Puzzles, arrangements & general mental ability

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

Answer: (c) Two

In reverse alphabetical order the 21 letters read TSSRPONNMLIIIIIHEEECB. The five I's fill positions 11 to 15, and the original word has I at positions 13 and 15, so exactly two positions stay the same.

1. Letters (21): I × 5, E × 3, N × 2, S × 2 and one each of B, C, H, L, M, O, P, R, T.
2. Reverse alphabetical order: T S S R P O N N M L I I I I I H E E E C B.
3. The I's now occupy positions 11 to 15.
4. In INCOMPREHENSIBILITIES, I stands at positions 1, 13, 15, 17 and 19 — so positions 13 and 15 match.
5. Check the rest one by one (1: I vs T, 11: N vs I, 16: L vs H, 20: E vs C …): no other match. Two positions stay unchanged.

Remember · For 'unchanged positions', write the new order with position numbers; check the block of the most repeated letter first — matches usually hide there.

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