Charts, Graphs and Visual Data Interpretation Previous Year Questions (PYQs) for CAT: 30+ Solved Questions with Step-by-Step Solutions

Solve 30+ Charts, Graphs and Visual Data Interpretation previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Charts, Graphs and Visual Data Interpretation

Chapter Roadmap

1. Research and Author Productivity Charts

Weightage: Moderate | Focus: Relational matrices, authorship counting, and logical constraints.

2. Financial and Firm Performance Charts

Weightage: High | Focus: Revenue, cost, profit after tax, and employee strength trends over time.

3. Travel, Sustainability and Demographic Charts

Weightage: High | Focus: Index values, pollution measures, and distinct categorical distributions.

4. Sales and Order Bar Charts

Weightage: Moderate-High | Focus: Decoding stacked, layered, or patterned bar representations.

Topic Hero: Research and Author Productivity Charts

Topic Hero: Research and Author Productivity

What is this topic?

These problems present data about collaborative outputs, typically research papers, articles, or projects. The data is split across two dimensions: Output Categories (e.g., single-author, two-author) and Contributors (e.g., Author A, B, C, D).

Why it matters

This is pure logical reasoning disguised as data interpretation. Charts rarely give the complete table. Instead, they provide fragments: total papers, breakdown by type, individual author totals, and specific constraints.

The Core Objective

To construct a complete Author by Paper-Type Matrix that satisfies all given row sums, column sums, and logical constraints.

Charts, Graphs and Visual Data Interpretation: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Data Interpretation and Logical Reasoning NAT

Common Description: The Sustainability Index (SI) of a country at a point in time is an integer between 1 and 100. This question is related to SI of six countries - A, B, C, D, E, and F - at three different points in time - 2016, 2020, and 2024. The plot represents the exact changes in their SI, with X-coordinate representing % increase in 2020 from 2016, i.e., (SI in 2020 minus SI in 2016) / (SI in 2016), and Y-coordinate representing % increase in 2024 from 2020. At any point in time, the country with highest SI is ranked 1, while the country with the lowest SI is ranked 6. The following additional facts are known.
1. In 2016, B, C, E, and A had ranks 1, 2, 3, and 4 respectively.
2. F had lower SI than any other country in 2016, 2020, and 2024.
3. In 2024, E was the only country with SI of 90.
4. The range of SI of the six countries was 60 in 2016 as well as in 2024.
What was the SI of E in 2016?

Correct Answer:

Step-by-Step Solution

Key idea: This is a coordinate-to-fraction reverse-engineering question, recognisable because scatter plot axes represent percentage changes and the underlying values must be integers.

Why this method applies: The chart encodes multiplicative relationships visually. Since the final SI is given as exactly 90 and all SIs are integers, the percentage increases must correspond to simple fractions that preserve integrality through two successive multiplications. Visual estimation narrows candidates; integer constraints select the unique valid value.

Step 1: Calibrate the axes from grid lines.

X-axis (% increase 2016 to 2020): 0% at x=219, 20% at x=284 $\to$ 65 pixels = 20%.

Y-axis (% increase 2020 to 2024): 0% at y=290, 10% at y=234 $\to$ 56 pixels = 10%.

Step 2: Estimate E's coordinates from the plot.

E is at approximately (322, 230).

X: $322 - 219 = 103$ pixels right of 0%. $103/65 \approx 1.585$ units of 20% $\to\approx 31.7%$ .

Y: $290 - 230 = 60$ pixels above 0%. $60/56 \approx 1.07$ units of 10% $\to\approx 10.7%$ .

Step 3: Test simple fractions near estimates.

For X $\approx 31.7%$ , try $1/3$ (33.3%). Multiplier = $4/3$ .

For Y $\approx 10.7%$ , try $1/8$ (12.5%). Multiplier = $9/8$ .

Step 4: Reverse-calculate from known $S I_{2024} (E) = 90$ .

$S I_{2020} = \frac{90}{1 + Y} = \frac{90}{9/8} = 80$

$S I_{2016} = \frac{80}{1 + X} = \frac{80}{4/3} = 60$

Both are integers. This confirms the fractions are correct.

Answer: 60

Question 2 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [43 - 46 ]
The chart below provides complete information about the number of countries visited by Dheeraj, Samantha and Nitesh, in Asia, Europe and the rest of the world (ROW).

Person	Asia	Europe	ROW
Dheeraj	3	7	1
Samantha	0	9	4
Nitesh	1	6	12

The following additional facts are known about the countries visited by them.
1. 32 countries were visited by at least one of them.
2. USA (in ROW) is the only country that was visited by all three of them.
3. China (in Asia) is the only country that was visited by both Dheeraj and Nitesh, but not by Samantha.
4. France (in Europe) is the only country outside Asia, which was visited by both Dheeraj and Samantha, but not by Nitesh.
5. Half of the countries visited by both Samantha and Nitesh are in Europe. How many countries in the ROW were visited by both Nitesh and Samantha?

Correct Answer:

Step-by-Step Solution

Key idea: This is a constrained three-set Venn diagram question, recognisable because we are given totals, a union count, and clues about specific overlaps.

Why this method applies: We need an unknown pairwise overlap (Samantha and Nitesh), but the question does not give it directly. The union count lets us find the total of all pairwise overlaps using inclusion-exclusion.

Step 1: Compute individual totals.

Dheeraj: $3 + 7 + 1 = 11$ .

Samantha: $0 + 9 + 4 = 13$ .

Nitesh: $1 + 6 + 12 = 19$ .

Sum of individual counts: $11 + 13 + 19 = 43$ .

Step 2: Use inclusion-exclusion.

Let $P$ be the sum of the three pairwise overlaps.

Union = 32, triple overlap = 1.

$32 = 43 - P + 1 ⟹ P = 12$ .

Step 3: Use the given pairwise clues.

Dheeraj and Samantha share France and USA, so their overlap is 2.

Dheeraj and Nitesh share China and USA, so their overlap is 2.

Therefore, Samantha and Nitesh overlap is $12 - 2 - 2 = 8$ .

Step 4: Split the Samantha-Nitesh overlap by region.

Half of these 8 are in Europe, so Europe contributes 4.

Samantha visited 0 Asian countries, so the Asia contribution is 0.

Therefore, ROW contribution is $8 - 4 - 0 = 4$ .

Answer: 4

Question 3 · Data Interpretation and Logical Reasoning NAT

Person	Asia	Europe	ROW
Dheeraj	3	7	1
Samantha	0	9	4
Nitesh	1	6	12

Correct Answer:

Step-by-Step Solution

Key idea: This is an exclusive-count question in a specific region, recognisable because it asks for countries visited "only" by one person in a particular subset (Europe), requiring set theory applied regionally.

Why this method applies: To get an only-count, we must remove all shared elements from Nitesh's total for that specific region. We first find the total pairwise overlaps using the union formula, then distribute those overlaps by region based on the clues.

Step 1: Compute individual totals and find the total pairwise overlap ( $P$ ).

Dheeraj (D) = 3 + 7 + 1 = 11.

Samantha (S) = 0 + 9 + 4 = 13.

Nitesh (N) = 1 + 6 + 12 = 19.

Total Union = 32. Triple overlap (D $\cap$ S $\cap$ N) = 1 (USA).

Formula: $Union = D + S + N - P + Triple$

$32 = 11 + 13 + 19 - P + 1 ⟹ 32 = 43 - P + 1 ⟹ P = 12$ .

Step 2: Find the Samantha-Nitesh total overlap ( $S \cap N$ ).

D $\cap$ S overlap is 2 (France, USA).

D $\cap$ N overlap is 2 (China, USA).

Since $P = (D \cap S) + (S \cap N) + (D \cap N) - 2 \times (Triple)$ is not the standard form, we use:

$Union = D + S + N - (D \cap S) - (S \cap N) - (D \cap N) + (D \cap S \cap N)$

$32 = 11 + 13 + 19 - 2 - (S \cap N) - 2 + 1$

$32 = 40 - (S \cap N) ⟹ S \cap N = 8$ .

Step 3: Place the $S \cap N$ overlap by region.

Half of these 8 are in Europe $⟹$ Europe has 4 countries visited by both Samantha and Nitesh.

Step 4: Check whether Nitesh shares any Europe country with Dheeraj.

China is in Asia, USA is in ROW, and France is only D $\cap$ S outside Asia. Therefore, there is no D $\cap$ N overlap in Europe (it is 0).

Step 5: Subtract the Europe overlap from Nitesh's Europe total.

Nitesh visited 6 European countries.

Countries visited only by Nitesh in Europe = $6 - 4 (shared with S) - 0 (shared with D) = 2$ .

Answer: 2

Question 4 · Data Interpretation and Logical Reasoning NAT

Common Description: 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?

Correct Answer:

Step-by-Step Solution

Key idea: This is a system of linear equations with integer constraints question, recognisable because we must deduce specific category counts from aggregate totals and distinctness constraints.

Why this method applies: We need Brajen's Two-author and Three-author paper counts. We can find these by locking in the minimum constraints (at least one of each) and using the distinctness of Single-author papers to force a single valid distribution.

Step 1: Verify total author-slots.

Total slots from authors = $5 + 8 + 12 + 10 = 35$ .

Total slots from paper types = $10 (1) + 4 (2) + 3 (3) + 2 (4) = 10 + 8 + 9 + 8 = 35$ . (Matches perfectly).

Step 2: Deduce Four-author and Single-author counts.

There are 2 Four-author papers, so every author gets exactly 2 slots from them.

There are 10 Single-author papers. Since each author wrote at least 1 and all 4 wrote different numbers, the counts must be exactly ${1, 2, 3, 4}$ .

Step 3: Analyze Arman and Brajen.

Arman's total is 5. Subtracting his 2 Four-author slots leaves 3 slots for Single, Two, and Three.

Since each must be $\geq 1$ , Arman MUST have exactly Single=1, Two=1, Three=1.

Brajen's total is 8. Subtracting 2 Four-author slots leaves 6 slots.

We are told Brajen's Single = Two. So $2 \times Single + Three = 6$ .

Since Three $\geq 1$ , Single can be 1 or 2. But Arman already has Single=1, and all Single counts are distinct, so Brajen's Single MUST be 2.

This gives Brajen: Single=2, Two=2, Three=2.

Step 4: Answer the specific question.

The total number of two-author and three-author papers written by Brajen is $2 + 2 = 4$ .

Answer: 4

Question 5 · Data Interpretation and Logical Reasoning MCQ

Common Description: The two most populous cities and the non-urban region (NUR) of each of three states, Whimshire, Fogglia, and Humbleset, are assigned Pollution Measures (PMs). These nine PMs are all distinct multiples of 10, ranging from 10 to 90. The six cities in increasing order of their PMs are: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo.
The Pollution Index (PI) of a state is a weighted average of the PMs of its NUR and cities, with a weight of 50% for the NUR, and 25% each for its two cities.
There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset.
The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively. Which pair of cities definitely belong to the same state?

A.
Mumpypore, Zingaloo
B.
Splutterville, Quackford
C.
Blusterburg, Mumpypore
D.
Noodleton, Quackford

Correct Answer:

Step-by-Step Solution

Key idea: This is an extremal principle logic puzzle combined with weighted averages, recognisable because we must distribute 9 distinct multiples of 10 into NUR and City groups such that a specific pairwise inequality holds exactly once, then determine which cities must share a state.

Why this method applies: The "only one NUR > City pair" condition severely restricts the placement of NUR values among the sorted list. Once the full assignment is locked, we can identify which city pairs are forced into the same state.

Step 1: Lock PM values for NURs and Cities.

The 9 PMs are {10, 20, 30, 40, 50, 60, 70, 80, 90}. For exactly one NUR > City pair, the three NURs must be placed so that only the largest NUR exceeds exactly one city. The sorted arrangement must be: NUR, NUR, City, NUR, City, City, City, City, City. This gives NURs = {10, 20, 40} and Cities = {30, 50, 60, 70, 80, 90}. The sole NUR > City pair is 40 > 30.

Step 2: Assign Humbleset.

The NUR > City pair belongs to Humbleset, so Humbleset gets NUR = 40 and City = 30 (Blusterburg). For integer PI, the second city must share parity with 30 (multiplier 3, odd). Testing: cities {30, 90} give PI = $0.5 (40) + 0.25 (120) = 50$ .

Step 3: Assign remaining cities and NURs.

Remaining cities: {50, 60, 70, 80} (multipliers 5, 6, 7, 8). Remaining NURs: {10, 20}.

Fogglia has the lowest PI. Testing Fogglia with NUR=10 and cities {50, 70} (both odd): PI = $5 + 30 = 35$ .

Whimshire gets NUR=20 and cities {60, 80} (both even): PI = $10 + 35 = 45$ .

PIs: Fogglia=35, Whimshire=45, Humbleset=50. All distinct integers. ✓

Step 4: Map cities to names.

Cities in order of PM: Blusterburg(30), Noodleton(50), Splutterville(60), Quackford(70), Mumpypore(80), Zingaloo(90).

Humbleset: {Blusterburg, Zingaloo} = {30, 90}.

Fogglia: {Noodleton, Quackford} = {50, 70}.

Whimshire: {Splutterville, Mumpypore} = {60, 80}.

Step 5: Check which pair definitely belongs to the same state.

Noodleton(50) and Quackford(70) are both in Fogglia. This is the only pairing among the options that is forced.

Checking other options: Mumpypore-Zingaloo are in different states. Splutterville-Quackford are in different states. Blusterburg-Mumpypore are in different states.

Answer: Option D (Noodleton, Quackford)

A question from this chapter

Question 1

Common Description: The Sustainability Index (SI) of a country at a point in time is an integer between 1 and 100. This question is related to SI of six countries - A, B, C, D, E, and F - at three different points in time - 2016, 2020, and 2024. The plot represents the exact changes in their SI, with X-coordinate representing % increase in 2020 from 2016, i.e., (SI in 2020 minus SI in 2016) / (SI in 2016), and Y-coordinate representing % increase in 2024 from 2020. At any point in time, the country with highest SI is ranked 1, while the country with the lowest SI is ranked 6. The following additional facts are known. 1. In 2016, B, C, E, and A had ranks 1, 2, 3, and 4 respectively. 2. F had lower SI than any other country in 2016, 2020, and 2024. 3. In 2024, E was the only country with SI of 90. 4. The range of SI of the six countries was 60 in 2016 as well as in 2024. 0% -10% -30% 10% 30% 50% -40% -20% 0% 20% 40% 60% 80% 100% A B C D E F % increase in index in 2020 from 2016 % increase in index in 2024 from 2020 What was the SI of E in 2016?

Question 2

Common Description: Instructions [43 - 46 ] The chart below provides complete information about the number of countries visited by Dheeraj, Samantha and Nitesh, in Asia, Europe and the rest of the world (ROW). Person Asia Europe ROW Dheeraj 371 Samantha 094 Nitesh 1612 The following additional facts are known about the countries visited by them. 1. 32 countries were visited by at least one of them. 2. USA (in ROW) is the only country that was visited by all three of them. 3. China (in Asia) is the only country that was visited by both Dheeraj and Nitesh, but not by Samantha. 4. France (in Europe) is the only country outside Asia, which was visited by both Dheeraj and Samantha, but not by Nitesh. 5. Half of the countries visited by both Samantha and Nitesh are in Europe. How many countries in the ROW were visited by both Nitesh and Samantha?

Question 3

Common Description: Instructions [43 - 46 ] The chart below provides complete information about the number of countries visited by Dheeraj, Samantha and Nitesh, in Asia, Europe and the rest of the world (ROW). Person Asia Europe ROW Dheeraj 371 Samantha 094 Nitesh 1612 The following additional facts are known about the countries visited by them. 1. 32 countries were visited by at least one of them. 2. USA (in ROW) is the only country that was visited by all three of them. 3. China (in Asia) is the only country that was visited by both Dheeraj and Nitesh, but not by Samantha. 4. France (in Europe) is the only country outside Asia, which was visited by both Dheeraj and Samantha, but not by Nitesh. 5. Half of the countries visited by both Samantha and Nitesh are in Europe. How many countries in Europe were visited only by Nitesh?

Question 4

Common Description: 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. Number of papers by each author 04812 581210 Arman Brajen Chintan Devon Number of papers by type 10432 Single-author Two-author Three-author Four-author 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?

Question 5

Common Description: The two most populous cities and the non-urban region (NUR) of each of three states, Whimshire, Fogglia, and Humbleset, are assigned Pollution Measures (PMs). These nine PMs are all distinct multiples of 10, ranging from 10 to 90. The six cities in increasing order of their PMs are: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo. The Pollution Index (PI) of a state is a weighted average of the PMs of its NUR and cities, with a weight of 50% for the NUR, and 25% each for its two cities. There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset. The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively. Which pair of cities definitely belong to the same state?

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Charts, Graphs and Visual Data Interpretation Previous Year Questions (PYQs) for CAT: 30+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Charts, Graphs and Visual Data Interpretation

Chapter Roadmap

Topic Hero: Research and Author Productivity Charts

Topic Hero: Research and Author Productivity

Why it matters

The Core Objective

Charts, Graphs and Visual Data Interpretation: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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