Understanding the set :- 1) We are given that there are three participants – Akhil, Bimal and Chatur. 2) Played a game for 5 days – where, i) Every day, each participant randomly picks up a ball numbered between 1 and 9. ii) number on the ball determines his score on that day 3) 2-day average = average of scores of that day and previous day 4) Participants are ranked each day, according to their scores (if tie same rank is given) In this question we will try to find the scores of every participant each day. —--

2 day average through day 2 to day 5.

Day 2	Day 3	Day 4	Day 5
15	15.5	16	17

From table 2, Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively. The table shows that, the average of two days, d1 and d2 is 15, then the average of two days d2 and d3 is 15.5, and so on (as given in the set). d1+d2 = 30 ... (1), d2+d3 = 31 ... (2), d3+d4 = 32 .... (3), d4+d5 = 34 ... (4) —--

Day 2	Day 3	Day 4	Day 5
15	15.5	16	17

clue 2 says, The total score on Day 3 is the same as the total score on Day 4. i.e., d3 = d4 And we know, d3 + d4 = 32 => d3 = d4 = 16 Now using previous equations, we know d2+ d3 = 31 => d2 = 15 and, d1 + d2 = 30 => d1 = 15 and , d4 + d5 = 34 => 18 —--

Clue 1 says, Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. therefore, chatur score on day 2 will be a multiple of 3 and also unique highest which can only be 9. (as it is given that each participant randomly picks up a ball numbered between 1 and 9, therefore, the highest score can't be greater then 9.) => No other person scored 9 or more then 9 on 5 days. —--

Continuing with Clue 1, Chatur's minimum score is observed only on Day 1, and it matches Akhil's score on Day 4. Now, we already know that his score is a multiple of 3 therefore, his minimum score will be 3. And the same will be Akhil's score on day 4. —--

Also question says that Chatur has a unique minimum and maximum score. which we found to be 3 and 9 So, the only multiple of 3 between them is 6, which will be his score on other days. From which we can find Bimal's score on day 4 to be Akhil + bimal + chatur = 16 (on day 4) => bimal = 16 - 6 - 3 => bimal = 7 —--

Ranks of participants on each day
	Day 1	Day 2	Day 3	Day 4	Day 5
Akhil	1	2	2	3	3
Bimal	2	3	2	1	1
Chatur	3	1	1	2	2

From the second table we know, we see that the rank of Bimal and Akhil is the same on day 3, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5 —--

Ranks of participants on each day
	Day 1	Day 2	Day 3	Day 4	Day 5
Akhil	1	2	2	3	3
Bimal	2	3	2	1	1
Chatur	3	1	1	2	2

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1 which implies the score obtained by Bimal will be more than Chatur. Chatur had score of 6, also we know that maximum score is 9 Therefore, Bimal's score on day 5 = 7/8 And accordingly subtracting both Chatur's and Bimal's score on day 5 we will get Akhil's score to be - 5/4 —--

Ranks of participants on each day
	Day 1	Day 2	Day 3	Day 4	Day 5
Akhil	1	2	2	3	3
Bimal	2	3	2	1	1
Chatur	3	1	1	2	2

Clue 3 says, Bimal's scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5 which implies The score obtained by Akhil is 7 on Day 1 —--

	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Akhil	7	4/5	5	3	5/4	23/24/25
Bimal	5	2/1	5	7	7/8	27/26/25
Chatur	3	9	6	6	6	30
Total Score	15	15	16	16	18

Ranks of participants on each day
	Day 1	Day 2	Day 3	Day 4	Day 5
Akhil	1	2	2	3	3
Bimal	2	3	2	1	1
Chatur	3	1	1	2	2

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2. Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b => a+b =6, and a> b Hence, the value of a can be 4/5, and the value of b can be 2/1.

Total Score

CAT (DILR) PYQs

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