CAT 2023 Slot 2 — DILR Question 11

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.

âââââââ

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.

What is Akhil's score on Day 1?

Answer & solution

Solution

Easy

One grid solves the whole set. Turn the 2-day averages into daily totals, then use the rank table plus the three extra facts to nail every score. We build the grid once and just read answers off it.

Given data (transcribed from the two tables).

Table 1 — 2-day averages. A day's 2-day average $=\dfrac{T_{\text{today}}+T_{\text{prev}}}{2}$ , where $T_d$ is that day's total of all three scores.

	Day 2	Day 3	Day 4	Day 5
2-day avg	15	15.5	16	17

Table 2 — daily ranks (rank 1 = highest score; ties share the best rank).

	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

Extra facts: (1) Chatur always scores a multiple of 3; his Day-2 score is the unique highest score of the whole competition; his minimum occurs only on Day 1 and equals Akhil's Day-4 score. (2) $T_3=T_4$ . (3) Bimal's Day-1 and Day-3 scores are equal.

Daily totals from the averages. Doubling each 2-day average gives the sum of two consecutive day-totals:

\begin{aligned} &T_1+T_2=30,\quad T_2+T_3=31,\quad T_3+T_4=32,\quad T_4+T_5=34. \end{aligned}

Fact 2 says $T_3=T_4$ , so $T_3+T_4=32\Rightarrow T_3=T_4=16$ . Back-substituting: $T_2=31-16=15$ , $T_1=30-15=15$ , $T_5=34-16=18$ .

T_1=15,\ T_2=15,\ T_3=16,\ T_4=16,\ T_5=18 \quad\Rightarrow\quad \text{grand total}=80.

Chatur's column (multiples of 3). On Day 2 Chatur is rank 1 and his score is the unique highest of the entire competition, so $C_2=9$ . On Day 3, Chatur (rank 1) beats Akhil = Bimal (both rank 2): $2A_3+C_3=16$ with $C_3>A_3$ a multiple of 3 forces $C_3=6,\ A_3=B_3=5$ . His minimum is only on Day 1 and is a multiple of 3 below 6, so $C_1=3$ .

Day 4 and Akhil's Day-4 score. Fact 1: Akhil's Day-4 score equals Chatur's minimum, so $A_4=3$ . Day-4 order is $B_4>C_4>A_4$ with total 16: $B_4+C_4=13$ , $C_4$ a multiple of 3 above 3 $\Rightarrow C_4=6,\ B_4=7$ .

Day 1 (Bimal link). Fact 3: $B_1=B_3=5$ . Day 1 order is $A_1>B_1>C_1$ with total 15 and $C_1=3$ : $A_1=15-5-3=7$ . So Akhil scores 7 on Day 1.

Day 5 finishes Chatur. Order $B_5>C_5>A_5$ , total 18, $C_5$ a multiple of 3 above 3. $C_5=9$ would need $B_5\ge10$ (impossible), so $C_5=6$ . Chatur's total is therefore

3+9+6+6+6=30\ \text{(fixed)}.

The two free cells. Only $A_2$ (Day 2) and $A_5$ (Day 5) remain loose. Day 2 needs $A_2>B_2$ with $A_2+B_2=6\Rightarrow A_2\in\{4,5\}$ . Day 5 needs $B_5>C_5=6$ i.e. $B_5\le9$ with $A_5+B_5=12$ , and since $C_2=9$ must stay the unique 9, $B_5\ne9\Rightarrow A_5\in\{4,5\}$ . The grid:

	Day 1	Day 2	Day 3	Day 4	Day 5
Akhil	7	$A_2$	5	3	$A_5$
Bimal	5	$6-A_2$	5	7	$12-A_5$
Chatur	3	9	6	6	6

Hence Akhil's total $=15+A_2+A_5$ and Bimal's total $=35-A_2-A_5$ , with $A_2,A_5\in\{4,5\}$ .

Akhil's score on Day 1 is $\mathbf{7}$ .