CAT 2025 Slot 2 DILR Set 5 Q19–Q22
A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2025 Slot 2, with 4 questions (Charts & Tables). Answers are CAT's official key.
The set
This set relies on a figure in the original paper, which is not reproduced here. The questions and answers are below.
The following charts depict details of research papers written by four authors, Arman, Brajen, Chintan, and Devon. The papers were of four types, single-author, two-author, three-author, and four-author, that is, written by one, two, three, or all four of these authors, respectively. No other authors were involved in writing these papers.
[Figure in the original paper]
The following additional facts are known.
1. Each of the authors wrote at least one of each of the four types of papers.
2. The four authors wrote different numbers of single-author papers.
3. Both Chintan and Devon wrote more three-author papers than Brajen.
4. The number of single-author and two-author papers written by Brajen were the same.
[Figure in the original paper]
The following additional facts are known.
1. Each of the authors wrote at least one of each of the four types of papers.
2. The four authors wrote different numbers of single-author papers.
3. Both Chintan and Devon wrote more three-author papers than Brajen.
4. The number of single-author and two-author papers written by Brajen were the same.
Question 19
What was the total number of two-author and three-author papers written by Brajen?
TITA: type in the answer. There are no options and no negative marking.
Answer: 4 Charts & Tables TITA
Solution
- Totals: Arman 5, Brajen 8, Chintan 12, Devon 10; papers: 10 single, 4 two-author, 3 three-author, 2 four-author. Each author is on both four-author papers.
- Single-author counts are distinct, at least 1, and sum to 10, so they are 1, 2, 3, 4. Arman has \(5 - 2 = 3\) other papers with at least one of each type: single 1, two 1, three 1.
- Three-author slots \(= 9\); Arman has 1, so Arman is left out of two papers. The third paper leaves out one of B, C, D; since C and D both beat Brajen, it leaves out Brajen: Brajen 2, Chintan 3, Devon 3. Papers: BCD, BCD, ACD.
- Brajen: \(s + t_2 = 8 - 2 - 2 = 4\) with \(s = t_2\), so \(s = t_2 = 2\). Chintan and Devon have singles 3 and 4 in some order and \(t_2\) summing to \(8 - 1 - 2 = 5\).
- Chintan \(s + t_2 = 7\), Devon \(s + t_2 = 5\): either (C: 3 single, 4 two; D: 4 single, 1 two) or (C: 4 single, 3 two; D: 3 single, 2 two). Both are realisable.
- Brajen: 2 two-author + 2 three-author \(= 4\).
Question 20
Which of the following statements is/are NECESSARILY true?
i. Chintan wrote exactly three two-author papers.
ii. Chintan wrote more single-author papers than Devon.
i. Chintan wrote exactly three two-author papers.
ii. Chintan wrote more single-author papers than Devon.
- ANeither i nor ii
- BOnly ii
- CBoth i and ii
- DOnly i
Answer: (A) Neither i nor ii Charts & Tables MCQ
Solution
- Totals: Arman 5, Brajen 8, Chintan 12, Devon 10; papers: 10 single, 4 two-author, 3 three-author, 2 four-author. Each author is on both four-author papers.
- Single-author counts are distinct, at least 1, and sum to 10, so they are 1, 2, 3, 4. Arman has \(5 - 2 = 3\) other papers with at least one of each type: single 1, two 1, three 1.
- Three-author slots \(= 9\); Arman has 1, so Arman is left out of two papers. The third paper leaves out one of B, C, D; since C and D both beat Brajen, it leaves out Brajen: Brajen 2, Chintan 3, Devon 3. Papers: BCD, BCD, ACD.
- Brajen: \(s + t_2 = 8 - 2 - 2 = 4\) with \(s = t_2\), so \(s = t_2 = 2\). Chintan and Devon have singles 3 and 4 in some order and \(t_2\) summing to \(8 - 1 - 2 = 5\).
- Chintan \(s + t_2 = 7\), Devon \(s + t_2 = 5\): either (C: 3 single, 4 two; D: 4 single, 1 two) or (C: 4 single, 3 two; D: 3 single, 2 two). Both are realisable.
- Both cases are valid: Chintan has 4 or 3 two-author papers, and Chintan's singles may be fewer or more than Devon's. Neither statement is forced.
Question 21
Which of the following statements is/are NECESSARILY true?
i. Arman wrote three-author papers only with Chintan and Devon.
ii. Brajen wrote three-author papers only with Chintan and Devon.
i. Arman wrote three-author papers only with Chintan and Devon.
ii. Brajen wrote three-author papers only with Chintan and Devon.
- ANeither i or ii
- BOnly ii
- COnly i
- DBoth i and ii
Answer: (D) Both i and ii Charts & Tables MCQ
Solution
- Totals: Arman 5, Brajen 8, Chintan 12, Devon 10; papers: 10 single, 4 two-author, 3 three-author, 2 four-author. Each author is on both four-author papers.
- Single-author counts are distinct, at least 1, and sum to 10, so they are 1, 2, 3, 4. Arman has \(5 - 2 = 3\) other papers with at least one of each type: single 1, two 1, three 1.
- Three-author slots \(= 9\); Arman has 1, so Arman is left out of two papers. The third paper leaves out one of B, C, D; since C and D both beat Brajen, it leaves out Brajen: Brajen 2, Chintan 3, Devon 3. Papers: BCD, BCD, ACD.
- Brajen: \(s + t_2 = 8 - 2 - 2 = 4\) with \(s = t_2\), so \(s = t_2 = 2\). Chintan and Devon have singles 3 and 4 in some order and \(t_2\) summing to \(8 - 1 - 2 = 5\).
- Chintan \(s + t_2 = 7\), Devon \(s + t_2 = 5\): either (C: 3 single, 4 two; D: 4 single, 1 two) or (C: 4 single, 3 two; D: 3 single, 2 two). Both are realisable.
- Arman's only three-author paper is ACD; Brajen's are both BCD. Both statements are true.
Question 22
If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?
TITA: type in the answer. There are no options and no negative marking.
Answer: 3 Charts & Tables TITA
Solution
- Totals: Arman 5, Brajen 8, Chintan 12, Devon 10; papers: 10 single, 4 two-author, 3 three-author, 2 four-author. Each author is on both four-author papers.
- Single-author counts are distinct, at least 1, and sum to 10, so they are 1, 2, 3, 4. Arman has \(5 - 2 = 3\) other papers with at least one of each type: single 1, two 1, three 1.
- Three-author slots \(= 9\); Arman has 1, so Arman is left out of two papers. The third paper leaves out one of B, C, D; since C and D both beat Brajen, it leaves out Brajen: Brajen 2, Chintan 3, Devon 3. Papers: BCD, BCD, ACD.
- Brajen: \(s + t_2 = 8 - 2 - 2 = 4\) with \(s = t_2\), so \(s = t_2 = 2\). Chintan and Devon have singles 3 and 4 in some order and \(t_2\) summing to \(8 - 1 - 2 = 5\).
- Chintan \(s + t_2 = 7\), Devon \(s + t_2 = 5\): either (C: 3 single, 4 two; D: 4 single, 1 two) or (C: 4 single, 3 two; D: 3 single, 2 two). Both are realisable.
- Devon with more than one two-author paper means the second case: Devon 2, so Chintan has 3.