CAT 2025 Slot 1 · DILR · Question 1

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 LevelMultiple choiceHard

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 was the occupancy factor for segment D – E?

A
35%
B
84%
C
77%
D
70%

Pick an option, then check.

Answer & solution

Answer: Option D

Only seven ticket types can exist once clue 5 removes A–B, B–D and D–E: A–C, A–D, A–E, B–C, B–E, C–D and C–E.

Write a for the number booked A–C, which by clue 4 also equals the number booked A–E.

Write d for A–D, c for C–D and e for C–E; clue 2 fixes B–C at 40 and B–E at 30.

A segment carries every ticket whose journey spans it, so segment D – E carries A–E, B–E and C–E, giving a load of a+30+e.

Seats on D – E booked from stations before C are those starting at A or B, that is a+30.

Clue 3 makes that four-sevenths of the whole segment load: 7\left( a+30 \right)=4\left( a+30+e \right).

Simplifying, 3\left( a+30 \right)=4e, so e=\frac{3\left( a+30 \right)}{4}.

Clue 6 keeps every count a multiple of 10 and clue 4 puts a above 30, so test a=40,50,60,\ldots

Only a=50 makes e a multiple of 10, giving e=60.

Segment C – D carries A–D, A–E, B–E, C–D and C–E, so d+50+30+c+60=190, that is d+c=50.

Segment B – C carries A–C, A–D, A–E, B–C and B–E, so its load is 50+d+50+40+30=170+d.

Clue 1 puts B – C above 190 and the capacity caps it at 200, so d=30 and then c=20.

Segment D – E therefore carries 50+30+60=140 seats.

As a percentage of 200 that is 70%, which is the fourth option.

Shortcut. Clue 3 alone pins the whole set down: it forces e=\frac{3\left( a+30 \right)}{4}, and the multiple-of-10 rule leaves a=50 as the only candidate above 30. Everything else follows by filling the two remaining segments.

CAT 2025 Slot 1 · DILR · Question 1

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