CAT 2019 Slot 1 — DILR Question 9
Answer the following question based on the information given below.
Six players – Tanzi, Umeza, Wangdu, Xyla, Yonita and Zeneca competed in an archery tournament. The tournament had three compulsory rounds, Rounds 1 to 3. In each round every player shot an arrow at a target. Hitting the centre of the target (called bull’s eye) fetched the highest score of 5. The only other possible scores that a player could achieve were 4, 3, 2 and 1. Every bull’s eye score in the first three rounds gave a player one additional chance to shoot in the bonus rounds, Rounds 4 to 6. The possible scores in Rounds 4 to 6 were identical to the first three. A player’s total score in the tournament was the sum of his/her scores in all rounds played by him/her. The table below presents partial information on points scored by the players after completion of the tournament. In the table, NP means that the player did not participate in that round, while a hyphen means that the player participated in that round and the score information is missing.
The following facts are also known.
- Tanzi, Umeza and Yonita had the same total score.
- Total scores for all players, except one, were in multiples of three.
- The highest total score was one more than double of the lowest total score.
- The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
- Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.
What was the highest total score?
Answer & solution
- A
23
- B
21
- C
24
25
Tanzi, Umeza, Xyla, Yonita and Zeneca got 1,2,3,1,2 chances to shoot in the bonus rounds respectively. Therefore, in the compulsory round, 9 bull’s eye were hit. From (4), it can be concluded that number of bull’s eye hit in Rounds 1, 2 and 3 were 3, 4 and 2 respectively.
Note that Xyla got three chances in the bonus round. So she must have hit three bull’s eye in the compulsory rounds. Therefore, Xyla’s minimum score is 5 × 4 + 1 + 1 = 22.
Zeneca’s maximum score could be 5 × 4 + 4 = 24. Her minimum score could be 5 × 4 + 1 = 21. If her score was the maximum score, it must be an odd number. So, 23 is the maximum score. Now using (3), the lowest score = 11. But both 11 and 23 are not divisible by 3. Using (2), we can conclude that 23 is not the maximum score. So, Xyla’s must have scored maximum. And Zeneca’s score was either 21(5 × 4 + 1) or 24(5 × 4 + 4).
Thus, Xyla scored 5 in each of the compulsory rounds and 4 in round 6.
Tanzi hit one bull’s eye either in round 1 or in round 3. So her minimum score = 5 + 4 + 1 + 5 = 15 and her maximum score = 5 + 4 + 4 + 5 = 18.
Yonita’s maximum score = 5 + 4 + 3 + 5 = 17.
So from (1) and (2), Tanzi, Umeza and Yonita each had total score 15. And hence, Wangdu scored least points i.e., 12 points. The only possible combination is 4 points in each of the compulsory rounds.
So, Tanzi scored 1 and 5 in rounds 1 and 3 in some order. Assume that she scored 1 in round 1 and 5 in round 3. From (5), Zeneca scored 1 in round 1. But then she must also have scored 5 in round 3 as she hit bull’s eye twice in the compulsory rounds. But this is contradiction to (5). So, Tanzi and Zeneca scored 5 in round 1. Tanzi scored 1 in round 3.
Thus, Tanzi, Zeneca and Xyla hit bull’s eye in round 1. Therefore Yonita (total score =15) must have hit bull’s eye in the second round and scored 2 points in the first round.
Umeza must have hit two bull’s eye in rounds 2 and 3. Also, she must have scored 2 points in the first round.
Thus, Umeza and Xyla hit the bull’s eye in the third round. Therefore, Zeneca hit bull’s eye in the second round and scored 4 points in the third round.
Thus, we have
So, the highest total score was 25.
Hence, option (d).