CAT 2025 Slot 1 · DILR · Question 4

CAT 2025 Slot 1 previous year question on Quant Based Puzzles and Optimisation, with the answer and a step-by-step solution.

Quant Based Puzzles and OptimisationLOGICAL REASONING - High LevelType in the answer (TITA)Medium

Set 1: Train Occupancy, Stations A to E — this question belongs to a set. Read the full set with all 5 questions →

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.

Segment C – D had an occupancy factor of 95%. Only segment B – C had a higher occupancy factor.

Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E.

Among the seats reserved on segment D – E, exactly four-sevenths were from stations before C.

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.

No tickets were booked from A to B, from B to D and from D to E.

The number of tickets booked for any segment was a multiple of 10.

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

Answer & solution

Answer: 40

The complete booking table follows from the earlier working.

A–C is 50, A–D is 30, A–E is 50, B–C is 40, B–E is 30, C–D is 20 and C–E is 60; the other three types are zero.

Tickets booked to Station C are those ending at C, namely A–C and B–C.

That is 50+40=90.

Tickets booked to Station D are those ending at D, namely A–D, B–D and C–D.

Clue 5 makes B–D zero, so that total is 30+0+20=50.

The difference is 90-50=40.

Shortcut. Clue 5 wipes out B–D entirely, so the count to D rests on just two numbers. Grouping the tickets by destination rather than by segment turns both totals into a single addition each.

CAT 2025 Slot 1 · DILR · Question 4

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