# CAT 2021 LRDI Slot 3 | Previous Year CAT Paper

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**Answer the next 4 questions based on the information given **

Each of the bottles mentioned in this question contains 50 ml of liquid. The liquid in any bottle can be 100% pure content (P) or can have certain amount of impurity (I). Visually it is not possible to distinguish between P and I. There is a testing device which detects impurity, as long as the percentage of impurity in the content tested is 10% or more.

For example, suppose bottle 1 contains only P, and bottle 2 contains 80% P and 20% I. If content from bottle 1 is tested, it will be found out that it contains only P. If content of bottle 2 is tested, the test will reveal that it contains some amount of I. If 10 ml of content from bottle 1 is mixed with 20 ml content from bottle 2, the test will show that the mixture has impurity, and hence we can conclude that at least one of the two bottles has I. However, if 10 ml of content from bottle 1 is mixed with 5 ml of content from bottle 2. the test will not detect any impurity in the resultant mixture.

**1. CAT 2021 LRDI Slot 3 | LR - Mathematical Reasoning**

5 ml of content from bottle A is mixed with 5 ml of content from bottle B. The resultant mixture, when tested, detects the presence of I. If it is known that bottle A contains only P, what BEST can be concluded about the volume of I in bottle B?

- A.
10 ml or more

- B.
Less than 1 ml

- C.
10 ml

- D.
1 ml

Answer: Option A

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**Explanation** :

Since impurity is detected in the final mixture, hence the mixture has at least 10% impurities.

∴ % impurity in the mixture = $\frac{5\times 0\%+5\times P\%}{5+5}$ ≥ 10

⇒ 5P ≥ 100

⇒ P ≥ 20%

∴ Volume of impurities in bottle B is at least 20% of 50 ml i.e., ≥ 10 ml.

Hence, option (a).

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**2. CAT 2021 LRDI Slot 3 | LR - Mathematical Reasoning**

There are four bottles. Each bottle is known to contain only P or only I. They will be considered to be “collectively ready for despatch” if all of them contain only P. In minimum how many tests, is it possible to ascertain whether these four bottles are “collectively ready for despatch”?

Answer: 1

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**Explanation** :

The bottles contain only P or only I.

Let us mix equal quantities from each of these bottles and check for impurities.

**Case 1**: All bottles contain only P

Concentration of I in final mixture = $\frac{0+0+0+0}{4}$ = 0%

∴ Impurities will not be detected.

**Case 2**: 3 bottles contain P and 1 contains I

Concentration of I in final mixture = $\frac{0+0+0+100}{4}$ = 25%

∴ Impurities will be detected.

**Case 3**: 2 bottles contain P and 2 contain I

Concentration of I in final mixture = $\frac{0+0+100+100}{4}$ = 50%

∴ Impurities will be detected.

**Case 4**: 1 bottle contains P and 3 contain I

Concentration of I in final mixture = $\frac{0+100+100+100}{4}$ = 75%

∴ Impurities will be detected.

**Case 5**: All bottles contain only I

Concentration of I in final mixture = $\frac{100+100+100+100}{4}$ = 100%

∴ Impurities will be detected.

There is only one case where impurities are not detected and that is when all the bottles have only P.

⇒ If we mix equal quantities of all 4 bottles and test for impurities, if impurity is not detected then we can safely accept all 4 bottles.

Only one test is required is such a situation.

Hence, 1.

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**3. CAT 2021 LRDI Slot 3 | LR - Mathematical Reasoning**

There are four bottles. It is known that three of these bottles contain only P, while the remaining one contains 80% P and 20% I. What is the minimum number of tests required to definitely identify the bottle containing some amount of I?

Answer: 2

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**Explanation** :

Let the bottles be A, B, C and D.

We first mix equal quantities of A and B and test the mixture.

**Case 1**: If any one of A or B contains 20% impurities, test will detect the impurity.

Then we check A for impurity. If impurity is detected then A contains I else B contains I.

∴ 2 tests are required to detect I.

**Case 2**: If none of A or B contains impurities, test will not detect the impurity.

This means one of C or D contains I.

Then we check C for impurity. If impurity is detected then C contains I else D contains I.

∴ 2 tests are required to detect I.

In both cases 2 tests are required to detect I.

Hence, 2.

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**4. CAT 2021 LRDI Slot 3 | LR - Mathematical Reasoning**

There are four bottles. It is known that either one or two of these bottles contain(s) only P, while the remaining ones contain 85% P and 15% I. What is the minimum number of tests required to ascertain the exact number of bottles containing only P?

- A.
3

- B.
2

- C.
4

- D.
1

Answer: Option D

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**Explanation** :

**Case 1**: Only bottle contains pure P.

Now if we mix equal quantities from each of the 4 bottles, the concentration of I in the mixture = $\frac{0+15+15+15}{4}$ = 11.25%

∴ Testing this mixture will detect the impurities.

**Case 2**: 2 bottles contains pure P.

Now if we mix equal quantities from each of the 4 bottles, the concentration of I in the mixture = $\frac{0+0+15+15}{4}$ = 7.5%

∴ Testing this mixture will not detect the impurities.

⇒ Mix equal quantities of all 4 bottles, it I is detected then only 1 bottle contains P else 2 bottles contain P.

Hence, option (d).

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**Answer the next 6 questions based on the information given below:**

Three reviewers Amal, Bimal, and Komal are tasked with selecting questions from a pool of 13 questions (Q01 to Q13). Questions can be created by external “subject matter experts” (SMEs) or by one of the three reviewers. Each of the reviewers either approves or disapproves a question that is shown to them. Their decisions lead to eventual acceptance or rejection of the question in the manner described below.

If a question is created by an SME, it is reviewed first by Amal, and then by Bimal. If both of them approve the question, then the question is accepted and is not reviewed by Komal. If both disapprove the question, it is rejected and is not reviewed by Komal. If one of them approves the question and the other disapproves it, then the question is reviewed by Komal. Then the question is accepted only if she approves it.

A question created by one of the reviewers is decided upon by the other two. If a question is created by Amal, then it is first reviewed by Bimal. If Bimal approves the question, then it is accepted. Otherwise, it is reviewed by Komal. The question is then accepted only if Komal approves it. A similar process is followed for questions created by Bimal, whose questions are first reviewed by Komal, and then by Amal only if Komal disapproves it. Questions created by Komal are first reviewed by Amal, and then, if required, by Bimal.

The following facts are known about the review process after its completion.

- Q02, Q06, Q09, Q11, and Q12 were rejected and the other questions were accepted.
- Amal reviewed only Q02, Q03, Q04, Q06, Q08, Q10, Q11, and Q13.
- Bimal reviewed only Q02, Q04, Q06 through Q09, Q12, and Q13.
- Komal reviewed only Q01 through Q05, Q07, Q08, Q09, Q11, and Q12.

**5. CAT 2021 LRDI Slot 3 | LR - Selection & Distribution**

How many questions were DEFINITELY created by Amal?

Answer: 3

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**Explanation** :

Based on the information given in the five points, we can make the following table.

Q1 is reviewed only by Komal, hence it must have been prepared by Bimal and Komal must have accepted it. [If Komal had rejected it the question would then have been reviewed by Amal which is not the case.]

Q2 is reviewed by all three, hence it must have been prepared by SMEs. Since it is reviewed by Komal, one of Amal and Bimal must have accepted and the other rejected this question. Finally Komal must also have rejected the question for it to be finally get rejected.

Q3 is reviewed by Komal and Amal, hence it was created by Bimal. Komal reviews it first and rejects it hence Bimal reviews it next and accepts it.

Similarly, we can complete the table:

Questions 7, 9 and 12 are definitely created by Amal.

Hence, 3.

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**6. CAT 2021 LRDI Slot 3 | LR - Selection & Distribution**

How many questions were DEFINITELY created by Komal?

Answer: 1

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**Explanation** :

Consider the solution to first question of this set.

Question 10 is definitely created by Komal.

Hence, 1.

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**7. CAT 2021 LRDI Slot 3 | LR - Selection & Distribution**

How many questions were DEFINITELY created by SMEs?

Answer: 3

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**Explanation** :

Consider the solution to first question of this set.

Questions 2, 4 and 8 are definitely created by SMEs.

Hence, 3.

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**8. CAT 2021 LRDI Slot 3 | LR - Selection & Distribution**

How many questions were DEFINITELY disapproved by Bimal?

- A.
7

- B.
5

- C.
4

- D.
3

Answer: Option C

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**Explanation** :

Consider the solution to first question of this set.

Bimal definitely rejected questions 6, 7, 9 and 12 i.e., 4 questions.

Hence, option (c).

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**9. CAT 2021 LRDI Slot 3 | LR - Selection & Distribution**

The approval ratio of a reviewer is the ratio of the number of questions (s)he approved to the number of questions (s)he reviewed. Which option best describes Amal’s approval ratio?

- A.
0.25

- B.
lies between 0.25 and 0.75

- C.
lies between 0.25 and 0.50

- D.
either 0.25 or 0.75

Answer: Option B

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**Explanation** :

Consider the solution to first question of this set.

Amal reviewed a total of 8 questions.

Out of these 8 questions Amal definitely accepted questions 2 questions and definitely rejected 2 questions.

∴ Minimum accepted questions for Amal = 2 while maximum accepted questions for Amal = 6.

⇒ Minimum approval ratio = 2/8 = 0.25

⇒ Maximum approval ratio = 6/8 = 0.75

∴ Approval ratio lies between 0.25 and 0.75.

Hence, option (b).

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**10. CAT 2021 LRDI Slot 3 | LR - Selection & Distribution**

How many questions created by Amal or Bimal were disapproved by at least one of the other reviewers?

- A.
4

- B.
7

- C.
2

- D.
5

Answer: Option D

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**Explanation** :

Consider the solution to first question of this set.

Amal or Bimal created questions 1, 3, 5, 7, 9, 11 and 12 i.e., 7 questions.

Out of these questions 3, 7, 9, 11 and 12 i.e., 5 questions were rejected by at least one of the other reviewers.

Hence, option (d).

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**Answer the next 6 questions based on the information given below:**

10 players – P1, P2, … , P10 - competed in an international javelin throw event. The number (after P) of a player reflects his rank at the beginning of the event, with rank 1 going to the topmost player. There were two phases in the event with the first phase consisting of rounds 1, 2, and 3, and the second phase consisting of rounds 4, 5, and 6. A throw is measured in terms of the distance it covers (in meters, up to one decimal point accuracy), only if the throw is a ‘valid’ one. For an invalid throw, the distance is taken as zero. A player’s score at the end of a round is the maximum distance of all his throws up to that round. Players are re-ranked after every round based on their current scores. In case of a tie in scores, the player with a prevailing higher rank retains the higher rank. This ranking determines the order in which the players go for their throws in the next round.

In each of the rounds in the first phase, the players throw in increasing order of their latest rank, i.e. the player ranked 1 at that point throws first, followed by the player ranked 2 at that point and so on. The top six players at the end of the first phase qualify for the second phase. In each of the rounds in the second phase, the players throw in decreasing order of their latest rank i.e. the player ranked 6 at that point throws first, followed by the player ranked 5 at that point and so on. The players ranked 1, 2, and 3 at the end of the sixth round receive gold, silver, and bronze medals respectively.

All the valid throws of the event were of distinct distances (as per stated measurement accuracy). The tables below show distances (in meters) covered by all valid throws in the first and the third round in the event.

Distances covered by all the valid throws in the first round

The following facts are also known.

- Among the throws in the second round, only the last two were valid. Both the throws enabled these players to qualify for the second phase, with one of them qualifying with the least score. None of these players won any medal.
- If a player throws first in a round AND he was also the last (among the players in the current round) to throw in the previous round, then the player is said to get a double. Two players got a double.
- In each round of the second phase, exactly one player improved his score. Each of these improvements was by the same amount.
- The gold and bronze medalists improved their scores in the fifth and the sixth rounds respectively. One medal winner improved his score in the fourth round.
- The difference between the final scores of the gold medalist and the silver medalist, as well as the difference between the final scores of the silver medalist and the bronze medalist was 1.0 m.

**11. CAT 2021 LRDI Slot 3 | LR - Puzzles**

Which two players got the double?

- A.
P1, P8

- B.
P2, P4

- C.
P1, P10

- D.
P8, P10

Answer: Option D

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**Explanation** :

Let us arrange the players in the order in which they throw in each round.

Round 1: Here the players throw in order of their initial seeds so the order is as follows:

P1 → P2 → P3 → P4 → P5 → P6 → P7 → P8 → P9 → P10

In the first round only 6 players had a valid throw: P1, P3, P5, P6, P7 and P9

∴ Ranking at the end of 1st round.

1 P7 87.2

2 P5 86.4

3 P9 84.1

4 P1 82.9

5 P6 82.5

6 P3 81.5

7 P2 -

8 P4 -

9 P8 -

10 P10 -

P7, P5, P9, P1, P6, P3 P2, P4, P8, P10

Now, in round 2, only last two throws i.e., of P8 and P10 were valid throws. Hence, their order will change at the start of Round 3, however, the remaining order stays the same. That is, P8 and P10 will move up in the table and occupy some higher places, whereas some of the others may move down consequently.

Round 3: In round 3, we can see that P1 improved his score from 82.9 to 88.6. The other 2 participants did not improve their scores. Also, after round 3, P8 and P10 qualify, where one of P8 or P10 is at the sixth position. So, at the end of round 3, we can say that P6, P3, P2 and P4 are at the bottom 4 positions. One of P8 or P10 is at the sixth positions. P1 > P7 > P5 > P9.

∴ Their ranks at the end of round 3.

1/2 P1 88.6

2/3 P7 87.2

3/4 P5 86.4

4/5 P9 84.1

6 P8/P10

The other person between P8 / P10 can go anywhere between rank 1 and 5.

Now let us consider the two players who got a double. Doubles happen in the transition between rounds.

Round 1 → Round 2: Double is not possible as P10 (last to throw in Round 1) did not have a valid throw.

Round 2 → Round 1: Possible if P10 (last to throw in Round 1) reaches Rank 1 in round 2.

Round 3 → Round 4: P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens.

Round 4 → Round 5: Not possible since only one player improves his/her rank in round 4.

Round 5 → Round 6: Not possible since only one player improves his/her rank in round 5.

∴ After Round 2 P10 definitely reaches top of the ranking i.e., P10 throws more than 87.2.

∴ Ranks at the end of round 3.

**Case 1**

1 P1 88.6

2 P10 ?

3 P7 87.2

4 P5 86.4

5 P9 84.1

6 P8

**Case 2**

1 P10 ?

2 P1 88.6

3 P7 87.2

4 P5 86.4

5 P9 84.1

6 P8

P10 and P8 do not win any medals.

**Case 2**: 3 of P1, P7, P5 and P9 need to score better than P10. Each of these 3 will improve their score by same amount say x.

Possible scores after improvement:

P1: 88.6 + x

P7: 87.2 + x

P5: 86.4 + x

P9: 84.1 + x

No value of x will satisfy the 5th condition given in the question. Hence, this case is rejected.

**Case 1**: 2 or 3 of P1, P7, P5 and P9 need to score better than P10. Each of these 3 will improve their score by same amount say x.

Possible scores after improvement:

P1: 88.6 + x

P7: 87.2 + x

P5: 86.4 + x

P9: 84.1 + x

If P1 does not win any medal, P7, P5 and P9 will have to improve their scores. But this will not satisfy conditions 3 and 5 simultaneously. Hence P1 has to be one of the medalists.

P10 has scored more than 87.2, hence the three medalists need to score more than 87.2.

The only way this is possible is when P7 improves his score twice by 1.2 while P5 improves his score once by 1.2.

Scores at the end of round 5.

P7: 89.6

P1: 88.6

P5: 87.6

∴ P8 and P10 get the doubles.

Hence, option (d).

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**12. CAT 2021 LRDI Slot 3 | LR - Puzzles**

Who won the silver medal?

- A.
P9

- B.
P7

- C.
P5

- D.
P1

Answer: Option D

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**Explanation** :

Consider the solution to first question of this set.

P1 won the silver medal.

Hence, option (d).

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**13. CAT 2021 LRDI Slot 3 | LR - Puzzles**

Who threw the last javelin in the event?

- A.
P7

- B.
P9

- C.
P10

- D.
P1

Answer: Option A

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**Explanation** :

Consider the solution to first question of this set.

Bronze medalist i.e., P5 improved his score in the 6^{th} round hence P7 must have improved his score in round 4 and 5.

At the end of round 5 P7 would be ranked 1 and hence will throw the last in round 6.

Hence, option (a).

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**14. CAT 2021 LRDI Slot 3 | LR - Puzzles**

What was the final score (in m) of the silver-medalist?

- A.
89.6

- B.
87.2

- C.
88.6

- D.
88.4

Answer: Option C

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**Explanation** :

Consider the solution to first question of this set.

Final score of silver medalist i.e., P1 is 88.6.

Hence, option (c).

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**15. CAT 2021 LRDI Slot 3 | LR - Puzzles**

Which of the following can be the final score (in m) of P8?

- A.
81.9

- B.
0

- C.
85.1

- D.
82.7

Answer: Option D

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**Explanation** :

Consider the solution to first question of this set.

P8 has to finish 6^{th} at the end of 3^{rd} round. Hence, he should score less than 84.1 but more than 81.5.

Hence, option (d).

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**16. CAT 2021 LRDI Slot 3 | LR - Puzzles**

By how much did the gold medalist improve his score (in m) in the second phase?

- A.
1.0

- B.
1.2

- C.
2.0

- D.
2.4

Answer: Option D

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**Explanation** :

Consider the solution to first question of this set.

P7 improved his score by 1.2 + 1.2 = 2.4 in the second phase.

Hence, option (d).

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**Answer the next 4 questions based on the information given **

The figure above shows the schedule of four employees – Abani, Bahni, Danni and Tinni – whom Dhoni supervised in 2020. Altogether there were five projects which started and concluded in 2020 in which they were involved. For each of these projects and for each employee, the starting day was at the beginning of a month and the concluding day was the end of a month, and these are indicated by the left and right end points of the corresponding horizontal bars. The number within each bar indicates the percentage of assigned work completed by the employee for that project, as assessed by Dhoni.

For each employee, his/her total project-month (in 2020) is the sum of the number of months (s)he worked across the five project, while his/her annual completion index is the weightage average of the completion percentage assigned from the different projects, with the weights being the corresponding number of months (s)he worked in these projects. For each project, the total employee-month is the sum of the number of months four employees worked in this project, while its completion index is the weightage average of the completion percentage assigned for the employees who worked in this project, with the weights being the corresponding number of months they worked in this project.

**17. CAT 2021 LRDI Slot 3 | DI - Tables & Graphs**

Which of the following statements is/are true?

I: The total project-month was the same for the four employees.

II: The total employee-month was the same for the five projects.

- A.
Both I and II

- B.
Neither I nor II

- C.
Only II

- D.
Only I

Answer: Option D

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**Explanation** :

Total project month for

Abani: 2 + 2 + 5 = 9

Bahni: 2 + 4 + 3 = 9

Danni: 3 + 3 + 2 + 1 = 9

Tinni: 2 + 2 + 3 + 2 = 9

∴ Total project month was same for all 4 employees.

Total employee month

Project 1: 2 + 2 +2 = 6

Project 2: 3 + 2 = 5

Project 3: 2 + 4 + 3 = 9

Project 4: 5 + 2 + 3 = 10

Project 5: 3 + 1 + 2 = 6

∴ Total employee month was not same for all projects.

Hence, option (d).

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**18. CAT 2021 LRDI Slot 3 | DI - Tables & Graphs**

Which employees did not work in multiple projects for any of the months in 2020?

- A.
Only Tinni

- B.
Only Abani, Bahni and Danni

- C.
Only Abani and Bahni

- D.
All four of them

Answer: Option B

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**Explanation** :

Only Tinni worked for project 4 and 5 during September.

No other employee worked for multiple projects at the same time.

Hence, option (b).

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**19. CAT 2021 LRDI Slot 3 | DI - Tables & Graphs**

The project duration, measured in terms of the number of months, is the time during which at least one employee worked in the project. Which of the following pairs of the projects had the same duration?

- A.
Project 4, Project 5

- B.
Project 3, Project 5

- C.
Project 1, Project 5

- D.
Project 3, Project 4

Answer: Option D

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**Explanation** :

Project duration for

Project 1: jan to mar i.e., 3 months

Project 2: feb to apr i.e., 3 months

Project 3: apr to aug i.e., 5 months

Project 4: jul to nov i.e., 5 months

Project 5: sep to dec i.e., 4 months

Hence, option (d).

Workspace:

**20. CAT 2021 LRDI Slot 3 | DI - Tables & Graphs**

The list of employees in decreasing order of annual completion index is:

- A.
Danni, Tinni, Bahni, Abani

- B.
Danni, Tinni, Abani, Bahni

- C.
Tinni, Danni, Abani, Bahni

- D.
Bahni, Abani, Tinni, Danni

Answer: Option B

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**Explanation** :

Annual completion index for

Abani = (1 × 2 + 1 × 2 + 0.8 × 5)/9 = 8/9

Bahni = (1 × 2 + 0.75 × 4 + 0.9 × 3)/9 = 7.7/9

Danni = (0.9 × 3 + 1 × 3 + 1 × 2 + 1 × 1)/9 = 8.7/9

Tinni = (0.8 × 2 + 1 × 2 + 1 × 3 + 1 × 2)/9 = 8.6/9

∴ Danni < Tinni < Abani < Bahni

Hence, option (b).

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