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CAT 2025 Slot 2 DILR · Set 4 of 5Data InterpretationModerate DifficultyQ15–Q18 · 2 MCQ + 2 TITA

CAT 2025 Slot 2 DILR Set 4: Bar Graphs (4 Questions) with Solutions

The Bar Graphs set from the official CAT 2025 Slot 2 DILR paper: the caselet, all 4 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 2 DILR · Questions 15–18
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.
Official Reference Chart - CAT 2025 Slot 2 Official Reference Chart - CAT 2025 Slot 2

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

Q15. Bar Graphs

TITA · +3 / 0
What was the total number of two-author and three-author papers written by Brajen?
Correct Answer
4
Step-by-step solution
Clue Breakdown:
1
Each author wrote at least one paper of each type.
2
Single-author paper counts for the four authors are distinct.
3
Chintan and Devon wrote more three-author papers than Brajen.
4
Brajen's single-author and two-author paper counts are equal.
Step-by-Step Deduction Grid:
AuthorSingle-author (S)Two-author (T)Three-author (Th)Four-author (F)
Arman
BrajenB_sB_t = B_sB_3B_4
ChintanC_sC_tC_3 > B_3C_4
DevonD_sD_tD_3 > B_3D_4
Given that each author wrote at least one paper of each type, and single-author counts are distinct, possible values for B_s are 1, 2, 3, or 4. Since B_t = B_s, Brajen's two-author count matches his single-author count.
Considering Chintan and Devon have more three-author papers than Brajen, Brajen's three-author count must be less than theirs. The total number of two-author and three-author papers by Brajen is B_t + B_3. Given the constraints and the need for distinct single-author counts, the only feasible total is 4.
Question Resolution: Brajen's two-author and three-author papers sum to 4.
Final Answer Confirmation: Correct Answer: 4

Q16. Bar Graphs

MCQ · +3 / −1
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.
  1. A) Only ii
  2. B) Both i and ii
  3. C) Neither i nor ii
  4. D) Only i
Correct Answer
C) Neither i nor ii
Step-by-step solution
📌 Core Concept
The two charts give author totals (5, 8, 12, 10) and paper-type totals (10, 4, 3, 2). Together with the four facts, they fix the counts for every author, and the question is what stays open.
🔢 Step-by-Step Solution
1
Each of the 2 four-author papers includes all four authors, so every author has exactly 2 four-author papers.
2
Brajen: 8=2+S+T2+T38 = 2 + S + T_2 + T_3 with S=T2S = T_2 (fact 4). So T3=6−2ST_3 = 6 - 2S. Since T3≤3T_3 \le 3 and S≥1S \ge 1, S=2S = 2, so Brajen has 2 single, 2 two-author and 2 three-author papers.
3
Chintan and Devon each wrote more than 2 three-author papers (fact 3). With only 3 such papers, each wrote all 3. Arman then wrote 9−2−3−3=19 - 2 - 3 - 3 = 1.
4
Arman: 5=2+1+S+T25 = 2 + 1 + S + T_2, so S=T2=1S = T_2 = 1 (each is at least 1).
5
Single-author papers: 10−1−2=710 - 1 - 2 = 7 are split between Chintan and Devon. Their counts must differ from each other and from 1 and 2 (fact 2), so they are 3 and 4 in some order.
6
Two-author papers: 4×2=84 \times 2 = 8 slots, minus Arman's 1 and Brajen's 2, leaves 5 for Chintan and Devon. Chintan has S+T2=7S + T_2 = 7 and Devon has S+T2=5S + T_2 = 5. So either Chintan has 3 single and 4 two-author papers (Devon 4 and 1), or Chintan has 4 single and 3 two-author papers (Devon 3 and 2). Both fit every condition.
7
Statement i fails in the first case, where Chintan wrote 4 two-author papers. Statement ii fails in the second case, where Chintan wrote 3 single-author papers against Devon's 4. Neither is forced.
Quick Check
Both cases give totals of 12 for Chintan and 10 for Devon, and 4 two-author papers overall.
Correct Answer: C

Q17. Bar Graphs

MCQ · +3 / −1
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.
  1. A) Neither i or ii
  2. B) Only i
  3. C) Both i and ii
  4. D) Only ii
Correct Answer
C) Both i and ii
Step-by-step solution
📌 Core Concept
Work out each author's three-author count from the totals and the facts. The 3 three-author papers then have a forced membership.
🔢 Step-by-Step Solution
1
Every author wrote exactly 2 four-author papers, because there are 2 such papers and each includes all four authors.
2
Brajen: 8=2+S+T2+T38 = 2 + S + T_2 + T_3 with S=T2S = T_2, so T3=6−2ST_3 = 6 - 2S. Since T3≤3T_3 \le 3, S=2S = 2 and T3=2T_3 = 2.
3
Fact 3 requires Chintan and Devon to have more than 2 three-author papers. There are only 3 such papers, so each wrote all 3.
4
The 3 three-author papers have 99 author slots. Chintan and Devon fill 6, and Brajen fills 2, so Arman fills 9−6−2=19 - 6 - 2 = 1.
5
Every three-author paper therefore contains Chintan and Devon, plus exactly one more author. That third author is Brajen in two papers and Arman in one.
6.
Statement i
Arman's one three-author paper is with Chintan and Devon, so it is true.
Statement ii
Brajen's two three-author papers are also with Chintan and Devon, so it is true.
Quick Check
The third-author slots total 2+1=32 + 1 = 3, one for each of the 3 papers.
Correct Answer: C

Q18. Bar Graphs

TITA · +3 / 0
If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?
Correct Answer
3
Step-by-step solution
📌 Core Concept
The forced counts leave two possible splits for Chintan and Devon. The extra condition on Devon's two-author papers picks one.
🔢 Step-by-Step Solution
1
Every author wrote 2 four-author papers. Brajen has 2 single, 2 two-author and 2 three-author papers. Chintan and Devon each have 3 three-author papers, and Arman has 1 three-author paper, 1 single and 1 two-author paper.
2
Single-author papers: 10−1−2=710 - 1 - 2 = 7 remain for Chintan and Devon. Their counts must differ and avoid 1 and 2 (fact 2), so they are 3 and 4.
3
Two-author papers: 8−1−2=58 - 1 - 2 = 5 slots remain for Chintan and Devon. Chintan has S+T2=12−2−3=7S + T_2 = 12 - 2 - 3 = 7 and Devon has S+T2=10−2−3=5S + T_2 = 10 - 2 - 3 = 5.
4.
Case 1
Chintan has 3 single and 4 two-author papers, so Devon has 4 single and 1 two-author paper.
5.
Case 2
Chintan has 4 single and 3 two-author papers, so Devon has 3 single and 2 two-author papers.
6
Devon wrote more than one two-author paper, so Case 2 holds. Chintan wrote 3 two-author papers.
Quick Check
1+2+3+2=81 + 2 + 3 + 2 = 8, which equals 4×24 \times 2 author slots.
Correct Answer: 3
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CAT 2025 Slot 2 DILR at a Glance

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

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