# CAT 2023 LRDI Slot 2 | Previous Year CAT Paper

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**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.

**1. CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning**

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

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**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:

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

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

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**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:

**3. CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning**

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

Answer: 17

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**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.

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**4. CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning**

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

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**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).

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**5. CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning**

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

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**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).

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

Anjali, Bipasha, and Chitra visited an entertainment park that has four rides. Each ride lasts one hour and can accommodate one visitor at one point. All rides begin at 9 am and must be completed by 5 pm except for Ride-3, for which the last ride has to be completed by 1 pm. Ride gates open every 30 minutes, e.g. 10 am, 10:30 am, and so on. Whenever a ride gate opens, and there is no visitor inside, the first visitor waiting in the queue buys the ticket just before taking the ride. The ticket prices are Rs. 20, Rs. 50, Rs. 30 and Rs. 40 for Rides 1 to 4, respectively. Each of the three visitors took at least one ride and did not necessarily take all rides. None of them took the same ride more than once. The movement time from one ride to another is negligible, and a visitor leaves the ride immediately after the completion of the ride. No one takes a break inside the park unless mentioned explicitly.

The following information is also known.

- Chitra never waited in the queue and completed her visit by 11 am after spending Rs. 50 to pay for the ticket(s).
- Anjali took Ride-1 at 11 am after waiting for 30 mins for Chitra to complete it. It was the only ride where Anjali waited.
- Bipasha began her first of three rides at 11:30 am. All three visitors incurred the same amount of ticket expense by 12:15 pm.
- The last ride taken by Anjali and Bipasha was the same, where Bipasha waited 30 mins for Anjali to complete her ride. Before standing in the queue for that ride, Bipasha took a 1-hour coffee break after completing her previous ride.

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

What was the total amount spent on tickets (in Rs.) by Bipasha?

- A.
110

- B.
100

- C.
90

- D.
120

Answer: Option A

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**Explanation** :

We get the following chart for the rides taken by the three visitors.

Bipasha took rides 1, 2 and 4.

Her total cost for tickets = 20 + 50 + 40 = Rs. 110.

Hence, option (a).

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

Which were all the rides that Anjali completed by 2:00 pm?

- A.
Ride-1 and Ride-3

- B.
Ride-1, Ride-2, and Ride-4

- C.
Ride-1 and Ride-4

- D.
Ride-1, Ride-2, and Ride-3

Answer: Option D

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**Explanation** :

We get the following chart for the rides taken by the three visitors.

By 2 pm Anjali took rides 1, 2 and 3.

Hence, option (d).

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

Which ride was taken by all three visitors?

- A.
Ride-1

- B.
Ride-3

- C.
Ride-2

- D.
Ride-4

Answer: Option A

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**Explanation** :

We get the following chart for the rides taken by the three visitors.

Ride 1 was taken by all thee visitors.

Hence, option (a).

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

How many rides did Anjali and Chitra take in total?

Answer: 6

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**Explanation** :

We get the following chart for the rides taken by the three visitors.

Anajli took rides 1, 2, 3 and 4 i.e., 4 rides

Chitra took rides 1 and 3 i.e., 2 rides

Anjali and Chitra together took 4 + 2 = 6 rides.

Hence, 6.

Workspace:

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

What was the total amount spent on tickets (in Rs.) by Anjali?

Answer: 140

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**Explanation** :

We get the following chart for the rides visited by the three persons.

Anjali took all 4 rides hence spent 20 + 50 + 30 + 40 = Rs. 140.

Hence, 140.

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**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.

**11. CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning**

What is Akhil's score on Day 1?

- A.
7

- B.
6

- C.
5

- D.
8

Answer: Option A

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**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).

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**12. CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning**

Who attains the maximum total score?

- A.
Bimal

- B.
Chatur

- C.
Cannot be determined

- D.
Akhil

Answer: Option B

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**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:

**13. CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning**

What is the minimum possible total score of Bimal?

Answer: 25

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**Explanation** :

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

Minimum total score of Bimal is 25.

Hence, 25.

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

If the total score of Bimal is a multiple of 3, what is the score of Akhil on Day 2?

- A.
Cannot be determined

- B.
5

- C.
6

- D.
4

Answer: Option D

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**Explanation** :

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

If Bimal's total score is a multiple of 3, it means Bimals get a total score of 27 hence Akhil gets a total score of 23.

This is possible when Akhil get 4 on Day 2 and 4 on Day 5.

Hence, option (d).

Workspace:

**15. CAT 2023 LRDI Slot 2 | LR - Mathematical Reasoning**

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

Answer: 26

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**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.

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

There are nine boxes arranged in a 3×3 array as shown in Tables 1 and 2. Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.

The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.

Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied.

- The minimum among the numbers of coins in the three sacks in the box is 1.
- The median of the numbers of coins in the three sacks is 1.
- The maximum among the numbers of coins in the three sacks in the box is 9.

**16. CAT 2023 LRDI Slot 2 | LR - Arrangements**

What is the total number of coins in all the boxes in the 3^{rd} row?

- A.
36

- B.
30

- C.
15

- D.
45

Answer: Option D

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**Explanation** :

Each box has three sacks. We will consider three values for each sacks representing the number of coins in each sack in ascending order.

**Table 1**: Medians of C1R2, C1R3, C2R1 and C3R1 are given. If numbers of coins in each sack is written in ascending order for all boxes, these medians will be the middle value.

Median for C2R1 is 9, hence the third value must be 9.

**Table 2**:

The number given represents the number of sacks having more than 5 coins. We can represent this information as follows:

C2R1: The sum of coins so far is 9 + 9 = 18.

*Exactly one condition is satisfied i.e., highest number coins in a sack is 9.

The first sack should have coins less than 5 and the average should be an integer.

This is possible only when the first sack has 3 coins.

∴ Average number of coins/sack = (3 + 9 + 9)/3 = 7

C2R3: Median has to be greater than 5, hence condition 2 cannot be satisfied, hence conditions 1 and 3 have to be satisfied.

The first and third sacks have 1 and 9 coins respectively.

The second sack has more than 5 coins, such that the average is an integer. This is only possible when it has 8 coins.

∴ Average number of coins/sack = (1 + 8 + 9)/3 = 6

C2R2: If median is one, then first sack will also have 1 coin which would satisfy 2 of the given conditions, hence median cannot be 1.

Since all the values are less than or equal to 5, condition 3 cannot be satisfied.

Hence, the only condition that can be satisfied is the first condition that lowest number of coins is 1.

C1R2: **At least 2 conditions should be satisfied.

The median cannot be 1, hence the conditions to be satisfied are 1st and 3rd.

∴ Frist sack has 1 coin and third sack has 9 coins.

∴ Average number of coins/sack = (1 + 2 + 9)/3 = 4

C1R3: All sacks have more than 5 coins, hence condition 1 and 2 cannot be satisfied.

Hence, the only condition that can be satisfied is that the third sack has 9 coins.

Now, the two sacks have combined 8 + 9 = 17 coins.

The total should be a multiple of 3 (average is an integer), and the first sack has more than 5 coins. This is possible only when the third sack has 7 coins.

∴ Average number of coins/sack = (7 + 8 + 9)/3 = 8

C1R1: Since 2 conditions need to be satisfied and the average has to be an integer, this is possible when number of coins is 1, 1, 7 or 1, 5, 9

**Case 1**: C1R1 has 1, 5, 9 coins.

Total coins in C1 = 1 + 5 + 9 + 1 + 2 +9 + 7 + 8 + 9 = 51

Now C2 so far has = 3 + 9 + 9 + 1 + 1 + 8 + 9 = 40 coins

∴ C2 needs 51 – 40 = 11 more coins from two sacks in C2R2 block. This is not possible since these blocks have less than or equal to 5 coins each in them.

Hence, this case is rejected.

**Case 2**: C1R1 has 1, 1, 7 coins.

Total coins in C1 = 1 + 1 + 7 + 1 + 2 + 9 + 7 + 8 + 9 = 45

∴ Number of coins in the two sacks in C2R2 = 45 – 40 = 5 coins.

⇒ Number of coins in each column = 45.

C2R2: Since first condition is satisfied, no other condition should be satisfied. Also, the sum of coins in second and third sack is 5. This is possible when the number of coins in them is 2 and 3 respectively.

C3R3: All sacks have ≤ 5 coins hence condition 3 cannot be satisfied. Hence, the first two conditions should be satisfied.

⇒ The first and the second sacks have 1 coin each.

Since the average has to be an integer, the third sack can be 1 or 4 coins.

With 4 coins the average will become (1 + 1 + 4)/3 = 2 which cannot be the case since C2R1 already have average of 2 and all blocks have distinct averages.

∴ Third sack has 1 coin.

∴ Average number of coins/sack = (1 + 1 + 1)/3 = 1

The averages so far have been, 1, 2, 3, 4, 6, 7, and 8. So C3R1 and C3R2 will have average 6 or 9 in any order.

For average to be 9, all sacks must have exactly 9 coins. This is possible only for C3R2.

∴ Average for C3R2 is 9 and that for C3R1 is 5.

C3R1: will have a total of 5 × 3 = 15 coins.

Second sack has 6 coins, hence the other 2 sacks will have 15 – 6 = 9 coins.

Since exactly one condition has to be satisfied, this is possible when first sack has 1 coin and third sack has 8 coins.

Hence, we get the following final table.

Total coins in third row = 7 + 8 + 9 + 1 + 8 + 9 + 1 + 1 + 1 = 45

Hence, option (d)

Workspace:

**17. CAT 2023 LRDI Slot 2 | LR - Arrangements**

How many boxes have at least one sack containing 9 coins?

- A.
3

- B.
8

- C.
5

- D.
4

Answer: Option C

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**Explanation** :

Consider the solution to first questions of this set.

C1R2, C1R3, C2R1, C2R3 and C3R2 i.e., 5 blocks have at least one sack containing 9 coins.

Hence, option (c).

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**18. CAT 2023 LRDI Slot 2 | LR - Arrangements**

For how many boxes are the average and median of the numbers of coins contained in the three sacks in that box the same?

Answer: 4

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**Explanation** :

Consider the solution to first questions of this set.

Average and median is same for C1R3, C2R2, C3R2 and C3R3 i.e., 4 blocks.

Hence, 4.

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**19. CAT 2023 LRDI Slot 2 | LR - Arrangements**

How many sacks have exactly one coin?

Answer: 9

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**Explanation** :

Consider the solution to first questions of this set.

9 sacks have exactly 1 coin.

Hence, 9.

Workspace:

**20. CAT 2023 LRDI Slot 2 | LR - Arrangements**

In how many boxes do all three sacks contain different numbers of coins?

Answer: 5

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**Explanation** :

Consider the solution to first questions of this set.

C1R2, C1R3, C2R2, C2R3 and C3R1 i.e., 5 blocks all sacks containing different number of coins.

Hence, 5.

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