): R is female, so cannot be U's husband; (c): S is married.

Remember · In family puzzles, fix the generations first, then use the male/female count to settle the remaining roles.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) North-West

A house facing East has its back to the West, so he first walks West. A right turn from West points North, and a left turn from North points West again. He ends 50 m West and 50 m North of the start — North-West.

1. House faces East, so walking down its backside means walking West: 25 m West.
2. Right turn while facing West → facing North: 50 m North.
3. Left turn while facing North → facing West: 25 m West.
4. Net displacement: 50 m West and 50 m North, so he is North-West of the starting point.

Remember · Fix the first direction carefully (the back of an east-facing house is West), then apply each turn relative to the current heading.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) Only Conclusion-I

A number divisible by both 2 and 3 is divisible by their LCM, 6, so Conclusion-I follows. Divisibility by 2 and 3 says nothing about a second factor of 2, so 4 need not divide the numbers — 6 itself is a counter-example.

1. 2 and 3 have no common factor, so a number divisible by both is divisible by 2 × 3 = 6: Conclusion-I follows.
2. Divisibility by 4 needs two factors of 2; the statements guarantee only one.
3. Counter-example: 6 is divisible by 2 and 3 but not by 4, so Conclusion-II does not follow.

Remember · Divisible by a and b means divisible by LCM(a, b) — no more. Test any stronger claim with the LCM itself.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) Neither Conclusion-I nor Conclusion-II

The statements only tell us that every cat is both a dog and black, so some dogs are black. The other dogs may be black or not: 'all dogs are black' and 'some dogs are not black' are each possible, but neither is forced, so neither follows.

1. Cats lie inside both 'dogs' and 'black', so some dogs (the cats) are black.
2. Diagram 1: dogs outside the cats circle are not black — then Conclusion-I fails.
3. Diagram 2: all dogs lie inside 'black' — then Conclusion-II fails.
4. Each conclusion fails in some valid diagram, so neither follows. (Exactly one of them is true in any given case, but the statements do not tell us which.)

Remember · Draw one diagram where a conclusion fails; if you can, it does not follow. Contradictory conclusions can both fail to follow.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 246

G, H, I are 7, 8, 9 and 789 × 2 = 1578; D, E, F are 4, 5, 6 and 456 × 2 = 912. So each code is the three letter numbers written together and doubled: ABC = 123 × 2 = 246.

1. GHI → 7, 8, 9 → 789; 789 × 2 = 1578.
2. DEF → 4, 5, 6 → 456; 456 × 2 = 912.
3. ABC → 1, 2, 3 → 123; 123 × 2 = 246.

Remember · In number codes, write the letter positions side by side first and look for a single operation that gives the code.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) NPJM

Going from ACPQ to BESU, the four letters move forward by 1, 2, 3 and 4 places. Applying the same shifts to MNGI gives N, P, J and M.

1. A → B (+1), C → E (+2), P → S (+3), Q → U (+4).
2. M + 1 = N, N + 2 = P, G + 3 = J, I + 4 = M.
3. The missing term is NPJM.

Remember · In letter analogies, write the shift for each position separately; the shifts often run 1, 2, 3, 4.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error