CAT 2024 Slot 2 DILR Set 2 Q5–Q9

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2024 Slot 2, with 5 questions (Quant-based Puzzles). Answers are CAT's official key.

The set

Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches - Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7, with 1 being the lowest rating and 7 the highest.

The following additional information is known.
1. Xena trained more players than Yuki.
2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
4. The average of the ratings of all the players was 4.
5. Player-2 got the highest rating.
6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
7. Player-4's rating was double of Player-8's and less than Player-5's.

Question 5

What best can be concluded about the number of players coached by Zara?
  1. A
    Exactly 3
  2. B
    Either 2 or 3 or 4
  3. C
    Exactly 2
  4. D
    Either 2 or 3

Answer: (C) Exactly 2 Quant-based Puzzles MCQ

Solution

  1. Player-2 is even (Xena or Yuki); if Xena trained 2, then 3 and 5 (odd) would both need Zara, contradicting 'all different'. So Yuki trained 2 and {3, 5} are split between Xena and Zara. Players 1 and 4 share a coach: 1 cannot be Yuki's and 4 cannot be Zara's, so both are Xena's.
  2. Ratings: 5 and 7 share a rating and the other six are distinct from it and each other, so the seven distinct values are exactly 1 to 7. Total \(= 8 \times 4 = 32 = 28 + r_5\), so \(r_5 = r_7 = 4\); Player-2 has 7; \(r_4 = 2r_8 < 4\) gives \(r_4 = 2, r_8 = 1\). Players 1, 3, 6 share 3, 5, 6.
  3. If Xena trained 5 (and 7), Zara's only possible player is 3, but each coach needs at least two. So Zara trained exactly 5 and 7 (average 4), and Xena trained 3. Then Yuki's average is 6 and Xena's is 3. Yuki = {2, 6} with \(r_6 = 5\) works (\((7+5)/2 = 6\)); adding 8 or using 8 instead fails. So Xena = {1, 3, 4, 8} with \(r_1 + r_3 = 3 + 6\): average \((9+2+1)/4 = 3\), as needed.
  4. So Zara coached exactly 2 players (5 and 7).

Question 6

What was the rating of Player-7?

TITA: type in the answer. There are no options and no negative marking.

Answer: 4 Quant-based Puzzles TITA

Solution

  1. Player-2 is even (Xena or Yuki); if Xena trained 2, then 3 and 5 (odd) would both need Zara, contradicting 'all different'. So Yuki trained 2 and {3, 5} are split between Xena and Zara. Players 1 and 4 share a coach: 1 cannot be Yuki's and 4 cannot be Zara's, so both are Xena's.
  2. Ratings: 5 and 7 share a rating and the other six are distinct from it and each other, so the seven distinct values are exactly 1 to 7. Total \(= 8 \times 4 = 32 = 28 + r_5\), so \(r_5 = r_7 = 4\); Player-2 has 7; \(r_4 = 2r_8 < 4\) gives \(r_4 = 2, r_8 = 1\). Players 1, 3, 6 share 3, 5, 6.
  3. Player-7 has the same rating as Player-5, which is 4.

Question 7

What was the rating of Player-6?

TITA: type in the answer. There are no options and no negative marking.

Answer: 5 Quant-based Puzzles TITA

Solution

  1. Player-2 is even (Xena or Yuki); if Xena trained 2, then 3 and 5 (odd) would both need Zara, contradicting 'all different'. So Yuki trained 2 and {3, 5} are split between Xena and Zara. Players 1 and 4 share a coach: 1 cannot be Yuki's and 4 cannot be Zara's, so both are Xena's.
  2. Ratings: 5 and 7 share a rating and the other six are distinct from it and each other, so the seven distinct values are exactly 1 to 7. Total \(= 8 \times 4 = 32 = 28 + r_5\), so \(r_5 = r_7 = 4\); Player-2 has 7; \(r_4 = 2r_8 < 4\) gives \(r_4 = 2, r_8 = 1\). Players 1, 3, 6 share 3, 5, 6.
  3. If Xena trained 5 (and 7), Zara's only possible player is 3, but each coach needs at least two. So Zara trained exactly 5 and 7 (average 4), and Xena trained 3. Then Yuki's average is 6 and Xena's is 3. Yuki = {2, 6} with \(r_6 = 5\) works (\((7+5)/2 = 6\)); adding 8 or using 8 instead fails. So Xena = {1, 3, 4, 8} with \(r_1 + r_3 = 3 + 6\): average \((9+2+1)/4 = 3\), as needed.
  4. So Player-6 got 5.

Question 8

For how many players the ratings can be determined with certainty?

TITA: type in the answer. There are no options and no negative marking.

Answer: 6 Quant-based Puzzles TITA

Solution

  1. Player-2 is even (Xena or Yuki); if Xena trained 2, then 3 and 5 (odd) would both need Zara, contradicting 'all different'. So Yuki trained 2 and {3, 5} are split between Xena and Zara. Players 1 and 4 share a coach: 1 cannot be Yuki's and 4 cannot be Zara's, so both are Xena's.
  2. Ratings: 5 and 7 share a rating and the other six are distinct from it and each other, so the seven distinct values are exactly 1 to 7. Total \(= 8 \times 4 = 32 = 28 + r_5\), so \(r_5 = r_7 = 4\); Player-2 has 7; \(r_4 = 2r_8 < 4\) gives \(r_4 = 2, r_8 = 1\). Players 1, 3, 6 share 3, 5, 6.
  3. If Xena trained 5 (and 7), Zara's only possible player is 3, but each coach needs at least two. So Zara trained exactly 5 and 7 (average 4), and Xena trained 3. Then Yuki's average is 6 and Xena's is 3. Yuki = {2, 6} with \(r_6 = 5\) works (\((7+5)/2 = 6\)); adding 8 or using 8 instead fails. So Xena = {1, 3, 4, 8} with \(r_1 + r_3 = 3 + 6\): average \((9+2+1)/4 = 3\), as needed.
  4. Known: Player-2 (7), 4 (2), 5 (4), 6 (5), 7 (4), 8 (1). Players 1 and 3 share 3 and 6 in either order. So 6 ratings are certain.

Question 9

Who all were the players trained by Xena?
  1. A
    Player-1, Player-3, Player-4, Player-6
  2. B
    Player-1, Player-4, Player-6, Player-8
  3. C
    Player-1, Player-3, Player-4, Player-8
  4. D
    Player-1, Player-3, Player-4

Answer: (C) Player-1, Player-3, Player-4, Player-8 Quant-based Puzzles MCQ

Solution

  1. Player-2 is even (Xena or Yuki); if Xena trained 2, then 3 and 5 (odd) would both need Zara, contradicting 'all different'. So Yuki trained 2 and {3, 5} are split between Xena and Zara. Players 1 and 4 share a coach: 1 cannot be Yuki's and 4 cannot be Zara's, so both are Xena's.
  2. Ratings: 5 and 7 share a rating and the other six are distinct from it and each other, so the seven distinct values are exactly 1 to 7. Total \(= 8 \times 4 = 32 = 28 + r_5\), so \(r_5 = r_7 = 4\); Player-2 has 7; \(r_4 = 2r_8 < 4\) gives \(r_4 = 2, r_8 = 1\). Players 1, 3, 6 share 3, 5, 6.
  3. If Xena trained 5 (and 7), Zara's only possible player is 3, but each coach needs at least two. So Zara trained exactly 5 and 7 (average 4), and Xena trained 3. Then Yuki's average is 6 and Xena's is 3. Yuki = {2, 6} with \(r_6 = 5\) works (\((7+5)/2 = 6\)); adding 8 or using 8 instead fails. So Xena = {1, 3, 4, 8} with \(r_1 + r_3 = 3 + 6\): average \((9+2+1)/4 = 3\), as needed.
  4. Xena trained Players 1, 3, 4 and 8.

Same concept in other years

See every question on Managers/coaches give ratings to members of their groups; deduce assignments.

More Quant-based Puzzles questions from CAT

All 127 Quant-based Puzzles questions asked since 2017.