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CSAT · 113 questions

Puzzles, arrangements & general mental ability

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Puzzles, arrangements & general mental ability questions per year: 2016: 19, 2017: 18, 2018: 17, 2019: 10, 2020: 3, 2021: 10, 2022: 5, 2023: 7, 2024: 8, 2025: 8, 2026: 8 Asked in 11 of 11 years · most in 2016 (19)
Timed · 30 minutes a set Practise Puzzles, arrangements & general mental ability as timed tests 6 sets of about 20 questions with UPSC's CSAT marking. Start practising

UPSC syllabus: “General mental ability;” Sub-topics in the detailed syllabus →

CSAT 2019 · Q10

Hard

एक ठोस घन को पीला, नीला और काला इस प्रकार रँगा गया है कि इसके विपरीत फलक एक ही रंग के हैं । तब इस घन को दो भिन्न आमापों के 36 घनों में इस प्रकार काटा गया है कि 32 घन छोटे हैं और अन्य 4 घन बड़े हैं । बड़े घनों का कोई भी फलक नीला नहीं रँगा गया है । कितने घनों में केवल एक फलक रँगा हुआ है ?

Read this question in English

A solid cube is painted yellow, blue and black such that opposite faces are of same colour. The cube is then cut into 36 cubes of two different sizes such that 32 cubes are small and the other four cubes are big. None of the faces of the bigger cubes is painted blue. How many cubes have only one face painted?

(a) 4 · (b) 6 · (c) 8 · (d) 10

उत्तर और व्याख्या · explanation in English

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.

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 ·

CSAT 2019 · Q12

Hard

श्रीमान ‘X’ के तीन बच्चे हैं । पहले बच्चे का जन्मदिन अप्रैल के पाँचवें सोमवार को पड़ता है, और दूसरे का नवम्बर के पाँचवें बृहस्पतिवार को पड़ता है । उसके तीसरे बच्चे का जन्मदिन किस दिन है, जो कि 20 दिसम्बर को पड़ता है ?

Read this question in English

Mr ‘X’ has three children. The birthday of the first child falls on the 5th Monday of April, that of the second one falls on the 5th Thursday of November. On which day is the birthday of his third child, which falls on 20th December?

(a) Monday · (b) Thursday · (c) Saturday · (d) Sunday

उत्तर और व्याख्या · explanation in English

Answer: (b) बृहस्पतिवार

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.

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 ·

CSAT 2019 · Q49

Medium

कोई दीवार-घड़ी प्रत्येक 24 घंटे में 10 मिनट तेज चलती है । इस घड़ी को सोमवार को पूर्वाह्न 8:00 बजे सही समय दिखाने के लिए सही किया गया । जब यह घड़ी बुधवार को अपराह्न 6:00 बजे का समय दिखाती है, तो सही समय क्या है ?

Read this question in English

A wall clock moves 10 minutes fast in every 24 hours. The clock was set right to show the correct time at 8:00 a.m. on Monday. When the clock shows the time 6:00 p.m. on Wednesday, what is the correct time?

(a) 5:36 p.m. · (b) 5:30 p.m. · (c) 5:24 p.m. · (d) 5:18 p.m.

उत्तर और व्याख्या · explanation in English

Answer: (a) अपराह्न 5:36

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.

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 ·

CSAT 2019 · Q54

Medium

किसी विन्यास का अनुसरण करने वाले निम्नलिखित अनुक्रम पर विचार कीजिए :

c_accaa_aa_bc_b

रिक्त स्थानों में आने वाले अक्षर कौन-से हैं ?

Read this question in English

Consider the following sequence that follows some arrangement:

c_accaa_aa_bc_b

The letters that appear in the gaps are

(a) abba · (b) cbbb · (c) bbbb · (d) cccc

उत्तर और व्याख्या · explanation in English

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.

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 ·

CSAT 2019 · Q59

Easy

किस वर्ष का कैलेंडर ठीक वैसा ही है जैसा 2009 का कैलेंडर है ?

Read this question in English

Which year has the same calendar as that of 2009?

(a) 2018 · (b) 2017 · (c) 2016 · (d) 2015

उत्तर और व्याख्या · explanation in English

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.

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 ·

CSAT 2019 · Q68

Easy

यदि B से आगे (B समेत) अंग्रेज़ी वर्णमाला का प्रत्येक एकांतर अक्षर छोटे अक्षरों (लोवर केस) में लिखा जाए और शेष अक्षरों को बड़े अक्षरों में लिखा जाए, तो वर्ष के उत्तरार्ध के प्रथम मास को कैसे लिखा जाएगा ?

Read this question in English

If every alternative letter of the English alphabet from B onwards (including B) is written in lower case (small letters) and the remaining letters are capitalized, then how is the first month of the second half of the year written?

(a) JuLY · (b) jULy · (c) jUly · (d) jUlY

उत्तर और व्याख्या · explanation in English

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.

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 ·

CSAT 2019 · Q70

Easy

अनुक्रम 1, 5, 7, 3, 5, 7, 4, 3, 5, 7 में, ऐसे कितने 5 हैं जिनके ठीक पहले 3 नहीं है किन्तु ठीक बाद 7 है ?

Read this question in English

In the sequence 1, 5, 7, 3, 5, 7, 4, 3, 5, 7, how many such 5s are there which are not immediately preceded by 3 but are immediately followed by 7?

(a) 1 · (b) 2 · (c) 3 · (d) None

उत्तर और व्याख्या · explanation in English

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.

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 ·

CSAT 2019 · Q76

Easy

आगे आने वाले 3 (तीन) प्रश्नांशों के लिए निर्देश : निम्नलिखित सूचना को पढ़िए और उनके बाद आने वाले तीन प्रश्नांशों के उत्तर दीजिए,:

छह विद्यार्थी A, B, C, D, E और F कई परीक्षाओं में बैठे । या तो C का या F का प्राप्तांक अधिकतम है । जब भी C का प्राप्तांक अधिकतम होता है, तब E का प्राप्तांक न्यूनतम होता है । जब भी F का प्राप्तांक अधिकतम होता है, B का प्राप्तांक न्यूनतम होता है ।

सभी परीक्षाओं में, उन्हें अलग-अलग अंक प्राप्त होते हैं; D का प्राप्तांक A के प्राप्तांक से अधिक है, लेकिन वे निकट प्रतिस्पर्धी हैं; A का प्राप्तांक B के प्राप्तांक से अधिक है; C का प्राप्तांक A के प्राप्तांक से अधिक है ।

यदि F योग्यता-क्रम में दूसरे स्थान पर आता है, तो B का स्थान क्या है ?

Read this question in English

Directions for the following 3 (three) items: Read the following information and answer the three items that follow:

Six students A, B, C, D, E and F appeared in several tests. Either C or F scores the highest. Whenever C scores the highest, then E scores the least. Whenever F scores the highest, B scores the least.

In all the tests they got different marks; D scores higher than A, but they are close competitors; A scores higher than B; C scores higher than A.

If F stands second in the ranking, then the position of B is

(a) Third · (b) Fourth · (c) Fifth · (d) Sixth

उत्तर और व्याख्या · explanation in English

Answer: (c) पाँचवाँ

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.

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 ·

CSAT 2019 · Q77

Medium

आगे आने वाले 3 (तीन) प्रश्नांशों के लिए निर्देश : निम्नलिखित सूचना को पढ़िए और उनके बाद आने वाले तीन प्रश्नांशों के उत्तर दीजिए,:

छह विद्यार्थी A, B, C, D, E और F कई परीक्षाओं में बैठे । या तो C का या F का प्राप्तांक अधिकतम है । जब भी C का प्राप्तांक अधिकतम होता है, तब E का प्राप्तांक न्यूनतम होता है । जब भी F का प्राप्तांक अधिकतम होता है, B का प्राप्तांक न्यूनतम होता है ।

सभी परीक्षाओं में, उन्हें अलग-अलग अंक प्राप्त होते हैं; D का प्राप्तांक A के प्राप्तांक से अधिक है, लेकिन वे निकट प्रतिस्पर्धी हैं; A का प्राप्तांक B के प्राप्तांक से अधिक है; C का प्राप्तांक A के प्राप्तांक से अधिक है ।

यदि B का प्राप्तांक न्यूनतम हो, तो C का स्थान क्या होगा ?

Read this question in English

Directions for the following 3 (three) items: Read the following information and answer the three items that follow:

Six students A, B, C, D, E and F appeared in several tests. Either C or F scores the highest. Whenever C scores the highest, then E scores the least. Whenever F scores the highest, B scores the least.

In all the tests they got different marks; D scores higher than A, but they are close competitors; A scores higher than B; C scores higher than A.

If B scores the least, the rank of C will be

(a) Second · (b) Third · (c) Fourth · (d) Second or third

उत्तर और व्याख्या · explanation in English

Answer: (d) दूसरा या तीसरा

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.

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 ·

CSAT 2019 · Q78

Medium

आगे आने वाले 3 (तीन) प्रश्नांशों के लिए निर्देश : निम्नलिखित सूचना को पढ़िए और उनके बाद आने वाले तीन प्रश्नांशों के उत्तर दीजिए,:

छह विद्यार्थी A, B, C, D, E और F कई परीक्षाओं में बैठे । या तो C का या F का प्राप्तांक अधिकतम है । जब भी C का प्राप्तांक अधिकतम होता है, तब E का प्राप्तांक न्यूनतम होता है । जब भी F का प्राप्तांक अधिकतम होता है, B का प्राप्तांक न्यूनतम होता है ।

सभी परीक्षाओं में, उन्हें अलग-अलग अंक प्राप्त होते हैं; D का प्राप्तांक A के प्राप्तांक से अधिक है, लेकिन वे निकट प्रतिस्पर्धी हैं; A का प्राप्तांक B के प्राप्तांक से अधिक है; C का प्राप्तांक A के प्राप्तांक से अधिक है ।

यदि E का योग्यता-क्रम तीसरा हो, तो निम्नलिखित में से कौन-सा एक सही है ?

Read this question in English

Directions for the following 3 (three) items: Read the following information and answer the three items that follow:

Six students A, B, C, D, E and F appeared in several tests. Either C or F scores the highest. Whenever C scores the highest, then E scores the least. Whenever F scores the highest, B scores the least.

In all the tests they got different marks; D scores higher than A, but they are close competitors; A scores higher than B; C scores higher than A.

If E is ranked third, then which one of the following is correct?

(a) E gets more marks than C · (b) C gets more marks than E · (c) A is ranked fourth · (d) D is ranked fifth

उत्तर और व्याख्या · explanation in English

Answer: (b) C का प्राप्तांक E के प्राप्तांक से अधिक होगा

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 ·

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