CAT 2021 Previous Year Questions: Tables Notes | Study Logical Reasoning (LR) and Data Interpretation (DI) - CAT
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Passage - 1
Ten objects o1, o2, …, o10 were distributed among Amar, Barat, Charles, Disha, and Elise. Each item went to exactly one person. Each person got exactly two of the items, and this pair of objects is called her/his bundle. The following table shows how each person values each object. The value of any bundle by a person is the sum of that person’s values of the objects in that bundle. A person X envies another person Y if X values Y’s bundle more than X’s own bundle. For example, hypothetically suppose Amar’s bundle consists of o1 and o2, and Barat’s bundle consists of o3 and o4. Then Amar values his own bundle at 4 + 9 = 13 and Barat’s bundle at 9 + 3 = 12. Hence Amar does not envy Barat. On the other hand, Barat values his own bundle at 7 + 5 = 12 and Amar’s bundle at 5 + 9 = 14. Hence Barat envies Amar. The following facts are known about the actual distribution of the objects among the five people. 1. If someone’s value for an object is 10, then she/he received that object. 2. Objects o1, o2, and o3 were given to three different people. 3. Objects o1 and o8 were given to different people. 4. Three people value their own bundles at 16. No one values her/his own bundle at a number higher than 16. 5. Disha values her own bundle at an odd number. All others value their own bundles at an even number. 6. Some people who value their own bundles less than 16 envy some other people who value their own bundle at 16. No one else envies others.
Question for CAT 2021 Previous Year Questions: Tables
Try yourself:What BEST can be said about object o8?
Explanation
We have the following table:
o10 is given to Elise and o9 is given to Bharat . Now as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7. Case 1 : o1 is given to Elise Now the total valuation of Elise = 12 Valuation of Disha is an odd number So we can say Amar , Bharat and Charles values their bundles at 16 . So for Bharat the valuation to be 16, o7 will be given to him so we get Bharat - o9 and o7 and Elise -o10 and o1 For charles to have valuation 16 the only way = 8+8 so we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together ) Now for Amart to have a valuation of 16 the only way possible = 9+7 Now so we can say Amar will receive either o2 or o3 and o5 . Now we are left with 04 and o6 So if Disha receives o4 and o6 The valuation of Disha will be 5+3 =8 which is not an odd number so this case is discarded
Case 2: Elise receives o5 or o7 . Now Valuation of Elise = 16 . And Elise receives o10 and o5/o7. Bharat received o9 and we know the evaluation of Bharat is an even number and the minimum even number possible for valuation of Bharat is 16 and no one can have evaluation more than 16 so Bharat received o7 . So Elise received o5 . So we have Bharat - o9 ,o7 Elise -o10,o5. Now as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha. Now As per Amar he values Bharat at 17 so he envy him So Amar will value his bundle less than 16 So the only possibility for Amar to value his bundle less than 16 = 12 =9+3. Now we can say Charu will have 16 as his own valuation so he will get 8+8 . Now o8 will be given to Charu, and he cannot have o1 , also he cannot have o2 because if he has o2 he will value Bharat’s bundle as 17 and will envy him which is not possible so Charu will have o3,o8 Now Amar will have o2 and Disha will have o1. Now Amar will not have o4 because in that case Charles will envy Amar and is not possible so we can say Amar will have o6 and Disha will have o4. So we have the following Amar - o2,o6 Bharat -o9,o7 Charu -o3,o8 Disha o1,o4 Elise -o10,o5 So o8 is given to Charu.
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*Answer can only contain numeric values
Question for CAT 2021 Previous Year Questions: Tables
Try yourself:What is Amar’s value for his own bundle?
Correct Answer : 12
Explanation
We have the following table:
o10 is given to Elise and o9 is given to Bharat . Now as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7. Case 1 : o1 is given to Elise Now the total valuation of Elise = 12 Valuation of Disha is an odd number So we can say Amar , Bharat and Charles values their bundles at 16 . So for Bharat the valuation to be 16, o7 will be given to him so we get Bharat - o9 and o7 and Elise -o10 and o1 For charles to have valuation 16 the only way = 8+8 so we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together ) Now for Amart to have a valuation of 16 the only way possible = 9+7 Now so we can say Amar will receive either o2 or o3 and o5 . Now we are left with 04 and o6 So if Disha receives o4 and o6 The valuation of Disha will be 5+3 =8 which is not an odd number so this case is discarded
Case 2: Elise receives o5 or o7 . Now Valuation of Elise = 16 . And Elise receives o10 and o5/o7. Bharat received o9 and we know the evaluation of Bharat is an even number and the minimum even number possible for valuation of Bharat is 16 and no one can have evaluation more than 16 so Bharat received o7 . So Elise received o5 . So we have Bharat - o9 ,o7 Elise -o10,o5. Now as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha. Now As per Amar he values Bharat at 17 so he envy him So Amar will value his bundle less than 16 So the only possibility for Amar to value his bundle less than 16 = 12 =9+3. Now we can say Charu will have 16 as his own valuation so he will get 8+8 . Now o8 will be given to Charu, and he cannot have o1 , also he cannot have o2 because if he has o2 he will value Bharat’s bundle as 17 and will envy him which is not possible so Charu will have o3,o8 Now Amar will have o2 and Disha will have o1. Now Amar will not have o4 because in that case Charles will envy Amar and is not possible so we can say Amar will have o6 and Disha will have o4. So we have the following Amar - o2,o6 Bharat -o9,o7 Charu -o3,o8 Disha o1,o4 Elise -o10,o5 Amar's own valuation = 9+3 =12
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:Object o5 was given to
Explanation
We have the following table:
o10 is given to Elise and o9 is given to Bharat . Now as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7. Case 1 : o1 is given to Elise Now the total valuation of Elise = 12 Valuation of Disha is an odd number So we can say Amar , Bharat and Charles values their bundles at 16 . So for Bharat the valuation to be 16, o7 will be given to him so we get Bharat - o9 and o7 and Elise -o10 and o1 For charles to have valuation 16 the only way = 8+8 so we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together ) Now for Amart to have a valuation of 16 the only way possible = 9+7 Now so we can say Amar will receive either o2 or o3 and o5 . Now we are left with 04 and o6 So if Disha receives o4 and o6 The valuation of Disha will be 5+3 =8 which is not an odd number so this case is discarded
Case 2: Elise receives o5 or o7 . Now Valuation of Elise = 16 . And Elise receives o10 and o5/o7. Bharat received o9 and we know the evaluation of Bharat is an even number and the minimum even number possible for valuation of Bharat is 16 and no one can have evaluation more than 16 so Bharat received o7 . So Elise received o5 . So we have Bharat - o9 ,o7 Elise -o10,o5. Now as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha. Now As per Amar he values Bharat at 17 so he envy him So Amar will value his bundle less than 16 So the only possibility for Amar to value his bundle less than 16 = 12 =9+3. Now we can say Charu will have 16 as his own valuation so he will get 8+8 . Now o8 will be given to Charu, and he cannot have o1 , also he cannot have o2 because if he has o2 he will value Bharat’s bundle as 17 and will envy him which is not possible so Charu will have o3,o8 Now Amar will have o2 and Disha will have o1. Now Amar will not have o4 because in that case Charles will envy Amar and is not possible so we can say Amar will have o6 and Disha will have o4. So we have the following Amar - o2,o6 Bharat -o9,o7 Charu -o3,o8 Disha o1,o4 Elise -o10,o5 o5 is given to Elise
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:Who among the following envies someone else?
Explanation
We have the following table:
o10 is given to Elise and o9 is given to Bharat . Now as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7. Case 1 : o1 is given to Elise Now the total valuation of Elise = 12 Valuation of Disha is an odd number So we can say Amar , Bharat and Charles values their bundles at 16 . So for Bharat the valuation to be 16, o7 will be given to him so we get Bharat - o9 and o7 and Elise -o10 and o1 For charles to have valuation 16 the only way = 8+8 so we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together ) Now for Amart to have a valuation of 16 the only way possible = 9+7 Now so we can say Amar will receive either o2 or o3 and o5 . Now we are left with 04 and o6 So if Disha receives o4 and o6 The valuation of Disha will be 5+3 =8 which is not an odd number so this case is discarded
Case 2: Elise receives o5 or o7 . Now Valuation of Elise = 16 . And Elise receives o10 and o5/o7. Bharat received o9 and we know the evaluation of Bharat is an even number and the minimum even number possible for valuation of Bharat is 16 and no one can have evaluation more than 16 so Bharat received o7 . So Elise received o5 . So we have Bharat - o9 ,o7 Elise -o10,o5. Now as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha. Now As per Amar he values Bharat at 17 so he envy him So Amar will value his bundle less than 16 So the only possibility for Amar to value his bundle less than 16 = 12 =9+3. Now we can say Charu will have 16 as his own valuation so he will get 8+8 . Now o8 will be given to Charu, and he cannot have o1 , also he cannot have o2 because if he has o2 he will value Bharat’s bundle as 17 and will envy him which is not possible so Charu will have o3,o8 Now Amar will have o2 and Disha will have o1. Now Amar will not have o4 because in that case Charles will envy Amar and is not possible so we can say Amar will have o6 and Disha will have o4. So we have the following: Amar - o2,o6 Bharat -o9,o7 Charu -o3,o8 Disha o1,o4 Elise -o10,o5 So Amar envies someone else
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:Object o4 was given to
Explanation
We have the following table:
o10 is given to Elise and o9 is given to Bharat . Now as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7. Case 1 : o1 is given to Elise Now the total valuation of Elise = 12 Valuation of Disha is an odd number So we can say Amar , Bharat and Charles values their bundles at 16 . So for Bharat the valuation to be 16, o7 will be given to him so we get Bharat - o9 and o7 and Elise -o10 and o1 For charles to have valuation 16 the only way = 8+8 so we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together ) Now for Amart to have a valuation of 16 the only way possible = 9+7 Now so we can say Amar will receive either o2 or o3 and o5 . Now we are left with 04 and o6 So if Disha receives o4 and o6 The valuation of Disha will be 5+3 =8 which is not an odd number so this case is discarded
Case 2: Elise receives o5 or o7 . Now Valuation of Elise = 16 . And Elise receives o10 and o5/o7. Bharat received o9 and we know the evaluation of Bharat is an even number and the minimum even number possible for valuation of Bharat is 16 and no one can have evaluation more than 16 so Bharat received o7 . So Elise received o5 . So we have Bharat - o9 ,o7 Elise -o10,o5. Now as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha. Now As per Amar he values Bharat at 17 so he envy him So Amar will value his bundle less than 16 So the only possibility for Amar to value his bundle less than 16 = 12 =9+3. Now we can say Charu will have 16 as his own valuation so he will get 8+8 . Now o8 will be given to Charu, and he cannot have o1 , also he cannot have o2 because if he has o2 he will value Bharat’s bundle as 17 and will envy him which is not possible so Charu will have o3,o8 Now Amar will have o2 and Disha will have o1. Now Amar will not have o4 because in that case Charles will envy Amar and is not possible so we can say Amar will have o6 and Disha will have o4. So we have the following: Amar - o2,o6 Bharat -o9,o7 Charu -o3,o8 Disha o1,o4 Elise -o10,o5 o4 is given to Disha
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:What BEST can be said about the distribution of object o1?
Explanation
We have the following table:
o10 is given to Elise and o9 is given to Bharat . Now as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7. Case 1 : o1 is given to Elise Now the total valuation of Elise = 12 Valuation of Disha is an odd number So we can say Amar , Bharat and Charles values their bundles at 16 . So for Bharat the valuation to be 16, o7 will be given to him so we get Bharat - o9 and o7 and Elise -o10 and o1 For charles to have valuation 16 the only way = 8+8 so we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together ) Now for Amart to have a valuation of 16 the only way possible = 9+7 Now so we can say Amar will receive either o2 or o3 and o5 . Now we are left with 04 and o6 So if Disha receives o4 and o6 The valuation of Disha will be 5+3 =8 which is not an odd number so this case is discarded
Case 2: Elise receives o5 or o7 . Now Valuation of Elise = 16 . And Elise receives o10 and o5/o7. Bharat received o9 and we know the evaluation of Bharat is an even number and the minimum even number possible for valuation of Bharat is 16 and no one can have evaluation more than 16 so Bharat received o7 . So Elise received o5 . So we have Bharat - o9 ,o7 Elise -o10,o5. Now as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha. Now As per Amar he values Bharat at 17 so he envy him So Amar will value his bundle less than 16 So the only possibility for Amar to value his bundle less than 16 = 12 =9+3. Now we can say Charu will have 16 as his own valuation so he will get 8+8 . Now o8 will be given to Charu, and he cannot have o1 , also he cannot have o2 because if he has o2 he will value Bharat’s bundle as 17 and will envy him which is not possible so Charu will have o3,o8 Now Amar will have o2 and Disha will have o1. Now Amar will not have o4 because in that case Charles will envy Amar and is not possible so we can say Amar will have o6 and Disha will have o4. So we have the following: Amar - o2,o6 Bharat -o9,o7 Charu -o3,o8 Disha o1,o4 Elise -o10,o5 So o1 is given to Disha
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Passage - 2
Ravi works in an online food-delivery company. After each delivery, customers rate Ravi on each of four parameters - Behaviour, Packaging, Hygiene, and Timeliness, on a scale from 1 to 9. If the total of the four rating points is 25 or more, then Ravi gets a bonus of ₹20 for that delivery. Additionally, a customer may or may not give Ravi a tip. If the customer gives a tip, it is either ₹30 or ₹50. One day, Ravi made four deliveries - one to each of Atal, Bihari, Chirag, and Deepak, and received a total of ₹120 in bonus and tips. He did not get both a bonus and a tip from the same customer. The following additional facts are also known.
1. In Timeliness, Ravi received a total of 21 points, and three of the customers gave him the same rating points in this parameter. Atal gave higher rating points than Bihari and Chirag in this parameter. 2. Ravi received distinct rating points in Packaging from the four customers adding up to 29 points. Similarly, Ravi received distinct rating points in Hygiene from the four customers adding up to 26 points. 3. Chirag gave the same rating points for Packaging and Hygiene. 4. Among the four customers, Bihari gave the highest rating points in Packaging, and Chirag gave the highest rating points in Hygiene. 5. Everyone rated Ravi between 5 and 7 in Behaviour. Unique maximum and minimum ratings in this parameter were given by Atal and Deepak respectively. 6. If the customers are ranked based on ratings given by them in individual parameters, then Atal’s rank based on Packaging is the same as that based on Hygiene. This is also true for Deepak.
*Answer can only contain numeric values
Question for CAT 2021 Previous Year Questions: Tables
Try yourself:What was the minimum rating that Ravi received from any customer in any parameter?
Correct Answer : 5
Explanation
Using condition 1 : Ravi had a total of 21 points in timeliness. 3 of the four customers among Atal, Bihari, Chirag, and Deepak gave him the same ratings. Atal gave the highest rating in timeliness in comparison with Bihari and Chirag. Hence he must have given the distinct rating.
Using condition 2 : The possibilities are A - 9, B - 4, C - 4, D - 4. A - 6, B - 5, C - 5, D - 5. Ravi received a total of 29 points in the Packaging and for this, the possibility of the four scores are (5, 7, 8, 9) awarded by the four customers. In Hygiene the sum of the ratings awarded was 26. This could possibly be awarded by considering the following cases : A-(4,6,7,9) , B-( 5,6,7,8), C-(4,5,8,9). Using condition 4: Bihari gave the highest rating in packaging and thus Bihari must have given a 9 rating in the packaging. Chirag gave the highest rating in Hygiene. In condition 3 it was mentioned that Chirag gave the same points for packaging and hygiene. Since 9 was rated by Bihari packaging it cannot be awarded by Chirag in packaging and hygiene. Since Chirag was awarded the highest in Hygiene. He must award 8 points in Hygiene and Packaging. Hence of the three possibilities among A, B, and C for Hygiene only B is the possible case with 8 as the maximum score. In condition 5 it was mentioned that everyone awarded Ravi between 5 and 7 in Behaviour. Unique maximum and minimum ratings in this parameter were given by Atal and Deepak respectively. Hence Atal must have awarded 7, Deepak 6, Bihari, and Chirag 6 each in Behaviour.
The two possible cases are : Case 1 :
Case 2 :
The ratings awarded by Atal and Deepak in Packaging are among 5 and 7.The ratings awarded by Atal, Bihari, Deepak are among 5,6, and 7. Atal individual ranking in Packaging and Hygiene are the same. The same is true for Deepak. Since Atal and Deepak can give the ranking among 3 and 4 in Packaging as Bihari is first and Chirag is second in this parameter. They can rank 3 or 4 in the Hygiene parameter also. Hence Bihari must rate 7 points in Hygiene. In both the possibilities Bihari and Chirag award a total of 26 points. Hence he wins 40 because the total ratings are greater than 25 received from Bihari and Chirag. Since he gets a total of 120 in bonuses and tips. He must have 80 from Atal and Deepak. This is possible if he gets a tip of 30 ad 50 from them respectively. In case 1 irrespective of Atal standing at rank 3 or rank 4 in Hygiene and Packaging Atal total rating is greater than 25 which implies Ravi gets a tip from Atal but this is not a possible case because Ravi needs a total of Rs 80 from Atal and Deepak. From Atal if he gets Rs 20 as a bonus he cannot get a total of Rs 120 and hence this case fails. Hence case 1 fails.
In case 2 there are two possibilities : Atal ranking 3 in both the parameters and Deepak 4th. Atal ranking 4th in both the parameters and Deepak 3rd In the case where Atal ranks 3rd in Packaging and Hygiene the total score is 26 and is not a feasible case. Case - 2A
Case - 2B :
Case - 2A fails because Atal's total rating is greater than 25 which should not be the case.The minimum rating awarded is 5. The minimum rating awarded is 5.
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*Answer can only contain numeric values
Question for CAT 2021 Previous Year Questions: Tables
Try yourself:What rating did Atal give on Timeliness?
Correct Answer : 6
Explanation
Using condition 1 : Ravi had a total of 21 points in timeliness. 3 of the four customers among Atal, Bihari, Chirag, and Deepak gave him the same ratings. Atal gave the highest rating in timeliness in comparison with Bihari and Chirag. Hence he must have given the distinct rating.
Using condition 2 : The possibilities are A - 9, B - 4, C - 4, D - 4. A - 6, B - 5, C - 5, D - 5. Ravi received a total of 29 points in the Packaging and for this, the possibility of the four scores are (5, 7, 8, 9) awarded by the four customers. In Hygiene the sum of the ratings awarded was 26. This could possibly be awarded by considering the following cases : A-(4,6,7,9) , B-( 5,6,7,8), C-(4,5,8,9). Using condition 4: Bihari gave the highest rating in packaging and thus Bihari must have given a 9 rating in the packaging. Chirag gave the highest rating in Hygiene. In condition 3 it was mentioned that Chirag gave the same points for packaging and hygiene. Since 9 was rated by Bihari packaging it cannot be awarded by Chirag in packaging and hygiene. Since Chirag was awarded the highest in Hygiene. He must award 8 points in Hygiene and Packaging. Hence of the three possibilities among A, B, and C for Hygiene only B is the possible case with 8 as the maximum score. In condition 5 it was mentioned that everyone awarded Ravi between 5 and 7 in Behaviour. Unique maximum and minimum ratings in this parameter were given by Atal and Deepak respectively. Hence Atal must have awarded 7, Deepak 6, Bihari, and Chirag 6 each in Behaviour.
The two possible cases are : Case 1 :
Case 2 :
The ratings awarded by Atal and Deepak in Packaging are among 5 and 7.The ratings awarded by Atal, Bihari, Deepak are among 5,6, and 7. Atal individual ranking in Packaging and Hygiene are the same. The same is true for Deepak. Since Atal and Deepak can give the ranking among 3 and 4 in Packaging as Bihari is first and Chirag is second in this parameter. They can rank 3 or 4 in the Hygiene parameter also. Hence Bihari must rate 7 points in Hygiene. In both the possibilities Bihari and Chirag award a total of 26 points. Hence he wins 40 because the total ratings are greater than 25 received from Bihari and Chirag. Since he gets a total of 120 in bonuses and tips. He must have 80 from Atal and Deepak. This is possible if he gets a tip of 30 ad 50 from them respectively. In case 1 irrespective of Atal standing at rank 3 or rank 4 in Hygiene and Packaging Atal total rating is greater than 25 which implies Ravi gets a tip from Atal but this is not a possible case because Ravi needs a total of Rs 80 from Atal and Deepak. From Atal if he gets Rs 20 as a bonus he cannot get a total of Rs 120 and hence this case fails. Hence case 1 fails.
In case 2 there are two possibilities : Atal ranking 3 in both the parameters and Deepak 4th. Atal ranking 4th in both the parameters and Deepak 3rd In the case where Atal ranks 3rd in Packaging and Hygiene the total score is 26 and is not a feasible case. Case - 2A
Case - 2B :
Case - 2A: fails because Atal's total rating is greater than 25 which should not be the case. Atal has given a rating of 6 in timeliness
Check
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:In which parameter did Atal give the maximum rating points to Ravi?
Explanation
Using condition 1 : Ravi had a total of 21 points in timeliness. 3 of the four customers among Atal, Bihari, Chirag, and Deepak gave him the same ratings. Atal gave the highest rating in timeliness in comparison with Bihari and Chirag. Hence he must have given the distinct rating.
Using condition 2 : The possibilities are A - 9, B - 4, C - 4, D - 4. A - 6, B - 5, C - 5, D - 5. Ravi received a total of 29 points in the Packaging and for this, the possibility of the four scores are (5, 7, 8, 9) awarded by the four customers. In Hygiene the sum of the ratings awarded was 26. This could possibly be awarded by considering the following cases : A-(4,6,7,9) , B-( 5,6,7,8), C-(4,5,8,9). Using condition 4: Bihari gave the highest rating in packaging and thus Bihari must have given a 9 rating in the packaging. Chirag gave the highest rating in Hygiene. In condition 3 it was mentioned that Chirag gave the same points for packaging and hygiene. Since 9 was rated by Bihari packaging it cannot be awarded by Chirag in packaging and hygiene. Since Chirag was awarded the highest in Hygiene. He must award 8 points in Hygiene and Packaging. Hence of the three possibilities among A, B, and C for Hygiene only B is the possible case with 8 as the maximum score. In condition 5 it was mentioned that everyone awarded Ravi between 5 and 7 in Behaviour. Unique maximum and minimum ratings in this parameter were given by Atal and Deepak respectively. Hence Atal must have awarded 7, Deepak 6, Bihari, and Chirag 6 each in Behaviour.
The two possible cases are : Case 1 :
Case 2 :
The ratings awarded by Atal and Deepak in Packaging are among 5 and 7.The ratings awarded by Atal, Bihari, Deepak are among 5,6, and 7. Atal individual ranking in Packaging and Hygiene are the same. The same is true for Deepak. Since Atal and Deepak can give the ranking among 3 and 4 in Packaging as Bihari is first and Chirag is second in this parameter. They can rank 3 or 4 in the Hygiene parameter also. Hence Bihari must rate 7 points in Hygiene. In both the possibilities Bihari and Chirag award a total of 26 points. Hence he wins 40 because the total ratings are greater than 25 received from Bihari and Chirag. Since he gets a total of 120 in bonuses and tips. He must have 80 from Atal and Deepak. This is possible if he gets a tip of 30 ad 50 from them respectively. In case 1 irrespective of Atal standing at rank 3 or rank 4 in Hygiene and Packaging Atal total rating is greater than 25 which implies Ravi gets a tip from Atal but this is not a possible case because Ravi needs a total of Rs 80 from Atal and Deepak. From Atal if he gets Rs 20 as a bonus he cannot get a total of Rs 120 and hence this case fails. Hence case 1 fails.
In case 2 there are two possibilities : Atal ranking 3 in both the parameters and Deepak 4th. Atal ranking 4th in both the parameters and Deepak 3rd In the case where Atal ranks 3rd in Packaging and Hygiene the total score is 26 and is not a feasible case. Case - 2A
Case - 2B :
Case - 2A: fails because Atal's total rating is greater than 25 which should not be the case. Atal has given the maximum rating in Behaviour.
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:The COMPLETE list of customers who gave the maximum total rating points to Ravi is
Explanation
Using condition 1 : Ravi had a total of 21 points in timeliness. 3 of the four customers among Atal, Bihari, Chirag, and Deepak gave him the same ratings. Atal gave the highest rating in timeliness in comparison with Bihari and Chirag. Hence he must have given the distinct rating.
Using condition 2 : The possibilities are A - 9, B - 4, C - 4, D - 4. A - 6, B - 5, C - 5, D - 5. Ravi received a total of 29 points in the Packaging and for this, the possibility of the four scores are (5, 7, 8, 9) awarded by the four customers. In Hygiene the sum of the ratings awarded was 26. This could possibly be awarded by considering the following cases : A-(4,6,7,9) , B-( 5,6,7,8), C-(4,5,8,9). Using condition 4: Bihari gave the highest rating in packaging and thus Bihari must have given a 9 rating in the packaging. Chirag gave the highest rating in Hygiene. In condition 3 it was mentioned that Chirag gave the same points for packaging and hygiene. Since 9 was rated by Bihari packaging it cannot be awarded by Chirag in packaging and hygiene. Since Chirag was awarded the highest in Hygiene. He must award 8 points in Hygiene and Packaging. Hence of the three possibilities among A, B, and C for Hygiene only B is the possible case with 8 as the maximum score. In condition 5 it was mentioned that everyone awarded Ravi between 5 and 7 in Behaviour. Unique maximum and minimum ratings in this parameter were given by Atal and Deepak respectively. Hence Atal must have awarded 7, Deepak 6, Bihari, and Chirag 6 each in Behaviour.
The two possible cases are : Case 1 :
Case 2 :
The ratings awarded by Atal and Deepak in Packaging are among 5 and 7.The ratings awarded by Atal, Bihari, Deepak are among 5,6, and 7. Atal individual ranking in Packaging and Hygiene are the same. The same is true for Deepak. Since Atal and Deepak can give the ranking among 3 and 4 in Packaging as Bihari is first and Chirag is second in this parameter. They can rank 3 or 4 in the Hygiene parameter also. Hence Bihari must rate 7 points in Hygiene. In both the possibilities Bihari and Chirag award a total of 26 points. Hence he wins 40 because the total ratings are greater than 25 received from Bihari and Chirag. Since he gets a total of 120 in bonuses and tips. He must have 80 from Atal and Deepak. This is possible if he gets a tip of 30 ad 50 from them respectively. In case 1 irrespective of Atal standing at rank 3 or rank 4 in Hygiene and Packaging Atal total rating is greater than 25 which implies Ravi gets a tip from Atal but this is not a possible case because Ravi needs a total of Rs 80 from Atal and Deepak. From Atal if he gets Rs 20 as a bonus he cannot get a total of Rs 120 and hence this case fails. Hence case 1 fails.
In case 2 there are two possibilities : Atal ranking 3 in both the parameters and Deepak 4th. Atal ranking 4th in both the parameters and Deepak 3rd In the case where Atal ranks 3rd in Packaging and Hygiene the total score is 26 and is not a feasible case. Case - 2A
Case - 2B :
Case - 2A : fails because Atal's total rating is greater than 25 which should not be the case. Bihari and Chirag has given the highest ratings
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:What BEST can be concluded about the tip amount given by Deepak?
Explanation
Using condition 1 : Ravi had a total of 21 points in timeliness. 3 of the four customers among Atal, Bihari, Chirag, and Deepak gave him the same ratings. Atal gave the highest rating in timeliness in comparison with Bihari and Chirag. Hence he must have given the distinct rating.
Using condition 2 : The possibilities are A - 9, B - 4, C - 4, D - 4. A - 6, B - 5, C - 5, D - 5. Ravi received a total of 29 points in the Packaging and for this, the possibility of the four scores are (5, 7, 8, 9) awarded by the four customers. In Hygiene the sum of the ratings awarded was 26. This could possibly be awarded by considering the following cases : A-(4,6,7,9) , B-( 5,6,7,8), C-(4,5,8,9). Using condition 4: Bihari gave the highest rating in packaging and thus Bihari must have given a 9 rating in the packaging. Chirag gave the highest rating in Hygiene. In condition 3 it was mentioned that Chirag gave the same points for packaging and hygiene. Since 9 was rated by Bihari packaging it cannot be awarded by Chirag in packaging and hygiene. Since Chirag was awarded the highest in Hygiene. He must award 8 points in Hygiene and Packaging. Hence of the three possibilities among A, B, and C for Hygiene only B is the possible case with 8 as the maximum score. In condition 5 it was mentioned that everyone awarded Ravi between 5 and 7 in Behaviour. Unique maximum and minimum ratings in this parameter were given by Atal and Deepak respectively. Hence Atal must have awarded 7, Deepak 6, Bihari, and Chirag 6 each in Behaviour.
The two possible cases are : Case 1 :
Case 2 :
The ratings awarded by Atal and Deepak in Packaging are among 5 and 7.The ratings awarded by Atal, Bihari, Deepak are among 5,6, and 7. Atal individual ranking in Packaging and Hygiene are the same. The same is true for Deepak. Since Atal and Deepak can give the ranking among 3 and 4 in Packaging as Bihari is first and Chirag is second in this parameter. They can rank 3 or 4 in the Hygiene parameter also. Hence Bihari must rate 7 points in Hygiene. In both the possibilities Bihari and Chirag award a total of 26 points. Hence he wins 40 because the total ratings are greater than 25 received from Bihari and Chirag. Since he gets a total of 120 in bonuses and tips. He must have 80 from Atal and Deepak. This is possible if he gets a tip of 30 ad 50 from them respectively. In case 1 irrespective of Atal standing at rank 3 or rank 4 in Hygiene and Packaging Atal total rating is greater than 25 which implies Ravi gets a tip from Atal but this is not a possible case because Ravi needs a total of Rs 80 from Atal and Deepak. From Atal if he gets Rs 20 as a bonus he cannot get a total of Rs 120 and hence this case fails. Hence case 1 fails.
In case 2 there are two possibilities : Atal ranking 3 in both the parameters and Deepak 4th. Atal ranking 4th in both the parameters and Deepak 3rd In the case where Atal ranks 3rd in Packaging and Hygiene the total score is 26 and is not a feasible case. Case - 2A
Case - 2B :
Case - 2A: fails because Atal's total rating is greater than 25 which should not be the case. Among Atal and Deepak, one of them gives a tip of 30 and the other gives a tip of 50. Hence 30 or 50 any case is possible
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:What rating did Deepak give on Packaging?
Explanation
Using condition 1 : Ravi had a total of 21 points in timeliness. 3 of the four customers among Atal, Bihari, Chirag, and Deepak gave him the same ratings. Atal gave the highest rating in timeliness in comparison with Bihari and Chirag. Hence he must have given the distinct rating.
Using condition 2 : The possibilities are A - 9, B - 4, C - 4, D - 4. A - 6, B - 5, C - 5, D - 5. Ravi received a total of 29 points in the Packaging and for this, the possibility of the four scores are (5, 7, 8, 9) awarded by the four customers. In Hygiene the sum of the ratings awarded was 26. This could possibly be awarded by considering the following cases : A-(4,6,7,9) , B-( 5,6,7,8), C-(4,5,8,9). Using condition 4: Bihari gave the highest rating in packaging and thus Bihari must have given a 9 rating in the packaging. Chirag gave the highest rating in Hygiene. In condition 3 it was mentioned that Chirag gave the same points for packaging and hygiene. Since 9 was rated by Bihari packaging it cannot be awarded by Chirag in packaging and hygiene. Since Chirag was awarded the highest in Hygiene. He must award 8 points in Hygiene and Packaging. Hence of the three possibilities among A, B, and C for Hygiene only B is the possible case with 8 as the maximum score. In condition 5 it was mentioned that everyone awarded Ravi between 5 and 7 in Behaviour. Unique maximum and minimum ratings in this parameter were given by Atal and Deepak respectively. Hence Atal must have awarded 7, Deepak 6, Bihari, and Chirag 6 each in Behaviour.
The two possible cases are : Case 1 :
Case 2 :
The ratings awarded by Atal and Deepak in Packaging are among 5 and 7.The ratings awarded by Atal, Bihari, Deepak are among 5,6, and 7. Atal individual ranking in Packaging and Hygiene are the same. The same is true for Deepak. Since Atal and Deepak can give the ranking among 3 and 4 in Packaging as Bihari is first and Chirag is second in this parameter. They can rank 3 or 4 in the Hygiene parameter also. Hence Bihari must rate 7 points in Hygiene. In both the possibilities Bihari and Chirag award a total of 26 points. Hence he wins 40 because the total ratings are greater than 25 received from Bihari and Chirag. Since he gets a total of 120 in bonuses and tips. He must have 80 from Atal and Deepak. This is possible if he gets a tip of 30 ad 50 from them respectively. In case 1 irrespective of Atal standing at rank 3 or rank 4 in Hygiene and Packaging Atal total rating is greater than 25 which implies Ravi gets a tip from Atal but this is not a possible case because Ravi needs a total of Rs 80 from Atal and Deepak. From Atal if he gets Rs 20 as a bonus he cannot get a total of Rs 120 and hence this case fails. Hence case 1 fails.
In case 2 there are two possibilities : Atal ranking 3 in both the parameters and Deepak 4th. Atal ranking 4th in both the parameters and Deepak 3rd In the case where Atal ranks 3rd in Packaging and Hygiene the total score is 26 and is not a feasible case. Case - 2A
Case - 2B :
Case - 2A : fails because Atal's total rating is greater than 25 which should not be the case. Deepak gives a rating of 7 in Packaging.
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Passage - 3
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 Distances covered by all the valid throws in the third round The following facts are also known. i. 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. ii. 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. iii. In each round of the second phase, exactly one player improved his score. Each of these improvements was by the same amount. iv. 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. v. 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.
Question for CAT 2021 Previous Year Questions: Tables
Try yourself:Which two players got the double?
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:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had 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(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9. So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5. Now let us consider the two players who got a double. Doubles happen in the transition between rounds. 1 -> 2 - Not possible 2 -> 3 - Possible if P10 reaches Rank 1 after round 2. 3 -> 4 - P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens. 4 - > 5 AND 5 - > 6 not possible. So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2. So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions' distances are in AP.P1 - 88.6 + x P7 - 87.2 + x P5 - 86.4 + x P9 - 84.1 + x The differences do not satisfy the condition. Hence, case 1 is invalidated. Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1. Let us take the cases where P1 is individually the G, S and B medallists. Case 1: P1 is a G medallist. P1 - 88.6 The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case. Case 2: P1 is the S medallist. P1 - 88.6 G - 89.6 B - 87.6 Now, if we see the differences 89.6 - 87.2 = 2.4 87.6 - 86.4 = 1.2 This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner. Hence, P1 - Silver P7 - Gold P5 - Bronze Hence, P8 and P10 got the doubles.
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:Who threw the last javelin in the event?
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:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had 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(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9. So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5. Now let us consider the two players who got a double. Doubles happen in the transition between rounds. 1 -> 2 - Not possible 2 -> 3 - Possible if P10 reaches Rank 1 after round 2. 3 -> 4 - P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens. 4 - > 5 AND 5 - > 6 not possible. So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2. So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions' distances are in AP.P1 - 88.6 + x P7 - 87.2 + x P5 - 86.4 + x P9 - 84.1 + x The differences do not satisfy the condition. Hence, case 1 is invalidated. Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1. Let us take the cases where P1 is individually the G, S and B medallists. Case 1: P1 is a G medallist. P1 - 88.6 The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case. Case 2: P1 is the S medallist. P1 - 88.6 G - 89.6 B - 87.6 Now, if we see the differences 89.6 - 87.2 = 2.4 87.6 - 86.4 = 1.2 This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner. Hence, P1 - Silver P7 - Gold P5 - Bronze P7 the gold winner had already received rank 1 at the end of Round 5. Hence, he was the last one to throw in the tournament.
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:Which of the following can be the final score (in m) of P8?
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:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had 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(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9. So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5. Now let us consider the two players who got a double. Doubles happen in the transition between rounds. 1 -> 2 - Not possible 2 -> 3 - Possible if P10 reaches Rank 1 after round 2. 3 -> 4 - P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens. 4 - > 5 AND 5 - > 6 not possible. So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2. So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions' distances are in AP.P1 - 88.6 + x P7 - 87.2 + x P5 - 86.4 + x P9 - 84.1 + x The differences do not satisfy the condition. Hence, case 1 is invalidated. Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1. Let us take the cases where P1 is individually the G, S and B medallists. Case 1: P1 is a G medallist. P1 - 88.6 The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case. Case 2: P1 is the S medallist. P1 - 88.6 G - 89.6 B - 87.6 Now, if we see the differences 89.6 - 87.2 = 2.4 87.6 - 86.4 = 1.2 This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner. Hence, P1 - Silver P7 - Gold P5 - Bronze P8 comes between P9 and P6. Hence, 82. 5 < P8 < 84.1 P8 = 82.7
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:Who won the silver medal?
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:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had 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(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9. So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5. Now let us consider the two players who got a double. Doubles happen in the transition between rounds. 1 -> 2 - Not possible 2 -> 3 - Possible if P10 reaches Rank 1 after round 2. 3 -> 4 - P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens. 4 - > 5 AND 5 - > 6 not possible. So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2. So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions' distances are in AP.P1 - 88.6 + x P7 - 87.2 + x P5 - 86.4 + x P9 - 84.1 + x The differences do not satisfy the condition. Hence, case 1 is invalidated. Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1. Let us take the cases where P1 is individually the G, S and B medallists. Case 1: P1 is a G medallist. P1 - 88.6 The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case. Case 2: P1 is the S medallist. P1 - 88.6 G - 89.6 B - 87.6 Now, if we see the differences 89.6 - 87.2 = 2.4 87.6 - 86.4 = 1.2 This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner. Hence, P1 - Silver P7 - Gold P5 - Bronze Hence, P1 won the silver medal.
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:What was the final score (in m) of the silver-medalist?
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:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had 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(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9. So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5. Now let us consider the two players who got a double. Doubles happen in the transition between rounds. 1 -> 2 - Not possible 2 -> 3 - Possible if P10 reaches Rank 1 after round 2. 3 -> 4 - P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens. 4 - > 5 AND 5 - > 6 not possible. So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2. So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions' distances are in AP.P1 - 88.6 + x P7 - 87.2 + x P5 - 86.4 + x P9 - 84.1 + x The differences do not satisfy the condition. Hence, case 1 is invalidated. Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1. Let us take the cases where P1 is individually the G, S and B medallists. Case 1: P1 is a G medallist. P1 - 88.6 The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case. Case 2: P1 is the S medallist. P1 - 88.6 G - 89.6 B - 87.6 Now, if we see the differences 89.6 - 87.2 = 2.4 87.6 - 86.4 = 1.2 This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner. Hence, P1 - Silver P7 - Gold P5 - Bronze P1 - 88.6 m.
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Question for CAT 2021 Previous Year Questions: Tables
Try yourself:By how much did the gold medalist improve his score (in m) in the second phase?
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:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had 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(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9. So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5. Now let us consider the two players who got a double. Doubles happen in the transition between rounds. 1 -> 2 - Not possible 2 -> 3 - Possible if P10 reaches Rank 1 after round 2. 3 -> 4 - P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens. 4 - > 5 AND 5 - > 6 not possible. So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2. So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions' distances are in AP.P1 - 88.6 + x P7 - 87.2 + x P5 - 86.4 + x P9 - 84.1 + x The differences do not satisfy the condition. Hence, case 1 is invalidated. Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1. Let us take the cases where P1 is individually the G, S and B medallists. Case 1: P1 is a G medallist. P1 - 88.6 The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case. Case 2: P1 is the S medallist. P1 - 88.6 G - 89.6 B - 87.6 Now, if we see the differences 89.6 - 87.2 = 2.4 87.6 - 86.4 = 1.2 This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner. Hence, P1 - Silver P7 - Gold P5 - Bronze P7 improved by a total of 2.4.