CAT 2025 Slot 2 · DILR · Set 2: Research Papers by Four Authors
The full set and all 4 questions (Q6–Q9) from CAT 2025 Slot 2, with answers and solutions.
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.
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 6 · Reasoning Based DI
What was the total number of two-author and three-author papers written by Brajen?
Answer & solution
Answer: 4
There are 2 four-author papers, and every author is on both, so each author has 2 four-author papers.
Arman wrote 5 papers and at least one of each type (fact 1): 1 single + 1 two-author + 1 three-author + 2 four-author = 5. So Arman's split is fixed as 1, 1, 1, 2.
The 3 three-author papers have 3 × 3 = 9 author slots. Arman takes 1, leaving 8 for Brajen, Chintan and Devon. Nobody can be on more than 3, and Chintan and Devon each have more than Brajen (fact 3). If Brajen had 1, Chintan + Devon = 7 > 6, impossible. So Brajen = 2, Chintan = 3, Devon = 3.
Brajen's total is 8 = single + two-author + 2 + 2, so single + two-author = 4. These are equal (fact 4), so each is 2.
Brajen's two-author + three-author papers = 2 + 2 = 4.
Shortcut: split the 9 three-author slots first (1, 2, 3, 3), then subtract Brajen's three-author and four-author papers from his total of 8.
Hence, the answer is 4.
Question 7 · Reasoning Based DI
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.
- A
Neither i nor ii
- B
Only i
- C
Only ii
- D
Both i and ii
Answer & solution
Answer: Option A
Known so far: Arman (single, two, three, four) = (1, 1, 1, 2); Brajen = (2, 2, 2, 2); Chintan and Devon each have 3 three-author and 2 four-author papers.
Single-author papers: 10 in total, split into four different positive numbers (fact 2). The only such split is 1 + 2 + 3 + 4. Arman has 1 and Brajen 2, so Chintan and Devon have 3 and 4 in some order.
Two-author papers: 4 papers give 8 author slots. Arman and Brajen take 1 + 2 = 3, so Chintan + Devon = 5.
Chintan: 12 − 3 − 2 = 7 single + two-author papers. Devon: 10 − 3 − 2 = 5.
Case 1: Chintan has 3 single and 4 two-author; Devon has 4 single and 1 two-author.
Case 2: Chintan has 4 single and 3 two-author; Devon has 3 single and 2 two-author.
Both cases can be realised (e.g. Case 1: pairs A–C, B–C, B–C, C–D; Case 2: pairs A–C, B–C, C–D, B–D).
Statement i holds only in Case 2, and statement ii holds only in Case 2. Neither is forced.
Shortcut: Chintan's single + two-author total is fixed at 7, but his single-author count can be 3 or 4, so no statement about either count is necessarily true.
Hence, the answer is option 1.
Question 8 · Reasoning Based DI
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.
- A
Only ii
- B
Both i and ii
- C
Only i
- D
Neither i or ii
Answer & solution
Answer: Option B
There are 3 three-author papers. Chintan and Devon are each on 3 of them, so both are on every three-author paper.
Each three-author paper is therefore Chintan + Devon + one more author. Arman's one paper is Arman–Chintan–Devon, and Brajen's two papers are both Brajen–Chintan–Devon.
Both statements are necessarily true.
Shortcut: two authors who appear on every three-author paper leave room for only one more name on each.
Hence, the answer is option 2.
Question 9 · Reasoning Based DI
If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?
Answer & solution
Answer: 3
The 4 two-author papers have 8 author slots. Arman has 1 and Brajen 2, so Chintan + Devon = 5.
Devon's single + two-author count is 5 and his single-author count is 3 or 4, so Devon has 1 or 2 two-author papers.
If Devon wrote more than one, Devon has 2, so Chintan has 5 − 2 = 3.
This is consistent: pairs A–C, B–C, C–D and B–D give Arman 1, Brajen 2, Chintan 3 and Devon 2, with single-author counts Chintan 4 and Devon 3, all distinct.
Shortcut: Chintan and Devon share 5 two-author slots; Devon with 2 leaves Chintan 3.
Hence, the answer is 3.
Take this paper as a timed test
Real exam timing, instant score, rank and topic-wise analysis.