Tables, Matrices and Structured Data Previous Year Questions (PYQs) for CAT: 17+ Solved Questions with Step-by-Step Solutions

Solve 17+ Tables, Matrices and Structured Data previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Tables, Matrices and Structured Data

Chapter Roadmap

1. Rating Tables and Score Matrices

High Weightage. Master integer reconstruction and cumulative averages.

2. Nutrition and Composition Tables

Moderate. Focus on percentage changes and mixture logic in grids.

3. Grid Matrices and Box Arrays

Moderate. Spatial logic, coin distributions, and 2D constraints.

4. Pandemic and Mortality Data Tables

Highest Weightage. Time-series data, rates, and overlapping conditions.

By the end of this chapter, you will be able to look at any structured data set, extract the hidden sums using averages, and reconstruct missing cells using integer constraints and logical bounds.

The Anatomy of a Rating Matrix

A rating table is a structured grid where:

Rows represent entities (e.g., Days, Restaurants, Products).

Columns represent categories (e.g., Rating scale 1 to 5, or specific Workers).

Cells contain either the count of ratings or the actual integer scores.

The Core Challenge

You are never asked to do complex arithmetic. You are asked to translate. You will be given summary statistics and must reverse-engineer the exact integer values hidden inside the grid.

The Golden Rule

Every summary statistic is just a disguised equation for the sum of that row or column.

Tables, Matrices and Structured Data: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [25 - 29]
There are nine boxes arranged in a

3 \times 3

array as shown in Tables 1 and 2. Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
Table 1
Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes.

	1st column	2nd column	3rd column
1st Row		9	6
2nd Row	2
3rd Row	8

Table 2
In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied.

	1st column	2nd column	3rd column
1st Row	1**	2*	2*
2nd Row	1**	0*	3*
3rd Row	3*	2**	0**

i) The minimum among the numbers of coins in the three sacks in the box is 1.
ii) The median of the numbers of coins in the three sacks is 1.
iii) The maximum among the numbers of coins in the three sacks in the box is 9. How many sacks have exactly one coin?

Correct Answer:

Step-by-Step Solution

Key idea: This is a Grid Matrix / Box Array reconstruction question, recognisable because there is a $3 \times 3$ array of boxes, equal row totals, equal column totals, and distinct integer box averages. To count sacks with exactly one coin, we must first reconstruct the triples in all nine boxes.

Step 1: Determine the box averages.

Each box has 3 sacks, and each box average is an integer. Since the nine averages are distinct, they must be $1, 2, 3, 4, 5, 6, 7, 8, 9$ .

Their total is $45$ . Since row totals are equal, each row's average-sum is $45/3 = 15$ . The same holds for columns. Thus the averages form a semi-magic square with constant $15$ .

Step 2: Use the strongest clues to fix some triples.

Sort each box as $a \leq b \leq c$ .

Conditions:

i) $a = 1$

ii) $b = 1$

iii) $c = 9$

R1C2 has median 9 and label $2 *$ . Its triple is $(a, 9, 9)$ . Condition iii is true, and exactly one condition is true, so $a \neq = 1$ . The sum must be divisible by 3, forcing $a = 3$ . So R1C2 is $(3, 9, 9)$ , average $7$ .

R2C1 has median 2 and label $1 * *$ . Its triple is $(a, 2, c)$ . Exactly one sack is greater than 5, so $c > 5$ . At least two conditions are true. Since the median is not 1, conditions i and iii must be true. So R2C1 is $(1, 2, 9)$ , average $4$ .

R3C1 has median 8 and label $3 *$ . All three sacks are greater than 5, and exactly one condition is true. Therefore $c = 9$ . The triple is $(a, 8, 9)$ , and the sum must be divisible by 3. The only valid $a$ is 7. So R3C1 is $(7, 8, 9)$ , average $8$ .

Step 3: Complete the magic square of averages.

Column 1 sum is 15. R1C1 + 4 + 8 = 15 $⟹$ R1C1 = 3.

Row 1 sum is 15. 3 + 7 + R1C3 = 15 $⟹$ R1C3 = 5.

The remaining averages for the grid are 1, 2, 6, 9.

Row 2: 4 + R2C2 + R2C3 = 15 $⟹$ R2C2 + R2C3 = 11.

The only pair from {1, 2, 6, 9} summing to 11 is 2 and 9.

Column 2: 7 + R2C2 + R3C2 = 15 $⟹$ R2C2 + R3C2 = 8.

If R2C2 = 9, R3C2 = -1 (impossible). So R2C2 = 2, R2C3 = 9.

Then R3C2 = 6.

Row 3: 8 + 6 + R3C3 = 15 $⟹$ R3C3 = 1.

The averages are:

3 7 5

4 2 9

8 6 1

Step 4: Reconstruct the remaining triples.

R1C1 (avg 3, 1**): 1 sack > 5. Conditions i, ii true $⟹ (1, 1, 7)$ .

R1C3 (avg 5, 2*, median 6): 2 sacks > 5. Condition i true $⟹ (1, 6, 8)$ .

R2C2 (avg 2, 0*): 0 sacks > 5. Condition i true $⟹ (1, 2, 3)$ .

R2C3 (avg 9, 3*): 3 sacks > 5. Condition iii true $⟹ (9, 9, 9)$ .

R3C2 (avg 6, 2**): 2 sacks > 5. Conditions i, iii true $⟹ (1, 8, 9)$ .

R3C3 (avg 1, 0**): 0 sacks > 5. Conditions i, ii true $⟹ (1, 1, 1)$ .

Step 5: Count sacks with exactly one coin.

Count the 1s in each triple:

R1C1: 2

R1C2: 0

R1C3: 1

R2C1: 1

R2C2: 1

R2C3: 0

R3C1: 0

R3C2: 1

R3C3: 3

Total = 2 + 0 + 1 + 1 + 1 + 0 + 0 + 1 + 3 = 9.

Answer: 9

Question 2 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [35 - 39]
Five restaurants, coded R1, R2, R3, R4 and R5 gave integer ratings to five gig workers - Ullas, Vasu, Waman, Xavier and Yusuf, on a scale of 1 to 5.
The means of the ratings given by R1, R2, R3, R4 and R5 were 3.4, 2.2, 3.8, 2.8 and 3.4 respectively. The summary statistics of these ratings for the five workers is given below.

	Ullas	Vasu	Waman	Xavier	Yusuf
Mean rating	2.2	3.8	3.4	3.6	2.6
Median rating	2	4	4	4	3
Modal rating	2	4	5	5	1 and 4
Range of rating*	3	3	4	4	3

* Range of ratings is defined as the difference between the maximum and minimum ratings awarded to a worker.
The following is partial information about ratings of 1 and 5 awarded by the restaurants to the workers.
(a) R1 awarded a rating of 5 to Waman, as did R2 to Xavier, R3 to Waman and Xavier, and R5 to Vasu.
(b) R1 awarded a rating of 1 to Ullas, as did R2 to Waman and Yusuf, and R3 to Yusuf. How many individual ratings cannot be determined from the above information?

Correct Answer:

Step-by-Step Solution

Key idea: This is a rating-matrix reconstruction question, recognisable because row means, column summary statistics, and a few fixed cells are given. The task is to determine whether any cells remain ambiguous after all constraints are applied.

Step 1: Convert means into sums.

Row sums (Restaurants, 5 workers each):

R1 = 17, R2 = 11, R3 = 19, R4 = 14, R5 = 17.

Column sums (Workers, 5 ratings each):

Ullas = 11, Vasu = 19, Waman = 17, Xavier = 18, Yusuf = 13.

Step 2: Reconstruct each worker's multiset.

Ullas: sum 11, median 2, mode 2, range 3. Unique set: ${1, 2, 2, 2, 4}$ .

Vasu: sum 19, median 4, mode 4, range 3. Unique set: ${2, 4, 4, 4, 5}$ .

Waman: sum 17, median 4, mode 5, range 4. Unique set: ${1, 2, 4, 5, 5}$ .

Xavier: sum 18, median 4, mode 5, range 4. Unique set: ${1, 3, 4, 5, 5}$ .

Yusuf: sum 13, median 3, modes 1 and 4, range 3. Unique set: ${1, 1, 3, 4, 4}$ .

Step 3: Place the given ratings.

Known 5s: R1W, R2X, R3W, R3X, R5V.

Known 1s: R1U, R2W, R2Y, R3Y.

Step 4: Use row and column constraints.

For R2, known ratings are R2W=1, R2X=5, R2Y=1. Sum is 7.

R2 sum is 11, so R2U + R2V = 4.

From remaining multisets, Ullas has ${2, 2, 4}$ and Vasu has ${2, 4, 4, 4}$ . The only way to sum to 4 is $2 + 2$ . So R2U=2, R2V=2.

For R1, known are R1U=1, R1W=5. Sum is 6. Need 11 from V, X, Y.

Vasu's remaining values are ${4, 4, 4}$ . So R1V=4.

Need 7 from R1X and R1Y. Xavier has ${1, 3, 4, 5}$ , Yusuf has ${3, 4, 4}$ . The only pairs summing to 7 are $(3, 4)$ or $(4, 3)$ .

For R3, known are R3W=5, R3X=5, R3Y=1. Sum is 11. Need 8 from U, V.

Vasu's remaining are ${4, 4}$ . So R3V=4, which forces R3U=4.

Now Ullas has ${2, 2}$ left for R4, R5. So R4U=2, R5U=2.

Vasu is fully placed except R4V, which must be 4.

Step 5: Resolve the remaining ambiguities.

R4 needs 8 from W, X, Y. R5 needs 10 from W, X, Y.

Waman's remaining values are ${2, 4}$ .

If R1X=4, R1Y=3: Xavier has ${1, 3}$ left, Yusuf has ${4, 4}$ left.

R4Y=4, R5Y=4. R4W+R4X=4. Since W is ${2, 4}$ , if R4W=4, R4X=0 (invalid). If R4W=2, R4X=2 (not in Xavier's remaining). This case fails.

If R1X=3, R1Y=4: Xavier has ${1, 4}$ left, Yusuf has ${3, 4}$ left.

Let R4Y=3, R5Y=4. Then R4W+R4X=5, R5W+R5X=6.

If R4W=4, R4X=1. Then R5W=2, R5X=4. This perfectly matches all remaining multisets!

Every cell is uniquely determined. 0 individual ratings cannot be determined.

Answer: 0

Question 3 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [25 - 29]
There are nine boxes arranged in a

3 \times 3

	1st column	2nd column	3rd column
1st Row		9	6
2nd Row	2
3rd Row	8

	1st column	2nd column	3rd column
1st Row	1**	2*	2*
2nd Row	1**	0*	3*
3rd Row	3*	2**	0**

Correct Answer:

Step-by-Step Solution

Key idea: This is a grid-matrix / box-array question, recognisable because there is a $3 \times 3$ array of boxes, equal row totals, equal column totals, and distinct integer box averages.

Step 1: Establish Global Constraints.

There are 9 boxes, each with 3 sacks. The averages are distinct integers.

Since the sacks contain values from 1 to 9, the possible averages for a box are 1 through 9.

With 9 distinct integer averages, they must be exactly ${1, 2, 3, 4, 5, 6, 7, 8, 9}$ .

Total sum of all averages = 45.

Since row totals and column totals are equal, each row and column of averages must sum to $45/3 = 15$ .

Step 2: Decode the Boxes using Conditions and Star Codes.

Conditions: i) Min=1, ii) Med=1, iii) Max=9.

means exactly 1 condition true. ** means 2 or more true.

R1C1: 1**, count>5 is 1. Med=1 (ii) and Min=1 (i) must be true. Max=9 (iii) is false.

Sorted: $(1, 1, c)$ with $c \leq 5$ . Sum $2 + c$ divisible by 3. $c = 7$ gives sum 9, avg 3. Sacks ${1, 1, 7}$ .

R1C2: 2*, Med=9. Sorted $(a, 9, 9)$ . $a \leq 5$ . Max=9 (iii) is true. So Min=1 (i) is false ( $a \neq = 1$ ).

Sum $a + 18$ divisible by 3. $a = 3$ gives sum 21, avg 7. Sacks ${3, 9, 9}$ .

R1C3: 2*, Med=6. Sorted $(a, 6, c)$ . $c > 5$ . Max=9 (iii) is false, so Min=1 (i) is true ( $a = 1$ ).

Sum $7 + c$ divisible by 3. $c = 8$ gives sum 15, avg 5. Sacks ${1, 6, 8}$ .

Row 1 averages: 3, 7, 5. Sum = 15. Valid.

R2C1: 1**, Med=2. Sorted $(a, 2, c)$ . $c > 5$ . Med=1 (ii) is false. So Min=1 (i) and Max=9 (iii) are true.

$a = 1, c = 9$ . Sum 12, avg 4. Sacks ${1, 2, 9}$ .

R2C2: 0*, count>5 is 0. All $\leq 5$ . Max=9 (iii) is false. Med=1 (ii) is false.

Thus Min=1 (i) must be true. Sorted $(1, b, c)$ with $b, c \leq 5$ .

Sum $1 + b + c$ divisible by 3. To get avg 2, sum=6. ${1, 2, 3}$ .

R2C3: 3*, count>5 is 3. All $> 5$ . Min=1 (i) and Med=1 (ii) are false. So Max=9 (iii) is true.

Sorted $(a, b, 9)$ with $a, b > 5$ . Sum $a + b + 9$ divisible by 3. To get avg 9, sum=27. ${9, 9, 9}$ .

Col 1 averages: 3, 4, 8. Sum = 15. So R3C1 avg = 8.

Col 2 averages: 7, 2, 6. Sum = 15. So R3C2 avg = 6.

Col 3 averages: 5, 9, 1. Sum = 15. So R3C3 avg = 1.

R3C1: 3*, Med=8. Sorted $(a, 8, c)$ . $a > 5$ . Min=1 (i) and Med=1 (ii) are false. So Max=9 (iii) is true ( $c = 9$ ).

Sum $a + 17$ divisible by 3. $a = 7$ gives sum 24, avg 8. Sacks ${7, 8, 9}$ .

R3C2: 2**, count>5 is 2. Sorted $(a, b, c)$ . Two $> 5$ . Min=1 (i) and Max=9 (iii) must be true.

$a = 1, c = 9$ . $b > 5$ . Sum $10 + b$ divisible by 3. $b = 8$ gives sum 18, avg 6. Sacks ${1, 8, 9}$ .

R3C3: 0**, count>5 is 0. All $\leq 5$ . Max=9 (iii) is false. So Min=1 (i) and Med=1 (ii) are true.

Sorted $(1, 1, 1)$ . Sum 3, avg 1. Sacks ${1, 1, 1}$ .

Step 3: Compare Average and Median for each box.

R1C1: ${1, 1, 7}$ , avg 3, med 1. (No)

R1C2: ${3, 9, 9}$ , avg 7, med 9. (No)

R1C3: ${1, 6, 8}$ , avg 5, med 6. (No)

R2C1: ${1, 2, 9}$ , avg 4, med 2. (No)

R2C2: ${1, 2, 3}$ , avg 2, med 2. (Yes)

R2C3: ${9, 9, 9}$ , avg 9, med 9. (Yes)

R3C1: ${7, 8, 9}$ , avg 8, med 8. (Yes)

R3C2: ${1, 8, 9}$ , avg 6, med 8. (No)

R3C3: ${1, 1, 1}$ , avg 1, med 1. (Yes)

Total matches = 4.

Answer: 4

Question 4 · Data Interpretation and Logical Reasoning NAT

	Ullas	Vasu	Waman	Xavier	Yusuf
Mean rating	2.2	3.8	3.4	3.6	2.6
Median rating	2	4	4	4	3
Modal rating	2	4	5	5	1 and 4
Range of rating*	3	3	4	4	3

Correct Answer:

Step-by-Step Solution

Key idea: This is a rating-matrix reconstruction question, recognisable because we have row means, worker summary statistics (mean, median, mode, range) and some fixed cells. For this sub-question, we only need to reconstruct the row for R2.

Step 1: Convert row means into row sums.

Each restaurant gives 5 ratings, so row sum = $5 \times mean$ .

For R2, mean = 2.2, so R2 sum = $5 \times 2.2 = 11$ .

Step 2: Reconstruct the relevant worker multisets.

We need Ullas and Vasu because the other three R2 ratings are partly known.

Ullas: mean 2.2 gives sum 11, median 2, mode 2, range 3.

Range 3 means max - min = 3. With median 2 and mode 2, the multiset must contain multiple 2s.

The only multiset of 5 integers from ${1, 2, 3, 4, 5}$ with sum 11, median 2, mode 2, range 3 is ${1, 2, 2, 2, 4}$ .

Vasu: mean 3.8 gives sum 19, median 4, mode 4, range 3.

Range 3 means max - min = 3. With median 4 and mode 4:

The only valid multiset is ${2, 4, 4, 4, 5}$ .

Step 3: Place the known R2 ratings.

From the given clues:

R2 gave Xavier a 5.

R2 gave Waman a 1.

R2 gave Yusuf a 1.

So R2 currently has: $5 + 1 + 1 = 7$ .

Step 4: Use the R2 row sum.

R2 total must be 11.

The remaining two ratings, for Ullas and Vasu, must sum to:

$11 - 7 = 4$ .

Step 5: Use the worker multisets to force the remaining ratings.

Ullas can contribute one of ${1, 2, 2, 2, 4}$ .

Vasu can contribute one of ${2, 4, 4, 4, 5}$ .

The only pair from these multisets that sums to 4 is:

Ullas = 2 and Vasu = 2.

Therefore R2's full row is:

Ullas = 2, Vasu = 2, Waman = 1, Xavier = 5, Yusuf = 1.

The question asks how many workers R2 gave a rating of 4.

Looking at the row, there are no 4s.

Answer: 0

Question 5 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [30 - 34]
There are only four neighbourhoods in a city - Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.
The following facts are also known:
1. There was at least one new case in every neighbourhood on Day 1.
2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2.
4. The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.
5. Kitmisto is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively. What BEST can be concluded about the total number of new cases in the city on Day 2?

A.
Either 7 or 8
B.
Exactly 7
C.
Either 6 or 7
D.
Exactly 8

Correct Answer:

Step-by-Step Solution

Key idea: This is a constrained pandemic-data puzzle, recognisable because daily case counts are limited to small integers, and several local and global conditions must all be satisfied. The best way is to work with the daily city totals first.

Step 1: Define the daily totals.

Let the total number of city cases on Days 1 to 5 be $D_{1}, D_{2}, D_{3}, D_{4}, D_{5}$ .

We are told the totals keep increasing:

$D_{1} < D_{2} < D_{3} < D_{4} < D_{5}$ .

Also $D_{3} = D_{2} + 1$ .

Step 2: Use the overall totals.

The five-day totals for the four neighbourhoods are:

Levmisto = 12, Tyhrmisto = 12, Pesmisto = 5, Kitmisto = 14.

Grand total = $12 + 12 + 5 + 14 = 43$ .

So:

$D_{1} + D_{2} + D_{3} + D_{4} + D_{5} = 43$ .

Since $D_{3} = D_{2} + 1$ :

$D_{1} + 2 D_{2} + D_{4} + D_{5} = 42$ .

Step 3: Find the maximum possible daily total.

Each neighbourhood can have at most 3 cases in a day.

But Pesmisto's maximum is 2, and this maximum happened only once.

Therefore on any day, the maximum city total is:

$3 + 3 + 3 + 2 = 11$ .

So $D_{5} \leq 11$ .

Step 4: Bound $D_{2}$ from above.

If $D_{2} \geq 9$ , then $D_{3} = D_{2} + 1 \geq 10$ .

Since totals are strictly increasing, $D_{4} \geq 11$ and $D_{5} \geq 12$ .

But $D_{5}$ cannot exceed 11. Contradiction.

Therefore $D_{2} \leq 8$ .

Step 5: Bound $D_{2}$ from below.

Since totals are strictly increasing:

$D_{1} \leq D_{2} - 1$ .

Also, because $D_{5} \leq 11$ and $D_{4} < D_{5}$ , the largest possible values are $D_{4} = 10$ and $D_{5} = 11$ .

Use the equation:

$42 = D_{1} + 2 D_{2} + D_{4} + D_{5}$ .

The maximum possible right-hand side is:

$42 \leq (D_{2} - 1) + 2 D_{2} + 10 + 11 = 3 D_{2} + 20$ .

$22 \leq 3 D_{2} ⟹ D_{2} \geq 7.33$ .

Since $D_{2}$ is an integer, $D_{2} \geq 8$ .

Step 6: Conclusion.

Combining the bounds, $D_{2}$ must be exactly 8.

Answer: D

Tables, Matrices and Structured Data Previous Year Questions (PYQs) for CAT: 17+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Tables, Matrices and Structured Data

Chapter Roadmap

The Anatomy of a Rating Matrix

The Anatomy of a Rating Matrix

The Core Challenge

The Golden Rule

Tables, Matrices and Structured Data: 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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