Puzzles, arrangements & general mental ability

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.

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Answer: (a) 100 m

After the first turn the two run in opposite directions, so they end up 60 + 40 = 100 m apart on one line. The next turns send both of them 50 m in the same direction, which does not change the gap.

1. Take the common first run as 50 m north, ending at a point O.
2. X turns right (east) and runs 60 m; Y turns left (west) and runs 40 m. They are now 60 + 40 = 100 m apart on an east–west line.
3. X, facing east, turns left and so runs north 50 m; Y, facing west, turns right and so also runs north 50 m.
4. Both moved 50 m north, so they are still 100 m apart.

Remember · Equal moves in the same direction never change the distance between two people; only the unequal or opposite legs count.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 7

Count odd days from the end of 2000: 31 December 2098 falls on a Wednesday, and January to May 2099 add 4 more odd days, so 31 May 2099 is a Sunday. One week later, 7 June 2099, is also a Sunday.

1. 2000 years contain 0 odd days, so 31 December 2000 was a Sunday.
2. From 2001 to 2098 there are 98 years, of which 24 are leap years (2004, 2008, …, 2096). Odd days = 98 + 24 = 122, and 122 ÷ 7 leaves 3.
3. So 31 December 2098 is Sunday + 3 = Wednesday.
4. 1 January to 31 May 2099 (2099 is not a leap year): 31 + 28 + 31 + 30 + 31 = 151 days; 151 ÷ 7 leaves 4.
5. 31 May 2099 is Wednesday + 4 = Sunday, so 1 June 2099 is a Monday and 7 June 2099 is a Sunday.
6. Check: 5 June 2022 (the day of this exam) was a Sunday. Up to 5 June 2099 there are 77 years and 19 leap days: 96 days, which leaves 5 on division by 7. So 5 June 2099 is a Friday and 7 June a Sunday.

Remember · Calendar items: 400 years give 0 odd days; add ordinary years plus leap years, reduce mod 7, then add month days.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 22

C is 12th from the front, so B is 4th or 20th, and A is then 9th, 15th or 25th. With 13 people behind A, the queue has A’s place + 13 people, which is smallest when A is 9th: 22.

1. 11 persons are ahead of C, so C is 12th from the front.
2. 7 persons between B and C means B is 8 places from C: 4th or 20th.
3. 4 persons between A and B means A is 5 places from B: if B is 4th, A is 9th; if B is 20th, A is 15th or 25th.
4. 13 persons are behind A, so the queue has (A’s place + 13) persons: 22, 28 or 38.
5. The minimum is 22, with B 4th, A 9th and C 12th.
6. Check: between B (4th) and A (9th) stand places 5–8, i.e. 4 persons; between B (4th) and C (12th) stand places 5–11, i.e. 7 persons; behind A (9th) stand places 10–22, i.e. 13 persons.

Remember · Queue items: ‘k persons between’ means a gap of k + 1 places; list the cases and take the minimum.

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Answer: (d) None of the above

The conditions allow more than one seating. In one valid seating C faces G but E does not face F; in another E faces F but C does not face G; and B can never sit beside A. So none of (a), (b) and (c) is definitely correct.

1. A sits between E and D, so E, A, D occupy three adjacent seats; F sits between B and H, so B, F, H occupy three adjacent seats. C and G take the two seats left.
2. A’s two neighbours are E and D, so B can never sit next to A — (a) is never true.
3. Valid seating 1 (clockwise): A, D, G, B, F, H, C, E. B and D are neither next to nor opposite C. Here C is opposite G, but E is opposite B, not F.
4. Valid seating 2 (clockwise): A, D, H, F, B, G, C, E. Again B and D are neither next to nor opposite C. Here E is opposite F, but C is opposite H, not G.
5. (b) fails in seating 2 and (c) fails in seating 1, so none of the statements is definitely correct.

Remember · For ‘definitely correct’, build two valid arrangements; any option that fails in either one is out.

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Answer: (c) Both 1 and 2

The hour and minute hands meet 16 4/11 minutes after 3 o’clock, i.e. between 3:16 and 3:17. The second hand catches the minute hand about 58.98 seconds after 4:58, i.e. before 4:59. Both statements are correct.

1. At 3:00 the hour hand is 90° ahead of the minute hand. The minute hand moves 6° a minute and the hour hand 0.5°, so the minute hand gains 5.5° a minute.
2. It catches up after 90 ÷ 5.5 = 180/11 = 16 4/11 minutes, about 3:16:22 — between 3:16 and 3:17. Statement 1 is correct.
3. t seconds after 4:58, the minute hand is at the 58 + t/60 minute mark and the second hand at the t mark. They meet when t = 58 + t/60, i.e. t = 58 × 60/59 ≈ 58.98 s.
4. That is before 4:59, so statement 2 is also correct.

• ✓ 1. The hands coincide at 3:16 4/11, inside the stated minute.
• ✓ 2. The second hand passes the minute hand about 58.98 seconds after 4:58, before 4:59.

Remember · Clock overlaps: gap ÷ relative speed. Minute hand gains 5.5° per minute on the hour hand.

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Answer: (b) 14

A capital letter looks unchanged in a mirror only if it is symmetric about a vertical line. Of the 21 consonants, only H, M, T, V, W, X and Y are, so 14 consonants look different.

1. The alphabet has 26 − 5 = 21 consonants (leaving out A, E, I, O, U).
2. A mirror flips left and right, so only letters symmetric about a vertical line look the same.
3. Such consonants: H, M, T, V, W, X, Y — 7 letters.
4. Consonants whose images look different: 21 − 7 = 14.
5. Check: B, C, D, F, G, J, K, L, N, P, Q, R, S, Z — 14 letters.

Remember · Mirror-symmetric capitals: A, H, I, M, O, T, U, V, W, X, Y — learn the list once.

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Answer: (c) Q and T

Write the three sets — blue covers (Q, R, T), old books (all but S and U) and Gazetteers (all but P, R and S) — and take what is common. Only Q and T are in all three.

1. Blue covers: Q, R, T.
2. Old books: all except S and U → P, Q, R, T, V.
3. Gazetteers: all except the law reports P, R, S → Q, T, U, V.
4. Common to all three: Q and T (R is a law report; U is new and red).

Remember · For multi-attribute questions, list each set and intersect; one mismatch eliminates an option.

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Answer: (a) 0

The odd-place terms rise by 4, 6, 8, 10, 12. The even-place terms follow the same steps only from 4 onwards, so their first term should be 4 − 4 = 0, not 2.

1. Odd places: 3, 7, 13, 21, 31, 43 — differences 4, 6, 8, 10, 12.
2. Even places: 2, 4, 10, 18, 28, 40 — differences 2, 6, 8, 10, 12.
3. Only the first even-place difference breaks the pattern; to match, the first even term must be 4 − 4 = 0.
4. So the incorrect term 2 should be replaced by 0.
5. Check: each even-place term is then 3 less than the odd-place term before it — 0, 4, 10, 18, 28, 40.

Remember · In an alternating series, split it into two sub-series and compare their differences; the misfit shows itself.

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Answer: (c) 3B

In each row the third number is the second minus the first, and the letters A, B and C each appear once. Row 3 therefore needs 13 − 10 = 3 with the letter B: 3B.

1. Numbers, row 1: 10 − 7 = 3; row 2: 9 − 3 = 6 — the third number is the second minus the first.
2. Row 3: 13 − 10 = 3.
3. Letters: each row uses A, B and C once; row 3 already has A and C, so it needs B.
4. The missing entry is 3B.

Remember · In matrix puzzles treat numbers and letters as separate patterns, and check each rule against every row.

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Answer: (c) 9.375

Sequence-I multiplies by 0.5, 1.5, 2.5, 3.5 and 4.5 in turn. Starting from 5, Sequence-II runs 2.5, 3.75, 9.375, so C = 9.375.

1. Sequence-I: 8 × 0.5 = 4, 4 × 1.5 = 6, 6 × 2.5 = 15, 15 × 3.5 = 52.5, 52.5 × 4.5 = 236.25 — the multiplier rises by 1 each time.
2. Sequence-II from 5: A = 5 × 0.5 = 2.5, B = 2.5 × 1.5 = 3.75.
3. C = 3.75 × 2.5 = 9.375.
4. Check: D = 9.375 × 3.5 = 32.8125 — option (d) is the next term, a trap; (a) is A.

Remember · When terms first fall and then shoot up, test multipliers; options often offer the neighbouring terms as traps.

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Answer: (b) T

After reversing each block of four, the 13th place holds P (block PONM). Four places to its right is the 17th place, which starts the next block TSRQ — so the letter is T.

1. Reverse each block of four: DCBA HGFE LKJI PONM TSRQ XWVU, and then Y Z become Z Y.
2. Places 1–12 hold D C B A H G F E L K J I, so the 13th letter is P.
3. The fourth letter to its right is in the 17th place: places 13–17 are P, O, N, M, T.
4. The answer is T.

Remember · Work with positions: in reversed blocks of four, place 4k − 3 holds the letter numbered 4k.

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Answer: (a) Sunday

Counting odd days gives 4 for the years 2001–2026 and 3 for 2027 up to 10 October — 7 in all, i.e. none left over, so the day is Sunday. A quicker route: this very paper was held on Sunday, 10 October 2021, and six years with one leap day add exactly 7 days.

1. Odd days up to the end of 2000: 0 (every 400 years has 0 odd days).
2. 2001–2026: 26 years with 6 leap years (2004, 2008, 2012, 2016, 2020, 2024), so 26 + 6 = 32 odd days; 32 = 4 × 7 + 4, leaving 4.
3. 2027 up to 10 October: Jan 3, Feb 0, Mar 3, Apr 2, May 3, Jun 2, Jul 3, Aug 3, Sep 2, Oct 3 — total 24, leaving 3.
4. Total = 4 + 3 = 7 odd days, i.e. 0, so 10 October 2027 is a Sunday.
5. Check: 10 October 2021 (the date of this paper) was a Sunday; 2022–2027 add 6 + 1 (for 29 February 2024) = 7 days — Sunday again.

Remember · Anchor on a date you know: each ordinary year moves the weekday on by 1 and each leap year by 2.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 607

Every term is three times the previous one plus 1. So X = 202 × 3 + 1 = 607, and 607 × 3 + 1 = 1822 confirms it.

1. 2 × 3 + 1 = 7, 7 × 3 + 1 = 22, 22 × 3 + 1 = 67, 67 × 3 + 1 = 202 — each term is 3 × previous + 1.
2. X = 202 × 3 + 1 = 607.
3. Check: 607 × 3 + 1 = 1822, the next term.

Remember · When terms roughly triple, test '× 3 ± k', and always verify with the term after the blank.

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Answer: (d) Between 7:05 hours and 7:10 hours

At 7:00 the hour hand is 210° ahead of the minute hand, and the minute hand gains 5.5° every minute. The gap falls to 180° after 30 ÷ 5.5 = 5 5/11 minutes — just after 7:05.

1. Angle between the hands = |30H − 5.5M| degrees (H = hour, M = minutes).
2. At 7:00 the angle is 30 × 7 = 210°, and it shrinks by 5.5° each minute.
3. 210 − 5.5M = 180 gives M = 30 ÷ 5.5 = 5 5/11 minutes.
4. The hands are opposite at 7:05 5/11, i.e. between 7:05 and 7:10.
5. Check: at 7:05 the angle is 210 − 27.5 = 182.5°, still above 180°.

Remember · Angle = |30H − 5.5M|; the minute hand gains 5.5° a minute on the hour hand.

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Answer: (d) ababab

Filling the blanks with a, b, a, b, a, b turns the series into abbaab abbaab abbaab — one six-letter block repeated three times. None of the other choices produces a repeating block.

1. The series has 18 places, so look for a repeating block of 6 (or 3, 9).
2. Fill the blanks with a, b, a, b, a, b: abbaababbaababbaab.
3. This splits as abbaab | abbaab | abbaab — the same block three times.
4. Check: bababa gives bbaabbaabbababaaab, which has no repeating block.

Remember · For letter-series gaps, count the length and test whether the filled series splits into equal repeating blocks.

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Answer: (c) 12th July is a Sunday if the year is a leap year.

From 12 January to 12 July is 181 days in an ordinary year and 182 days in a leap year. 182 is exactly 26 weeks, so in a leap year 12 July falls on the same day, Sunday.

1. Days from 12 January to 12 July: January 31, February 28 or 29, March 31, April 30, May 31, June 30.
2. Ordinary year: 181 days = 25 weeks + 6 days, so 12 July is Saturday.
3. Leap year: 182 days = 26 weeks exactly, so 12 July is Sunday.
4. 15 July is then Tuesday (ordinary year) or Wednesday (leap year), never Sunday — (a), (b) and (d) fail.

Remember · Count days between the two dates and divide by 7; the remainder is the number of weekdays to move forward.

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Answer: (b) 6

Scan the 26 digits once and mark every place where an odd digit is immediately followed by another odd digit. There are six such places: 5-1, 7-3, 3-9, 5-7, 3-1 and 1-5.

1. Pairs in order: 5-1 ✓, 1-4, 4-7, 7-3 ✓, 3-9 ✓, 9-8, 8-5, 5-7 ✓, 7-2, 2-6, 6-3, 3-1 ✓, 1-5 ✓, 5-8 …
2. After 5-8 the odd digits are 3, 5, 3 and 9, and each is followed by an even digit (8, 2, 4, 6).
3. Total odd-odd pairs = 6.

Remember · For adjacency counts, scan pairs left to right in one pass and tick each hit; do not restart counting.

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Answer: (b) 41

G is 11 places to the right of A, so G is 27th from the left. G is also 3 places to the right of V, so V is 24th from the left. V is 18th from the right, so the row has 24 + 18 − 1 = 41 boys.

1. A is 16th from the left; G is 11th to A's right, so G is 16 + 11 = 27th from the left.
2. G is 3rd to V's right, so V is 27 − 3 = 24th from the left.
3. V is 18th from the right, so total = 24 + 18 − 1 = 41.
4. Check: G is then 41 − 27 + 1 = 15th from the right, which is 3 places right of V (18th).

Remember · Convert every position to one end, then total = position from left + position from right − 1.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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

Answer: (b) 20

The solid is a regular icosahedron, and every one of its faces is a triangle. Its edges close into no triangles other than the faces, so the count is simply the number of faces: 20.

1. Identify the solid: every face is a triangle and five faces meet at each vertex — a regular icosahedron.
2. Count the faces in three layers: 5 around the top vertex, 5 around the bottom vertex and a band of 10 in the middle.
3. 5 + 10 + 5 = 20 triangular faces.
4. No three edges outside a face close into a triangle, so there are no extra triangles to add.
5. Check: Euler's formula V − E + F = 2 with 12 vertices and 30 edges gives F = 20.

Remember · For a solid made of triangular faces, count faces layer by layer and confirm with Euler's formula V − E + F = 2.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 1 Oct 2026 (how we verify). Permalink · Report an error