CAT 2021 | DILR Set 1
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.
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.
1. Which two players got the double?
- P1, P10
- P2, P4
- P8, P10
- P1, P8
Among round 1, only P1, P3, P5, P6, P7 and P9 have valid throws, so P2, P4, P8 and P10 must be scoring zero distance in round 1. Among round 2, only last two throws were valid and after re-ranking that two must be P8 and P10, which makes them qualify for the 2nd phase as well.
Now we will check for the rounds in which double is possible
In round 1, double never happened
In round 2, double is possible if P10 ranks 1st after round 2
In round 3, double is possible as P8, the other qualifier having least scores in 1st phase and he will be the 1st one to throw in round 4 as per decreasing order of their latest rank
In round 4 and round 5, double is again not possible as exactly one player improved his score.
Therefore, two players that scored a double must be P10 from round 2 to round 3 and P8 from round 3 to round 4.
P10 must be ranked 1 in round 2 with score more than 87.2 m (top scorer P7) to score a double and P8, the other qualifier having least score, must have scored more than 82.5 m (P6) and less than 82.9 m (P1) such that it has least score to qualify and scored a double from round 3 to round 4.
Also P1 has increased his score to 88.6 m in round 3, so it may be at 1st position surpassing the score of P10 or may be at 2nd position less than P10 score.
Given, in 2nd phase, exactly one player improved his score and also P8 and P10 did not win any medal, so P10 cannot finish at 1st in round 3. Therefore, P1 scores 1st rank and P10 scores 2nd rank after round 3 such that scores of P10 is more than 87.2 m but less than 88.6 m.
Now, in 2nd phase, in each round exactly one player improved his score (with same amount) such that all of them are medalist and also final scores of gold, silver and bronze medalist have common difference of 1.0 m. Also, P8 and 10 are not one of the medalists. All three P9, P5 and P7 cannot be the medalists as if they improved their scores with same amount, their common difference is not 1.0 m.
Therefore, among P9, P5 and P7, two of them surpass P10 to become medalist and P1 is definitely a medalist. Now if P1 improves his scores, still the difference among the scores of P9, P5 and P7 cannot be the same, so it is not possible. Now let us consider following cases.
Case I – Let P1 is a gold medalist with score 88.6 m
So, silver medalists score must be 87.6 m and bronze medalists score must be 86.6 m
This case is not possible as P10 scores more than 87.2 m and can’t be a medalist
Case II – Let P1 is a silver medalist with score 88.6 m
So, the score of gold medalist be 89.6 m and score of bronze medalist be 87.6 m
This case is possible if P7 improves his scores twice with an amount of 1.2 m each and P5 improves his score with 1.2 m once such that the final score of P7 becomes 87.6 + 2.4 = 89.6 m and final score of P5 becomes 86.4 + 1.2 = 87.6 m
Case III – Let P1 is a bronze medalist with score 88.6 m
So, silver medalists score must be 89.6 m and gold medalist score must be 90.6 m
Now, in case of P7, 90.6 – 87.2 = 3.4 m, so each improved score = 3.4/2 = 1.7 m
But for P5, 89.6 – 86.4 = 3.2 m ≠ 1.7 m
So, this case is also not possible.
Therefore according to Case – II, P7 improves his score in round 4 and round 5 and P5 improves his scores in round 6 and the rest of the information can be gathered as follows
Hence, the two players who got the doubles are P8 and P10
2. Who won the silver medal?
- P9
- P5
- P7
- P1
Among round 1, only P1, P3, P5, P6, P7 and P9 have valid throws, so P2, P4, P8 and P10 must be scoring zero distance in round 1. Among round 2, only last two throws were valid and after re-ranking that two must be P8 and P10, which makes them qualify for the 2nd phase as well.
Now we will check for the rounds in which double is possible
In round 1, double never happened
In round 2, double is possible if P10 ranks 1st after round 2
In round 3, double is possible as P8, the other qualifier having least scores in 1st phase and he will be the 1st one to throw in round 4 as per decreasing order of their latest rank
In round 4 and round 5, double is again not possible as exactly one player improved his score.
Therefore, two players that scored a double must be P10 from round 2 to round 3 and P8 from round 3 to round 4.
P10 must be ranked 1 in round 2 with score more than 87.2 m (top scorer P7) to score a double and P8, the other qualifier having least score, must have scored more than 82.5 m (P6) and less than 82.9 m (P1) such that it has least score to qualify and scored a double from round 3 to round 4.
Also P1 has increased his score to 88.6 m in round 3, so it may be at 1st position surpassing the score of P10 or may be at 2nd position less than P10 score.
Given, in 2nd phase, exactly one player improved his score and also P8 and P10 did not win any medal, so P10 cannot finish at 1st in round 3. Therefore, P1 scores 1st rank and P10 scores 2nd rank after round 3 such that scores of P10 is more than 87.2 m but less than 88.6 m.
Now, in 2nd phase, in each round exactly one player improved his score (with same amount) such that all of them are medalist and also final scores of gold, silver and bronze medalist have common difference of 1.0 m. Also, P8 and 10 are not one of the medalists. All three P9, P5 and P7 cannot be the medalists as if they improved their scores with same amount, their common difference is not 1.0 m.
Therefore, among P9, P5 and P7, two of them surpass P10 to become medalist and P1 is definitely a medalist. Now if P1 improves his scores, still the difference among the scores of P9, P5 and P7 cannot be the same, so it is not possible. Now let us consider following cases.
Case I – Let P1 is a gold medalist with score 88.6 m
So, silver medalists score must be 87.6 m and bronze medalists score must be 86.6 m
This case is not possible as P10 scores more than 87.2 m and can’t be a medalist
Case II – Let P1 is a silver medalist with score 88.6 m
So, the score of gold medalist be 89.6 m and score of bronze medalist be 87.6 m
This case is possible if P7 improves his scores twice with an amount of 1.2 m each and P5 improves his score with 1.2 m once such that the final score of P7 becomes 87.6 + 2.4 = 89.6 m and final score of P5 becomes 86.4 + 1.2 = 87.6 m
Case III – Let P1 is a bronze medalist with score 88.6 m
So, silver medalists score must be 89.6 m and gold medalist score must be 90.6 m
Now, in case of P7, 90.6 – 87.2 = 3.4 m, so each improved score = 3.4/2 = 1.7 m
But for P5, 89.6 – 86.4 = 3.2 m ≠ 1.7 m
So, this case is also not possible.
Therefore according to Case – II, P7 improves his score in round 4 and round 5 and P5 improves his scores in round 6 and the rest of the information can be gathered as follows
Hence, P1 won the silver medal
3. Who threw the last javelin in the event?
- P7
- P1
- P10
- P9
Among round 1, only P1, P3, P5, P6, P7 and P9 have valid throws, so P2, P4, P8 and P10 must be scoring zero distance in round 1. Among round 2, only last two throws were valid and after re-ranking that two must be P8 and P10, which makes them qualify for the 2nd phase as well.
Now we will check for the rounds in which double is possible
In round 1, double never happened
In round 2, double is possible if P10 ranks 1st after round 2
In round 3, double is possible as P8, the other qualifier having least scores in 1st phase and he will be the 1st one to throw in round 4 as per decreasing order of their latest rank
In round 4 and round 5, double is again not possible as exactly one player improved his score.
Therefore, two players that scored a double must be P10 from round 2 to round 3 and P8 from round 3 to round 4.
P10 must be ranked 1 in round 2 with score more than 87.2 m (top scorer P7) to score a double and P8, the other qualifier having least score, must have scored more than 82.5 m (P6) and less than 82.9 m (P1) such that it has least score to qualify and scored a double from round 3 to round 4.
Also P1 has increased his score to 88.6 m in round 3, so it may be at 1st position surpassing the score of P10 or may be at 2nd position less than P10 score.
Given, in 2nd phase, exactly one player improved his score and also P8 and P10 did not win any medal, so P10 cannot finish at 1st in round 3. Therefore, P1 scores 1st rank and P10 scores 2nd rank after round 3 such that scores of P10 is more than 87.2 m but less than 88.6 m.
Now, in 2nd phase, in each round exactly one player improved his score (with same amount) such that all of them are medalist and also final scores of gold, silver and bronze medalist have common difference of 1.0 m. Also, P8 and 10 are not one of the medalists. All three P9, P5 and P7 cannot be the medalists as if they improved their scores with same amount, their common difference is not 1.0 m.
Therefore, among P9, P5 and P7, two of them surpass P10 to become medalist and P1 is definitely a medalist. Now if P1 improves his scores, still the difference among the scores of P9, P5 and P7 cannot be the same, so it is not possible. Now let us consider following cases.
Case I – Let P1 is a gold medalist with score 88.6 m
So, silver medalists score must be 87.6 m and bronze medalists score must be 86.6 m
This case is not possible as P10 scores more than 87.2 m and can’t be a medalist
Case II – Let P1 is a silver medalist with score 88.6 m
So, the score of gold medalist be 89.6 m and score of bronze medalist be 87.6 m
This case is possible if P7 improves his scores twice with an amount of 1.2 m each and P5 improves his score with 1.2 m once such that the final score of P7 becomes 87.6 + 2.4 = 89.6 m and final score of P5 becomes 86.4 + 1.2 = 87.6 m
Case III – Let P1 is a bronze medalist with score 88.6 m
So, silver medalists score must be 89.6 m and gold medalist score must be 90.6 m
Now, in case of P7, 90.6 – 87.2 = 3.4 m, so each improved score = 3.4/2 = 1.7 m
But for P5, 89.6 – 86.4 = 3.2 m ≠ 1.7 m
So, this case is also not possible.
Therefore according to Case – II, P7 improves his score in round 4 and round 5 and P5 improves his scores in round 6 and the rest of the information can be gathered as follows
Hence, the last javelin is thrown by gold medalist P7
4. What was the final score (in m) of the silver-medalist?
- 87.2
- 89.6
- 88.6
- 88.4
Among round 1, only P1, P3, P5, P6, P7 and P9 have valid throws, so P2, P4, P8 and P10 must be scoring zero distance in round 1. Among round 2, only last two throws were valid and after re-ranking that two must be P8 and P10, which makes them qualify for the 2nd phase as well.
Now we will check for the rounds in which double is possible
In round 1, double never happened
In round 2, double is possible if P10 ranks 1st after round 2
In round 3, double is possible as P8, the other qualifier having least scores in 1st phase and he will be the 1st one to throw in round 4 as per decreasing order of their latest rank
In round 4 and round 5, double is again not possible as exactly one player improved his score.
Therefore, two players that scored a double must be P10 from round 2 to round 3 and P8 from round 3 to round 4.
P10 must be ranked 1 in round 2 with score more than 87.2 m (top scorer P7) to score a double and P8, the other qualifier having least score, must have scored more than 82.5 m (P6) and less than 82.9 m (P1) such that it has least score to qualify and scored a double from round 3 to round 4.
Also P1 has increased his score to 88.6 m in round 3, so it may be at 1st position surpassing the score of P10 or may be at 2nd position less than P10 score.
Given, in 2nd phase, exactly one player improved his score and also P8 and P10 did not win any medal, so P10 cannot finish at 1st in round 3. Therefore, P1 scores 1st rank and P10 scores 2nd rank after round 3 such that scores of P10 is more than 87.2 m but less than 88.6 m.
Now, in 2nd phase, in each round exactly one player improved his score (with same amount) such that all of them are medalist and also final scores of gold, silver and bronze medalist have common difference of 1.0 m. Also, P8 and 10 are not one of the medalists. All three P9, P5 and P7 cannot be the medalists as if they improved their scores with same amount, their common difference is not 1.0 m.
Therefore, among P9, P5 and P7, two of them surpass P10 to become medalist and P1 is definitely a medalist. Now if P1 improves his scores, still the difference among the scores of P9, P5 and P7 cannot be the same, so it is not possible. Now let us consider following cases.
Case I – Let P1 is a gold medalist with score 88.6 m
So, silver medalists score must be 87.6 m and bronze medalists score must be 86.6 m
This case is not possible as P10 scores more than 87.2 m and can’t be a medalist
Case II – Let P1 is a silver medalist with score 88.6 m
So, the score of gold medalist be 89.6 m and score of bronze medalist be 87.6 m
This case is possible if P7 improves his scores twice with an amount of 1.2 m each and P5 improves his score with 1.2 m once such that the final score of P7 becomes 87.6 + 2.4 = 89.6 m and final score of P5 becomes 86.4 + 1.2 = 87.6 m
Case III – Let P1 is a bronze medalist with score 88.6 m
So, silver medalists score must be 89.6 m and gold medalist score must be 90.6 m
Now, in case of P7, 90.6 – 87.2 = 3.4 m, so each improved score = 3.4/2 = 1.7 m
But for P5, 89.6 – 86.4 = 3.2 m ≠ 1.7 m
So, this case is also not possible.
Therefore according to Case – II, P7 improves his score in round 4 and round 5 and P5 improves his scores in round 6 and the rest of the information can be gathered as follows
Hence, the final score of silver medalist is 88.6 m
5. Which of the following can be the final score (in m) of P8?
- 85.1
- 0
- 81.9
- 82.7
Among round 1, only P1, P3, P5, P6, P7 and P9 have valid throws, so P2, P4, P8 and P10 must be scoring zero distance in round 1. Among round 2, only last two throws were valid and after re-ranking that two must be P8 and P10, which makes them qualify for the 2nd phase as well.
Now we will check for the rounds in which double is possible
In round 1, double never happened
In round 2, double is possible if P10 ranks 1st after round 2
In round 3, double is possible as P8, the other qualifier having least scores in 1st phase and he will be the 1st one to throw in round 4 as per decreasing order of their latest rank
In round 4 and round 5, double is again not possible as exactly one player improved his score.
Therefore, two players that scored a double must be P10 from round 2 to round 3 and P8 from round 3 to round 4.
P10 must be ranked 1 in round 2 with score more than 87.2 m (top scorer P7) to score a double and P8, the other qualifier having least score, must have scored more than 82.5 m (P6) and less than 82.9 m (P1) such that it has least score to qualify and scored a double from round 3 to round 4.
Also P1 has increased his score to 88.6 m in round 3, so it may be at 1st position surpassing the score of P10 or may be at 2nd position less than P10 score.
Given, in 2nd phase, exactly one player improved his score and also P8 and P10 did not win any medal, so P10 cannot finish at 1st in round 3. Therefore, P1 scores 1st rank and P10 scores 2nd rank after round 3 such that scores of P10 is more than 87.2 m but less than 88.6 m.
Now, in 2nd phase, in each round exactly one player improved his score (with same amount) such that all of them are medalist and also final scores of gold, silver and bronze medalist have common difference of 1.0 m. Also, P8 and 10 are not one of the medalists. All three P9, P5 and P7 cannot be the medalists as if they improved their scores with same amount, their common difference is not 1.0 m.
Therefore, among P9, P5 and P7, two of them surpass P10 to become medalist and P1 is definitely a medalist. Now if P1 improves his scores, still the difference among the scores of P9, P5 and P7 cannot be the same, so it is not possible. Now let us consider following cases.
Case I – Let P1 is a gold medalist with score 88.6 m
So, silver medalists score must be 87.6 m and bronze medalists score must be 86.6 m
This case is not possible as P10 scores more than 87.2 m and can’t be a medalist
Case II – Let P1 is a silver medalist with score 88.6 m
So, the score of gold medalist be 89.6 m and score of bronze medalist be 87.6 m
This case is possible if P7 improves his scores twice with an amount of 1.2 m each and P5 improves his score with 1.2 m once such that the final score of P7 becomes 87.6 + 2.4 = 89.6 m and final score of P5 becomes 86.4 + 1.2 = 87.6 m
Case III – Let P1 is a bronze medalist with score 88.6 m
So, silver medalists score must be 89.6 m and gold medalist score must be 90.6 m
Now, in case of P7, 90.6 – 87.2 = 3.4 m, so each improved score = 3.4/2 = 1.7 m
But for P5, 89.6 – 86.4 = 3.2 m ≠ 1.7 m
So, this case is also not possible.
Therefore according to Case – II, P7 improves his score in round 4 and round 5 and P5 improves his scores in round 6 and the rest of the information can be gathered as follows
Among options, 82.7 is the only possible score in the range 82.5 < P8 < 82.9
6. By how much did the gold medalist improve his score (in m) in the second phase?
- 2.0
- 1.2
- 1.0
- 2.4
Among round 1, only P1, P3, P5, P6, P7 and P9 have valid throws, so P2, P4, P8 and P10 must be scoring zero distance in round 1. Among round 2, only last two throws were valid and after re-ranking that two must be P8 and P10, which makes them qualify for the 2nd phase as well.
Now we will check for the rounds in which double is possible
In round 1, double never happened
In round 2, double is possible if P10 ranks 1st after round 2
In round 3, double is possible as P8, the other qualifier having least scores in 1st phase and he will be the 1st one to throw in round 4 as per decreasing order of their latest rank
In round 4 and round 5, double is again not possible as exactly one player improved his score.
Therefore, two players that scored a double must be P10 from round 2 to round 3 and P8 from round 3 to round 4.
P10 must be ranked 1 in round 2 with score more than 87.2 m (top scorer P7) to score a double and P8, the other qualifier having least score, must have scored more than 82.5 m (P6) and less than 82.9 m (P1) such that it has least score to qualify and scored a double from round 3 to round 4.
Also P1 has increased his score to 88.6 m in round 3, so it may be at 1st position surpassing the score of P10 or may be at 2nd position less than P10 score.
Given, in 2nd phase, exactly one player improved his score and also P8 and P10 did not win any medal, so P10 cannot finish at 1st in round 3. Therefore, P1 scores 1st rank and P10 scores 2nd rank after round 3 such that scores of P10 is more than 87.2 m but less than 88.6 m.
Now, in 2nd phase, in each round exactly one player improved his score (with same amount) such that all of them are medalist and also final scores of gold, silver and bronze medalist have common difference of 1.0 m. Also, P8 and 10 are not one of the medalists. All three P9, P5 and P7 cannot be the medalists as if they improved their scores with same amount, their common difference is not 1.0 m.
Therefore, among P9, P5 and P7, two of them surpass P10 to become medalist and P1 is definitely a medalist. Now if P1 improves his scores, still the difference among the scores of P9, P5 and P7 cannot be the same, so it is not possible. Now let us consider following cases.
Case I – Let P1 is a gold medalist with score 88.6 m
So, silver medalists score must be 87.6 m and bronze medalists score must be 86.6 m
This case is not possible as P10 scores more than 87.2 m and can’t be a medalist
Case II – Let P1 is a silver medalist with score 88.6 m
So, the score of gold medalist be 89.6 m and score of bronze medalist be 87.6 m
This case is possible if P7 improves his scores twice with an amount of 1.2 m each and P5 improves his score with 1.2 m once such that the final score of P7 becomes 87.6 + 2.4 = 89.6 m and final score of P5 becomes 86.4 + 1.2 = 87.6 m
Case III – Let P1 is a bronze medalist with score 88.6 m
So, silver medalists score must be 89.6 m and gold medalist score must be 90.6 m
Now, in case of P7, 90.6 – 87.2 = 3.4 m, so each improved score = 3.4/2 = 1.7 m
But for P5, 89.6 – 86.4 = 3.2 m ≠ 1.7 m
So, this case is also not possible.
Therefore according to Case – II, P7 improves his score in round 4 and round 5 and P5 improves his scores in round 6 and the rest of the information can be gathered as follows
The gold medalist improve his score by 2.4 m in the second phase
CAT 2021 | DILR Set 2
DIRECTIONS for the question: Read the information given below and answer the question that follows.
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.
7. How many questions were DEFINITELY created by Amal? (in numerical value)
Given that Q02, Q06, Q09, Q11, and Q12 were rejected and Q01, Q03, Q04, Q05, Q07, Q08, Q10, and Q13 were accepted.
Q01 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q02 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is rejected, so it must be disapproved by Komal as well
Q03 is reviewed by Amal and Komal, so it must be created by Bimal and Komal disapproves it but Amal approves it as it is accepted
Q04 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Similar to Q01, Q05 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q06 is reviewed by Amal and Bimal, so it may be created by either SME or Komal and since it is rejected, so both Amal and Bimal must have disapproved
Q07 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is accepted, so Bimal disapproves it but Komal approves it
Similar to Q04, Q08 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Q09 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q10 is reviewed by Amal only and it is accepted as well, so it must be created by Komal and Amal approves it
Q11 is reviewed by Amal and Komal, so it must be created by Bimal and since it is rejected, so both Komal and Amal disapproves it
Similar to Q09, Q12 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q13 is reviewed by Amal and Bimal, so it may be created by either SME or Komal
Now if it is created by SME, both Amal and Bimal approves it to be accepted
And if it is created by Komal, then Amal disapproves it but Bimal approves it to be accepted
The rest of the information can be gathered as follows-
✓ means approved and × means disapproved
Amal definitely created 3 questions Q07, Q09, Q12
8. How many questions were DEFINITELY created by Komal? (in numerical value)
Given that Q02, Q06, Q09, Q11, and Q12 were rejected and Q01, Q03, Q04, Q05, Q07, Q08, Q10, and Q13 were accepted.
Q01 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q02 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is rejected, so it must be disapproved by Komal as well
Q03 is reviewed by Amal and Komal, so it must be created by Bimal and Komal disapproves it but Amal approves it as it is accepted
Q04 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Similar to Q01, Q05 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q06 is reviewed by Amal and Bimal, so it may be created by either SME or Komal and since it is rejected, so both Amal and Bimal must have disapproved
Q07 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is accepted, so Bimal disapproves it but Komal approves it
Similar to Q04, Q08 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Q09 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q10 is reviewed by Amal only and it is accepted as well, so it must be created by Komal and Amal approves it
Q11 is reviewed by Amal and Komal, so it must be created by Bimal and since it is rejected, so both Komal and Amal disapproves it
Similar to Q09, Q12 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q13 is reviewed by Amal and Bimal, so it may be created by either SME or Komal
Now if it is created by SME, both Amal and Bimal approves it to be accepted
And if it is created by Komal, then Amal disapproves it but Bimal approves it to be accepted
The rest of the information can be gathered as follows-
✓ means approved and × means disapproved
Komal definitely created 1 question Q10
9. How many questions were DEFINITELY created by the SMEs? (in numerical value)
Given that Q02, Q06, Q09, Q11, and Q12 were rejected and Q01, Q03, Q04, Q05, Q07, Q08, Q10, and Q13 were accepted.
Q01 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q02 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is rejected, so it must be disapproved by Komal as well
Q03 is reviewed by Amal and Komal, so it must be created by Bimal and Komal disapproves it but Amal approves it as it is accepted
Q04 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Similar to Q01, Q05 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q06 is reviewed by Amal and Bimal, so it may be created by either SME or Komal and since it is rejected, so both Amal and Bimal must have disapproved
Q07 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is accepted, so Bimal disapproves it but Komal approves it
Similar to Q04, Q08 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Q09 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q10 is reviewed by Amal only and it is accepted as well, so it must be created by Komal and Amal approves it
Q11 is reviewed by Amal and Komal, so it must be created by Bimal and since it is rejected, so both Komal and Amal disapproves it
Similar to Q09, Q12 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q13 is reviewed by Amal and Bimal, so it may be created by either SME or Komal
Now if it is created by SME, both Amal and Bimal approves it to be accepted
And if it is created by Komal, then Amal disapproves it but Bimal approves it to be accepted
The rest of the information can be gathered as follows-
✓ means approved and × means disapproved
SME definitely created 3 questions Q02, Q04 and Q08
10. How many questions were DEFINITELY disapproved by Bimal? (in numerical value)
Given that Q02, Q06, Q09, Q11, and Q12 were rejected and Q01, Q03, Q04, Q05, Q07, Q08, Q10, and Q13 were accepted.
Q01 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q02 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is rejected, so it must be disapproved by Komal as well
Q03 is reviewed by Amal and Komal, so it must be created by Bimal and Komal disapproves it but Amal approves it as it is accepted
Q04 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Similar to Q01, Q05 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q06 is reviewed by Amal and Bimal, so it may be created by either SME or Komal and since it is rejected, so both Amal and Bimal must have disapproved
Q07 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is accepted, so Bimal disapproves it but Komal approves it
Similar to Q04, Q08 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Q09 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q10 is reviewed by Amal only and it is accepted as well, so it must be created by Komal and Amal approves it
Q11 is reviewed by Amal and Komal, so it must be created by Bimal and since it is rejected, so both Komal and Amal disapproves it
Similar to Q09, Q12 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q13 is reviewed by Amal and Bimal, so it may be created by either SME or Komal
Now if it is created by SME, both Amal and Bimal approves it to be accepted
And if it is created by Komal, then Amal disapproves it but Bimal approves it to be accepted
The rest of the information can be gathered as follows-
✓ means approved and × means disapproved
Bimal definitely disapproved 4 questions Q06, Q07, Q09 and Q12
11. 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?
- lies between 0.25 and 0.75
- 0.25
- either 0.25 or 0.75
- lies between 0.25 and 0.50
Given that Q02, Q06, Q09, Q11, and Q12 were rejected and Q01, Q03, Q04, Q05, Q07, Q08, Q10, and Q13 were accepted.
Q01 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q02 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is rejected, so it must be disapproved by Komal as well
Q03 is reviewed by Amal and Komal, so it must be created by Bimal and Komal disapproves it but Amal approves it as it is accepted
Q04 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Similar to Q01, Q05 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q06 is reviewed by Amal and Bimal, so it may be created by either SME or Komal and since it is rejected, so both Amal and Bimal must have disapproved
Q07 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is accepted, so Bimal disapproves it but Komal approves it
Similar to Q04, Q08 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Q09 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q10 is reviewed by Amal only and it is accepted as well, so it must be created by Komal and Amal approves it
Q11 is reviewed by Amal and Komal, so it must be created by Bimal and since it is rejected, so both Komal and Amal disapproves it
Similar to Q09, Q12 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q13 is reviewed by Amal and Bimal, so it may be created by either SME or Komal
Now if it is created by SME, both Amal and Bimal approves it to be accepted
And if it is created by Komal, then Amal disapproves it but Bimal approves it to be accepted
The rest of the information can be gathered as follows-
✓ means approved and × means disapproved
Amal reviewed 8 questions, among them he approved at least 2 questions Q03, Q10 and maximum Amal can approve 6 questions Q02, Q03, Q04, Q08, Q10 and Q13
Approval ratio of Amal lies between 2/8 = 0.25 and 6/8 = 0.75
12. How many questions created by Amal or Bimal were disapproved by at least one of the other reviewers?
- 5
- 2
- 4
- 7
Given that Q02, Q06, Q09, Q11, and Q12 were rejected and Q01, Q03, Q04, Q05, Q07, Q08, Q10, and Q13 were accepted.
Q01 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q02 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is rejected, so it must be disapproved by Komal as well
Q03 is reviewed by Amal and Komal, so it must be created by Bimal and Komal disapproves it but Amal approves it as it is accepted
Q04 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Similar to Q01, Q05 is only reviewed by Komal and is accepted, so it must be created by Bimal
Q06 is reviewed by Amal and Bimal, so it may be created by either SME or Komal and since it is rejected, so both Amal and Bimal must have disapproved
Q07 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is accepted, so Bimal disapproves it but Komal approves it
Similar to Q04, Q08 is reviewed by all three Amal, Bimal and Komal, so it must be created by SME and either Amal or Bimal disapproves it and since it is accepted, so it must be approved by Komal
Q09 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q10 is reviewed by Amal only and it is accepted as well, so it must be created by Komal and Amal approves it
Q11 is reviewed by Amal and Komal, so it must be created by Bimal and since it is rejected, so both Komal and Amal disapproves it
Similar to Q09, Q12 is reviewed by both Bimal and Komal, so it must be created by Amal and since it is rejected, so both Bimal and Komal disapproves it
Q13 is reviewed by Amal and Bimal, so it may be created by either SME or Komal
Now if it is created by SME, both Amal and Bimal approves it to be accepted
And if it is created by Komal, then Amal disapproves it but Bimal approves it to be accepted
The rest of the information can be gathered as follows-
✓ means approved and × means disapproved
For Amal, Q07, Q09 and Q12 or for Bimal Q03 and Q11 are disapproved by at least one of the other reviewers
Total = 5 questions
CAT 2021 | DILR Set 3
DIRECTIONS for the question: Analyse the graph/s given below and answer the question that follows.
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 weight age 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 weight age 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.
13. 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.
- Only II
- Only I
- Neither I nor II
- Both I and II
The given information can be gathered as follows
From the above table, it is clear that
I: The total project-month was the same for the four employees = 9 is true
II: The total employee-month was the same for the five projects is not true
Only I is true
14. Which employees did not work in multiple projects for any of the months in 2020?
- Only Abani and Bahni
- Only Tinni
- All four of them
- Only Abani, Bahni and Danni
The given information can be gathered as follows
By observation only Tinni worked in multiple projects, so Abani, Bahni and Danni did not work in multiple projects
15. 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?
- Project 4, Project 5
- Project 3, Project 5
- Project 3, Project 4
- Project 1, Project 5
The given information can be gathered as follows
Again, by observation, project duration of
Project 1 = Jan to Mar = 3 months
Project 2 = Feb to Apr = 3 months
Project 3 = Apr to Aug = 5 months
Project 4 = Jul to Nov = 5 months
Project 5 = Sep to Dec = 4 months
Hence, project duration as per options, Project 3 and Project 4 is same
16. The list of employees in decreasing order of annual completion index is:
- Danni, Tinni, Bahni, Abani
- Danni, Tinni, Abani, Bahni
- Tinni, Danni, Abani, Bahni
- Bahni, Abani, Tinni, Danni
The given information can be gathered as follows
From the above table, it is clear that in terms of Annual Completion Index
Danni > Tinni > Abani > Bahni
CAT 2021 | DILR Set 4
DIRECTIONS for the question: Read the information given below and answer the question that follows.
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.
17. 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?
- 10 ml or more
- Less than 1 ml
- 1 ml
- 10 ml
Given, A = 0% I and 100% P Let B = x% I and (100 – x)% P Since they are mixed in equal quantity to detect the presence of I. (0% I + x% I)/2 ≥ 10% Total, x% I ≥ 20% Total Hence, x% I ≥ 20% of 50 ml = 10 ml
18. 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”? (in numerical value)
Since each bottle contains only P or only I Take equal quantity from each bottle say 10 ml and mix them Now, if at least any one of them will contain only I, then I % = 10/40 × 100 = 25% > 10%, so impurity will be detected And if all four bottles contain only P, then 0% I will be detected Hence, minimum of 1 test required to ascertain that all of them contain only P
19. 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? (in numerical value)
Since three bottles contains only P and one contains 80% P and 20% I Taking equal quantity from each bottle say 10 ml and mixing them I % = 2/40 × 100 = 5% < 10%, so only one test is not sufficient to detect the bottle Let the bottles be B1, B2, B3 and B4 in any order Now consider any two bottles (say B1 and B2) and take equal quantity from each bottle say 10 ml and mix them, Case I, if I% = 0, then both the bottles are only P and now take any one bottle from either B1 or B2 and mix with either B3 or B4 if I% is still = 0, then the remaining bottle contains 20% I and if I% = (0 + 20)/2 = 10, then the newly mixed bottle contains 20% I Case II, if I% = 10, then remaining B3 and B4 are only P, now take either B1 or B2 and mix with B3 or B4 If I% = 0, then the other bottle among B1 or B2 contains 20% I And if I% is still = 10, then the bottle taken among B1 or B2 contains 20% I Hence, minimum of 2 tests required in either of the cases
20. 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?
- 2
- 1
- 3
- 4
Case I, if only one bottle contains only P and remaining three bottles contains 15% I Take equal quantity from each bottle and mix them I% = (0 + 15 + 15 + 15)/4 = 11.25 > 10, so impurity will be detected Case II, if two bottles contain only P and remaining two bottles contain 15% I Take equal quantity from each bottle and mix them I% = (0 + 0 + 15 + 15)/4 = 7.5 < 10, so impurity will not be detected Hence, minimum of only 1 test ascertain the exact number of bottles containing only P
