# LR - Mathematical Reasoning - Previous Year CAT/MBA Questions

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Here we bring you all previous year LR - Mathematical Reasoning CAT questions along with detailed solutions.

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**Answer the following questions based on the information given below:**

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.

* 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.

**CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning CAT Question**

How many individual ratings cannot be determined from the above information?

Answer: 0

**Text Explanation** :

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 sum of rating given by R1, R2, R3, R4 and R5 were 17, 11, 19, 14 and 17 respectively. (Multiplied with 5)

Similarly, we can find the sum of rating of all the 5 people.

Filling the data given in point 1 and 2, we get the following table:

**Range of ratings:**

Ullas: lowest is given as 1, hence his highest = 1 + 3 = 4.

Vasu: highest is given as 5, hence his lowest = 5 – 3 = 2.

Waman: highest is 5 and lowest is 1.

Xavier: highest is given as 5, hence his lowest = 5 – 4 = 1.

Yusuf: lowest is given as 1, hence his highest = 1 + 3 = 4.

**Ullas**: Highest = 4, Lowest = 1, Median = 2 and Mode = 2

∴ Since median is 2, his reading in ascending order can be

Sum of reading of Ullas = 11, and his mode is 2

This is possible only when his readings in ascending order must be 1, 2, 2, 2, 4.

**Vasu**: Highest = 5, Lowest = 2, Median = 4 and Mode = 4.

∴ Since median is 4, his reading in ascending order can be

Sum of reading of Vasu = 19, and his mode is 2

This is possible when his readings in ascending order must be 2, 4, 4, 4, 5.

**Waman**: Highest = 5, Lowest = 1, Median = 4 and Mode = 5.

∴ Since median is 4, his reading in ascending order can be

Sum of reading of Waman = 17, and his mode is 5

This is possible when his readings in ascending order must be 1, 2, 4, 5, 5.

**Xavier**: Highest = 5, Lowest = 1, Median = 4 and Mode = 5.

∴ Since median is 4, his reading in ascending order can be

Sum of reading of Xavier = 18, and his mode is 5

This is possible when his readings in ascending order must be 1, 3, 4, 5, 5.

**Yusuf**: Highest = 4, Lowest = 1, Median = 3 and Mode = 1 and 4.

∴ Since median is 4, his reading in ascending order can be

Sum of reading of Yusuf = 17, and his mode is 1 and 4

This is possible when his readings in ascending order must be 1, 1, 3, 4, 4.

Now, we have the following table.

For R3, sum of ratings is 19, hence R3’s sum of rating for Ullas and Vasu = 19 – 5 – 5 – 1 = 8.

This is possible only when R3 gives a rating of 4 to both Ullas and Vasu.

For Ullas, the remaining 3 2’s would be given by R2, R4 and R5.

For R2, rating given to Vasu = 11 – 2 – 1 – 5 – 1 = 2.

∴ For Vasu the remaining 2 4s would be given by R1 and R4.

Xavier receives a rating of 1. This can only be given by R4. If R1 or R5 give a rating of 1 to Xavier, their sum of 17 each could not be achieved.

Now, the sum of ratings given by R4 to Waman and Yusuf = 14 – 2 – 4 – 1 = 7.

This is only possible when Waman gets a rating of 4 from R4 and Yusuf gets a rating of 3 from R4.

∴ The remaining rating of 2 for Waman must have come from R5.

∴ The remaining 2 4s for Yusuf must have come from R1 and R5.

For R1, rating given to Xavier = 17 – 1 – 4 – 5 – 4 = 3

⇒ The remaining rating of 4 for Xavier must have been given by R5.

Ratings of all can be uniquely determined.

Hence, 0.

Workspace:

**CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning CAT Question**

To how many workers did R2 give a rating of 4?

Answer: 0

**Text Explanation** :

Consider the solution to first question of this set.

R2 gave a rating of 4 to no one.

Hence, 0.

Workspace:

**CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning CAT Question**

What rating did R1 give to Xavier?

Answer: 3

**Text Explanation** :

Consider the solution to first question of this set.

R1 gave a rating of 3 to Xavier.

Hence, 3.

Workspace:

**CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning CAT Question**

What is the median of the ratings given by R3 to the five workers?

Answer: 4

**Text Explanation** :

Consider the solution to first question of this set.

R3 gave ratings of 1, 4, 4, 5, 5.

∴ Median rating is 4.

Hence, 4.

Workspace:

**CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning CAT Question**

Which among the following restaurants gave its median rating to exactly one of the workers?

- (a)
R5

- (b)
R4

- (c)
R2

- (d)
R3

Answer: Option B

**Text Explanation** :

Consider the solution to first question of this set.

Median rating for:

R1 is 4, given to 2 persons.

R2 is 2, given to 2 persons.

R3 is 4, given to 2 persons.

R4 is 3, given to only 1 person.

R5 is 4, given to 2 persons.

∴ R4 gave median rating to only 1 person.

Hence, option (b).

Workspace:

**Answer the following questions based on the information given below:**

Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later.

Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.

The table below provides partial information about the five firms.

**CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning CAT Question**

For which firm(s) can the amounts raised by them be concluded with certainty in each year?

- (a)
Only Czechy

- (b)
Only Bzygoo and Czechy and Drjbna

- (c)
Only Drjbna

- (d)
Only Czechy and Drjbna

Answer: Option D

**Text Explanation** :

We can represent the money raised each year for each company as follows.

For each or these companies we will try to make cases satifying the given conditions.

We will write the amount of money raised by them in consecutive years from starting from their 1st year till their last year.

**For A**, the sum is 21 where first and last values must be 1.

This is possible for the following order of numbers.

Case 1: 1 → 2 → 3 → 4 → 5 → 3 → 2 → 1

Case 2: 1 → 2 → 3 → 5 → 4 → 3 → 2 → 1

**For B**, the only two possible order of numbers is

Case 1: 1 → 2 → 3 → 1

Case 2: 1 → 3 → 2 → 1

**For C**, the sum is 9 where first and last values must be 1.

This is only possible when the order of numbers is

1 → 2 → 3 → 2 → 1

**For D**, the sum is 10 where first and last values must be 1.

This is only possible when the order of numbers is

1 → 2 → 4 → 2 → 1

**For E**, the sum is 13 where first and last values must be 1.

This is possible for the following order of numbers.

Case 1: 1 → 2 → 4 → 3 → 2 → 1

Case 2: 1 → 2 → 3 → 4 → 2 → 1

Case 3: 1 → 3 → 5 → 3 → 1

Hence, we get the following table with all possible cases.

For C and D we can definitely determine the amounts raised by them each year.

Hence, option (d).

Workspace:

**CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning CAT Question**

What best can be concluded about the total amount of money raised in 2015?

- (a)
It is either Rs. 7 crores or Rs. 8 crores or Rs. 9 crores.

- (b)
It is either Rs. 7 crores or Rs. 8 crores.

- (c)
It is exactly Rs. 8 crores.

- (d)
It is either Rs. 8 crores or Rs. 9 crores.

Answer: Option B

**Text Explanation** :

Consider the solution for the first questions of this set.

Amount of money raised in 2015 by:

A = 2

B = 1

C = 3

D = 1

E = 1 or 0

∴ Total money raised by them in 2015 = 2 + 1 + 3 + 1 + (1 or 0) = 8 or 7 crores.

Hence, option (b).

Workspace:

**CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning CAT Question**

What is the largest possible total amount of money (in Rs. crores) that could have been raised in 2013?

Answer: 17

**Text Explanation** :

Consider the solution for the first questions of this set.

Highest amoun of money that can be raised in 2013 by:

A = 5

B = 3

C = 1

D = 4

E = 4

∴ Highest total amount of money that can be raised by them in 2013 = 5 + 3 + 1 + 4 + 4 = 17 crores.

Hence, 17.

Workspace:

**CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning CAT Question**

If Elavalaki raised Rs. 3 crores in 2013, then what is the smallest possible total amount of money (in Rs. crores) that could have been raised by all the companies in 2012?

- (a)
10

- (b)
9

- (c)
12

- (d)
11

Answer: Option D

**Text Explanation** :

Consider the solution for the first questions of this set.

Smallest amount of money that can be raised in 2012 by:

A = 4

B = 1

C = 1

D = 2

E = 3

∴ Smallest total amount of money that can be raised by them in 2012 = 4 + 1 + 1 + 2 + 3 = 11 crores.

Hence, option (d).

Workspace:

**CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning CAT Question**

If the total amount of money raised in 2014 is Rs. 12 crores, then which of the following is not possible?

- (a)
Alfloo raised the same amount of money as Drjbna in 2013.

- (b)
Bzygoo raised more money than Elavalaki in 2014.

- (c)
Bzygoo raised the same amount of money as Elavalaki in 2013.

- (d)
Alfloo raised the same amount of money as Bzygoo in 2014.

Answer: Option C

**Text Explanation** :

Consider the solution for the first question of this set.

Amount of money raised in 2014 by

A = 3

B = 2 or 3

C = 2

D = 2

E = 1 or 2

If the total amount of money raised by them in 2014 is 12 crores this is possible only when B raised 3 crores i.e., case 1 for B and E raises 2 crores i.e., case 1 or 2 for E.

We have the following cases left.

Option (a): A and D can both raise 4 crores in 2013, hence option (a) is possible.

Option (b): B raises 3 crores and E raises 2 crores in 2014, hence option (b) is possible.

Option (c): B cannot raise the same amount as E in 2013, hence option (c) is not possible.

Option (d): A and B can both raise 3 crores in 2014, hence option (d) is possible.

Hence, option (c).

Workspace:

**Answer the following questions based on the information given below:**

Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants’ scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for

Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.

The following information is also known.

- Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.
- The total score on Day 3 is the same as the total score on Day 4.
- Bimal’s scores are the same on Day 1 and Day 3.

**CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning CAT Question**

What is Akhil's score on Day 1?

- (a)
7

- (b)
6

- (c)
5

- (d)
8

Answer: Option A

**Text Explanation** :

We get the following table for their score on each of the 5 days.

Akhil's score on Day 1 is 7.

Hence, option (a).

Workspace:

**CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning CAT Question**

Who attains the maximum total score?

- (a)
Bimal

- (b)
Chatur

- (c)
Cannot be determined

- (d)
Akhil

Answer: Option B

**Text Explanation** :

We get the following table for their score on each of the 5 days.

Chatur gets the maximum total score of 30.

Hence, option (b).

Workspace:

**CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning CAT Question**

What is the minimum possible total score of Bimal?

Answer: 25

**Text Explanation** :

We get the following table for their score on each of the 5 days.

Minimum total score of Bimal is 25.

Hence, 25.

Workspace:

**CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning CAT Question**

If Akhil attains a total score of 24, then what is the total score of Bimal?

Answer: 26

**Text Explanation** :

We get the following table for their score on each of the 5 days.

If Akil gets a total score of 24, Bimal will get a total score of 26.

Hence, 26.

Workspace:

**Answer the following questions based on the information given below:**

In a coaching class, some students register online, and some others register offline. No student registers both online and offline; hence the total registration number is the sum of online and offline registrations. The following facts and table pertain to these registration numbers for the five months – January to May of 2023. The table shows the minimum, maximum, median registration numbers of these five months, separately for online, offline and total number of registrations. The following additional facts are known.

1. In every month, both online and offline registration numbers were multiples of 10.

2. In January, the number of offline registrations was twice that of online registrations.

3. In April, the number of online registrations was twice that of offline registrations.

4. The number of online registrations in March was the same as the number of offline registrations in February.

5. The number of online registrations was the largest in May.

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

What was the total number of registrations in April?

Answer: 120

**Text Explanation** :

We get the following table based on the conditions given.

Total registrations in April is 120.

Hence, 120.

Workspace:

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

What was the number of online registrations in January?

Answer: 40

**Text Explanation** :

We get the following table based on the conditions given.

Number of online registrations in January is 40.

Hence, 40.

Workspace:

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

Which of the following statements can be true?

I. The number of offline registrations was the smallest in May.

II. The total number of registrations was the smallest in February.

- (a)
Only II

- (b)
Only I

- (c)
Both I and II

- (d)
Neither I nor II

Answer: Option B

**Text Explanation** :

We get the following table based on the conditions given.

Offline registrations is smallest in May. Hence, I is correct.

Total registrations was highest in Feb. Hence, II is incorrect.

Hence, option (b).

Workspace:

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

What best can be concluded about the number of offline registrations in February?

- (a)
50

- (b)
30 or 50 or 80

- (c)
50 or 80

- (d)
80

Answer: Option A

**Text Explanation** :

We get the following table based on the conditions given.

Number of offline registrations in February is 50.

Hence, option (a).

Workspace:

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

Which pair of months definitely had the same total number of registrations?

I. January and April

II. February and May

- (a)
Neither I nor II

- (b)
Only II

- (c)
Only I

- (d)
Both I and II

Answer: Option D

**Text Explanation** :

We get the following table based on the conditions given.

Jan and April both have 120 total registrations.

Feb and May both have 130 total registrations.

Hence, optiono (d).

Workspace:

**Answer the following questions based on the information given below:**

There are only three female students – Amala, Koli and Rini – and only three male students – Biman, Mathew and Shyamal – in a course. The course has two evaluation components, a project and a test. The aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1.

The projects are done in groups of two, with each group consisting of a female and a male student. Both the group members obtain the same score in the project.

The following additional facts are known about the scores in the project and the test.

1. The minimum, maximum and the average of both project and test scores were identical – 40, 80 and 60, respectively.

2. The test scores of the students were all multiples of 10; four of them were distinct and the remaining two were equal to the average test scores.

3. Amala’s score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Yet Amala had the highest aggregate score.

4. Shyamal scored the second highest in the test. He scored two more than Koli, but two less than Amala in the aggregate.

5. Biman scored the second lowest in the test and the lowest in the aggregate.

6. Mathew scored more than Rini in the project, but less than her in the test.

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

What was Rini’s score in the project?

Answer: 60

**Text Explanation** :

We get the following table from conditions given in the question.

Rani's score in project is 60.

Hence, 60.

Workspace:

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

What was the weight of the test component?

- (a)
0.75

- (b)
0.60

- (c)
0.40

- (d)
0.50

Answer: Option B

**Text Explanation** :

We get the following table from conditions given in the question.

Weight of test componenet is 0.6.

Hence, option (b).

Workspace:

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

What was the maximum aggregate score obtained by the students?

- (a)
66

- (b)
80

- (c)
62

- (d)
68

Answer: Option D

**Text Explanation** :

We get the following table from conditions given in the question.

Aggregate score was maximum for Amal.

Amala's maximum score = 80 × 0.4 + 60 × 0.6 = 32 + 36 = 68.

Hence, option (d).

Workspace:

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

What was Mathew’s score in the test?

Answer: 40

**Text Explanation** :

We get the following table from conditions given in the question.

Mathew's score in test is 40.

Hence, 40.

Workspace:

**CAT 2023 LRDI Slot 3 | LR - Mathematical Reasoning CAT Question**

Which of the following pairs of students were part of the same project team?

i) Amala and Biman

ii) Koli and Mathew

- (a)
Neither i) nor ii)

- (b)
Only ii)

- (c)
Only i)

- (d)
Both i) and ii)

Answer: Option A

**Text Explanation** :

We get the following table from conditions given in the question.

A and B are not in same team.

K and M are not in same team.

Neither i) nor ii) is correct.

Hence, option (a).

Workspace:

**Answer the next 5 questions based on the information given below:**

Adhara, Bithi, Chhaya, Dhanavi, Esther, and Fathima are the interviewers in a process that awards funding for new initiatives. Every interviewer individually interviews each of the candidates individually and awards a token only if she recommends funding. A token has a face value of 2, 3, 5, 7, 11, or 13. Each interviewer awards tokens of a single face value only.

Once all six interviews are over for a candidate, the candidate receives a funding that is Rs.1000 times the product of the face values of all the tokens. For example, if a candidate has tokens with face values 2, 5, and 7, then they get a funding of Rs.1000 × (2 × 5 × 7) = Rs.70,000.

Pragnyaa, Qahira, Rasheeda, Smera, and Tantra were five candidates who received funding. The funds they received, in descending order, were Rs.390,000, Rs.210,000, Rs.165,000, Rs.77,000, and Rs.66,000.

The following additional facts are known:

1. Fathima awarded tokens to everyone except Qahira, while Adhara awarded tokens to no one except Pragnyaa.

2. Rashida received the highest number of tokens that anyone received, but she did not receive one from Esther.

3. Bithi awarded a token to Smera but not to Qahira, while Dhanavi awarded a token to Qahira but not to Smera.

**CAT 2022 LRDI Slot 1 | LR - Mathematical Reasoning CAT Question**

How many tokens did Qahira receive?

Answer: 2

**Text Explanation** :

The funds they received, in descending order, were Rs.390,000, Rs.210,000, Rs.165,000, Rs.77,000, and Rs.66,000.

∴ The product of tokes = 390, 210, 165, 77, 66

⇒ Break of tokens received

390 = 2 × 3 × 5 × 13

210 = 2 × 3 × 5 × 7

165 = 3 × 5 × 11

77 = 7 × 11

66 = 2 × 3 × 11

From (1): Fathima awarded tokens to everyone except Qahira,

∴ Fatima must have awarded token number 3 to everyone and Qahira received tokens whose product is 77.

From (1): Adhara awarded tokens to no one except Pragnyaa

∴ Adhara must have awared token 13 and Pragnyaa received tokens whose product is 390.

From (2): Rashida received the highest number of tokens that anyone received, but she did not receive one from Esther.

∴ Rashida must have received 4 tokes whose product can only be 210. Also, since Esther did not give her the token, hence Esther must have distributed token number 11.

From (3): Dhanavi awarded a token to Qahira but not to Smera.

∴ Dhanavi must have awared token number 7.

From (3): Bithi awarded a token to Smera but not to Qahira

∴ Bithi/Chhaya could have awared token 2/5 in any order.

If Bithi gives token number 2 to S/T, then Chhaya will given token number 5 to T/S and vice-a-versa.

∴ Qahira received two tokens i.e., 7 and 11.

Hence, 2.

Workspace:

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