CAT 2025 Slot 1 · DILR · Set 1: Train Occupancy, Stations A to E

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.

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.

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.

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.

A ticket spans exactly one segment when it runs between consecutive stations.

The four such types are A–B, B–C, C–D and D–E.

Clue 5 sets both A–B and D–E to zero.

Clue 2 gives B–C as 40.

The C–D count was found to be 20 while solving the 190-seat equation for segment C – D.

So the single-segment tickets number 0+40+20+0=60.

Shortcut. Of the four consecutive-station journeys, clue 5 kills two outright and clue 2 hands over a third. Only the C–D figure needs to be worked out, so this question is almost free once the grid is built.