# CAT 2022 LRDI Slot 3 | Previous Year CAT Paper

Pulak, Qasim, Ritesh, and Suresh participated in a tournament comprising of eight rounds. In each round, they formed two pairs, with each of them being in exactly one pair. The only restriction in the pairing was that the pairs would change in successive rounds. For example, if Pulak formed a pair with Qasim in the first round, then he would have to form a pair with Ritesh or Suresh in the second round. He would be free to pair with Qasim again in the third round. In each round, each pair decided whether to play the game in that round or not. If they decided not to play, then no money was exchanged between them. If they decided to play, they had to bet either ₹1 or ₹2 in that round. For example, if they chose to bet ₹2, then the player winning the game got ₹2 from the one losing the game.

At the beginning of the tournament, the players had ₹10 each. The following table shows partial information about the amounts that the players had at the end of each of the eight rounds. It shows every time a player had ₹10 at the end of a round, as well as every time, at the end of a round, a player had either the minimum or the maximum amount that he would have had across the eight rounds. For example, Suresh had ₹10 at the end of Rounds 1, 3, and 8 and not after any of the other rounds. The maximum amount that he had at the end of any round was ₹13 (at the end of Round 5), and the minimum amount he had at the end of any round was ₹8 (at the end of Round 2). At the end of all other rounds, he must have had either ₹9, ₹11, or ₹12.

It was also known that Pulak and Qasim had the same amount of money with them at the end of Round 4.

**1. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

What BEST can be said about the amount of money that Ritesh had with him at the end of Round 8?

- A.
Exactly ₹5

- B.
₹5 or ₹6

- C.
₹4 or ₹5

- D.
Exactly ₹6

Answer: Option D

**Explanation** :

Whenever a player has Rs. 10, it is already mentioned in the table. Their can be no other entry of Rs. 10.

Also, Highest and Lowest amounts of a player are also already mentioned.

E.g.: Pulak’s highest and lowest amount is Rs. 13 and Rs. 10 respectively in rounds 2, 8, 1 and 5. . Pulak cannot have Rs. 13 or more and Rs. 10 or less in any other round. Hence, the only amount Pulak can have at the end of any round is Rs. 11 or Rs. 12.

Allowed entries for

Pulak: 11 or 12

Qasim: 9 or 11

Ritesh: 5 or 6 or 7 or 8 or 9

Suresh: 9 or 11 or 12

Total amount at the end of any round should be same as the total amount initially i.e., Rs. 40.

**Round 1:** Amount with Pulak = 40 – (8 + 10 + 10) = Rs. 12

Ritesh and Suresh did not lose/gain, hence they paired up and decided not to play.

while, Qasim lost Rs. 2 to Pulak.

**Round 2:** Amount with Ritesh = 40 – (13 + 10 + 8) = Rs. 9

Pulak gained Rs. 1 while Ritesh lost Rs. 1. Hence, Pulak and Ritesh paired and bet Rs. 1.

while, Suresh paired up with Qasim for Rs. 2.

**Round 5:** Amount with Ritesh = 40 – (10 + 10 + 13) = Rs. 7

**Round 7:** Amount with Pulak and Suresh together = 40 – 12 – 4 = Rs. 24

Amount with Suresh and Pulak in Round 7 should be less than 13.

Only way this is possible is that both Pulak and Suresh have Rs. 12.

**Round 8:** Suresh loses Rs. 2. The only one who could gain Rs. 2 is Ritesh. Hence, Suresh paired with Ritesh and Pulak with Qasim.

**Round 4:** Pulak and Qasim had same amount. This amount can only be Rs. 11.

Now Pulak and Qasim lose Rs. 1 each Round 5, hence Ritesh and Suresh must gain Rs. 1 in round 5.

⇒ Ritesh and Suresh had Rs. 1 less in round 4 than round 5.

**Round 3:** Since Suresh paired with Qasim in round 2, he will pair with either Pulak or Ritesh in round 3.

If Suresh pairs with Ritesh, Ritesh will have Rs. 7 in round 3. Hence, Pulak + Qasim will have Rs. 23. This is possible when Pulak has Rs. 12 and Qasim has Rs. 11.

But that would mean that only Qasim doesn’t bet in round 4 which is not possible.

∴ Suresh pairs with Pulak in round 3. Hence, Pulak will have Rs. 12 in round 3.

⇒ Qasim and Ritesh have together Rs. 18. This is possible when Qasim has Rs. and Ritesh has Rs. .

⇒ Pulak paired with Qasim in round 4 and Ritesh with Suresh.

**Round 6:** Highest amount with Pulak, Qasim and Suresh in round 6 can be 12 + 11 + 12 = 35.

In that case Ritesh will have Rs. 5.

Had the amount with Ritesh been Rs. 6 at the end of round 6. Other three will have a total of Rs. 34. In this case, we would not get proper arrangement for round 7.

Therefore, the final table is as follows:

∴ The amount of money that Ritesh had with him at the end of Round 8 is Rs. 6.

Hence, option (d).

Workspace:

**2. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

What BEST can be said about the amount of money that Pulak had with him at the end of Round 6?

- A.
₹11 or ₹12

- B.
₹12 or ₹13

- C.
Exactly ₹12

- D.
Exactly ₹11

Answer: Option C

**Explanation** :

Consider the solution to first questions of this set.

∴ The amount of money that Pulak had with him at the end of Round 6 is Rs. 12.

Hence, option (c).

Workspace:

**3. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

How much money (in ₹) did Ritesh have at the end of Round 4?

Answer: 6

**Explanation** :

Consider the solution to first questions of this set.

∴ The amount of money (in ₹) Ritesh has at the end of Round 4 is Rs. 6.

Hence, 6.

Workspace:

**4. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

How many games were played with a bet of ₹2?

Answer: 6

**Explanation** :

Consider the solution to first questions of this set.

Number of games with bet of Rs. 2 are:

P vs. Q in round 1

Q vs. S in round 2

P vs. S in round 3

R vs S in round 4

P vs. R in round 6

R vs S in round 8

∴ There are 6 games with bet of Rs. 2.

Hence, 6.

Workspace:

**5. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

Which of the following pairings was made in Round 5?

- A.
Pulak and Suresh

- B.
Pulak and Qasim

- C.
Qasim and Suresh

- D.
Pulak and Ritesh

Answer: Option A

**Explanation** :

Consider the solution to first questions of this set.

Pulak paired with Suresh while Qasim paired with Ritesh in round 5.

Hence, option (as).

Workspace:

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

There are only four neighbourhoods in a city - Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.

The following facts are also known:

1. There was at least one new case in every neighbourhood on Day 1.

2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.

3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2.

4. The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.

5. Kitmisto is the only place to have 3 new cases on Day 2.

6. The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.

**6. CAT 2022 LRDI Slot 3 | LR - Arrangements**

What BEST can be concluded about the total number of new cases in the city on Day 2?

- A.
Exactly 7

- B.
Either 7 or 8

- C.
Either 6 or 7

- D.
Exactly 8

Answer: Option D

**Explanation** :

We can make the following table:

1. There was at least one new case in every neighborhood on Day 1.

2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.

3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2.

5. Kitmisto is the only place to have 3 new cases on Day 2.

From (6): The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.

Total cases during the 5 days = 12 + 12 + 5 + 14 = 43

On day 5, neighborhoods L, T and K can have maximum 3 cases each, while P can have maximum 2 cases. Hence, maximum possible total cases on day 5 = 3 + 2 + 3 + 3 = 11.

⇒ Maximum possible cases on day 4 = 10.

∴ Maximum possible cases on day 4 + day 5 = 10 + 11 = 21.

Since on day 1 there will be at least 5 cases in the city and every day total cases increase, hence on day 2 total cases must be 6 or more.

Case 1: Total cases on Day 2 = 6

⇒ Total cases on Day 3 = 7

⇒ Total cases on Day 1 = 5

∴ Total cases on day 4 + day 5 = 43 – (5 + 6 + 7) = 25

This is not possible as maximum cases possible on day 4 + day 5 is 21.

Case 2: Total cases on Day 2 = 7

⇒ Total cases on Day 3 = 8

⇒ Total cases on Day 1 = 5 or 6

∴ Total cases on day 4 + day 5 = 43 – (5/6 + 7 + 8) = 23 or 22

This is not possible as maximum cases possible on day 4 + day 5 is 21.

Case 3: Total cases on Day 2 = 8

⇒ Total cases on Day 3 = 9

Since total cases on day 4 cannot be more than 10, hence total cases on day 4 = 10.

Since total cases on day 5 cannot be more than 11, hence total cases on day 5 = 11.

⇒ Total cases on Day 1 = 43 – (8 + 9 + 10 + 11) = 5

Only possibility for total cases of 11 on day 5 is: L – 3, T – 3, P – 2, K – 3.

Only possibility for total cases of 11 on day 5 is: L – 3, T – 3, P – 1, K – 3.

[Note: There is only 1 day when Pesmisto had 2 cases. On remaining days P will have less than 2 cases.]

Total cases in Kitmisto is 14. This is possible when there are 3 cases on 4 days each and 2 cases on the remaining 5th day.

Levmisto and Tyhrmisto have total 12 cases.

Total cases on day 2 + day 3 for these two neighborhoods = 12 – (1 + 3 + 3) = 5

5 cases in 2 days are possible when there are 2 and 3 cases in these 2 days.

On day 2 only Kitmisto had 3 cases, hence we can fill the table accordingly.

Now the remaining 2 slots can be filled.

∴ The total number of new cases in the city on Day 2 = 8

Hence, option (d).

Workspace:

**7. CAT 2022 LRDI Slot 3 | LR - Arrangements**

What BEST can be concluded about the number of new cases in Levmisto on Day 3?

- A.
Either 2 or 3

- B.
Exactly 3

- C.
Either 0 or 1

- D.
Exactly 2

Answer: Option B

**Explanation** :

Consider the solution to first question of this set.

∴ The number of new cases in Levmisto on Day 3 = 3.

Hence, option (b).

Workspace:

**8. CAT 2022 LRDI Slot 3 | LR - Arrangements**

On which day(s) did Pesmisto not have any new case?

- A.
Both Day 2 and Day 3

- B.
Only Day 2

- C.
Only Day 3

- D.
Both Day 2 and Day 4

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

∴ The number of new cases in Levmisto on Day 3 = 3.

Hence, option (b).

Workspace:

**9. CAT 2022 LRDI Slot 3 | LR - Arrangements**

Which of the two statements below is/are necessarily false?

Statement A: There were 2 new cases in Tyhrmisto on Day 3.

Statement B: There were no new cases in Pesmisto on Day 2.

- A.
Statement B only

- B.
Statement A only

- C.
Both Statement A and Statement B

- D.
Neither Statement A nor Statement B

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

Both Statement A and Statement B are false.

Hence, option (c).

Workspace:

**10. CAT 2022 LRDI Slot 3 | LR - Arrangements**

On how many days did Levmisto and Tyhrmisto have the same number of new cases?

- A.
5

- B.
4

- C.
3

- D.
2

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

Levmisto and Tyhrmisto have the same number of new cases on all 5 days.

Hence, option (c).

Workspace:

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

In the following, a year corresponds to 1^{st} of January of that year.

A study to determine the mortality rate for a disease began in 1980. The study chose 1000 males and 1000 females and followed them for forty years or until they died, whichever came first. The 1000 males chosen in 1980 consisted of 250 each of ages 10 to less than 20, 20 to less than 30, 30 to less than 40, and 40 to less than 50. The 1000 females chosen in 1980 also consisted of 250 each of ages 10 to less than 20, 20 to less than 30, 30 to less than 40, and 40 to less than 50.

The four figures below depict the age profile of those among the 2000 individuals who were still alive in 1990, 2000, 2010, and 2020. The blue bars in each figure represent the number of males in each age group at that point in time, while the pink bars represent the number of females in each age group at that point in time. The numbers next to the bars give the exact numbers being represented by the bars. For example, we know that 230 males among those tracked and who were alive in 1990 were aged between 20 and 30.

**11. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

In 2000, what was the ratio of the number of dead males to dead females among those being tracked?

- A.
109 : 107

- B.
41 : 43

- C.
129 : 131

- D.
71 : 69

Answer: Option D

**Explanation** :

Total number of males/females in 1980 = 1000 each.

Total number males in 2000 = 180 + 205 + 160 + 100 = 645

∴ Number of dead males = 1000 – 645 = 355

Total number females in 2000 = 210 + 175 + 150 + 120 = 655

∴ Number of dead males = 1000 – 655 = 345

⇒ Ratio of dead males to females = 355 : 345 = 71 : 69.

Hence, option (d).

Workspace:

**12. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

How many people who were being tracked and who were between 30 and 40 years of age in 1980 survived until 2010?

- A.
190

- B.
310

- C.
90

- D.
110

Answer: Option A

**Explanation** :

Those who were in the age range 30-40 in 1980 will be in the age range 60-70 in 2010.

Number of people alive in age range 60-70 in 2010 = 90 + 100 = 190

Hence, option (a).

Workspace:

**13. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

How many individuals who were being tracked and who were less than 30 years of age in 1980 survived until 2020?

- A.
240

- B.
470

- C.
580

- D.
230

Answer: Option B

**Explanation** :

Those who were less than 30 years old in 1980 will be less than 70 years old in 2020.

∴ Number of people alive below 70 in 2020 = 140 + 100 + 125 + 105 = 470

Hence, option (b).

Workspace:

**14. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

How many of the males who were being tracked and who were between 20 and 30 yearsof age in 1980 died in the period 2000 to 2010?

Answer: 40

**Explanation** :

1980 age range 20-30 (500 males) is same as

2000 age range 40-50 (205 males) is same as

2010 age range 50-60 (165 males)

∴ Number of required males dying between 2000 and 2010 = 205 – 165 = 40.

Hence, 40.

Workspace:

**15. CAT 2022 LRDI Slot 3 | DI - Tables & Graphs**

How many of the females who were being tracked and who were between 20 and 30 years of age in 1980 died between the ages of 50 and 60?

[**Note**: There is an ambiguity in this question and hence was discarded by IIM Bangalore.]

Answer: 30

**Explanation** :

1980 age range 20-30 (500) is same as

2000 age range 40-50 (175) is same as

2010 age range 50-60 (145 females)

30 people from this category died from 2000 to 2010, but it is not necessary that they were between 50-60 years old.

Someone who was 41 years old in 2000 may have died before he/she attains the age of 50.

The answer to this question should be cannot be determined.

Hence, this question was discarded.

**Note**: Answer provided in Candidate response sheet is 30.

Workspace:

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

All the first-year students in the computer science (CS) department in a university take both the courses (i) AI and (ii) ML. Students from other departments (non-CS students) can also take one of these two courses, but not both. Students who fail in a course get an F grade; others pass and are awarded A or B or C grades depending on their performance. The following are some additional facts about the number of students who took these two courses this year and the grades they obtained.

1. The numbers of non-CS students who took AI and ML were in the ratio 2 : 5.

2. The number of non-CS students who took either AI or ML was equal to the number of CS students.

3. The numbers of non-CS students who failed in the two courses were the same and their total is equal to the number of CS students who got a C grade in ML.

4. In both the courses, 50% of the students who passed got a B grade. But, while the numbers of students who got A and C grades were the same for AI, they were in the ratio 3 : 2 for ML.

5. No CS student failed in AI, while no non-CS student got an A grade in AI.

6. The numbers of CS students who got A, B and C grades respectively in AI were in the ratio 3 : 5 : 2, while in ML the ratio was 4 : 5 : 2.

7. The ratio of the total number of non-CS students failing in one of the two courses to the number of CS students failing in one of the two courses was 3 : 1.

8. 30 students failed in ML.

**16. CAT 2022 LRDI Slot 3 | LR - Rank & Ordering**

How many students took AI?

- A.
90

- B.
60

- C.
210

- D.
270

Answer: Option D

**Explanation** :

From (1) and (2): Let the number of non-CS students who took AI and ML are 2x and 5x respectively.

⇒ Number of CS students = 2x + 5x = 7x.

From (3): Let the number of non-CS students failing in two courses is ‘a’ each and CS students who got a C in ML = ‘2a’

From (6): Let number of CS students who got A, B and C grades respectively in AI are 3y, 5y, 2y.

While number of CS students who got A, B and C grades respectively in ML are 4a, 5a, 2a.

From (5): No CS student failed in AI, while no non-CS student got an A grade in AI.

From (4): The numbers of students who got A and C grades were the same for AI

⇒ 3y + 0 = 2y + non-CS students in AI with C grade

⇒ non-CS students in AI with C grade = 3y – 2y = y

From (8): 30 students failed in ML hence CS students failing in ML = 30 – a.

From (7):

⇒ $\frac{a+a}{30-a}$ = $\frac{3}{1}$

⇒ 2a = 90 – 3a

⇒ a = 18

⇒ 7x = 72 + 90 + 36 + 12 = 210

⇒ x = 30

⇒ 210 = 3y + 5y + 2y

⇒ y = 21

⇒ 60 = 0 + (non-CS student in AI getting B grade) + 21 + 18

∴ non-CS student in AI getting B grade = 60 – 39 = 21

From (4): In both the courses, 50% of the students who passed got a B grade

Number of students passing in ML = 72 + 90 + 36 + (150 - 18) = 330

⇒ Number of students receiving B grade in ML = 330/2 = 165.

⇒ non-CS students with B grade in ML = 165 – 90 = 75

Let number of non-CS students with A grade in ML = c, hence number of non-CS students with C grade in ML = 150 – 75 – 18 – c = 57 – c

From (4): the numbers of students who got A and C grades were in the ratio 3 : 2 for ML.

⇒ $\frac{72+c}{36+57-c}$ = $\frac{3}{2}$

⇒ 144 + 2c = 279 – 3c

⇒ 5c = 135

⇒ c = 27

∴ Number of students taking AI = 210 (CS students) + 60 (non-CS students) = 270.

Hence, option (d).

Workspace:

**17. CAT 2022 LRDI Slot 3 | LR - Rank & Ordering**

How many CS students failed in ML?

Answer: 12

**Explanation** :

Consider the solution to first question of this set.

Number of CS students failing in ML = 12.

Hence, 12.

Workspace:

**18. CAT 2022 LRDI Slot 3 | LR - Rank & Ordering**

How many non-CS students got A grade in ML?

Answer: 27

**Explanation** :

Consider the solution to first question of this set.

Number of non-CS students who got A grade in ML = 27.

Hence, 27.

Workspace:

**19. CAT 2022 LRDI Slot 3 | LR - Rank & Ordering**

How many students got A grade in AI?

- A.
84

- B.
63

- C.
99

- D.
42

Answer: Option B

**Explanation** :

Consider the solution to first question of this set.

Number of students who got A grade in AI = 63 + 0 = 63

Hence, option (b).

Workspace:

**20. CAT 2022 LRDI Slot 3 | LR - Rank & Ordering**

How many non-CS students got B grade in ML?

- A.
75

- B.
165

- C.
25

- D.
90

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

Number of non-CS students who got B grade in ML = 75

Hence, option (a).

Workspace:

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