CAT 2026 DILR Practice Questions with Solutions: High-Weightage DI & LR Sets for Exam Readiness

CAT 2026 DILR Practice Questions

Solving a lot of CAT DILR questions is the most effective CAT 2026 preparation strategy for the DILR section. The candidates should solve as many CAT DILR practice questions as possible, and this includes various CAT preparation resources such as the CAT mock test 2026 and CAT sample papers. A few of the solved CAT 2026 DILR practice questions are provided below for the reference of the candidates.

CAT 2026 DILR Practice Questions: Data Interpretation

1. Comprehension:
Over the top (OTT) subscribers of a platform are segregated into three categories: i) Kid, ii) Elder, and iii) Others. Some of the subscribers used one app and the others used multiple apps to access the platform. The figure below shows the percentage of the total number of subscribers in 2023 and 2024 who belong to the ‘Kid’ and ‘Elder’ categories.

The following additional facts are known about the number of subscribers.

1. The total number of subscribers increased by 10% from 2023 to 2024.
2. In 2024, 12 of the subscribers from the 'Kid' category and 23 of the subscribers from the 'Elder' category use one app.
3. In 2023, the number of subscribers from the 'Kid' category who used multiple apps was the same as the number of subscribers from the 'Elder' category who used one app.
4. 10,000 subscribers from the 'Kid' category used one app, and 15,000 subscribers from the 'Elder' category used multiple apps in 2023.

Question: What could be the minimum percentage of subscribers who used multiple apps in 2024?

1. 22.00%

2. 16.5%

3. 20.0%

4. 10.0%

Solution

After reading the bar graphs, we can find the distribution as:

Take Clue 1, assume that the total number of subscribers will be 100x in 2023 and 110x in 2024.

This provides the division of subscribers as:

2023: 100x

2024: 110x → further divided in Clue 2

Based on Clue 2, of the 55x subscribers in 2024:

11x=half of 22x, and

23 of 33x=22x had utilised a single app.

Therefore, the total usage of a single app in 2024 = 11x+22x=33x

Now with Clues 3 and 4:

By Clue 4:

Ten thousand out of 15x Kids used one app

Of 20x Elders, 15000 use multiple apps

Clue 3:

Children multi-tasking with several applications = 15x−10000, equal to Older adults using one app = 20x−15000

Equating these (as per Clue 3):

15x−10,000=20x−15,000

Solution: 5x=5000

It means x=1000

And, Clue 2 makes us know that, of the 55000 kids and elders, 33000 use one app, thus 22000 use more than one app.

To minimize the total number of multiple app users, we can assume that all 55000 "other" subscribers use only one app.

Therefore, the number of multiple app users = 22000

Percentage = 22000110,000×100=20%

Hence, the third option is correct.

2. Comprehension:
Over the top (OTT) subscribers of a platform are segregated into three categories: i) Kid, ii) Elder, and iii) Others. Some of the subscribers used one app, and the others used multiple apps to access the platform. The figure below shows the percentage of the total number of subscribers in 2023 and 2024 who belong to the ‘Kid’ and ‘Elder’ categories.

The following additional facts are known about the number of subscribers.

1. The total number of subscribers increased by 10% from 2023 to 2024.
2. In 2024, 12 of the subscribers from the 'Kid' category and 23 of the subscribers from the 'Elder' category use one app.
3. In 2023, the number of subscribers from the 'Kid' category who used multiple apps was the same as the number of subscribers from the 'Elder' category who used one app.
4. 10,000 subscribers from the 'Kid' category used one app, and 15,000 subscribers from the 'Elder' category used multiple apps in 2023.

Question: What was the percentage increase in the number of subscribers in the 'Elder' category from 2023 to 2024?

1. 65%

2. 50%

3. 60%

4. 40%

Solution

After reading the bar graphs, we can find the distribution as:

Take Clue 1, assume that the total number of subscribers will be 100x in 2023 and 110x in 2024.

This provides the division of subscribers as:

2023: 100x

2024: 110x → further divided in Clue 2

Based on Clue 2, of the 55x subscribers in 2024:

11x=half of 22x, and

23 of 33x=22x had utilised a single app.

Therefore, the total usage of a single app in 2024 = 11x+22x=33x

Now with Clues 3 and 4:

By Clue 4:

Ten thousand out of 15x Kids used one app

Of 20x Elders, 15,000 use multiple apps

Clue 3:

Children multi-tasking with several applications = 15x−10000, equal to Older adults using one app = 20x−15000

Equating these (as per Clue 3):

15x−10000=20x−15000

Solution: 5x=5000

It means x=1000

In 2023, there were 20000 elders, and in 2024, there were 33000 elders

Thus, the percentage increase in the number of subscribers in the 'Elder' category from 2023 to 2024 = 33000−2000020000×100=65%

Hence, the first option is correct.

3. Comprehension:
Over-the-top (OTT) subscribers of a platform are segregated into three categories: i) Kid, ii) Elder, and iii) Others. Some of the subscribers used one app, and the others used multiple apps to access the platform. The figure below shows the percentage of the total number of subscribers in 2023 and 2024 who belong to the ‘Kid’ and ‘Elder’ categories.

The following additional facts are known about the number of subscribers.

1. The total number of subscribers increased by 10% from 2023 to 2024.
2. In 2024, 12 of the subscribers from the 'Kid' category and 23 of the subscribers from the 'Elder' category use one app.
3. In 2023, the number of subscribers from the 'Kid' category who used multiple apps was the same as the number of subscribers from the 'Elder' category who used one app.
4. 10,000 subscribers from the 'Kid' category used one app, and 15,000 subscribers from the 'Elder' category used multiple apps in 2023.

Question: What was the percentage increase in the number of subscribers in the 'Elder' category from 2023 to 2024?

1. 65%

2. 50%

3. 60%

4. 40%

Solution

After reading the bar graphs, we can find the distribution as:

Take Clue 1, assume that the total number of subscribers will be 100x in 2023 and 110x in 2024.

This provides the division of subscribers as:

2023: 100x

2024: 110x → further divided in Clue 2

Based on Clue 2, of the 55x subscribers in 2024:

11x=half of 22x, and

23 of 33x=22x had utilised a single app.

Therefore, the total usage of a single app in 2024 = 11x+22x=33x

Now with Clues 3 and 4:

By Clue 4:

Ten thousand out of 15x Kids used one app

Of 20x Elders, 15000 use multiple apps

Clue 3:

Children multi-tasking with several applications = 15x−10000, equal to Older adults using one app = 20x−15000

Equating these (as per Clue 3):

15x−10000=20x−15000

Solution: 5x=5000

It means x=1000

In 2023, there were 20000 elders, and in 2024, there were 33000 elders

Thus, the percentage increase in the number of subscribers in the 'Elder' category from 2023 to 2024 = 33000−2000020000×100=65%

Hence, the first option is correct.

4. Comprehension:
The table given below shows the amount, in grams, of carbohydrate, protein, fat, and all other nutrients per 100 grams of nutrients in seven foodgrains. The first column shows the foodgrain category and the second column its codename. The table has some missing values.

The following additional facts are known.

1. Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any millet.
2. Both the cereals had higher amounts of carbohydrate than any pseudo-cereal.
3. All the missing values of carbohydrate amounts (in grams) for all the foodgrains are nonzero multiples of 5.
4. All the missing values of protein, fat, and other nutrients (in grams) for all the foodgrains are non-zero multiples of 4.
5. P1 contains double the amount of protein that M3 contains.

Question: What is the median of the number of grams of protein in 100 grams of nutrients among these food grains?

Solution

The initial clue in the puzzle is that the rows have to add up to 100 grams.

From Clue 3, we know that all missing values in the Carbs column must be multiples of 5, and from Clue 4, all missing values in the other three columns (Proteins, Fats, and Others) must be multiples of 4.

Next, take Clue 1, which says that the amount of Carbs in C1 and C2 should be higher than in any pseudo cereal. Because P1 is a pseudo cereal, and it contains 66g of Carbs, it means that C1 and C2 should contain over 66g of Carbs.

Attempt to fill in the values of C1:
The possible Carb values (multiples of 5 greater than 66) are:
70, 75, 80, 85, 90, 95

We are told that 12g of the total 100g is already occupied by 'Other nutrients'. Therefore, the remaining 88g has to be shared by Carbs, Proteins, and Fats.

Suppose we take:

90g: Carbs 90+12=102>total, remove

95g: Carbs 95+12=107>total, remove

85g: Carbs 85+12=97→leaves 3g of Protein, 3 is not a multiple of 4→eliminate

75g: Carbs 75+12=87→leaves 13g of Protein, 13 is not a multiple of 4→eliminate

80g: Carbs 80+12=92→leaves 8g for Protein, 8 is a multiple of 4

→ Therefore, C1 contains: 80g Carbs, 8g Protein, 12g Others

Next, C2:

We’re told that 13g is already allocated (presumably to Fats or Others).

Therefore, out of 100g, 100−13=87g is left over to be divided between Carbs and Protein.

Possible Carbs values to be tested:

90g: 90+13=103>total, remove

95g→total exceeds, discard

85g: 85+13=98→leaves 2g Protein, 2 is not a multiple of 4→eliminate

80g: 80+13=93→leaves 7g, 7 is not a multiple of 4→discard

70g: 70+13=83→leaves 17g, 17 is not a multiple of 4→discard

75g Carbs: 75+13=88→leaves 12g, 12 is a multiple of 4

→ C2 therefore contains: 75g Carbs, 12g Protein, 13g Other

With Clues 1 and 2, we can conclude that the Carbs in P2 should be over 62 and below 75. The only values that are valid are 65 and 70.

If Carbs = 65g, then Protein = 100−65−14−8=13g→ Invalid (not a multiple of 4)

If Carbs = 70g, then Protein = 100−70−14−8=8g

Clue 1: Protein in M2 should be lower than any pseudo-cereal, which is 14g at the maximum.

So, Protein in M2 (and M3) can be: 12,8,4,0

Fats = 7g, Others = 16g given:

Carbs + Protein = 100−7−16=77g

Valid only combination: Protein = 12g, Carbs = 65g

The protein in M3 can be 0,4,8, or 12 (Clue 4).

From Clue 5, we know that P1's protein is double that of M3, so possible values for P1 protein are 0,8,16, or 24.

However, Clue 1 also states that P1's protein must be more than that of M1 and M2, so 0 and 8 are not valid.

This reduces to 16 and 24.

The protein and fats in P1 should now total:

100−66 (carbs)−10 (others)=24g

If protein = 24g, then fats = 0g, which violates Clue 4 (fats must be a non-zero multiple of 4) → Invalid

If protein = 16g, then fats = 8g

So, P1 has 16g protein and 8g fats, meaning M3 must have 8g protein (half of P1's), and with known carbs and fats, M3's others = 24g.

The proteins in the food grains, when arranged in ascending order, become: 8, 8, 10, 12, 12, 14, 16, giving the median as 12.

Hence, 12 is the correct answer.

5. Comprehension:
The table given below shows the amount, in grams, of carbohydrate, protein, fat, and all other nutrients, per 100 grams of nutrients in seven foodgrains. The first column shows the foodgrain category and the second column its codename. The table has some missing values.

The following additional facts are known.

1. Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any millet.
2. Both the cereals had higher amounts of carbohydrate than any pseudo-cereal.
3. All the missing values of carbohydrate amounts (in grams) for all the foodgrains are nonzero multiples of 5.
4. All the missing values of protein, fat, and other nutrient amounts (in grams) for all the foodgrains are non-zero multiples of 4.
5. P1 contains double the amount of protein that M3 contains.

Question: How many grams of other nutrients were there in 100 grams of nutrients in M3?

Solution

The initial clue in the puzzle is that the rows have to add up to 100 grams.

From Clue 3, we know that all missing values in the Carbs column must be multiples of 5, and from Clue 4, all missing values in the other three columns (Proteins, Fats, and Others) must be multiples of 4.

Next, take Clue 1, which says that the amount of Carbs in C1 and C2 should be higher than in any pseudo cereal. Because P1 is a pseudo cereal, and it contains 66g of Carbs, it means that C1 and C2 should contain over 66g of Carbs.

Attempt to fill in the values of C1:
The possible Carb values (multiples of 5 greater than 66) are:
70, 75, 80, 85, 90, 95

We are told that 12g of the total 100g is already occupied by 'Other nutrients'. Therefore, the remaining 88g has to be shared by Carbs, Proteins, and Fats.

Suppose we take:

90g: Carbs90+12=102>total, remove

95g: Carbs 95+12=107>total, remove

85g: Carbs 85+12=97→leaves 3g of Protein, 3 is not a multiple of 4→eliminate

75g: Carbs 75+12=87→leaves 13g of Protein, 13 is not a multiple of 4→eliminate

80g: Carbs 80+12=92→leaves 8g for Protein, 8 is a multiple of 4

→ Therefore, C1 contains: 80g Carbs, 8g Protein, 12g Others

Next, C2:

We’re told that 13g is already allocated (presumably to Fats or Others).

Therefore, out of 100g, 100−13=87g is left over to be divided between Carbs and Protein.

Possible Carbs values to be tested:

90g: 90+13=103>total, remove

95g→total exceeds, discard

85g: 85+13=98→leaves 2g Protein, 2 is not a multiple of 4→eliminate

80g: 80+13=93, 7g remaining. 93 is not a multiple of 4; therefore, discard

70g: 70+13=83, leaves 17g, 17 is not a multiple of 4; discard

75g Carbs: 75+13=88→leaves 12g, 12 is a multiple of 4

→ C2 therefore contains: 75g Carbs, 12g Protein, 13g Other

With Clues 1 and 2, we can conclude that the Carbs in P2 should be over 62 and below 75. The only values that are valid are 65 and 70.

If Carbs = 65g, then Protein = 100−65−14−8=13g→ Invalid (not a multiple of 4)

If Carbs = 70g, then Protein = 100−70−14−8=8g

Clue 1: Protein in M2 should be lower than any pseudo-cereal, which is 14g at the maximum.

So, Protein in M2 (and M3) can be: 12,8,4,0

Fats = 7g, Others = 16g given:

Carbs + Protein = 100−7−16=77g

Valid only combination: Protein = 12g, Carbs = 65g

The protein in M3 can be 0,4,8, or 12 (Clue 4).

From Clue 5, we know that P1's protein is double that of M3, so possible values for P1 protein are 0,8,16, or 24.

However, Clue 1 also states that P1's protein must be more than that of M1 and M2, so 0 and 8 are not valid.

This reduces to 16 and 24.

The protein and fats in P1 should now total:

100−66 (carbs)−10 (others)=24g

If protein = 24g, then fats = 0g, which violates Clue 4 (fats must be a non-zero multiple of 4) → Invalid

If protein = 16g, then fats = 8g

So, P1 has 16g protein and 8g fats, meaning M3 must have 8g protein (half of P1's), and with known carbs and fats, M3 others = 24g.

Therefore, we can see that there were 24 grams of other nutrients in M3.

Hence, 24 is the correct answer.

6. Comprehension:
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).

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 that 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.

Question: How many countries in Europe were visited by exactly one of Dheeraj, Samantha, and Nitesh?

1. 12

2. 14

3. 10

4. 5

Solution

Converting the data in tabular form, we get:

To find the answer to this, let us draw Venn diagrams of Asia, Europe, and the Rest of the World (ROW) depending on the given conditions.

It is given that half of the countries visited by Samantha and Nitesh are in Europe.

Note that those visited by the two of them in Asia are 0; this means the remaining half is in the rest of the world, and let x be the countries visited by Samantha and Nitesh only.

Therefore, the countries visited by Samantha and Nitesh are in Europe, be: (x+1) [as one country, USA, is visited by all three in ROW].

Using this, we can deduce:

In ROW, the number of countries visited only by Samantha is (3−x), and those visited only by Nitesh is (11−x).

In Europe, the number of countries visited only by Samantha is (7−x), and those visited only by Nitesh is (5−x).

It is said that 32 countries were visited by at least one of them.

⇒15−x+3+20−x=32⇒x=3

Replacing the value of x, we get the following Venn diagrams:

Therefore, the number of countries in Europe that were visited by exactly one of Dheeraj, Samantha, and Nitesh =6+2+4=12.

Hence, the first option is correct.

7. Comprehension:
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).

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 that 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.

Question: How many countries in the ROW were visited by both Nitesh and Samantha?

1. 4

2. 6

3. 7

4. 9

Solution

Converting the data in tabular form, we get:

To find the answer to this, let us draw Venn diagrams of Asia, Europe, and the Rest of the World (ROW) depending on the given conditions.

It is given that half of the countries visited by Samantha and Nitesh are in Europe.

Note that those visited by the two of them in Asia are 0; this means the remaining half is in the rest of the world, and let x be the countries visited by Samantha and Nitesh only.

Therefore, the countries visited by Samantha and Nitesh are in Europe, be: (x+1) [as one country, USA, is visited by all three in ROW].

Using this, we can deduce:

In ROW, the number of countries visited only by Samantha is (3−x), and those visited only by Nitesh is (11−x).

In Europe, the number of countries visited only by Samantha is (7−x), and those visited only by Nitesh is (5−x).

It is said that 32 countries were visited by at least one of them.

⇒15−x+3+20−x=32⇒x=3

Replacing the value of x, we get the following Venn diagrams:

Therefore, four countries in the Rest of the world were visited by both Nitesh and Samantha.

8. Comprehension: 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).

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.

Question: How many countries in Europe were visited only by Nitesh?

1. 2

2. 3

3. 4

4. 5

Solution

Converting the data in tabular form, we get:

To find the answer to this, let us draw Venn diagrams of Asia, Europe, and the Rest of the World (ROW) depending on the given conditions.

It is given that half of the countries visited by Samantha and Nitesh are in Europe.

Note that those visited by the two of them in Asia is 0, this means the remaining half is in the rest of the world, and let x be the countries visited by Samantha and Nitesh only.

Therefore, the countries visited by Samantha and Nitesh in Europe are: (x+1)

[as one country, the USA, is visited by all three in ROW].

Using this, we can deduce:

In ROW, the number of countries visited only by Samantha is (3−x), and those visited only by Nitesh is (11−x).

In Europe, the number of countries visited only by Samantha is (7−x), and those visited only by Nitesh is (5−x).

It is said that 32 countries were visited by at least one of them.

⇒15−x+3+20−x=32⇒x=3

Replacing the value of x, we get the following Venn diagrams:

Therefore, exactly two countries in Europe were visited only by Nitesh.

9. Comprehension: 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).

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.

Question: How many countries in Asia were visited by at least one of Dheeraj, Samantha, and Nitesh?

1. 3

2. 4

3. 5

4. 6

Solution

Converting the data into tabular form, we get:

To find the answer to this, let us draw Venn diagrams of Asia, Europe and the Rest of the World (ROW) depending on the given conditions.

It is given that half of the countries visited by Samantha and Nitesh are in Europe.

Note that those visited by the two of them in Asia is 0, this means the remaining half is in the rest of the world, and let x be the countries visited by Samantha and Nitesh only.

Therefore, the countries visited by Samantha and Nitesh are in Europe, be: (x+1) [as one country, USA, is visited by all three in ROW].

Using this, we can deduce:

In ROW, the number of countries visited only by Samantha is (3−x), and those visited only by Nitesh is (11−x).

In Europe, the number of countries visited only by Samantha is (7−x), and those visited only by Nitesh is (5−x).

It is said that 32 countries were visited by at least one of them. 15−x+3+20−x=32⇒x=3 Replacing the value of x, we get the following Venn diagrams:

Therefore, three countries in Asia were visited by at least one of them.

CAT 2026 DILR Practice Questions: Logical Reasoning

1. Comprehension:
The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km ) between two adjacent intersections. Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.

The following additional information is known.
1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
2. The road distance between the ATM with the second-highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km.

Question: What is the number of ATMs whose locations and cash requirements can both be uniquely determined?

1. 3

2. 4

3. 5

4. 6

Solution

It is indicated to us that, of the 9 intersections in the figure, 6 contain ATMs. 3 of these intersections are vacant.
It is also given to us that both the highest capacity and the lowest capacity ATMs are on the same road, the highest capacity of 15L and the lowest of 7L.
This information not only provides us with hints of the location of these two ATMs, but also, we now have the upper and lower limits of cash in the six ATMs with different cash amounts.
The next set of information, which is provided,d is that the road distance between the ATM having the second highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km. Since we can only traverse the roads, from (RC, V3), we have to either traverse the 5km road or the 7km road. It can only sum up to 12 in one way, 5+7. That means, the ATM with the second-highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are informed that 15 and 7 are on the same road. We are provided with the total capacities on the roads; therefore, we have to find which roads have a capacity greater than or equal to 22.
The only two choices are R-A or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20, and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not possible since there cannot be an ATM with 5L capacity.

Case 1: 15L ATM is on the intersection (RA, V3).

In order to get a cumulative sum of 26L in column V3, there is a need to have an ATM containing 11L of cash at any of the intersections. This is because placing two ATMs in V3 is not viable, the minimum ATM capacity is 7L, and using two such machines would exceed 26L (since 7L + 7L = 14L, requiring a third ATM with 12L, which is not allowed).

The 11L ATM can only be placed at either (RB, V3) or (RC, V3).

Suppose we situate the 11L ATM in one of these intersections, then the second largest ATM, namely 9L, needs to be placed so that it does not break any row or column totals.

Where could 9L place?:

It cannot be placed at (RC, V1) or (RB, V1), because V1 totals 15L, and placing a 9L ATM there would require a 6L ATM to complete the sum, which is invalid since 6L is not an available capacity.

It also cannot be at (RB, V2), because we’ve already placed the 11L ATM, and V2 cannot accommodate both the highest (11L) and the second-highest (9L) values, as this would unbalance the totals and available combinations.

This means the only feasible position for the 9L ATM is at (RC, V2).

Through this placement, it is possible to logically infer and fill in the remaining values using the constraints of the row and column totals. This brings us to the final and coherent ATM configuration of Case-1.

Case 2: When the 15L ATM is placed at (RA, V1).

With 15L at (RA, V1), no other ATM can be placed in column V1, as the column total is already fully accounted for.

Now, the 7L ATM, which must be placed somewhere in row RA, cannot go to (RA, V2). The reason is that the sum of V2 is 21L, and to put 7L ATM at that point would need:

Another 7L ATM (which isn’t allowed, as ATM values are unique), or

A 14L ATM (which exceeds the row total limit of 20L for both RB and RC).

Hence, the 7L ATM must be placed in (RA, V3).

That leaves 19L to fill in column V3 because V3 contains a total of 26L, and we have already put 7L. A single ATM of 19L is not possible (not among the allowed capacities), so two additional ATMs must collectively sum to 19L.

After trying out the possible combinations and abiding by all the row and column restrictions, we can conclude about the location of the remaining ATMs and get the valid solution of Case 2.

Using the two cases, we can answer the given questions.
ATMs that can be uniquely determined are the ATMs with cash 9L, 11L, and 12L. Hence, the answer is 3.

10. Comprehension:
The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C, and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km ) between two adjacent intersections. Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.

The following additional information is known.
1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
2. The road distance between the ATM with the second-highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km.

Question: What can be best said about the road distance (in km) between the ATMs having the second-highest and the second-lowest cash requirements?

1. 7 km

2. 4 km

3. Either 4 km or 7 km

4. 5 km

Solution

The network consists of three horizontal roads, R-A, R-B, R-C, and three vertical roads V1, V2, V3. The intersections form a 3×3 grid, with distances given between adjacent intersections.

Six ATMs are placed at six of the nine intersections. Let the cash requirements of the six ATMs be distinct integers in Rs. Lakhs. The total cash requirement of ATMs on each road is given. For example, on R-A, the total cash requirement = 22 Lakhs.

Given information:

1. The minimum and maximum cash requirements are Rs. 7 Lakhs and Rs. 15 Lakhs, placed on the same road.

2. The road distance between the ATM with the second highest cash requirement and the ATM at intersection (R-C,V3) is 12 km.

Identify possible locations of minimum and maximum ATMs

Let the road containing both minimum (7 Lakhs) and maximum (15 Lakhs) ATMs be R-B.

Determine possible distances between second highest and second lowest ATMs

Let the second-highest cash requirement = 14 Lakhs, and the second-lowest = 8 Lakhs.

The road distance between 14 Lakhs ATM and ATM at (R-C,V3) = 12 km. Compute possible distances between second highest and second lowest ATMs

Considering the grid layout and distances between intersections, the second highest and second lowest ATMs can be either 4 km or 7 km apart.

The road distance between the ATMs with the second highest and second lowest cash requirements = Either 4 km or 7 km.

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

1. 6

2. 7

3. 8

4. 9

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

(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 = 3813, (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.

12. 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: What was the rating of Player-6?

1. 3

2. 4

3. 5

4. 8

Solution

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

(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 = 3813, (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, the rating of player 6 is 5.

Hence, 5 is the correct answer.

CAT 2026 DILR Practice Questions PDF Link

Considering the vastness of the CAT DILR section, it is necessary for the candidates to shortlist the most important topics and solve a lot of CAT 2026 DILR practice questions under those topics. Careers360 has shortlisted a set of topics under which the CAT 2026 DILR practice question are repeatedly asked in the previous examinations. The candidates are encouraged to download the CAT practice questions DILR of the CAT important topics using the links given below.

CAT 2026 DILR Topic-wise Weightage and Difficulty

Understanding the topic-wise weightage and difficulty of the CAT 2026 DILR section helps streamline your preparation. Based on past trends, Data Interpretation and Logical Reasoning carry equal importance. Candidates must solve CAT 2026 DILR practice questions to improve accuracy and speed across varying difficulty levels. These can be checked on the CAT Mock Test 2026.

CAT 2026 DILR Practice Questions: Previous Year Trends and Analysis

Analysing previous year CAT DILR papers helps you understand the changing patterns in topic weightage, difficulty level, and question types. Whether you're aiming for CAT 99 percentile or starting your CAT preparation without coaching, reviewing trends from previous year. This helps you fine-tune your CAT 2026 DILR preparation strategy. Practise with CAT 2026 DILR practice questions aligned to the latest CAT syllabus 2026 for maximum impact.

Slot-Wise CAT DILR Difficulty Trends from Previous Years

Analysing slot-wise trends of the DILR section in previous CAT exams helps aspirants identify patterns, expected difficulty, and high-weightage question types. Understanding these trends alongside CAT 2026 DILR practice questions enables better time management. Candidates must use the CAT sample papers 2026 to create an edge in the exam.

CAT 2024 DILR Analysis

• Slot 1: Moderate difficulty, with 3–4 high-weightage DI sets and 2–3 LR puzzles. Quick attempts in familiar topics like tables and bar charts could fetch 12–14 marks.

• Slot 2: Slightly tougher, featuring multi-variable LR puzzles and mixed DI-LR sets. Time management was crucial; practising CAT 2026 DILR practice questions beforehand would have helped improve speed.

• Slot 3: Most challenging slot. Complex caselets, seating arrangements, and calculation-heavy DI sets dominated. High-confidence attempts with shortcut techniques could have maximised marks.

CAT 2023 DILR Analysis

• Slot 1: Moderate section with simpler DI sets and small LR puzzles. Candidates familiar with CAT 2026 DILR practice questions could attempt 10–12 questions confidently.

• Slot 2: Moderate to difficult, with mixed DI-LR sets requiring multi-step reasoning. Regular practice improved accuracy in calculating answers quickly.

• Slot 3: Difficult, lengthy section. Puzzles involving arrangement, grouping, and integration of DI data were common. Aspirants using time-efficient strategies gained an advantage.

CAT 2022 DILR Analysis

• Slot 1: Mostly moderate, featuring bar charts, tables, and small LR puzzles. About 8–10 questions were manageable for well-prepared aspirants.

• Slot 2: Moderate difficulty, with DI sets requiring calculation and LR puzzles needing logical deduction. Practising CAT 2026 DILR practice questions would have built confidence in handling multi-step reasoning.

• Slot 3: Slightly tougher, with complex caselets and multi-variable arrangement puzzles. Effective shortcut techniques and prior practice helped aspirants attempt high-scoring questions under time pressure.

Expected CAT 2026 DILR Practice Questions Types

Preparing for CAT 2026 requires familiarity with the types of DILR practice questions likely to appear. Knowing the CAT exam pattern 2026 helps aspirants focus on high-weightage areas, manage time efficiently, and improve accuracy. Regular practice with CAT 2026 DILR practice questions ensures confidence in handling multi-step reasoning and data interpretation problems.

Table-Based Data Interpretation Sets

Tables presenting numerical data are common in CAT 2026 DILR practice questions. Candidates need to compare, calculate percentages, ratios, or totals, and draw conclusions. Practising these improves speed in identifying key trends and relationships quickly.

Bar and Line Chart Analysis

Graphs and charts frequently appear in CAT 2026 DILR practice questions. Understanding scales, trends, and proportionality is essential. Regular practice enhances the ability to extract insights without unnecessary calculations.

Caselet Sets

Caselets with a short paragraph describing relationships or data are high-return in CAT 2026 DILR practice questions. Breaking information into smaller facts and forming tables or diagrams helps solve multi-variable problems effectively.

Puzzle-Based Logical Reasoning

Seating arrangements, scheduling, and grouping puzzles are key CAT 2026 DILR practice questions. Recognising patterns, using the process of elimination, and mapping data visually improve accuracy under time pressure.

Combination of DI and LR

Some complex sets mix data interpretation with logical reasoning. Practising integrated CAT 2026 DILR practice questions trains aspirants to handle multi-step reasoning efficiently, ensuring no marks are lost due to misinterpretation.

Venn Diagrams and Set Theory

Questions involving sets, overlaps, or categories often appear in CAT 2026 DILR practice questions. Using Venn diagrams to visualise relationships simplifies calculations and reduces errors in complex reasoning sets.

Shortcut Techniques for Complex CAT 2026 DILR Practice Questions

In CAT 2026 DILR, complex sets demand speed with precision. Using shortcut techniques helps reduce cognitive load, avoid brute-force calculations, and quickly identify workable paths. Focus on structure recognition, constraint pruning, and efficient representation of data. Smart shortcuts allow you to convert lengthy data into solvable formats, eliminate non-viable cases early

Constraint Mapping

Constraint mapping involves converting every condition in the set into a structured visual format such as a grid, table, or flowchart. Instead of rereading long paragraphs, all relationships (greater than, equal to, before/after, inclusion/exclusion) are placed in one consolidated view. This helps quickly detect conflicts, missing links, and forced placements. In CAT-level DILR, this technique significantly reduces reprocessing time and prevents logical errors caused by scattered information.

Option Elimination First

Rather than solving the entire set, scan the questions and answer options to test feasibility. Many CAT DILR questions are designed so that 2–3 options violate at least one constraint immediately. By plugging options back into key conditions, you can reject them without full calculations. This approach is especially useful in high-complexity sets where complete case construction is time-consuming.

Anchor Variable Fixing

An anchor variable is the element with the maximum number of constraints attached to it (for example, a person with multiple conditions or a time slot with fixed boundaries). Fixing this variable first drastically reduces permutations. Once anchored, other variables adjust naturally around it. This shortcut transforms multi-case problems into one or two manageable scenarios, improving speed and logical clarity.

Ratio-to-Value Conversion

When ratios are given (e.g., A:B:C = 2:3:5), assume base values that satisfy the ratio instead of solving algebraically. Choose the smallest possible multiplier that keeps numbers integral and aligned with constraints. This simplifies calculations and allows faster comparison across cases. Final answers usually remain unaffected because CAT questions test relative values, not absolute magnitudes.

Extreme Case Testing

Extreme case testing means checking minimum and maximum possible values allowed by constraints. If a case fails at an extreme, all derived cases automatically fail. This helps quickly eliminate entire branches of reasoning. It is particularly powerful in selection, distribution, and optimization-based DILR sets where limits are clearly defined.

Data Compression

Lengthy data should be reduced into symbols, abbreviations, or coded variables (for example, P1–P6 for persons, T1–T5 for time slots). This reduces reading load and speeds up logical deductions. Efficient data compression allows you to focus on relationships rather than text, which is crucial for maintaining accuracy under CAT’s strict time pressure.

Does Practising CAT 2026 DILR Practice Questions Maximise Score?

Yes, regular practice of CAT 2026 DILR questions plays a major role in maximising your overall score. The DILR section checks how well you can handle complex data and think logically under time pressure. The more you practise, the faster you recognise question patterns, improve decision-making, and reduce mistakes, key skills needed for a high percentile in CAT 2026.

Importance of Regular DILR Practice

Consistent practice helps you understand different question formats like arrangements, puzzles, and data interpretation sets. The CAT DILR section changes in difficulty each year, so regular exposure to diverse problems prepares you for surprises. Practising every day builds familiarity, helping you approach each set with clarity and confidence during the exam.

Strengthening Logical and Analytical Thinking

Practising DILR questions for CAT 2026 improves your logical reasoning and analytical skills. When you solve new problem types daily, your brain learns to connect patterns faster. Over time, you begin to spot shortcuts, eliminate wrong approaches, and build structured thinking—all of which directly boost accuracy and speed in the exam.

Improving Time Management

The DILR section is often time-consuming, so practising under timed conditions teaches you how to manage 40 minutes wisely. You learn to identify easy sets first and skip tricky ones. With regular timed practice, you develop the skill to attempt 2–3 correct sets accurately, enough to reach a 95–99 percentile range.

Reducing Errors Through Practice

Many students lose marks due to small calculation mistakes or missing key data points. Frequent practice helps reduce such errors by training you to read data carefully and double-check steps. Analysing every mock test also helps you find weak areas, refine your approach, and prevent repeated mistakes in the CAT 2026 DILR section.

CAT 2026 DILR Practice Strategy for High Percentile

The DILR section in CAT 2026 tests your logical thinking, pattern recognition, and data interpretation skills. To score a high percentile, you must practise the right sets, improve speed, and develop accuracy through regular mock analysis. A clear strategy helps you select doable sets, manage time, and perform consistently across varying difficulty levels.

Understanding the DILR Section Pattern

The CAT 2026 DILR section usually contains 20 questions divided into four sets. Each set has 4–6 questions based on data charts, graphs, puzzles, or logical arrangements. The difficulty level varies, so your goal is to identify two or three doable sets quickly. Understanding question types and practising across different patterns ensures better performance during the exam.

Building Strong Fundamentals

Before attempting complex DILR sets, strengthen your basics. Learn how to read tables, pie charts, and bar graphs efficiently. Practise topics like arrangements, team selection, and binary logic daily. Focus on understanding relationships between data points instead of memorising steps. A strong foundation helps you solve even unfamiliar question types confidently in the CAT 2026 exam.

Choosing the Right Sets During Practice

Set selection decides your DILR score. While practising CAT mock tests, learn to identify easy-to-moderate sets within the first few minutes. Skip sets that seem lengthy or confusing and move to ones where you can visualise a clear path. Developing this habit through regular timed practice saves crucial minutes during the real test.

Improving Speed and Accuracy

Speed in DILR comes from smart practice, not guesswork. Use a timer while solving every set and gradually reduce the time limit. Focus on accuracy by avoiding random assumptions. Once a set is solved, review your approach, check if a quicker method exists. Balancing speed and precision helps you handle the pressure of the CAT 2026 DILR section effectively.

Analysing Mock Tests and Tracking Progress

After every CAT 2026 mock test, spend time reviewing all DILR sets, both attempted and skipped. Identify why some sets took longer or led to errors. Keep a log of question types that trouble you and revise them regularly. Consistent analysis helps you track improvement, build confidence, and steadily move toward a high percentile in CAT 2026.

Best Books for CAT DILR Preparation

The candidates can refer to the following best CAT 2026 books for the DILR section to enhance their CAT DILR preparation. These books will provide the candidates with even more CAT 2026 DILR practice questions to enhance their CAT DILR preparation.

CAT DILR Preparation Materials by Careers360

Candidates can access a wide range of CAT 2026 DILR practice questions and preparation materials, carefully curated by Careers360 subject experts. These resources are designed to help aspirants strengthen their logical reasoning and data interpretation skills, improve accuracy, and boost speed.