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CAT 2025 Slot 1 DILR · Set 3 of 5Logical ReasoningHard DifficultyQ9–Q13 · 1 MCQ + 4 TITA

CAT 2025 Slot 1 DILR Set 3: Math Puzzles (5 Questions) with Solutions

The Math Puzzles set from the official CAT 2025 Slot 1 DILR paper: the caselet, all 5 questions, the official answer key and a step-by-step solution for each. 100% free, no sign-up.

The Set: Read the Caselet First

CAT 2025 Slot 1 DILR · Questions 9–13
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.

In the exam this set competed with 4 others for 40 minutes. Budget about 8 minutes if you attempt it.

Q9. Math Puzzles

MCQ · +3 / −1Solve on its own page
What was the occupancy factor for segment D – E?
  1. A) 84%
  2. B) 70%
  3. C) 35%
  4. D) 77%
Correct Answer
B) 70%
Step-by-step solution
📌 Core Concept
Count the tickets that cover each segment, then use the four-sevenths condition on segment D – E to fix the unknown ticket numbers.
🔢 Step-by-Step Solution
1
Clue 4 makes A–C and A–E equal, say xx each. Clue 5 gives zero tickets for A-B, B-D and D–E. Clue 2 gives B–E = 30. Clue 4 says x>30x > 30, and Clue 6 makes xx a multiple of 10.
2
The tickets covering D - E are A–E, B–E and C–E, because D–E has none. Those from stations before C are A–E and B–E, which total x+30x + 30.
3
Let C-E be yy. Clue 3 gives x+30=47(x+30+y)x + 30 = \frac{4}{7}(x + 30 + y), so y=34(x+30)y = \frac{3}{4}(x + 30).
4
For yy to be a multiple of 10, x+30x + 30 must be a multiple of 40. With x>30x > 30, the options are x=50x = 50 or x=90x = 90.
5
x=90x = 90 gives 120+90=210120 + 90 = 210 seats on D - E, which exceeds 200. So x=50x = 50 and y=60y = 60.
6
Seats reserved on D - E = 50+30+60=14050 + 30 + 60 = 140, so the occupancy factor is 140200=70%\frac{140}{200} = 70\%.
Why Not the Others
A) 84% would mean 168 seats, but D – E has 140.
C) 35% would mean 70 seats, which is too few.
D) 77% would mean 154 seats, which does not match 140.
Correct Answer: B

Q10. Math Puzzles

TITA · +3 / 0
How many tickets were booked from Station A to Station E?
Correct Answer
50
Step-by-step solution
📌 Core Concept
The four-sevenths condition on segment D – E, together with the capacity limit of 200, fixes the number of A–E tickets.
🔢 Step-by-Step Solution
1
Clue 4 makes A–C and A–E equal, say xx each. Clue 2 gives B-E = 30, and Clue 4 says x>30x > 30. Clue 6 makes xx a multiple of 10.
2
D - E carries A–E, B–E and C–E, because Clue 5 gives zero D–E tickets. The seats from stations before C are A–E and B–E, which total x+30x + 30.
3
Let C-E be yy. Clue 3 gives x+30=47(x+30+y)x + 30 = \frac{4}{7}(x + 30 + y), so y=34(x+30)y = \frac{3}{4}(x + 30).
4
yy must be a multiple of 10, so x+30x + 30 must be a multiple of 40. With x>30x > 30, the options are x=50x = 50 or x=90x = 90.
5
For x=90x = 90, D - E would hold 120+90=210120 + 90 = 210 seats, which exceeds 200. So x=50x = 50.
6
Then y=60y = 60, and D - E holds 50+30+60=14050 + 30 + 60 = 140 seats, which is within capacity.
Quick Check
47×140=80=50+30\frac{4}{7} \times 140 = 80 = 50 + 30, as Clue 3 requires.
Correct Answer: 50

Q11. Math Puzzles

TITA · +3 / 0
How many tickets were booked from Station C?
Correct Answer
80
Step-by-step solution
📌 Core Concept
First find C–E from the four-sevenths condition. Then find C–D from the occupancy of segment C – D, using the capacity limit on B – C.
🔢 Step-by-Step Solution
1
Clue 4 makes A–C and A–E equal, say xx each. B-E = 30 (Clue 2), and x>30x > 30. D-E, A–B and B–D have no tickets (Clue 5).
2
Clue 3 on D – E gives x+30=47(x+30+C-E)x + 30 = \frac{4}{7}(x + 30 + \text{C-E}), so C-E = 34(x+30)\frac{3}{4}(x + 30).
3
C-E must be a multiple of 10, so x+30x + 30 must be a multiple of 40. With x>30x > 30, x=90x = 90 would put 210 seats on D - E. So x=50x = 50 and C-E = 60.
4
Clue 1 puts C – D at 95%, which is 190 seats. Its tickets are A–D, A–E, B–E, C–D and C–E, so A-D+50+30+C–D+60=190\text{A-D} + 50 + 30 + \text{C–D} + 60 = 190. This gives A-D+C-D=50\text{A-D} + \text{C-D} = 50.
5
Segment B – C holds A–C, A–D, A–E, B–C and B–E, which is 50+A-D+50+40+30=170+A-D50 + \text{A-D} + 50 + 40 + 30 = 170 + \text{A-D}.
6
Clue 1 says B – C is higher than 190, so A-D≥30\text{A-D} \ge 30. Capacity gives 170+A-D≤200170 + \text{A-D} \le 200, so A-D≤30\text{A-D} \le 30. Hence A–D = 30 and C–D = 20.
7
Tickets booked from Station C = C–D + C–E = 20+60=8020 + 60 = 80.
Correct Answer: 80

Q12. Math Puzzles

TITA · +3 / 0
What is the difference between the number of tickets booked to Station C and the number of tickets booked to Station D?
Correct Answer
40
Step-by-step solution
📌 Core Concept
Solve for every ticket count using segments D – E, C – D and B – C. Then compare the tickets ending at C with those ending at D.
🔢 Step-by-Step Solution
1
Clue 4 makes A–C and A–E equal, say xx each. B-E = 30 (Clue 2), and x>30x > 30. Clue 5 gives zero for A-B, B–D and D–E.
2
Clue 3 on D – E gives C–E = 34(x+30)\frac{3}{4}(x + 30). For this to be a multiple of 10, x+30x + 30 must be a multiple of 40. Capacity (200) on D - E rules out x=90x = 90, so x=50x = 50 and C-E = 60.
3
Clue 1 puts C – D at 190 seats: A-D+50+30+C–D+60=190\text{A-D} + 50 + 30 + \text{C–D} + 60 = 190, so A-D+C-D=50\text{A-D} + \text{C-D} = 50.
4
B – C holds 50+A-D+50+40+30=170+A-D50 + \text{A-D} + 50 + 40 + 30 = 170 + \text{A-D}. It must exceed 190 (Clue 1) and not exceed 200, so A–D = 30. Then C–D = 20.
5
Tickets to C = A–C + B–C = 50+40=9050 + 40 = 90.
6
Tickets to D = A-D + C–D = 30+20=5030 + 20 = 50, since B-D = 0.
7
Difference = 90−50=4090 - 50 = 40.
Correct Answer: 40

Q13. Math Puzzles

TITA · +3 / 0
How many tickets were booked to travel in exactly one segment?
Correct Answer
60
Step-by-step solution
📌 Core Concept
A ticket uses exactly one segment only if it joins neighbouring stations. So the count needed is A–B + B–C + C–D + D–E.
🔢 Step-by-Step Solution
1
Clue 5 gives A–B = 0 and D–E = 0. Clue 2 gives B–C = 40. Only C–D remains to be found.
2
Clue 4 makes A–C and A–E equal, say xx each. B-E = 30, and x>30x > 30.
3
Clue 3 on D - E gives C–E = 34(x+30)\frac{3}{4}(x + 30), so x+30x + 30 must be a multiple of 40. Since x=90x = 90 would put 210 seats on D - E, x=50x = 50 and C-E = 60.
4
C – D has 190 seats (Clue 1): A-D+50+30+C–D+60=190\text{A-D} + 50 + 30 + \text{C–D} + 60 = 190, so A-D+C-D=50\text{A-D} + \text{C-D} = 50.
5
B – C holds 50+A-D+50+40+30=170+A-D50 + \text{A-D} + 50 + 40 + 30 = 170 + \text{A-D}. Clue 1 needs this above 190, and capacity needs it at most 200. So A–D = 30 and C–D = 20.
6
Single-segment tickets = 0+40+20+0=600 + 40 + 20 + 0 = 60.
Correct Answer: 60
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CAT 2025 Slot 1 DILR at a Glance

Sets in this section5
Questions in this section22
Section time40 minutes
Convenor IIMIIM Kozhikode

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