Puzzles, arrangements & general mental ability

Answer: (b) 9

64 identical pieces means a 4 × 4 × 4 arrangement. Each direction needs 3 cuts to make 4 slices, so the least number is 3 + 3 + 3 = 9.

1. With a, b and c straight cuts in the three directions, the number of pieces is (a + 1)(b + 1)(c + 1).
2. We need (a + 1)(b + 1)(c + 1) = 64 with a + b + c as small as possible; the total is least when the three factors are equal: 4 × 4 × 4.
3. So a = b = c = 3, and the number of cuts = 3 + 3 + 3 = 9.
4. Check: other splits cost more, e.g. 8 × 4 × 2 needs 7 + 3 + 1 = 11 cuts.

Remember · Pieces = (a + 1)(b + 1)(c + 1) for a, b, c cuts in three directions; cuts are fewest when the factors are equal.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 0

Zero itself can be reached: 5 − 4 − 3 + 2 × 1 = 0, using minus twice and plus and times once each. No non-negative value can be smaller than 0.

1. The smallest possible non-negative value is 0, so first try to make 0.
2. 5 − 4 − 3 = −2, so the last part must give +2; and 2 × 1 = 2.
3. 5 − 4 − 3 + 2 × 1 = 5 − 4 − 3 + 2 = 0 (multiplication is done first).
4. Check the rule: − is used twice, + once and × once, so no symbol is used more than two times.

Remember · For a ‘smallest value’ question, test the smallest option first; one valid construction settles it.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) A, A, C, D

Split the sequence into blocks of five letters. Each block is ABCD with one letter doubled, and the doubled letter moves along: AABCD, ABBCD, ABCCD, ABCDD. The blanks are A, A, C, D.

1. Write it in blocks of five: A_BCD | _BBCD | ABC_D | ABC_D.
2. The second block has B doubled in ABCD, so each block is ABCD with one letter doubled.
3. The doubled letter moves A → B → C → D: AABCD, ABBCD, ABCCD, ABCDD.
4. So the blanks are A (block 1), A (block 2), C (block 3) and D (block 4).
5. Check: the completed series is AABCD ABBCD ABCCD ABCDD, a clean pattern.

Remember · In letter series with blanks, split the string into equal blocks (try the length of the repeating unit plus one).

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 3 times

The hands coincide once every 12/11 hours, and in this window that happens at about 10:54:33, 12:00 and 1:05:27. There is no separate meeting in the 11 o’clock hour, because the meeting after 11 falls exactly at 12:00.

1. The hands coincide every 12/11 hours = 65 5/11 minutes, counting from 12:00.
2. Working back and forward from 12:00: 10:54 6/11, 12:00 and 1:05 5/11; the next one, 2:10 10/11, is after 2:00 p.m.
3. So the hands coincide 3 times between 10:00 a.m. and 2:00 p.m.
4. Check: the hands meet only 11 times in 12 hours, so one hour-slot (11 to 12) has no meeting of its own.

Remember · Hands coincide 11 times in 12 hours; the meetings of the 11 o’clock and 12 o’clock hours are the same one, at 12:00.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 2031

2025 is an ordinary year. Adding odd days year by year (2025: 1, 2026: 1, 2027: 1, 2028: 2, 2029: 1, 2030: 1) reaches 7 at the end of 2030, so 2031 begins on the same weekday. It is also an ordinary year, so its calendar matches 2025.

1. An ordinary year has 1 odd day; a leap year has 2.
2. Odd days from the start of 2025: 1 + 1 + 1 + 2 (2028 is a leap year) + 1 + 1 = 7, a whole number of weeks, by the end of 2030.
3. So 1st January 2031 falls on the same weekday as 1st January 2025 (a Wednesday).
4. 2031, like 2025, is not a leap year, so the whole calendar repeats.
5. Check: by the start of 2029 only 5 odd days have gathered, and by the start of 2033 there are 10, which leaves 3; neither gives a match.

Remember · For an ordinary year, add odd days until the total is a multiple of 7, then confirm the target year is also ordinary.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 17.5°

At 4:25 the minute hand is at 25 × 6° = 150° from 12, and the hour hand at 4 × 30° + 25 × 0.5° = 132.5°. The angle between them is 17.5°.

1. Minute hand moves 6° per minute: 25 × 6 = 150° from the 12.
2. Hour hand moves 30° per hour plus 0.5° per minute: 4 × 30 + 25 × 0.5 = 132.5°.
3. Angle = 150 − 132.5 = 17.5°.
4. Check with the formula |30H − 5.5M| = |120 − 137.5| = 17.5°.

Remember · Angle between the hands = |30H − 5.5M|; if the result is more than 180°, subtract it from 360°.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) −25

Replace every sign at once: 10 + 30 − 100 × 50 ÷ 25 becomes 10 − 30 × 100 ÷ 50 + 25. By BODMAS, 30 × 100 ÷ 50 = 60, so the value is 10 − 60 + 25 = −25.

1. Swap the signs: + becomes −, − becomes ×, × becomes ÷, ÷ becomes +.
2. 10 + 30 − 100 × 50 ÷ 25 becomes 10 − 30 × 100 ÷ 50 + 25.
3. 30 × 100 ÷ 50 = 3000 ÷ 50 = 60.
4. 10 − 60 + 25 = −25.

Remember · Rewrite the whole expression with the new signs first, then apply BODMAS; never substitute and calculate at the same time.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 2

Q − S = 2S means Q = 3S, so (Q, S) is (15, 5) or (12, 4). Only (15, 5) lets T = R + S + 3 work, with R = 4 and T = 12, which leaves P = 10. So P + R − T = 10 + 4 − 12 = 2.

1. Q − S = 2S gives Q = 3S. From 4, 5, 10, 12, 15, the options are (Q, S) = (15, 5) or (12, 4).
2. Case Q = 12, S = 4: T = R + 7 with R and T from 5, 10, 15; no pair works.
3. Case Q = 15, S = 5: T = R + 8 with R and T from 4, 10, 12; R = 4, T = 12 works, leaving P = 10.
4. P + R − T = 10 + 4 − 12 = 2.

Remember · Use the tightest equation first to list the cases, then eliminate cases with the second equation.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error