CAT 2025 DILR Topic-wise Questions, Syllabus, Exam Pattern, Important Topics & Practice

CAT DILR Section – Exam Pattern

Before starting the candidate’s CAT DILR preparation, it is essential to thoroughly understand the CAT DILR exam pattern. The CAT exam pattern of the DILR section went through a significant change in the CAT 2024 examination, as 2 additional questions were added in 2024. Hence, the CAT DILR exam pattern can be considered dynamic, and there can also be a change in the CAT 2025 question paper as well. This year, the CAT exam will be conducted by IIM Kozhikode. The expected CAT DILR exam pattern after analysing the CAT previous year question paper is provided in the table below. Aspirants must remember that their performance in DILR not only impacts their overall percentile but also plays a role in clearing sectional cut-offs. This becomes crucial after result day when shortlists are released for tier 1 MBA colleges in India.

CAT DILR 2025 Syllabus

The CAT DILR syllabus for 2025 includes a lot of topics from both the data interpretation and logical reasoning sections. As per the CAT's previous year analysis, the data interpretation section of the CAT DILR section is expected to outweigh the CAT logical reasoning section. The candidates can expect 22 questions in the CAT 2025 DILR section. This will also include CAT TITA questions, which will not include any options. According to the CAT experts, the difficulty level of the CAT 2025 DILR section is expected to be moderate. The extensive syllabus of the CAT DILR section is given below. The extensive syllabus of the CAT DILR section is given below. By covering every topic thoroughly, candidates increase their chances of scoring well. High scores in DI-based sets, for example, are often considered indicators of strong analytical ability, which is valued by tier 1 MBA colleges in India during interviews.

CAT 2025 Topic-wise Practice Questions: DILR

As discussed before, the candidates can expect CAT DILR questions from a wide range of topics. Hence, it is essential for the candidates to solve as many CAT DILR questions as possible to get acquainted with the different CAT DILR question types that can be potentially asked in the CAT DILR section. The experts of Careers360 have designed a set of CAT topic-wise practice questions for improving the CAT 2025 preparation of the candidates under the CAT DILR section. The answers to the CAT DILR questions are written elaborately with adequate explanations and examples.

Regular practice helps aspirants build speed and accuracy, two critical factors for success in the CAT exam. It also ensures that after result, candidates are not just dependent on one section but have a balanced profile, which is highly valued by tier 1 MBA colleges in India. In fact, top institutes featuring prominently in IIM rankings 2025 often look for well-rounded candidates with consistent performance across QA, VARC, and DILR — making this practice all the more important. The candidates can access the CAT topic-wise questions DILR using the links provided in the table below.

CAT DILR Sample Questions

A few of the CAT DILR sample questions are provided below for the reference of the candidates. The candidates can expect these types of questions to be asked in the actual CAT examination.

Comprehension:
Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches - Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7 , with 1 being the lowest rating and 7 the highest.

The following additional information is known.
1. Xena trained more players than Yuki.
2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
4. The average of the ratings of all the players was 4.
5. Player-2 got the highest rating.
6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
7. Player-4's rating was double of Player-8's and less than Player-5's.

Question: Who all were the players trained by Xena?

1. Player-1, Player-3, Player-4, Player-6

2. Player-1, Player-3, Player-4, Player-8

3. Player-1, Player-3, Player-4

4. Player-1, Player-4, Player-6, Player-8

Solution:

From conditions 3 and 4, only two players got the same ratings. Thus, let $K$ be the repetitive number, then:

$\frac{(1+2+3+4+5+6+7+K)}{8} = 4$

It means $K$ = 4

Thus, players 5 and 7 got the same rating. As Xena trained more players than Yuki. So, there are two possibilities, which are as follows:

Possibility 1, X = $\frac{38}{13}$, (Not possible)

Possibility 2, X = 3, (Possible)

Step 2:

From condition 5, player 2 got the rating of 7. Also, player 4 got the rating of 2, and player 8 got the rating of 1.

So, we get:

Yuki has 2 players, and their total score is 12. The only available combination that works is 7 + 5. Because player-2 got a score of 7, the other score can only be that of player-6, that is 5. This is also consistent with the fact that the only even-numbered player left is player-6, and Yuki trained only even-numbered players.
Therefore, Yuki trains player-2 and player-6.

The overall score of Zara is 8. The correct combinations used in 8 are 2+6 and 4+4.
The point of 2 is player 4, but Zara trained only those players who have an odd number; therefore, it is not possible.
The only valid move is 4 + 4, which can be given to player-5 and player-7, both belonging to odd-numbered.
Therefore, Zara has to be above player-5 and player-7.

Thus, the table of the players and their trainers is as follows:

Thus, Player-1, player-3, player-4 and player-8 were trained by Xena.

Hence, the second option is correct.

2. Question: For how many players the ratings can be determined with certainty?

Solution:

From conditions 3 and 4, only two players got the same ratings. Thus, let $K$ be the repetitive number, then:

$\frac{(1+2+3+4+5+6+7+K)}{8} = 4$

It means $K$ = 4

Thus, players 5 and 7 got the same rating. As Xena trained more players than Yuki. So, there are two possibilities, which are as follows:

Possibility 1, X = $\frac{38}{13}$, (Not possible)

Possibility 2, X = 3, (Possible)

Step 2:

From condition 5, player 2 got the rating of 7. Also, player 4 got the rating of 2, and player 8 got the rating of 1.

So, we get:

Yuki has 2 players, and their total score is 12. The only available combination that works is 7 + 5. Because player-2 got a score of 7, the other score can only be that of player-6, that is 5. This is also consistent with the fact that the only even-numbered player left is player-6, and Yuki trained only even-numbered players.
Therefore, Yuki trains player-2 and player-6.

The overall score of Zara is 8. The correct combinations used in 8 are 2+6 and 4+4.
The point of 2 is player 4, but Zara trained only those players who have an odd number; therefore, it is not possible.
The only valid move is 4 + 4, which can be given to player-5 and player-7, both belonging to odd-numbered.
Therefore, Zara has to be above player-5 and player-7.

Thus, the table of the players and their trainers is as follows:

Thus, for 6 players, the ratings can be determined with certainty.

Hence, 6 is the correct answer.

3. Comprehension:
An online e-commerce firm receives daily integer product ratings from 1 through 5 given by buyers. The daily average is the average of the ratings given on that day. The cumulative average is the average of all ratings given on or before that day.
The rating system began on Day 1, and the cumulative averages were 3 and 3.1 at the end of Day 1 and Day 2, respectively. The distribution of ratings on Day 2 is given in the figure below.

The following information is known about ratings on Day 3.

1. 100 buyers gave product ratings on Day 3.
2. The modes of the product ratings were 4 and 5.
3. The numbers of buyers giving each product rating are non-zero multiples of 10.
4. The same number of buyers gave product ratings of 1 and 2, and that number is half the number of buyers who gave a rating of 3.

Question: Which of the following is true about the cumulative average ratings of Day 2 and Day 3?

1. The cumulative average of Day 3 increased by a percentage between $5 \%$ and $8 \%$ from Day 2.

2. The cumulative average of Day 3 increased by more than $8 \%$ from Day 2 .

3. The cumulative average of Day 3 decreased from Day 2.

4. The cumulative average of Day 3 increased by less than $5 \%$ from Day 2 .

Solution:

We are given:

Cumulative average at the end of Day 2 = 3.1

Total ratings by the end of Day 2 = 200

Average rating on Day 3 = 3.6

Total ratings on Day 3 = 100

Therefore, the cumulative average at the end of Day 3 is:

$
\frac{(3.1 \times 200) + (3.6 \times 100)}{200 + 100}
= \frac{620 + 360}{300}
= \frac{980}{300}
= 3.266\overline{6}
$

The increase in the cumulative average from day 2 to day 3 can be calculated as:

$\frac{3.266 - 3.1}{3.1} \times 100 ≈ 5.34\%$

Hence, the first option is correct.

4. Question: What is the median of all ratings given on Day 3?

Solution:

As deduced earlier, there were 10-1 ratings, 10-2 ratings, 20-3 ratings, 30-4 ratings, and 30-5 ratings. The median would be the average of the 50th and 51st ratings when arranged in ascending or descending order. Both of which would be 4.

Hence, 4 is the correct answer.

5. Comprehension:
The numbers $1,2,3,4,5,6,7,8,9$, and 10 are placed in ten slots of the following grid based on the conditions below.

1. Numbers in any row appear in an increasing order from left to right.
2. Numbers in any column appear in a decreasing order from top to bottom.
3. 1 is placed either in the same row or in the same column as 10 .
4. Neither 2 nor 3 is placed in the same row or in the same column as 10.
5. Neither 7 nor 8 is placed in the same row or in the same column as 9 .
6. 4 and 6 are placed in the same row.

Question: What is the sum of the numbers placed in Column 4?

Solution:

We are provided with the following clues: the numbers grow across the column (Clue 1 ) and diminish across the rows (Clue 2).
The main conclusion is that 10 can only be put in Row 1, Column 4, as there is no other position where it can be put and still a larger figure to its right or above would be put.

In Clue 3, 1 is in either the same row or the same column as 10.
Using the same rationale as on 10, 1 will have to be in either Row 1, Column 1 or Row 4, Column 4. At this moment, we cannot know which one.

And we also hear that 2 and 3 do not assign to the same row or column as 10, so they can only be in two of the following:

Row 2, Column 2

Row 2, Column 3

Column 3, Row 3

But 2 and 3 must go in Row 2, Column 2 or Row 3, Column 3, as there is no smaller number with which we can write a smaller number in any other position.

In Clue 6, we know that 4 and 6 share the same row, which may be Row 1, Row 2, or Row 3.

Clue 5 offers a good place to get started.

With 10 in place, the moves we can make with 9 are only Row 1 and Column 3 or Row 2 and Column 4.

If 9 is placed in R2C4, 7 and 8 must be placed in row 1. Clue 6 says that 4 and 6 are in the same row, but with 7 and 8 in row 1, no rows are left with two spaces. Hence, 9 can not be in R2C4, and the arrangement must be:

7 and 8 must be in column 4, occupying R3C4 and R2C4, respectively.

Hence, 4, 6 must be in row 1:

The sum of numbers in the 4th column is 10 + 8 + 7 + 1 = 26.

Hence, 26 is the correct answer.

6. Question: For how many slots in the grid, placement of numbers CANNOT be determined with certainty?

Solution:

We are provided with the following clues: the numbers grow across the column (Clue 1 ) and diminish across the rows (Clue 2).
The main conclusion is that 10 can only be put in Row 1, Column 4, as there is no other position where it can be put and still a larger figure to its right or above would be put.

In Clue 3, 1 is in either the same row or the same column as 10.
Using the same rationale as on 10, 1 will have to be in either Row 1, Column 1 or Row 4, Column 4. At this moment, we cannot know which one.

And we also hear that 2 and 3 do not assign to the same row or column as 10, so they can only be in two of the following:

Row 2, Column 2

Row 2, Column 3

Column 3, Row 3

But 2 and 3 must go in Row 2, Column 2 or Row 3, Column 3, as there is no smaller number with which we can write a smaller number in any other position.

In Clue 6, we know that 4 and 6 share the same row, which may be Row 1, Row 2, or Row 3.

Clue 5 offers a good place to get started.

With 10 in place, the moves we can make with 9 are only Row 1 and Column 3 or Row 2 and Column 4.

If 9 is placed in R2C4, 7 and 8 must be placed in row 1. Clue 6 says that 4 and 6 are in the same row, but with 7 and 8 in row 1, no rows are left with two spaces. Hence, 9 can not be in R2C4, and the arrangement must be:

7 and 8 must be in column 4, occupying R3C4 and R2C4, respectively.

Hence, 4, 6 must be in row 1:

Placements of all the numbers except 2 and 3 can be determined.

Hence, 2 is the correct answer.

7. Directions: An air conditioner (AC) company has four dealers - D1, D2, D3 and D4 in a city. It evaluates the sales performances of these dealers. The company sells two variants of ACs Window and Split. Both these variants can be either Inverter type or Non-inverter type. It is known that of the total number of ACs sold in the city, 25% were of the Window variant, while the rest were of the Split variant. Among the Inverter ACs sold, 20% were of the Window variant.

The following information is also known:

1. Every dealer sold at least two window ACs.

2. D1 sold 13 inverter ACs, while D3 sold 5 Non-inverter ACs.

3. A total of six Window Non-inverter ACs and 36 Split Inverter ACs were sold in the city.

4. The number of Split ACs sold by D1 was twice the number of Window ACs sold by it.

5. D3 and D4 sold an equal number of Window ACs and this number was one-third of the number of similar ACs sold by D2.

6. D2 and D3 were the only ones who sold Window Non-inverter ACs. The number of these ACs sold by D2 was twice the number of these ACs sold by D3.

7. D3 and D4 sold an equal number of Split Inverter ACs. This number was half the number of similar ACs sold by D2.

Question: How many Split Inverter ACs did D2 sell?

1. 14

2. 15

3. 16

4. 17

Solution:

As per the statements given in the question:
In inverter AC, 80% are split = 36 (given in point 3)
So, inverter window AC = 20% of total inverter AC = 9
Also, non-inverter window AC = 6 (given)
Total Window AC = 9 + 6 = 15 which is 25% of total AC sold.
So, the number of total ACs sold = 60

It is given: that D2 and D3 were the only ones who sold Window Non-inverter ACs. The number of these ACs sold by D2 was twice the number of these ACs sold by D3.
So, divide 6 in the ratio of 2: 1, D2 = 4, D3 = 2 [in case of non-inverter window AC]
Since D3 sold 5 non-inverter AC (point 2), So, D3 will sell 3 (5-2=3) split non-inverter AC.
D3 and D4 sold an equal number of Window ACs and this number was one-third of the number of similar ACs sold by D2.
Let D3 = D4 = b, then D2 = 3b and let D1 = a.
a + b + b + 3b = 15
Also, 3b is a multiple of 3. Possible values are 3, 6, 9, 12
If 3b = 12, b = 3, Sum becomes greater than 15.
Similarly, for 3b = 9, Sum is greater than 15.
3b = 6, b = 2, then a = 15 – 6 – 2 – 2 = 5. Now using this we can find the number of window inverter AC for each dealer.

D1 sold 13 inverter ACs. So, the Inverter split AC sold by D1 = 13 – 5 = 8.
The number of Split ACs sold by D1 was twice the number of Window ACs sold by it. So, D1 sells 10 split AC.
Split non-inverter AC sold by D1 = 10 -8 = 2

D3 and D4 sold an equal number of Split Inverter ACs. This number was half the number of similar ACs sold by D2.
Let D3 = D4 = x, then D2 = 2x
So, x + x + 2x + 8 = 36
So, x = 7

From the above table,
The number of Split Inverter ACs sold by D2 is 14.

Hence, the first option is correct.

8. Question: What percentage of ACs sold were of non-inverter type?

1. 75%

2. 20%

3. 25%

4. 4.33%

Solution:

According to statements given in the question:
In inverter AC, 80% are split = 36 (given in point 3)
So, inverter window AC = 20% of total inverter AC = 9
Also, non-inverter window AC = 6 (given)
Total Window AC = 9 + 6 = 15 which is 25% of total AC sold.
So, the number of total ACs sold = 60

From the above diagram,
Total number of AC= 60
Total number of non-inverter AC= 15
Percentage of non-inverter AC = (15/60) × 100 = 25%.

Hence, the third option is the correct answer.

CAT DILR Important Topics and Previous Year Questions

The CAT DILR syllabus is very broad, and the candidates require a lot of practice to solve the CAT DILR questions. Apart from solving the CAT DILR topic-wise questions, the candidates should also be aware of the various CAT DILR important topics and solve as many questions from these topics as they have a greater chance of being repeated in the CAT examination. By going through the CAT previous year question papers, solving the CAT sample papers and attempting the CAT mock tests, the candidates can score better in the CAT DILR section. The list of the CAT DILR important topics and the previous year's CAT questions asked under them are listed below.

CAT Preparation Books for DILR

The set of best CAT books that the candidates can choose for the CAT data interpretation and logical reasoning section is provided below. The candidates can solve more CAT topic-wise practice questions by referring to these selected CAT preparation books.

CAT 2025 Preparation Resources by Careers360

The candidates can download and study the various CAT preparation resources designed by Careers30 for CAT DILR preparation, as well as the other CAT preparation of other sections, such as the quantitative aptitude and logical reasoning, using the links given below.