CAT 2025 Slot 1 DILR Set 1 Q1–Q5

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2025 Slot 1, with 5 questions (Routes & Networks). Answers are CAT's official key.

The set

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.
1. Segment C – D had an occupancy factor of 95%. Only segment B – C had a higher occupancy factor.
2. Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E.
3. Among the seats reserved on segment D – E, exactly four-sevenths were from stations before C.
4. 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.
5. No tickets were booked from A to B, from B to D and from D to E.
6. The number of tickets booked for any segment was a multiple of 10.

Question 1

What was the occupancy factor for segment D – E?
  1. A
    35%
  2. B
    84%
  3. C
    77%
  4. D
    70%

Answer: (D) 70% Routes & Networks MCQ

Solution

  1. Segments: \(AB = AC + AD + AE\), \(BC = AB + 40 + 30\), \(CD = AD + AE + BE + CD + CE = 190\) (95% of 200), \(DE = AE + BE + CE\) (no A–B, B–D or D–E tickets).
  2. Only \(BC\) exceeds 190 and every count is a multiple of 10, so \(BC = 200\) and \(AB = 130\).
  3. \(AE + BE = \frac47 DE\), so \(DE\) is a multiple of 70: 70 or 140. \(DE = 70\) gives \(AE = 10\), but \(AE = AC > BE = 30\). So \(DE = 140\), \(AE = 80 - 30 = 50\) and \(CE = 60\).
  4. Then \(AC = 50\), \(AD = 130 - 100 = 30\), and from \(CD\): \(30 + 50 + 30 + CD + 60 = 190 \Rightarrow CD = 20\).
  5. \(DE = 140\) seats, i.e. \(\frac{140}{200} = 70\%\).

Question 2

How many tickets were booked from Station A to Station E?

TITA: type in the answer. There are no options and no negative marking.

Answer: 50 Routes & Networks TITA

Solution

  1. Segments: \(AB = AC + AD + AE\), \(BC = AB + 40 + 30\), \(CD = AD + AE + BE + CD + CE = 190\) (95% of 200), \(DE = AE + BE + CE\) (no A–B, B–D or D–E tickets).
  2. Only \(BC\) exceeds 190 and every count is a multiple of 10, so \(BC = 200\) and \(AB = 130\).
  3. \(AE + BE = \frac47 DE\), so \(DE\) is a multiple of 70: 70 or 140. \(DE = 70\) gives \(AE = 10\), but \(AE = AC > BE = 30\). So \(DE = 140\), \(AE = 80 - 30 = 50\) and \(CE = 60\).
  4. Then \(AC = 50\), \(AD = 130 - 100 = 30\), and from \(CD\): \(30 + 50 + 30 + CD + 60 = 190 \Rightarrow CD = 20\).
  5. Tickets A to E \(= 50\).

Question 3

How many tickets were booked from Station C?

TITA: type in the answer. There are no options and no negative marking.

Answer: 80 Routes & Networks TITA

Solution

  1. Segments: \(AB = AC + AD + AE\), \(BC = AB + 40 + 30\), \(CD = AD + AE + BE + CD + CE = 190\) (95% of 200), \(DE = AE + BE + CE\) (no A–B, B–D or D–E tickets).
  2. Only \(BC\) exceeds 190 and every count is a multiple of 10, so \(BC = 200\) and \(AB = 130\).
  3. \(AE + BE = \frac47 DE\), so \(DE\) is a multiple of 70: 70 or 140. \(DE = 70\) gives \(AE = 10\), but \(AE = AC > BE = 30\). So \(DE = 140\), \(AE = 80 - 30 = 50\) and \(CE = 60\).
  4. Then \(AC = 50\), \(AD = 130 - 100 = 30\), and from \(CD\): \(30 + 50 + 30 + CD + 60 = 190 \Rightarrow CD = 20\).
  5. From C: \(CD + CE = 20 + 60 = 80\).

Question 4

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

TITA: type in the answer. There are no options and no negative marking.

Answer: 40 Routes & Networks TITA

Solution

  1. Segments: \(AB = AC + AD + AE\), \(BC = AB + 40 + 30\), \(CD = AD + AE + BE + CD + CE = 190\) (95% of 200), \(DE = AE + BE + CE\) (no A–B, B–D or D–E tickets).
  2. Only \(BC\) exceeds 190 and every count is a multiple of 10, so \(BC = 200\) and \(AB = 130\).
  3. \(AE + BE = \frac47 DE\), so \(DE\) is a multiple of 70: 70 or 140. \(DE = 70\) gives \(AE = 10\), but \(AE = AC > BE = 30\). So \(DE = 140\), \(AE = 80 - 30 = 50\) and \(CE = 60\).
  4. Then \(AC = 50\), \(AD = 130 - 100 = 30\), and from \(CD\): \(30 + 50 + 30 + CD + 60 = 190 \Rightarrow CD = 20\).
  5. To C: \(AC + BC = 50 + 40 = 90\). To D: \(AD + BD + CD = 30 + 0 + 20 = 50\). Difference \(= 40\).

Question 5

How many tickets were booked to travel in exactly one segment?

TITA: type in the answer. There are no options and no negative marking.

Answer: 60 Routes & Networks TITA

Solution

  1. Segments: \(AB = AC + AD + AE\), \(BC = AB + 40 + 30\), \(CD = AD + AE + BE + CD + CE = 190\) (95% of 200), \(DE = AE + BE + CE\) (no A–B, B–D or D–E tickets).
  2. Only \(BC\) exceeds 190 and every count is a multiple of 10, so \(BC = 200\) and \(AB = 130\).
  3. \(AE + BE = \frac47 DE\), so \(DE\) is a multiple of 70: 70 or 140. \(DE = 70\) gives \(AE = 10\), but \(AE = AC > BE = 30\). So \(DE = 140\), \(AE = 80 - 30 = 50\) and \(CE = 60\).
  4. Then \(AC = 50\), \(AD = 130 - 100 = 30\), and from \(CD\): \(30 + 50 + 30 + CD + 60 = 190 \Rightarrow CD = 20\).
  5. One-segment tickets: \(AB + BC + CD + DE = 0 + 40 + 20 + 0 = 60\).

More Routes & Networks questions from CAT

All 41 Routes & Networks questions asked since 2017.