Test Index

CAT 2023 Slot 2 Solved Paper

Instructions [35 - 39 ]

Three participants - Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9 . The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants' scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20 , then the 2 -day average on Day 2 is calculated as 22.5 . Table 1 gives the 2 -day averages for Days 2 through 5 .

Table 1: 2-day averages for Days 2 through 5
Day 2	Day 3	Day 4	Day 5
15	15.5	16	17

Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8,7 and 7 respectively, then their ranks will be 1,2 and 2 respectively on that day. These ranks are given in Table 2.

Table 2: 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 following information is also known.
1. Chatur always scores in multiples of 3 . His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1 , and it matches Akhil's score on Day 4.
2. The total score on Day 3 is the same as the total score on Day 4 .
3. Bimal's scores are the same on Day 1 and Day 3.

Question : 37

Total: 66

What is the minimum possible total score of Bimal?

Your Answer:

Validate

Solution:

Table 1: 2-day averages for Days 2 through 5
Day 2	Day 3	Day 4	Day 5
15	15.5	16	17

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 d1+d2=30... eq (1), d2+d3=31... eq ( 2 ), d3+d4=32.... eq(3), d4+d5=34... eq(4)
It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.
It is given that the total score on Day 3 is the same as the total score on Day 4.
Therefore, d3=d4⇒d3=d4=16, which implies d2=15,d5=18, and d1=15.
The day-wise score is given below:

	Day 1	Day 2	Day 3	Day 4	Day 5
Akhil
Bimal
Chatur
Total Score	15	15	16	16	18

It is known that Chatur always scores in multiples of 3 . His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil's score on Day 4.
Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1 . Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.
Hence, we get the following table:

	Day 1	Day 2	Day 3	Day 4	Day 5
Akhil				3
Bimal				7
Chatur	3	9	6	6	6
Total Score	15	15	16	16	18

From Table 2, we see that the rank of Bimal and Akhil is the same, 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
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. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4 .

Day 1

Day 2

Day 3

Day 4

Day 5

Akhil

Bimal

Chatur

Total Score

It is given that 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.
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
Therefore, the final table is given below:

Day 1

Day 2

Day 3

Day 4

Day 5

Akhil

Bimal

Chatur

Total Score

From the table, we can see that the minimum score obtained by Bimal is 25.

Go to Question:

Prev Question

Next Question