# CAT 2023 LRDI Slot 1 | Previous Year CAT Paper

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

A visa processing office (VPO) accepts visa applications in four categories – US, UK, Schengen, and Others. The applications are scheduled for processing in twenty 15-minute slots starting at 9:00 am and ending at 2:00 pm. Ten applications are scheduled in each slot.

There are ten counters in the office, four dedicated to US applications, and two each for UK applications, Schengen applications and Others applications. Applicants are called in for processing sequentially on a first-come-first-served basis whenever a counter gets freed for their category. The processing time for an application is the same within each category. But it may vary across the categories. Each US and UK application requires 10 minutes of processing time. Depending on the number of applications in a category and time required to process an application for that category, it is possible that an applicant for a slot may be processed later.

On a particular day, Ira, Vijay and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot but entered the VPO at 9:20 am. When they entered the office, exactly six out of the ten counters were either processing applications, or had finished processing one and ready to start processing the next.

Mahira and Osman were scheduled in the 9:30 am slot on that day for visa processing in the Others category.

The following additional information is known about that day.

1. All slots were full.

2. The number of US applications was the same in all the slots. The same was true for the other three categories.

3. 50% of the applications were US applications.

4. All applicants except Ira, Vijay and Nandini arrived on time.

5. Vijay was called to a counter at 9:25 am.

**1. CAT 2023 LRDI Slot 1 | LR - Puzzles**

How many UK applications were scheduled on that day?

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**2. CAT 2023 LRDI Slot 1 | LR - Puzzles**

What is the maximum possible value of the total time (in minutes, nearest to its integer value) required to process all applications in the Others category on that day?

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**3. CAT 2023 LRDI Slot 1 | LR - Puzzles**

Which of the following is the closest to the time when Nandini’s application process got over?

- A.
9:35 am

- B.
9:37 am

- C.
9:45 am

- D.
9:50 am

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**4. CAT 2023 LRDI Slot 1 | LR - Puzzles**

Which of the following statements is false?

- A.
The application process of Mahira started after Nandini’s.

- B.
The application process of Osman was completed before Vijay’s.

- C.
The application process of Osman was completed before 9:45 am.

- D.
The application process of Mahira was completed before Nandini’s.

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**5. CAT 2023 LRDI Slot 1 | LR - Puzzles**

When did the application processing for all US applicants get over on that day?

- A.
3:40 pm

- B.
2:05 pm

- C.
2:25 pm

- D.
2:00 pm

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

Faculty members in a management school can belong to one of four departments – Finance and Accounting (F&A), Marketing and Strategy (M&S), Operations and Quants (O&Q) and Behaviour and Human Resources (B&H). The numbers of faculty members in F&A, M&S, O&Q and B&H departments are 9, 7, 5 and 3 respectively.

Prof. Pakrasi, Prof. Qureshi, Prof. Ramaswamy and Prof. Samuel are four members of the school's faculty who were candidates for the post of the Dean of the school. Only one of the candidates was from O&Q.

Every faculty member, including the four candidates, voted for the post. In each department, all the faculty members who were not candidates voted for the same candidate. The rules for the election are listed below.

1. There cannot be more than two candidates from a single department.

2. A candidate cannot vote for himself/herself.

3. Faculty members cannot vote for a candidate from their own department.

After the election, it was observed that Prof. Pakrasi received 3 votes, Prof. Qureshi received 14 votes, Prof. Ramaswamy received 6 votes and Prof. Samuel received 1 vote. Prof. Pakrasi voted for Prof. Ramaswamy, Prof. Qureshi for Prof. Samuel, Prof. Ramaswamy for Prof. Qureshi and Prof. Samuel for Prof. Pakrasi.

**6. CAT 2023 LRDI Slot 1 | LR - Puzzles**

Which two candidates can belong to the same department?

- A.
Prof. Pakrasi and Prof. Qureshi

- B.
Prof. Qureshi and Prof. Ramaswamy

- C.
Prof. Pakrasi and Prof. Samuel

- D.
Prof. Ramaswamy and Prof. Samuel

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**7. CAT 2023 LRDI Slot 1 | LR - Puzzles**

Which of the following can be the number of votes that Prof. Qureshi received from a single department?

- A.
7

- B.
9

- C.
6

- D.
8

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**8. CAT 2023 LRDI Slot 1 | LR - Puzzles**

If Prof. Samuel belongs to B&H, which of the following statements is/are true?

Statement A: Prof. Pakrasi belongs to M&S.

Statement B: Prof. Ramaswamy belongs to O&Q.

- A.
Only statement A

- B.
Only statement B

- C.
Neither statement A not statement B

- D.
Both statements A and B

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**9. CAT 2023 LRDI Slot 1 | LR - Puzzles**

What best can be concluded about the candidate from O&Q?

- A.
It was Prof. Samuel.

- B.
It was either Prof. Ramaswamy or Prof. Samuel.

- C.
It was Prof. Ramaswamy.

- D.
It was either Prof. Pakrasi or Prof. Qureshi.

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**10. CAT 2023 LRDI Slot 1 | LR - Puzzles**

Which of the following statements is/are true?

Statement A: Non-candidates from M&S voted for Prof. Qureshi.

Statement B: Non-candidates from F&A voted for Prof. Qureshi.

- A.
Neither statement A nor statement B

- B.
Only statement A

- C.
Only statement B

- D.
Both statements A and B

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

The schematic diagram below shows 12 rectangular houses in a housing complex. House numbers are mentioned in the rectangles representing the houses. The houses are located in six columns – Column-A through Column-F, and two rows – Row-1 and Row-2. The houses are divided into two blocks - Block XX and Block YY. The diagram also shows two roads, one passing in front of the houses in Row-2 and another between the two blocks.

Some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.

The road adjacency value of a house is the number of its sides adjacent to a road. For example, the road adjacency values of C2, F2, and B1 are 2, 1, and 0, respectively. The neighbour count of a house is the number of sides of that house adjacent to occupied houses in the same block. For example, E1 and C1 can have the maximum possible neighbour counts of 3 and 2, respectively.

The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbour count).

The following information is also known.

1. The maximum quoted price of a house in Block XX is Rs. 24 lakhs. The minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column-E.

2. Row-1 has two occupied houses, one in each block.

3. Both houses in Column-E are vacant. Each of Column-D and Column-F has at least one occupied house.

4. There is only one house with parking space in Block YY.

**11. CAT 2023 LRDI Slot 1 | LR - Arrangements**

How many houses are vacant in Block XX?

Answer: 3

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**Explanation** :

**[Note**: There might be discrepancy in one of the questions of this set. We will update the solution once the objections window closes and answers are updated by IIM Lucknow.]

Price of a house = (base price) + 5 × (road adjacency value) + 3 × (neighbor count)

**Block XX:**

Maximum price of a house in XX is 24

∴ 24 = (10 or 12) + 5 × (0 or 1 or 2) + 3 × (0 or 1 or 2 or 3)

This is only possible when:

base price = 10 lakhs, road adjacency = 1 and neighbor count = 3.

⇒ Only house B2 satisfies the given criteria i.e., road adjacency = 1 and neighbor count = 3.

∴ B2 is vacant and its price is 24 lakhs.

⇒ A2, B1 and C2 are occupied houses.

There is only one occupied house in Row 1 and block XX.

Since B1 is occupied, ⇒ A1 and C1 are vacant.

**Block YY:**

Both E1 and E2 are vacant and one of them costs 15 lakhs.

Let us focus on E1

For E1 neighbor count = 1 (exactly one of D1 or F1 is occupied)

For E1 road adjacency = 0

∴ Cost of E1 = (10 or 12) + 5 × 0 + 3 × 1 = 13 or 15 lakhs

Since 15 lakhs is the least cost of a house in block YY, E1 must cost 15 lakhs.

⇒ E1 is the only house in YY which has a parking space.

Given, Row-1 has two occupied houses, one in each block.

∴ Exactly one of D1 or F1 is occupied.

If F1 is vacant, let’s calculate its price.

It should not have parking space as only 1 house has parking space in block YY. Hence, it’s base price is 10 lakhs. Now even if F2 is occupied, F1’s price will be 10 + 0 + 3 = 13 lakhs.

This is not possible as least price for a house in YY is 15 lakhs.

∴ F1 should be occupied.

⇒ D1 is vacant.

Also, at least one house in column D is occupied, hence D2 must be occupied.

Now, F2 may be vacant or occupied, both cases are possible.

∴ 3 houses are vacant in block XX.

Hence, 3.

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**12. CAT 2023 LRDI Slot 1 | LR - Arrangements**

Which of the following houses is definitely occupied?

- A.
D2

- B.
F2

- C.
B1

- D.
A1

Answer: Option C

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**Explanation** :

Consider the solution to first question of this set.

B1 and D2 are definitely occupied.

[There might be discrepancy in this question. We will update the solution once the objections window closes and answers are updated by IIM Lucknow]

Hence, option (c).

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**13. CAT 2023 LRDI Slot 1 | LR - Arrangements**

Which of the following options best describes the number of vacant houses in Row-2?

- A.
Exactly 3

- B.
Exactly 2

- C.
Either 3 or 4

- D.
Either 2 or 3

Answer: Option D

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**Explanation** :

Consider the solution to first question of this set.

In row 2, B2 and E2 are definitely vacant.

Out of D2 and F2, at least one is occupied, hence either 1 is vacant or 0 are vacant.

**Case 1**: One of D2 or F2 is vacant.

∴ We have B2, F2 and (D2 or F2), i.e., 3 houses vacant in row 2.

**Case 2**: None of D2 or F2 is vacant.

∴ We have B2 and F2, i.e., 2 houses vacant in row 2.

∴ In row 2, either only 2 houses are vacant or 3 houses are vacant.

Hence, option (d).

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**14. CAT 2023 LRDI Slot 1 | LR - Arrangements**

What is the maximum possible quoted price (in lakhs of Rs.) for a vacant house in Column-E?

Answer: 21

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**Explanation** :

Consider the solution to first question of this set.

E1 costs 15 laks.

Let us calculate the maximum possible price of E2.

E2's base price is 10 lakh as it cannot have a parking space. (Only 1 house in YY i.e., E1 has a parking space.)

Road adjancecy for E2 = 1 and maximum neighbor count of E2 will be 2 (Both D2 and F2 are occupied and E1 is vacant)

∴ E2's price = 10 + 5 × 1 + 3 × 2 = 21 lakhs

Hence, 21.

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**15. CAT 2023 LRDI Slot 1 | LR - Arrangements**

Which house in Block YY has parking space?

- A.
F2

- B.
E2

- C.
E1

- D.
F1

Answer: Option C

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**Explanation** :

Consider the solution to first question of this set.

Only 1 house in YY i.e., E1 has a parking space.

Hence, option (c).

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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.

**16. CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning**

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

Answer: 0

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**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.

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**17. CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning**

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

Answer: 0

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**Explanation** :

Consider the solution to first question of this set.

R2 gave a rating of 4 to no one.

Hence, 0.

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**18. CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning**

What rating did R1 give to Xavier?

Answer: 3

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**Explanation** :

Consider the solution to first question of this set.

R1 gave a rating of 3 to Xavier.

Hence, 3.

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**19. CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning**

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

Answer: 4

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**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.

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**20. CAT 2023 LRDI Slot 1 | LR - Mathematical Reasoning**

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

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**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).

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