Logical & analytical reasoning

Answer: (a) CCIDD

Each letter of MUMBAI is moved back by 1, 2, 3, 4, 5 and 6 places in turn, wrapping from A round to Z. Moving the letters of DELHI back by 1 to 5 gives CCIDD.

1. M − 1 = L, U − 2 = S, M − 3 = J, B − 4 = X (B → A → Z → Y → X), A − 5 = V, I − 6 = C.
2. DELHI: D − 1 = C, E − 2 = C, L − 3 = I, H − 4 = D, I − 5 = D.
3. Code: CCIDD.
4. Check: (d) CCIFE agrees in the first three letters but fails at H − 4 = D.

Remember · In letter codes, write the shift for each position; a growing shift (−1, −2, −3 …) is common, and letters wrap from A to Z.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 92323

Each letter stands for a fixed digit. From the two words, H = 9, A = 2 and M = 3, so HAMAM reads 9-2-3-2-3.

1. RAMON = 12345: R = 1, A = 2, M = 3, O = 4, N = 5.
2. DINESH = 675849: D = 6, I = 7, N = 5, E = 8, S = 4, H = 9.
3. HAMAM = H, A, M, A, M = 9, 2, 3, 2, 3 = 92323.

Remember · For letter-to-digit codes, map each letter once and read the new word letter by letter; repeated letters repeat their digit.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) लेखनी < पुस्तक

Pen < Pencil < Book is an unbroken chain, so Pen < Book always. Cap is tied only through Book > Cap, which says nothing about how Cap compares with Pen or Pencil.

1. Join the first two relations: Pen < Pencil < Book, so Pen < Book.
2. Book > Cap points the other way; Cap could be above or below Pen and Pencil.
3. So (a), (c) and (d) are not always true, while (b) is.

Remember · A comparison is certain only if an unbroken chain of signs points the same way between the two items.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 6

Every boy wears at least a shirt or pants, so adding shirts and pants counts the boys in full uniform twice. 11 + 14 − 19 = 6 boys wear both.

1. Every one of the 19 boys wears a shirt, pants or both.
2. Shirts + pants = 11 + 14 = 25, which counts the boys wearing both twice.
3. Both = 25 − 19 = 6.
4. Check: 5 with shirts only + 8 with pants only + 6 with both = 19.

Remember · For two overlapping groups that together cover everyone: both = first group + second group − total.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error