CATSprint

CAT 2025Slot 1DILRQues & Sol

Logical ReasoningHard

Data Set

A train travels from Station A to Station E, passing through stations B, C, and D, in that order. The train has a seating capacity of 200. A ticket may be booked from any station to any other station ahead on the route, but not to any earlier station.

A ticket from one station to another reserves one seat on every intermediate segment of the route. For example, a ticket from B to E reserves a seat in the intermediate segments B - C, C - D, and D - E.

The occupancy factor for a segment is the total number of seats reserved in the segment as a percentage of the seating capacity. The total number of seats reserved for any segment cannot exceed 200.

The following information is known.
1. Segment C - D had an occupancy factor of 95%. Only segment B - C had a higher occupancy factor.
2. Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E.
3. Among the seats reserved on segment D - E, exactly four-sevenths were from stations before C.
4. The number of tickets booked from A to C was equal to that booked from A to E, and it was higher than that from B to E.
5. No tickets were booked from A to B, from B to D and from D to E.
6. The number of tickets booked for any segment was a multiple of 10.

Question 1

What was the occupancy factor for segment D - E?

Solution

Setup: We assign variables to all unknown ticket counts and build segment occupancy expressions. Using the constraints on C–D occupancy (95%), B–C being the highest occupancy segment, the four-sevenths condition for D–E, and the multiples-of-10 rule, we solve for each variable and then compute the D–E occupancy factor.

Steps:

  1. Let tickets from A→C = a, A→D = b, A→E = c, C→D = x, C→E = y. From clue 4, a = c. From clue 2, B→C = 40, B→E = 30. From clue 5, no tickets for A→B, B→D, D→E. Express each segment's reserved seats: A–B = a + b + c = 2a + b; B–C = 2a + b + 40 + 30 = 2a + b + 70; C–D = a + b + x + y + 30; D–E = a + y + 30 (since c = a).
  2. C–D occupancy = 95% → 190 seats. B–C has strictly higher occupancy; the only multiple of 10 above 190 that doesn't exceed 200 is 200. So 2a + b + 70 = 200 → 2a + b = 130 ... (1). From C–D: a + b + x + y = 160 ... (2).
  3. From clue 3, seats on D–E from stations before C = a + 30 (tickets A→E and B→E). These are 4/7 of total D–E seats: (4/7)(a + y + 30) = a + 30 → a + y + 30 = (7/4)(a + 30) ... (3). For this to be an integer, (a + 30) must be divisible by 4. From clue 4, a > 30. Valid multiples of 10 satisfying (1): a = 40, 50, 60. Only a = 50 gives (50 + 30) = 80, divisible by 4. So a = 50, b = 30 from (1).
  4. Substituting a = 50 into (3): 50 + y + 30 = (7/4)(80) = 140 → y = 60. From (2): 50 + 30 + x + 60 = 160 → x = 20. D–E total = a + y + 30 = 50 + 60 + 30 = 140 seats.
  5. Occupancy factor for D–E = (140/200) × 100 = 70%.

Final Answer: 70% (Option B)

Question 2

How many tickets were booked from Station A to Station E?

Solution

Setup: Using the same variable framework established for the set — where tickets A→C = a, A→D = b, A→E = c = a — we solve the system of equations derived from the occupancy constraints to find the exact value of c (= a).

Steps:

  1. From the core solution: 2a + b = 130 (B–C at 200 seats) and a + b + x + y = 160 (C–D at 190 seats). The four-sevenths condition on D–E forces (a + 30) to be divisible by 4, and a > 30 (from clue 4). Among valid candidates a = 40, 50, 60, only a = 50 satisfies divisibility.
  2. Since a = c (from clue 4: tickets A→C = tickets A→E), the number of tickets from A to E equals c = a = 50.
  3. Cross-check: b = 130 − 2(50) = 30 ✓; y = 60, x = 20 — all are positive multiples of 10 ✓.

Final Answer: 50

Question 3

How many tickets were booked from Station C?

Solution

Setup: Tickets 'booked from Station C' means tickets whose origin is C — these are C→D (= x) and C→E (= y). Using the solved values of x and y, we sum them up.

Steps:

  1. From the complete solution: a = 50, b = 30, x = 20, y = 60. Segment values confirm: C–D = 50 + 30 + 20 + 60 + 30 = 190 ✓; D–E = 50 + 60 + 30 = 140 ✓.
  2. Tickets originating at Station C are those with source = C: C→D = x = 20 and C→E = y = 60. Note that no C→… ticket types are excluded by clue 5 (which only restricts A→B, B→D, D→E).
  3. Total tickets booked from Station C = 20 + 60 = 80.

Final Answer: 80

Question 4

What is the difference between the number of tickets booked to Station C and the number of tickets booked to Station D?

Solution

Setup: 'Tickets booked to Station C' means tickets whose destination is C; 'tickets booked to Station D' means tickets whose destination is D. We identify all valid O-D pairs ending at C and D respectively, then compute using the solved variable values.

Steps:

  1. From the solution: a = 50, b = 30, x = 20, y = 60, B→C = 40, B→E = 30. Recall clue 5 eliminates A→B, B→D, D→E. No ticket can go backward, so all possible destinations for C are: A→C = 50 and B→C = 40. Tickets to Station C = 50 + 40 = 90.
  2. Tickets whose destination is D: A→D = b = 30 and C→D = x = 20. (B→D is 0 by clue 5.) Tickets to Station D = 30 + 20 = 50.
  3. Difference = 90 − 50 = 40.

Final Answer: 40

Question 5

How many tickets were booked to travel in exactly one segment?

Solution

Setup: A ticket covers exactly one segment if the origin and destination are adjacent stations on the route (A–B, B–C, C–D, or D–E). We check each such pair against the known ticket counts and clue 5.

Steps:

  1. The four single-segment ticket types are: A→B, B→C, C→D, and D→E. From clue 5: A→B = 0 and D→E = 0. From clue 2: B→C = 40. From the solution: C→D = x = 20.
  2. Single-segment tickets = A→B + B→C + C→D + D→E = 0 + 40 + 20 + 0 = 60.
  3. Verify: B→C spans only segment B–C ✓; C→D spans only segment C–D ✓. All other ticket types (A→C, A→D, A→E, B→E, C→E) cover 2 or more segments, so they are correctly excluded.

Final Answer: 60