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Train Seat Occupancy Across Route Segments

CAT 2025 Slot 1 · Data Interpretation & Logical Reasoning · Hard · Data Interpretation

This is a hard Data Interpretation & Logical Reasoning question set from the CAT 2025 Slot 1 paper. It tests Data Interpretation. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Passage / 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 of 5

What was the occupancy factor for segment D – E?

  1. A.

    35%

  2. B.

    70%

  3. C.

    84%

  4. D.

    77%

Answer

B

Explanation

Segment C–D occupancy factor is 95%×200=19095\% \times 200 = 190 seats. From station before C for segment D–E: tickets booked from A to E and B to E contribute. Since 4/74/7 of D–E total seats are from before C, total seats on D–E must be a multiple of 7, which gives 140 seats out of 200, yielding an occupancy factor of 140/200=70%140/200 = 70\%.

Question 2 of 5

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

Answer

50

Explanation

From the constraints and multiples of 10, tickets booked from B to E = 30. Tickets from A to E are higher than 30 and equal to tickets from A to C. Solving the system yields A to E = 50.

Question 3 of 5

How many tickets were booked from Station C?

Answer

60

Explanation

By balancing total tickets originating at C to D and E, taking into account total reservations on C-D (190) and B-C (>190, max 200), we get tickets booked from Station C equal to 60.

Question 4 of 5

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

Answer

30

Explanation

Tickets ending at C = (A to C) + (B to C) = 50 + 40 = 90. Tickets ending at D = (A to D) + (B to D) + (C to D) = 60. The difference is 90−60=3090 - 60 = 30.

Question 5 of 5

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

Answer

100

Explanation

Single segment tickets are A to B (0), B to C (40), C to D (60), and D to E (0). Total = 0+40+60+0=1000 + 40 + 60 + 0 = 100.

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