CAT 2023 Slot 2 — DILR Question 14

Mixed PracticeEasy

Passage / Data

Answer the following questions based on the information given below:

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

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.

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The following information is also known.

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.
The total score on Day 3 is the same as the total score on Day 4.
Bimal’s scores are the same on Day 1 and Day 3.

If the total score of Bimal is a multiple of 3, what is the score of Akhil on Day 2?

Answer & solution

A
Cannot be determined
B
5
C
6
4

Solution

Easy

Bimal's total can only be $25$ , $26$ , or $27$ . Force it to be a multiple of 3, which pins the free cells and reveals Akhil's Day-2 score.

From the set deduction: Bimal's total $=35-(A_2+A_5)$ with $A_2,A_5\in\{4,5\}$ , so $A_2+A_5\in\{8,9,10\}$ and Bimal's total $\in\{25,26,27\}$ .

Pick the multiple of 3. Among $\{25,26,27\}$ , only $27$ is a multiple of 3. So Bimal $=27$ , which means $A_2+A_5=35-27=8$ .

Solve the free cells. With each of $A_2,A_5\in\{4,5\}$ , the only way their sum is $8$ is $A_2=4$ and $A_5=4$ .

A_2=4,\qquad A_5=4.

Read the answer. Akhil's Day-2 score is $A_2=4$ (and then Akhil's total $=15+8=23$ , a neat cross-check against Bimal $=27$ , Chatur $=30$ , summing to $80$ ).

Akhil's score on Day 2 is $\mathbf{4}$ .