CAT 2025 Slot 1 · DILR · Question 2

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 A to Station E?

Answer & solution

Answer: 50

Clue 4 says the A–C and A–E bookings are equal, so call each of them a.

Segment D – E carries the A–E, B–E and C–E tickets, a load of a+30+e where e is the C–E count.

The seats on that segment originating before C are the A–E and B–E tickets, numbering a+30.

Clue 3 makes those four-sevenths of the load, so 7\left( a+30 \right)=4\left( a+30+e \right).

That reduces to 4e=3\left( a+30 \right).

Clue 6 requires both a and e to be multiples of 10, and clue 4 requires a>30.

Testing upward, a=40 gives e=52.5 and a=60 gives e=67.5, neither admissible.

At a=50, e=\frac{3\times 80}{4}=60, which works.

Larger values fail the capacity limit on segment B – C, whose load is 170+d and cannot exceed 200.

So 50 tickets were booked from A to E.

Shortcut. Reading clue 3 as "three-sevenths of the D – E load came from C" gives e:\left( a+30 \right)=3:4 directly, so a+30 must be a multiple of 4 as well as of 10 — that is, a multiple of 20. The first such value above 60 is 80, giving a=50 at once.

CAT 2025 Slot 1 · DILR · Question 2

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