Minimalist IAS
CSAT

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)

UPSC syllabus: “General mental ability;” See the full syllabus →

CSAT 2026 · Q14

Medium Provisional key

Eight persons P, Q, R, S, T, U, V and W sit around a round table in eight different seats placed with equal distance between any two consecutive seats. Both P and R are adjacent to Q. Both T and R are adjacent to S. Both U and W are adjacent to V. S and W are on opposite chairs. If while going in the clockwise direction around the table from P, one meets R before T, then how many persons shall Q cross while moving in the clockwise direction around the table before meeting W?

Answer & explanation

Answer: (a) 5

The clues chain into the block P-Q-R-S-T, leaving three seats for U-V-W with V in the middle. W must be opposite S, which fixes the full circle as P, Q, R, S, T, U, V, W in clockwise order, so Q passes R, S, T, U and V before W.

  1. P and R sit on both sides of Q, and R and T on both sides of S: this gives the block P-Q-R-S-T (5 seats).
  2. The remaining 3 seats go to U, V, W with V in the middle (U and W both next to V).
  3. Number the seats 1 to 8: P1, Q2, R3, S4, T5. The seat opposite S (4) is seat 8, so W8, V7, U6.
  4. From P, going the way P → Q → R → S → T meets R before T, so this direction is clockwise.
  5. Clockwise from Q: R, S, T, U, V, then W. Q crosses 5 persons.

Remember · In circular seating, first join 'adjacent' clues into blocks; the 'opposite' clue then fixes the rest.

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

CSAT 2026 · Q18

Medium Provisional key

A pattern formed by two characters a and b is repeated more than once in the following string:

× b × a × a × × a × a × bab

What is × × in the 7th and 8th positions from the left in the above string?

Answer & explanation

Answer: (d) bb

The string has 15 places, so the repeating block has length 3 or 5. Length 3 clashes (place 2 is b but place 11 is a); length 5 fits every known letter and gives the block 'abbab', so places 7 and 8 are b and b.

  1. Number the 15 places: 1 ×, 2 b, 3 ×, 4 a, 5 ×, 6 a, 7 ×, 8 ×, 9 a, 10 ×, 11 a, 12 ×, 13 b, 14 a, 15 b.
  2. Block length 3 fails: places 2 and 11 should match but are b and a.
  3. Block length 5: places 1, 6, 11 → a; 2, 7, 12 → b; 3, 8, 13 → b; 4, 9, 14 → a; 5, 10, 15 → b.
  4. Block = abbab, repeated three times: abbab abbab abbab.
  5. Places 7 and 8 = 2nd and 3rd letters of the block = b, b.

Remember · For a repeated pattern, try block lengths that divide the string length; letters one block apart must match.

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

CSAT 2026 · Q24

Hard Provisional key

A cut on a solid object divides the object into two parts where the new surfaces thus produced are plane. On the other hand, one single cut can be used to cut more than one object at a time. In an experiment, the total number of pieces produced by applying n cuts is denoted by xₙ. The experiment is performed on a solid cube where pieces remain unmoved after each cut. In this experiment, if after the third cut, the pieces are identical, then which of the following is not a possible value for x₄?

Answer & explanation

Answer: (a) 16

With pieces left unmoved, four plane cuts can give at most 15 pieces (1 + 4 + 6 + 4), so 16 is impossible. The other values arise easily from identical pieces after three cuts: 5 and 8 from four equal slabs, 12 from eight small cubes.

  1. Identical pieces after three cuts: 4 equal slabs (three parallel cuts), 6 equal blocks (two parallel + one perpendicular), or 8 small cubes (three mutually perpendicular cuts through the centre).
  2. Four plane cuts on an unmoved solid give at most 1 + 4 + 6 + 4 = 15 pieces; equivalently, the fourth plane can cross at most 7 of the 8 small cubes, giving 8 + 7 = 15.
  3. So x₄ = 16 is not possible.
  4. Check the rest: 4 slabs + a fourth parallel cut through one slab = 5; 4 slabs + a perpendicular cut through all four = 8; 8 cubes + a cut parallel to a face through four of them = 12.
  • ✗ (a) Not possible: four plane cuts on unmoved pieces give at most 15 pieces.
  • ✓ (b) Possible: after three central perpendicular cuts (8 cubes), a fourth cut parallel to one face through one layer splits 4 cubes → 12.
  • ✓ (c) Possible: after three parallel cuts (4 equal slabs), a fourth cut perpendicular to them splits all 4 → 8.
  • ✓ (d) Possible: after 4 equal slabs, a fourth cut parallel to them inside one slab → 5.

Remember · Unmoved pieces: plane cuts give at most 2, 4, 8 and 15 pieces for one to four cuts.

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

CSAT 2026 · Q33

Easy Provisional key

P, Q, R, S and T are ranked 1 to 5 (not necessarily in that order). The rank of P is 4, the rank of Q is not 5, the rank of R is 1, the rank of S is not 2, the rank of T is not 3. Then which of the following is/are correct?

  1. I.If the rank of S is 3, then that of T is 2.
  2. II.If the rank of Q is 3, then that of T is 5.

Select the answer using the code given below.

Answer & explanation

Answer: (d) Neither I nor II

With P = 4 and R = 1, the ranks 2, 3 and 5 remain for Q, S and T, and only two orders fit the conditions: (Q, S, T) = (2, 3, 5) or (3, 5, 2). In each case the 'then' part of the statement is the opposite of what actually happens.

  1. P = 4, R = 1, so Q, S, T share ranks 2, 3 and 5.
  2. Q ≠ 5 → Q is 2 or 3; S ≠ 2 → S is 3 or 5; T ≠ 3 → T is 2 or 5.
  3. Q = 2: then T ≠ 3 forces T = 5 and S = 3.
  4. Q = 3: then S ≠ 2 forces S = 5 and T = 2.
  5. S = 3 happens only with T = 5, so I is wrong; Q = 3 happens only with T = 2, so II is wrong.
  • ✗ I S = 3 occurs only in the order Q = 2, S = 3, T = 5, so T is 5, not 2.
  • ✗ II Q = 3 occurs only in the order Q = 3, S = 5, T = 2, so T is 2, not 5.

Remember · List all arrangements that satisfy the restrictions; there are usually very few, then test each 'if…then' against them.

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

CSAT 2026 · Q34

Hard Provisional key

Two identical straight rods are painted in five distinct colours so that each of them gets divided into five equal parts along the length. In one of them, the portions are marked P1, P2, P3, P4 and P5 (not necessarily in that order) whereas in the other, they are marked Q1, Q2, Q3, Q4 and Q5 (not necessarily in that order). When the rods are kept parallel to each other side by side, P1 and Q3 match, P4 matches Q1 or Q2, and Q4 matches P3 or P5. If Q3 and Q5 are adjacent, which of the following is/are possible?

  1. I.Q3 is marked at the middle portion of the straight rod.
  2. II.P2 is marked at one of the extreme portions of the straight rod.

Select the answer using the code given below.

Answer & explanation

Answer: (c) Both I and II

For 'possible', one valid layout is enough. The layout Q4, Q5, Q3, Q1, Q2 beside P3, P5, P1, P4, P2 meets every condition, and it has Q3 in the middle and P2 at an end, so both are possible.

  1. Number the portions 1 to 5 along the rods. Try Q3 (and so P1) at portion 3, with Q5 next to it at portion 2.
  2. Put Q4 at portion 1 with P3 beside it (Q4 must match P3 or P5).
  3. Put Q1 at portion 4 with P4 beside it (P4 must match Q1 or Q2), and Q2 at portion 5.
  4. The remaining P5 goes to portion 2 and P2 to portion 5.
  5. Layout: Q-rod Q4, Q5, Q3, Q1, Q2; P-rod P3, P5, P1, P4, P2. All conditions hold.
  6. Here Q3 is in the middle (I possible) and P2 is at an end (II possible).
  • ✓ I Possible: in the layout Q4, Q5, Q3, Q1, Q2, Q3 occupies the middle portion and every condition holds.
  • ✓ II Possible: in the same layout, P2 sits at portion 5, an extreme end.

Remember · 'Is possible' needs just one example that satisfies every condition; 'must be' needs proof for all cases.

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

CSAT 2026 · Q35

Easy Provisional key

Seven persons A, B, C, D, E, F and G travel by three cars X, Y, Z. A and another two of them travel by X. Only E travels with G. C travels by Z, but B does not travel by Y. Besides, A and B do not travel by the same car. Then which of the following are correct?

  1. I.No one travels alone.
  2. II.Only D travels with F.
  3. III.Only C travels with B

Select the answer using the code given below.

Answer & explanation

Answer: (b) I and III only

B cannot be in Y or with A in X, so B joins C in Z. E and G ride alone together, which leaves them car Y; D and F then fill X with A. So cars are X: A, D, F; Y: E, G; Z: B, C — D is not the only one with F, since A is there too.

  1. B is not in Y and not in X (A is in X and A, B are apart), so B is in Z with C.
  2. E and G travel together with nobody else; X has A and Z has B, C, so E and G take Y.
  3. X has A and two others: the only ones left are D and F.
  4. Cars: X = A, D, F; Y = E, G; Z = B, C.
  5. I true (no car has one person); II false (A also travels with F); III true (Z has only B and C).
  • ✓ I Each car carries two or three persons.
  • ✗ II F shares car X with both A and D, so D is not the only one.
  • ✓ III Car Z carries only B and C.

Remember · Place the person with the most 'not' conditions first; the rest usually falls into place.

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

CSAT 2026 · Q51

Easy Provisional key

Seven cubes are identical in shape. Out of these, the weight of each of the six cubes is equal and the weight of the remaining cube is less than the weight of any other cube. A balance is used to identify the lightest cube. What is the minimum number of attempts required to distinguish the odd cube with certainty?

Answer & explanation

Answer: (a) 2

Each weighing has three outcomes (left lighter, right lighter, balance), so split the cubes into three groups: 3, 3 and 1. One weighing narrows the light cube to a group of at most three, and a second weighing of one against one finds it.

  1. Weigh 3 cubes against 3. If they balance, the 7th cube is the light one.
  2. If one side rises, the light cube is among those 3.
  3. Weigh 1 against 1 from that group: the lighter pan shows it; if balanced, it is the third cube.
  4. So 2 weighings always suffice; 1 cannot, since one weighing separates at most 3 possibilities, not 7.

Remember · A balance gives three outcomes: with k weighings you can find one odd (known-lighter) item among up to 3ᵏ items.

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

CSAT 2026 · Q57

Easy Provisional key

In a sequence of numbers, each number other than the first two is the sum of the two immediately preceding numbers from it. If the first two numbers in the sequence are 4 and 7, then the sixth number is

Answer & explanation

Answer: (d) 47

Build the sequence by adding the last two terms each time: 4, 7, 11, 18, 29, 47. The sixth term is 47.

  1. Term 3 = 4 + 7 = 11.
  2. Term 4 = 7 + 11 = 18.
  3. Term 5 = 11 + 18 = 29.
  4. Term 6 = 18 + 29 = 47.

Remember · Short sequences: just write the terms out; count positions carefully (option 29 is the fifth term).

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

Other CSAT skills