CAT 2024 Slot 3 · DILR · Set 1: ATMs on a Road Network
The full set and all 5 questions (Q1–Q5) from CAT 2024 Slot 3, with answers and solutions.
The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km) between two adjacent intersections. Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.
The following additional information is known.
1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
2. The road distance between the ATM with the second highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km.
Question 1 · Caselet DI
- AThere is no ATM placed at the (R-C, V2) intersection.
- BThe ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.
- CThe ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 8 Lakhs.
- DThe cash requirement of the ATM placed at the (R-C, V2) intersection cannot be uniquely determined.
Answer & solution
Answer: Option B
Solving the set
Step 1 – Totals. Rows: 22 + 20 + 20 = 62; columns: 15 + 21 + 26 = 62. So the six distinct requirements add up to 62. The minimum is 7 and the maximum is 15, so every value lies between 7 and 15.
Step 2 – Number of ATMs on each road. One ATM gives at most 15; three ATMs give at least 7 + 8 + 9 = 24.
Road | Total (Rs L) | Possible number of ATMs | Reason |
R-A | 22 | 2 | 1 ATM ≤ 15 < 22; 3 ATMs ≥ 7 + 8 + 9 = 24 > 22 |
R-B | 20 | 2 | 1 ATM ≤ 15 < 20; 3 ATMs ≥ 24 > 20 |
R-C | 20 | 2 | Same as R-B |
V1 | 15 | 1 (15) or 2 (7 + 8) | 3 ATMs ≥ 24 > 15 |
V2 | 21 | 2 | 1 ATM ≤ 15 < 21; 3 ATMs ≥ 24 > 21 |
V3 | 26 | 2 or 3 | 1 ATM ≤ 15 < 26 |
So each row has exactly two ATMs, V2 has two, and V1 and V3 share the remaining four: either V1 = 2 and V3 = 2, or V1 = 1 and V3 = 3.
Step 3 – Fact 2 says there is an ATM at (R-C, V3).
Step 4 – Case I: V1 has two ATMs. 15 = 7 + 8 is the only possibility. 15 cannot be on V2 (its partner would be 6), so 15 is on V3 with 11 (26 = 15 + 11). 7 and 15 are on the same road (fact 1), so they share a row; 7 + 15 = 22 means that row is R-A. V2's two ATMs (sum 21) must avoid 7, 8, 11 and 15, so they are 9 and 12. The ATM at (R-C, V3) must be 11, so R-C = 11 + 9 and R-B = 8 + 12.
Step 5 – Case II: V1 has one ATM, 15. Then V3 has three ATMs adding to 26 and including 7 (the 7 must share a row with 15, and that row is R-A since 7 + 15 = 22). V3 = 7, 8, 11 or 7, 9, 10.
• V3 = 7, 8, 11: V2 = 9 and 12. Rows B and C each need one V2 ATM and one V3 ATM adding to 20: 12 + 8 and 9 + 11. Fact 2: the second highest (12) must be 12 km from (R-C, V3). If 12 were in R-C, the distance would be only 7 km; so 12 is at (R-B, V2), 8 at (R-B, V3), 9 at (R-C, V2), 11 at (R-C, V3). Distance = 5 + 7 = 12 km ✓.
• V3 = 7, 9, 10: V2 = 8 and 13, but no V2 + V3 pair makes 20. Not possible.
Case | V1 | V3 | V2 (sum 21) | Rows B and C (sum 20 each) | Result |
I | 7, 8 | 15, 11 | 9, 12 | R-B = 8 + 12, R-C = 9 + 11 | Valid (12 to R-C/V3 = 5 + 7 = 12 km) |
II (a) | 15 | 7, 8, 11 | 9, 12 | R-B = 12 + 8, R-C = 9 + 11 | Valid (12 to R-C/V3 = 12 km) |
II (a′) | 15 | 7, 8, 11 | 9, 12 | R-B = 9 + 11, R-C = 12 + 8 | Invalid (12 to R-C/V3 = 7 km) |
II (b) | 15 | 7, 9, 10 | 8, 13 | No pair adds to 20 | Invalid |
Step 6 – The two valid arrangements:
Arrangement 1 | V1 | V2 | V3 | Row total |
R-A | 7 | — | 15 | 22 |
R-B | 8 | 12 | — | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
Arrangement 2 | V1 | V2 | V3 | Row total |
R-A | 15 | — | 7 | 22 |
R-B | — | 12 | 8 | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
This question
In both arrangements (Step 6), R-C has 9 at V2 and 11 at V3. So the ATM at (R-C, V2) needs Rs. 9 Lakhs.
Hence, the answer is option 2.
Question 2 · Caselet DI
Answer & solution
Answer: 3
Solving the set
Step 1 – Totals. Rows: 22 + 20 + 20 = 62; columns: 15 + 21 + 26 = 62. So the six distinct requirements add up to 62. The minimum is 7 and the maximum is 15, so every value lies between 7 and 15.
Step 2 – Number of ATMs on each road. One ATM gives at most 15; three ATMs give at least 7 + 8 + 9 = 24.
Road | Total (Rs L) | Possible number of ATMs | Reason |
R-A | 22 | 2 | 1 ATM ≤ 15 < 22; 3 ATMs ≥ 7 + 8 + 9 = 24 > 22 |
R-B | 20 | 2 | 1 ATM ≤ 15 < 20; 3 ATMs ≥ 24 > 20 |
R-C | 20 | 2 | Same as R-B |
V1 | 15 | 1 (15) or 2 (7 + 8) | 3 ATMs ≥ 24 > 15 |
V2 | 21 | 2 | 1 ATM ≤ 15 < 21; 3 ATMs ≥ 24 > 21 |
V3 | 26 | 2 or 3 | 1 ATM ≤ 15 < 26 |
So each row has exactly two ATMs, V2 has two, and V1 and V3 share the remaining four: either V1 = 2 and V3 = 2, or V1 = 1 and V3 = 3.
Step 3 – Fact 2 says there is an ATM at (R-C, V3).
Step 4 – Case I: V1 has two ATMs. 15 = 7 + 8 is the only possibility. 15 cannot be on V2 (its partner would be 6), so 15 is on V3 with 11 (26 = 15 + 11). 7 and 15 are on the same road (fact 1), so they share a row; 7 + 15 = 22 means that row is R-A. V2's two ATMs (sum 21) must avoid 7, 8, 11 and 15, so they are 9 and 12. The ATM at (R-C, V3) must be 11, so R-C = 11 + 9 and R-B = 8 + 12.
Step 5 – Case II: V1 has one ATM, 15. Then V3 has three ATMs adding to 26 and including 7 (the 7 must share a row with 15, and that row is R-A since 7 + 15 = 22). V3 = 7, 8, 11 or 7, 9, 10.
• V3 = 7, 8, 11: V2 = 9 and 12. Rows B and C each need one V2 ATM and one V3 ATM adding to 20: 12 + 8 and 9 + 11. Fact 2: the second highest (12) must be 12 km from (R-C, V3). If 12 were in R-C, the distance would be only 7 km; so 12 is at (R-B, V2), 8 at (R-B, V3), 9 at (R-C, V2), 11 at (R-C, V3). Distance = 5 + 7 = 12 km ✓.
• V3 = 7, 9, 10: V2 = 8 and 13, but no V2 + V3 pair makes 20. Not possible.
Case | V1 | V3 | V2 (sum 21) | Rows B and C (sum 20 each) | Result |
I | 7, 8 | 15, 11 | 9, 12 | R-B = 8 + 12, R-C = 9 + 11 | Valid (12 to R-C/V3 = 5 + 7 = 12 km) |
II (a) | 15 | 7, 8, 11 | 9, 12 | R-B = 12 + 8, R-C = 9 + 11 | Valid (12 to R-C/V3 = 12 km) |
II (a′) | 15 | 7, 8, 11 | 9, 12 | R-B = 9 + 11, R-C = 12 + 8 | Invalid (12 to R-C/V3 = 7 km) |
II (b) | 15 | 7, 9, 10 | 8, 13 | No pair adds to 20 | Invalid |
Step 6 – The two valid arrangements:
Arrangement 1 | V1 | V2 | V3 | Row total |
R-A | 7 | — | 15 | 22 |
R-B | 8 | 12 | — | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
Arrangement 2 | V1 | V2 | V3 | Row total |
R-A | 15 | — | 7 | 22 |
R-B | — | 12 | 8 | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
This question
Both arrangements use the values 7, 8, 9, 11, 12 and 15. Those of 10 or more are 11, 12 and 15.
Hence, the answer is 3.
Question 3 · Caselet DI
Which of the following two statements is/are DEFINITELY true?
Statement A: Each of R-A, R-B, and R-C has two ATMs.
Statement B: Each of V1, V2, and V3 has two ATMs.
- AOnly Statement A
- BNeither Statement A nor Statement B
- COnly Statement B
- DBoth Statement A and Statement B
Answer & solution
Answer: Option A
Solving the set
Step 1 – Totals. Rows: 22 + 20 + 20 = 62; columns: 15 + 21 + 26 = 62. So the six distinct requirements add up to 62. The minimum is 7 and the maximum is 15, so every value lies between 7 and 15.
Step 2 – Number of ATMs on each road. One ATM gives at most 15; three ATMs give at least 7 + 8 + 9 = 24.
Road | Total (Rs L) | Possible number of ATMs | Reason |
R-A | 22 | 2 | 1 ATM ≤ 15 < 22; 3 ATMs ≥ 7 + 8 + 9 = 24 > 22 |
R-B | 20 | 2 | 1 ATM ≤ 15 < 20; 3 ATMs ≥ 24 > 20 |
R-C | 20 | 2 | Same as R-B |
V1 | 15 | 1 (15) or 2 (7 + 8) | 3 ATMs ≥ 24 > 15 |
V2 | 21 | 2 | 1 ATM ≤ 15 < 21; 3 ATMs ≥ 24 > 21 |
V3 | 26 | 2 or 3 | 1 ATM ≤ 15 < 26 |
So each row has exactly two ATMs, V2 has two, and V1 and V3 share the remaining four: either V1 = 2 and V3 = 2, or V1 = 1 and V3 = 3.
Step 3 – Fact 2 says there is an ATM at (R-C, V3).
Step 4 – Case I: V1 has two ATMs. 15 = 7 + 8 is the only possibility. 15 cannot be on V2 (its partner would be 6), so 15 is on V3 with 11 (26 = 15 + 11). 7 and 15 are on the same road (fact 1), so they share a row; 7 + 15 = 22 means that row is R-A. V2's two ATMs (sum 21) must avoid 7, 8, 11 and 15, so they are 9 and 12. The ATM at (R-C, V3) must be 11, so R-C = 11 + 9 and R-B = 8 + 12.
Step 5 – Case II: V1 has one ATM, 15. Then V3 has three ATMs adding to 26 and including 7 (the 7 must share a row with 15, and that row is R-A since 7 + 15 = 22). V3 = 7, 8, 11 or 7, 9, 10.
• V3 = 7, 8, 11: V2 = 9 and 12. Rows B and C each need one V2 ATM and one V3 ATM adding to 20: 12 + 8 and 9 + 11. Fact 2: the second highest (12) must be 12 km from (R-C, V3). If 12 were in R-C, the distance would be only 7 km; so 12 is at (R-B, V2), 8 at (R-B, V3), 9 at (R-C, V2), 11 at (R-C, V3). Distance = 5 + 7 = 12 km ✓.
• V3 = 7, 9, 10: V2 = 8 and 13, but no V2 + V3 pair makes 20. Not possible.
Case | V1 | V3 | V2 (sum 21) | Rows B and C (sum 20 each) | Result |
I | 7, 8 | 15, 11 | 9, 12 | R-B = 8 + 12, R-C = 9 + 11 | Valid (12 to R-C/V3 = 5 + 7 = 12 km) |
II (a) | 15 | 7, 8, 11 | 9, 12 | R-B = 12 + 8, R-C = 9 + 11 | Valid (12 to R-C/V3 = 12 km) |
II (a′) | 15 | 7, 8, 11 | 9, 12 | R-B = 9 + 11, R-C = 12 + 8 | Invalid (12 to R-C/V3 = 7 km) |
II (b) | 15 | 7, 9, 10 | 8, 13 | No pair adds to 20 | Invalid |
Step 6 – The two valid arrangements:
Arrangement 1 | V1 | V2 | V3 | Row total |
R-A | 7 | — | 15 | 22 |
R-B | 8 | 12 | — | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
Arrangement 2 | V1 | V2 | V3 | Row total |
R-A | 15 | — | 7 | 22 |
R-B | — | 12 | 8 | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
This question
Statement A: from Step 2, every row must have exactly two ATMs. True.
Statement B: in Arrangement 2, V1 has one ATM (15) and V3 has three (7, 8, 11). So it is not definitely true.
Hence, the answer is option 1.
Question 4 · Caselet DI
- A4 km
- B7 km
- C5 km
- DEither 4 km or 7 km
Answer & solution
Answer: Option D
Solving the set
Step 1 – Totals. Rows: 22 + 20 + 20 = 62; columns: 15 + 21 + 26 = 62. So the six distinct requirements add up to 62. The minimum is 7 and the maximum is 15, so every value lies between 7 and 15.
Step 2 – Number of ATMs on each road. One ATM gives at most 15; three ATMs give at least 7 + 8 + 9 = 24.
Road | Total (Rs L) | Possible number of ATMs | Reason |
R-A | 22 | 2 | 1 ATM ≤ 15 < 22; 3 ATMs ≥ 7 + 8 + 9 = 24 > 22 |
R-B | 20 | 2 | 1 ATM ≤ 15 < 20; 3 ATMs ≥ 24 > 20 |
R-C | 20 | 2 | Same as R-B |
V1 | 15 | 1 (15) or 2 (7 + 8) | 3 ATMs ≥ 24 > 15 |
V2 | 21 | 2 | 1 ATM ≤ 15 < 21; 3 ATMs ≥ 24 > 21 |
V3 | 26 | 2 or 3 | 1 ATM ≤ 15 < 26 |
So each row has exactly two ATMs, V2 has two, and V1 and V3 share the remaining four: either V1 = 2 and V3 = 2, or V1 = 1 and V3 = 3.
Step 3 – Fact 2 says there is an ATM at (R-C, V3).
Step 4 – Case I: V1 has two ATMs. 15 = 7 + 8 is the only possibility. 15 cannot be on V2 (its partner would be 6), so 15 is on V3 with 11 (26 = 15 + 11). 7 and 15 are on the same road (fact 1), so they share a row; 7 + 15 = 22 means that row is R-A. V2's two ATMs (sum 21) must avoid 7, 8, 11 and 15, so they are 9 and 12. The ATM at (R-C, V3) must be 11, so R-C = 11 + 9 and R-B = 8 + 12.
Step 5 – Case II: V1 has one ATM, 15. Then V3 has three ATMs adding to 26 and including 7 (the 7 must share a row with 15, and that row is R-A since 7 + 15 = 22). V3 = 7, 8, 11 or 7, 9, 10.
• V3 = 7, 8, 11: V2 = 9 and 12. Rows B and C each need one V2 ATM and one V3 ATM adding to 20: 12 + 8 and 9 + 11. Fact 2: the second highest (12) must be 12 km from (R-C, V3). If 12 were in R-C, the distance would be only 7 km; so 12 is at (R-B, V2), 8 at (R-B, V3), 9 at (R-C, V2), 11 at (R-C, V3). Distance = 5 + 7 = 12 km ✓.
• V3 = 7, 9, 10: V2 = 8 and 13, but no V2 + V3 pair makes 20. Not possible.
Case | V1 | V3 | V2 (sum 21) | Rows B and C (sum 20 each) | Result |
I | 7, 8 | 15, 11 | 9, 12 | R-B = 8 + 12, R-C = 9 + 11 | Valid (12 to R-C/V3 = 5 + 7 = 12 km) |
II (a) | 15 | 7, 8, 11 | 9, 12 | R-B = 12 + 8, R-C = 9 + 11 | Valid (12 to R-C/V3 = 12 km) |
II (a′) | 15 | 7, 8, 11 | 9, 12 | R-B = 9 + 11, R-C = 12 + 8 | Invalid (12 to R-C/V3 = 7 km) |
II (b) | 15 | 7, 9, 10 | 8, 13 | No pair adds to 20 | Invalid |
Step 6 – The two valid arrangements:
Arrangement 1 | V1 | V2 | V3 | Row total |
R-A | 7 | — | 15 | 22 |
R-B | 8 | 12 | — | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
Arrangement 2 | V1 | V2 | V3 | Row total |
R-A | 15 | — | 7 | 22 |
R-B | — | 12 | 8 | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
This question
Second highest = 12 at (R-B, V2) in both arrangements. Second lowest = 8:
• Arrangement 1: 8 at (R-B, V1), distance along R-B = 4 km.
• Arrangement 2: 8 at (R-B, V3), distance along R-B = 7 km.
So the distance is either 4 km or 7 km.
Hence, the answer is option 4.
Question 5 · Caselet DI
Answer & solution
Answer: 3
Solving the set
Step 1 – Totals. Rows: 22 + 20 + 20 = 62; columns: 15 + 21 + 26 = 62. So the six distinct requirements add up to 62. The minimum is 7 and the maximum is 15, so every value lies between 7 and 15.
Step 2 – Number of ATMs on each road. One ATM gives at most 15; three ATMs give at least 7 + 8 + 9 = 24.
Road | Total (Rs L) | Possible number of ATMs | Reason |
R-A | 22 | 2 | 1 ATM ≤ 15 < 22; 3 ATMs ≥ 7 + 8 + 9 = 24 > 22 |
R-B | 20 | 2 | 1 ATM ≤ 15 < 20; 3 ATMs ≥ 24 > 20 |
R-C | 20 | 2 | Same as R-B |
V1 | 15 | 1 (15) or 2 (7 + 8) | 3 ATMs ≥ 24 > 15 |
V2 | 21 | 2 | 1 ATM ≤ 15 < 21; 3 ATMs ≥ 24 > 21 |
V3 | 26 | 2 or 3 | 1 ATM ≤ 15 < 26 |
So each row has exactly two ATMs, V2 has two, and V1 and V3 share the remaining four: either V1 = 2 and V3 = 2, or V1 = 1 and V3 = 3.
Step 3 – Fact 2 says there is an ATM at (R-C, V3).
Step 4 – Case I: V1 has two ATMs. 15 = 7 + 8 is the only possibility. 15 cannot be on V2 (its partner would be 6), so 15 is on V3 with 11 (26 = 15 + 11). 7 and 15 are on the same road (fact 1), so they share a row; 7 + 15 = 22 means that row is R-A. V2's two ATMs (sum 21) must avoid 7, 8, 11 and 15, so they are 9 and 12. The ATM at (R-C, V3) must be 11, so R-C = 11 + 9 and R-B = 8 + 12.
Step 5 – Case II: V1 has one ATM, 15. Then V3 has three ATMs adding to 26 and including 7 (the 7 must share a row with 15, and that row is R-A since 7 + 15 = 22). V3 = 7, 8, 11 or 7, 9, 10.
• V3 = 7, 8, 11: V2 = 9 and 12. Rows B and C each need one V2 ATM and one V3 ATM adding to 20: 12 + 8 and 9 + 11. Fact 2: the second highest (12) must be 12 km from (R-C, V3). If 12 were in R-C, the distance would be only 7 km; so 12 is at (R-B, V2), 8 at (R-B, V3), 9 at (R-C, V2), 11 at (R-C, V3). Distance = 5 + 7 = 12 km ✓.
• V3 = 7, 9, 10: V2 = 8 and 13, but no V2 + V3 pair makes 20. Not possible.
Case | V1 | V3 | V2 (sum 21) | Rows B and C (sum 20 each) | Result |
I | 7, 8 | 15, 11 | 9, 12 | R-B = 8 + 12, R-C = 9 + 11 | Valid (12 to R-C/V3 = 5 + 7 = 12 km) |
II (a) | 15 | 7, 8, 11 | 9, 12 | R-B = 12 + 8, R-C = 9 + 11 | Valid (12 to R-C/V3 = 12 km) |
II (a′) | 15 | 7, 8, 11 | 9, 12 | R-B = 9 + 11, R-C = 12 + 8 | Invalid (12 to R-C/V3 = 7 km) |
II (b) | 15 | 7, 9, 10 | 8, 13 | No pair adds to 20 | Invalid |
Step 6 – The two valid arrangements:
Arrangement 1 | V1 | V2 | V3 | Row total |
R-A | 7 | — | 15 | 22 |
R-B | 8 | 12 | — | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
Arrangement 2 | V1 | V2 | V3 | Row total |
R-A | 15 | — | 7 | 22 |
R-B | — | 12 | 8 | 20 |
R-C | — | 9 | 11 | 20 |
Column total | 15 | 21 | 26 | 62 |
This question
Comparing the two arrangements: 12 at (R-B, V2), 9 at (R-C, V2) and 11 at (R-C, V3) are the same in both. 7, 8 and 15 move. So 3 ATMs are fully determined.
Hence, the answer is 3.
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