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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)

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

CSAT 2017 · Q11

Easy

If second and fourth Saturdays and all the Sundays are taken as only holidays for an office, what would be the minimum number of possible working days of any month of any year?

Answer & explanation

Answer: (b) 22

Working days are fewest in a short month with many holidays. Every month has exactly two holiday Saturdays, so only the Sundays vary: a 28-day February has exactly 4 Sundays, giving 6 holidays and 22 working days. Longer months add more days than holidays.

  1. Every month has a 2nd and a 4th Saturday, so there are always exactly 2 holiday Saturdays; only the number of Sundays varies.
  2. February (28 days): exactly 4 Sundays → 4 + 2 = 6 holidays → 28 − 6 = 22 working days.
  3. February in a leap year (29 days): at most 5 Sundays → 7 holidays → 29 − 7 = 22 working days.
  4. 30-day month: at most 5 Sundays → 7 holidays → 23 working days. 31-day month: at most 5 Sundays → 7 holidays → 24.
  5. Minimum = 22.
  6. Check: no month can have 6 Sundays, since that needs at least 36 days.

Remember · For 'minimum working days', test the shortest month with the most holidays — February, including the leap-year case.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q13

Easy

Four tests—Physics, Chemistry, Mathematics and Biology are to be conducted on four consecutive days, not necessarily in the same order. The Physics test is held before the test which is conducted after Biology. Chemistry is conducted exactly after two tests are held. Which is the last test held?

Answer & explanation

Answer: (c) Mathematics

Chemistry is on day 3. Biology cannot be last, since a test follows it, and cannot be first, since Physics must come before the test after Biology. So Biology is on day 2, Physics on day 1, and Mathematics is left for day 4.

  1. 'Exactly after two tests are held' puts Chemistry on day 3.
  2. Some test is held after Biology, so Biology is not on day 4.
  3. If Biology were on day 1, the test after it would be on day 2, and Physics would have to be on day 1 — already taken. So Biology is not on day 1.
  4. So Biology is on day 2; the test after it is Chemistry (day 3), and Physics is before that — day 1.
  5. Order: Physics, Biology, Chemistry, Mathematics. The last test is Mathematics.

Remember · Fix the absolute clue first (a definite day), then test the remaining slots against each relative clue.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q14

Medium

The sum of income of A and B is more than that of C and D taken together. The sum of income of A and C is the same as that of B and D taken together. Moreover, A earns half as much as the sum of the income of B and D. Whose income is the highest?

Answer & explanation

Answer: (b) B

A is half of B + D, and A + C equals B + D, so C = A. Then A + B > C + D gives B > D, and since A is the average of B and D, B is above A. B has the highest income.

  1. A = (B + D)/2, so B + D = 2A.
  2. A + C = B + D = 2A, so C = A.
  3. A + B > C + D becomes A + B > A + D, so B > D.
  4. A is the average of B and D with B > D, so B > A > D. With C = A, the order is B > A = C > D.
  5. B has the highest income.

Remember · Turn each sentence into an equation or inequality and substitute; one equality (here C = A) usually unlocks the ranking.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q18

Easy

15 students failed in a class of 52. After removing the names of failed students, a merit order list has been prepared in which the position of Ramesh is 22nd from the top. What is his position from the bottom?

Answer & explanation

Answer: (c) 16th

Only the 52 − 15 = 37 students who passed are in the merit list. With 21 above Ramesh, 37 − 22 = 15 are below him, so he is 16th from the bottom.

  1. Students in the merit list = 52 − 15 = 37.
  2. Position from bottom = total − position from top + 1 = 37 − 22 + 1 = 16.
  3. Check: 21 above + Ramesh + 15 below = 37.

Remember · Rank from top + rank from bottom = total + 1 — but first count only those actually in the list.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q29

Easy

In a group of six women, there are four tennis players, four postgraduates in Sociology, one postgraduate in Commerce and three bank employees. Vimala and Kamla are the bank employees while Amala and Komala are unemployed. Komala and Nirmala are among the tennis players. Amala, Kamla, Komala and Nirmala are postgraduates in Sociology of whom two are bank employees. If Shyamala is a postgraduate in Commerce, who among the following is both a tennis player and a bank employee?

Answer & explanation

Answer: (c) Nirmala

Two of the four Sociology postgraduates are bank employees. Amala and Komala are unemployed, so the two are Kamla and Nirmala. Nirmala is also a tennis player, so she is both.

  1. Sociology postgraduates: Amala, Kamla, Komala, Nirmala — two of them are bank employees.
  2. Amala and Komala are unemployed, so those two are Kamla and Nirmala.
  3. Bank employees (three): Vimala, Kamla and Nirmala.
  4. Tennis players include Komala and Nirmala; of these only Nirmala is a bank employee.
  5. Check the other options: Amala and Komala are unemployed, and Shyamala is not one of the three bank employees.

Remember · In grouping puzzles, start with the clue naming a small closed set ('of whom two are…') and eliminate using the negative facts.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q31

Medium

A watch loses 2 minutes in every 24 hours while another watch gains 2 minutes in every 24 hours. At a particular instant, the two watches showed an identical time. Which of the following statements is correct if 24-hour clock is followed?

Answer & explanation

Answer: (d) None of the above statements is correct.

Each day the two watches drift 4 minutes further apart. On a 24-hour dial they show the same time again only when the gap is a full 24 hours (1440 minutes), which takes 360 days. None of 30, 90 or 120 days works.

  1. One watch loses 2 min and the other gains 2 min a day, so the gap between them grows by 4 min a day.
  2. On a 24-hour clock the two dials match again when the gap reaches 24 × 60 = 1440 min.
  3. Days needed = 1440 ÷ 4 = 360.
  4. Check: after 120 days the gap is only 480 min (8 hours), so options (a) to (c) all fail.

Remember · Two clocks agree again when their total gap equals one full dial: 720 minutes on a 12-hour clock, 1440 minutes on a 24-hour clock.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q33

Easy

A clock strikes once at 1 o’clock, twice at 2 o’clock and thrice at 3 o’clock, and so on. If it takes 12 seconds to strike at 5 o’clock, what is the time taken by it to strike at 10 o’clock?

Answer & explanation

Answer: (b) 24 seconds

Why not the tempting option · UPSC's key, 24 seconds, treats the time as proportional to the number of strikes (2.4 s a strike). The textbook method counts the gaps between strikes — 4 gaps in 12 s, so 9 gaps at 10 o'clock make 27 s — but 27 is not among the options, so the paper plainly intends the simple proportion. In the exam, if the gap method gives no option, switch to proportion; (b) is the only option that fits the data.

UPSC's key takes the striking time as proportional to the number of strikes: 5 strikes take 12 seconds, so each strike accounts for 2.4 seconds and 10 strikes take 24 seconds. This is the only option that fits the data.

  1. Read the data as 'each strike takes the same time': 12 s ÷ 5 strikes = 2.4 s a strike.
  2. At 10 o'clock the clock strikes 10 times: 10 × 2.4 = 24 s, option (b).
  3. Check the alternative: counting only the gaps between strikes (4 gaps of 3 s, then 9 gaps) gives 27 s, which is not offered, so that reading cannot be what the paper intends.

Remember · In a clock-strike question, first try the gap method (strikes − 1); if the result is not among the options, use simple proportion, as UPSC did here.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 1 Oct 2026 (how we verify). Permalink ·

CSAT 2017 · Q36

Easy

Six boys A, B, C, D, E and F play a game of cards. Each has a pack of 10 cards. F borrows 2 cards from A and gives away 5 to C who in turn gives 3 to B while B gives 6 to D who passes on 1 to E. Then the number of cards possessed by D and E is equal to the number of cards possessed by

Answer & explanation

Answer: (b) B, C and F

After all the transfers A has 8, B 7, C 12, D 15, E 11 and F 7 cards. D and E together hold 26, the same as B, C and F together (7 + 12 + 7).

  1. Start: everyone has 10 cards.
  2. F borrows 2 from A: A = 8, F = 12. F gives 5 to C: F = 7, C = 15.
  3. C gives 3 to B: C = 12, B = 13. B gives 6 to D: B = 7, D = 16. D gives 1 to E: D = 15, E = 11.
  4. D + E = 15 + 11 = 26.
  5. B + C + F = 7 + 12 + 7 = 26. (A + B + C = 27, A + B + F = 22, A + C + F = 27.)
  6. Check: total = 8 + 7 + 12 + 15 + 11 + 7 = 60, the same as at the start.

Remember · Track transfers in a small table and confirm the total stays unchanged — a quick guard against slips.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q40

Medium

In a test, Randhir obtained more marks than the total marks obtained by Kunal and Debu. The total marks obtained by Kunal and Shankar are more than those of Randhir. Sonal obtained more marks than Shankar. Neha obtained more marks than Randhir. Who amongst them obtained highest marks?

Answer & explanation

Answer: (d) Data are inadequate

Neha scores above Randhir, and Sonal above Shankar, but no clue compares Neha with Sonal. Either of them can come out on top, so the data are inadequate.

  1. Clues: Randhir > Kunal + Debu; Kunal + Shankar > Randhir; Sonal > Shankar; Neha > Randhir.
  2. Randhir is below Neha, and Kunal and Debu are below Randhir; Shankar is below Sonal. So only Neha or Sonal can be highest.
  3. No clue links Neha and Sonal.
  4. Example: Kunal 10, Debu 5, Randhir 20, Shankar 15, Sonal 16, Neha 25 — Neha is highest. Change Sonal to 30 and Sonal is highest. All clues hold both times.

Remember · In ranking puzzles, find who can still be on top; if two candidates are never compared, the data are inadequate.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q58

Easy

The outer surface of a 4 cm × 4 cm × 4 cm cube is painted completely in red. It is sliced parallel to the faces to yield sixty four 1 cm × 1 cm × 1 cm small cubes. How many small cubes do not have painted faces?

Answer & explanation

Answer: (a) 8

Only the small cubes that touch no outer face stay unpainted. Peeling a 1 cm layer off every side leaves a 2 × 2 × 2 core, so 8 small cubes have no painted face.

  1. A 4 × 4 × 4 cube gives 64 small cubes.
  2. Cubes with no painted face form the inner block: (4 − 2) × (4 − 2) × (4 − 2).
  3. (4 − 2)³ = 2³ = 8.
  4. Check: 8 corner cubes + 24 edge cubes + 24 face-centre cubes = 56 painted, and 56 + 8 = 64.

Remember · For an n × n × n painted cube, the unpainted small cubes number (n − 2)³.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q59

Medium

Consider the following:

A, B, C, D, E, F, G and H are standing in a row facing North.

B is not neighbour of G.

F is to the immediate right of G and neighbour of E.

G is not at the extreme end.

A is sixth to the left of E.

H is sixth to the right of C.

Which one of the following is correct in respect of the above?

Answer & explanation

Answer: (c) G is to the immediate right of D.

'Sixth to the left/right' in a row of eight forces the pairs A–E and C–H into places 1 & 7 and 2 & 8. F must stand next to E, which works only with A at 1 and E at 7. The row is A C B D G F E H, so G is to the immediate right of D.

  1. Number the places 1–8 from left to right; facing North, a person's right is the higher number.
  2. A is sixth to the left of E: (A, E) = (1, 7) or (2, 8). H is sixth to the right of C: (C, H) = (1, 7) or (2, 8). The two pairs must take different sets.
  3. Case C = 1, H = 7, A = 2, E = 8: E's only neighbour is place 7 (H), so F cannot be next to E — rejected.
  4. So A = 1, C = 2, E = 7, H = 8. F must be at 6 (next to E) and G at 5 (immediately left of F); G is not at an end.
  5. B and D fill places 3 and 4; B is not next to G (5), so B = 3 and D = 4.
  6. Row: A C B D G F E H. G (5) is immediately right of D (4).
  7. Check the others: C is to the right of A, not the left; D's neighbours are B and G; A and E are at 1 and 7, not both ends.

Remember · In linear seating, 'k-th to the left' clues with a large k allow few places — list them first, then test the adjacency clues.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q69

Medium

Directions for the following 3 (three) items: Consider the given information and answer the three items that follow.

A, B, C, D, E, F and G are Lecturers from different cities—Hyderabad, Delhi, Shillong, Kanpur, Chennai, Mumbai and Srinagar (not necessarily in the same order) who participated in a conference. Each one of them is specialized in a different subject, viz., Economics, Commerce, History, Sociology, Geography, Mathematics and Statistics (not necessarily in the same order). Further

  1. 1.Lecturer from Kanpur is specialized in Geography
  2. 2.Lecturer D is from Shillong
  3. 3.Lecturer C from Delhi is specialized in Sociology
  4. 4.Lecturer B is specialized in neither History nor Mathematics
  5. 5.Lecturer A who is specialized in Economics does not belong to Hyderabad
  6. 6.Lecturer F who is specialized in Commerce belongs to Srinagar
  7. 7.Lecturer G who is specialized in Statistics belongs to Chennai

Who is specialized in Geography?

Answer & explanation

Answer: (a) B

Once the direct clues are placed, History, Geography and Mathematics are left for B, D and E. Clue 4 rules out History and Mathematics for B, so B is the Geography lecturer (and hence the one from Kanpur).

  1. Direct clues: C – Delhi – Sociology; D – Shillong; F – Srinagar – Commerce; G – Chennai – Statistics; A – Economics.
  2. Cities left for A, B and E: Hyderabad, Kanpur, Mumbai. A is not from Hyderabad (clue 5), and Kanpur's lecturer teaches Geography while A teaches Economics — so A is from Mumbai.
  3. Subjects left for B, D and E: History, Geography, Mathematics. B teaches neither History nor Mathematics (clue 4), so B teaches Geography.
  4. Geography goes with Kanpur (clue 1), so B is from Kanpur, and the last city, Hyderabad, goes to E.
  5. D and E share History and Mathematics in some order — the data do not decide which.
  6. Geography: B.

Remember · In matching grids, place every direct clue first, then use 'neither/nor' and 'does not' clues to eliminate; the last blanks fall into place.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q70

Medium

Directions for the following 3 (three) items: Consider the given information and answer the three items that follow.

A, B, C, D, E, F and G are Lecturers from different cities—Hyderabad, Delhi, Shillong, Kanpur, Chennai, Mumbai and Srinagar (not necessarily in the same order) who participated in a conference. Each one of them is specialized in a different subject, viz., Economics, Commerce, History, Sociology, Geography, Mathematics and Statistics (not necessarily in the same order). Further

  1. 1.Lecturer from Kanpur is specialized in Geography
  2. 2.Lecturer D is from Shillong
  3. 3.Lecturer C from Delhi is specialized in Sociology
  4. 4.Lecturer B is specialized in neither History nor Mathematics
  5. 5.Lecturer A who is specialized in Economics does not belong to Hyderabad
  6. 6.Lecturer F who is specialized in Commerce belongs to Srinagar
  7. 7.Lecturer G who is specialized in Statistics belongs to Chennai

To which city does the Lecturer specialized in Economics belong?

Answer & explanation

Answer: (b) Mumbai

The Economics lecturer is A. Only Hyderabad, Kanpur and Mumbai are free; clue 5 rules out Hyderabad, and Kanpur's lecturer teaches Geography, so A must be from Mumbai.

  1. Direct clues: C – Delhi – Sociology; D – Shillong; F – Srinagar – Commerce; G – Chennai – Statistics; A – Economics.
  2. Cities left for A, B and E: Hyderabad, Kanpur, Mumbai. A is not from Hyderabad (clue 5), and Kanpur's lecturer teaches Geography while A teaches Economics — so A is from Mumbai.
  3. Subjects left for B, D and E: History, Geography, Mathematics. B teaches neither History nor Mathematics (clue 4), so B teaches Geography.
  4. Geography goes with Kanpur (clue 1), so B is from Kanpur, and the last city, Hyderabad, goes to E.
  5. D and E share History and Mathematics in some order — the data do not decide which.
  6. Economics is A's subject, and A is from Mumbai.

Remember · When a city is tied to a subject (Kanpur–Geography), anyone with a different subject is ruled out of that city at once.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q71

Medium

Directions for the following 3 (three) items: Consider the given information and answer the three items that follow.

A, B, C, D, E, F and G are Lecturers from different cities—Hyderabad, Delhi, Shillong, Kanpur, Chennai, Mumbai and Srinagar (not necessarily in the same order) who participated in a conference. Each one of them is specialized in a different subject, viz., Economics, Commerce, History, Sociology, Geography, Mathematics and Statistics (not necessarily in the same order). Further

  1. 1.Lecturer from Kanpur is specialized in Geography
  2. 2.Lecturer D is from Shillong
  3. 3.Lecturer C from Delhi is specialized in Sociology
  4. 4.Lecturer B is specialized in neither History nor Mathematics
  5. 5.Lecturer A who is specialized in Economics does not belong to Hyderabad
  6. 6.Lecturer F who is specialized in Commerce belongs to Srinagar
  7. 7.Lecturer G who is specialized in Statistics belongs to Chennai

Who of the following belongs to Hyderabad?

Answer & explanation

Answer: (b) E

A (Economics) must be from Mumbai, and B, the Geography lecturer, must be from Kanpur. The only city left, Hyderabad, therefore belongs to E.

  1. Direct clues: C – Delhi – Sociology; D – Shillong; F – Srinagar – Commerce; G – Chennai – Statistics; A – Economics.
  2. Cities left for A, B and E: Hyderabad, Kanpur, Mumbai. A is not from Hyderabad (clue 5), and Kanpur's lecturer teaches Geography while A teaches Economics — so A is from Mumbai.
  3. Subjects left for B, D and E: History, Geography, Mathematics. B teaches neither History nor Mathematics (clue 4), so B teaches Geography.
  4. Geography goes with Kanpur (clue 1), so B is from Kanpur, and the last city, Hyderabad, goes to E.
  5. D and E share History and Mathematics in some order — the data do not decide which.
  6. Hyderabad: E.

Remember · Solve the whole grid once; later questions on the same set then take seconds.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q72

Easy

In a school, there are five teachers A, B, C, D and E. A and B teach Hindi and English. C and B teach English and Geography. D and A teach Mathematics and Hindi. E and B teach History and French. Who teaches maximum number of subjects?

Answer & explanation

Answer: (b) B

Collect each teacher's subjects, counting a repeated subject once. B teaches Hindi, English, Geography, History and French — five subjects. A teaches three, and C, D and E teach two each.

  1. A: Hindi, English (with B) and Mathematics, Hindi (with D) → Hindi, English, Mathematics = 3.
  2. B: Hindi, English; English, Geography; History, French → Hindi, English, Geography, History, French = 5.
  3. C: English, Geography = 2. D: Mathematics, Hindi = 2. E: History, French = 2.
  4. B teaches the most subjects.

Remember · List each person's subjects as a set, counting a subject once even if it appears in two statements.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q76

Medium

Directions for the following 3 (three) items: Consider the given information and answer the three items that follow.

Eight railway stations A, B, C, D, E, F, G and H are connected either by two-way passages or one-way passages. One-way passages are from C to A, E to G, B to F, D to H, G to C, E to C and H to G. Two-way passages are between A and E, G and B, F and D, and E and D.

While travelling from C to H, which one of the following stations must be passed through?

Answer & explanation

Answer: (b) E

UPSC's official answer: (b) · the answer UPSC accepted, and the one that counts in the exam

Also defensible: (c)

  • Exits from each station: A → E; B → F, G; C → A; D → H, F, E; E → G, C, A, D; F → D; G → C, B; H → G.
  • Working forward: C's only exit is A, and A's only exit is E, so every route from C passes through E. That is UPSC's (b).
  • Working backward: the only passage into H is D → H, so every route must also pass through D. The two complete routes are C → A → E → D → H and C → A → E → G → B → F → D → H; both contain E and D.
  • G and F lie only on the longer route, so (a) and (d) are wrong; but (b) and (c) are both forced, and the item has two correct options.

UPSC's key is (b) E, which every route must pass through; D is equally unavoidable, so (c) is also correct. In the exam, trace the forced moves from the start station first, as the key does, and give E.

This box is Minimalist IAS's analysis, with its sources; it does not change UPSC's answer.

From C the only exit is to A, and from A the only onward move is to E, so every journey from C to H has to pass through E. G and F lie only on the longer of the two routes, so they are not compulsory.

  1. Exits from each station: A → E; B → F, G; C → A; D → H, F, E; E → G, C, A, D; F → D; G → C, B; H → G.
  2. From C the only move is to A, and from A the only move is to E — so E is on every route out of C.
  3. Routes to H: C → A → E → D → H, or C → A → E → G → B → F → D → H; both pass through E.
  4. G and F appear only on the longer route, so they are not compulsory; E is the station UPSC's key accepts.

Remember · For 'must pass through', trace the forced moves from the start: a station with a single exit forces the next station onto every route.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 1 Oct 2026 (how we verify). Permalink ·

CSAT 2017 · Q77

Medium

Directions for the following 3 (three) items: Consider the given information and answer the three items that follow.

Eight railway stations A, B, C, D, E, F, G and H are connected either by two-way passages or one-way passages. One-way passages are from C to A, E to G, B to F, D to H, G to C, E to C and H to G. Two-way passages are between A and E, G and B, F and D, and E and D.

In how many different ways can a train travel from F to A without passing through any station more than once?

Answer & explanation

Answer: (d) 4

From F the only exit is D, and from D the train can go to E or H. Through E it reaches A directly, via C, or via G and C (3 routes); through H it must go to G, then C, then A (1 route). That makes 4 routes.

  1. Exits from each station: A → E; B → F, G; C → A; D → H, F, E; E → G, C, A, D; F → D; G → C, B; H → G.
  2. F's only exit is D.
  3. D → E: then E → A (F-D-E-A), E → C → A (F-D-E-C-A), or E → G → C → A (F-D-E-G-C-A). E → G → B leads only back to F or G, already used.
  4. D → H: H → G, then G → C → A (F-D-H-G-C-A); G → B leads back to F — a dead end.
  5. Total routes without repeating a station = 3 + 1 = 4.

Remember · Count routes with a tree: branch at every station and prune any branch that revisits a station or reaches a dead end.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q78

Medium

Directions for the following 3 (three) items: Consider the given information and answer the three items that follow.

Eight railway stations A, B, C, D, E, F, G and H are connected either by two-way passages or one-way passages. One-way passages are from C to A, E to G, B to F, D to H, G to C, E to C and H to G. Two-way passages are between A and E, G and B, F and D, and E and D.

If the route between G and C is closed, which one of the following stations need not be passed through while travelling from H to C?

Answer & explanation

Answer: (c) A

H's only exit is G, and with G to C closed, G must go to B, then F, then D. From D the train reaches C only through E. The route H-G-B-F-D-E-C passes E, D and B but never needs A.

  1. Exits from each station: A → E; B → F, G; C → A; D → H, F, E; E → G, C, A, D; F → D; G → C, B; H → G.
  2. H → G is the only exit from H.
  3. G → C is closed, so G → B; then B → F (B → G goes back) and F → D.
  4. From D: D → H goes back, so D → E; from E go straight to C.
  5. Route: H → G → B → F → D → E → C. It passes E, D and B but not A.
  6. A can be entered only from C or E and leads only back to E, so it is never needed.

Remember · When a link closes, trace the forced path one exit at a time; stations off that path need not be passed.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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