CAT 2025 Slot 1 · DILR · Question 3

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.

How many tickets were booked from Station C?

Answer & solution

Answer: 80

Tickets booked from Station C are those going C–D and C–E, since no station lies beyond E.

From the earlier working the C–E count is 60 and the A–C and A–E counts are 50 each.

Segment C – D has an occupancy factor of 95%, so it carries 0.95\times 200=190 seats.

That segment is spanned by the A–D, A–E, B–E, C–D and C–E tickets.

So d+50+30+c+60=190, where d is the A–D count and c the C–D count.

That gives d+c=50.

Segment B – C carries A–C, A–D, A–E, B–C and B–E, a load of 170+d.

Clue 1 says only B – C beats C – D, so 170+d>190 and hence d>20.

The capacity limit gives 170+d\le 200, so d\le 30.

With d a multiple of 10, d=30 and therefore c=20.

Tickets from Station C number c+e=20+60=80.

Shortcut. Segment B – C is squeezed between 190 and 200, and its load is 170+d — so d has just one admissible multiple of 10. The C–D count then falls out of the 190-seat equation without any further case work.

CAT 2025 Slot 1 · DILR · Question 3

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