CAT 2025 Slot 2 · DILR · Question 6
CAT 2025 Slot 2 previous year question on Reasoning Based DI, with the answer and a step-by-step solution.
Set 2: Research Papers by Four Authors — this question belongs to a set. Read the full set with all 4 questions →
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.
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.
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