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Six players- Tanzi, Umeza, Wangdu, Xyla, Yonita and Zeneca completed in an archery tournament. The tournament had three compulsory rounds, Rounds $1$ and $3$. 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.

  Round-1 Round-2 Round-3 Round-4 Round-5 Round-6
Tanzi - 4 - 5 NP NP
Umeza - - - 1 2 NP
Wangdu - 4 - NP NP NP
Xyla - - - 1 5 -
Yonita - - 3 5 NP NP
Zeneca - - - 5 5 NP

The following facts are also known.

  1. Tanzi, Umeza and Yonita had the same total score.
  2. total scores for all players, except one, were in multiples of three.
  3. The highest total score was one more than double of the lowest score.
  4. The number of players hitting bull’s eye in Round $2$ was double of that in Round $3$.
  5. Tanzi and Zeneca had the same score in Round $1$ but different scores in Round $3$.

What was Zeneca’s total score?

  1. $22$
  2. $23$
  3. $21$
  4. $24$
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  Round1 Round2 Round 3 Round 4 Round 5 Round 6
Tanzi 5 4 1 5 NP NP
Umeza 2 5 5 1 2 NP
Wangdu 4 4 4 NP NP NP
Xyka 5 5 5 1 5 4
Yonita 2 5 3 5 NP NP
Zeneca 5 5 4 5 5 NP

A typical or methodical approach of these constraint based LR questions is to pin-point that clue which reduces the search space substantially. I’ll list down the clues which helped me to fill the table :-

  1. “Every bull’s eye hit in the compulsory rounds gave a player an additional chance in the bonus rounds”. So, as Xyka participated in all 3 bonus rounds, her cores in all 3 compulsory rounds should be 5. Tanzi will have a 5 either in Round1 or Round 2. Umeza will be having 2 5’s in the compulsory rounds. Wangdu will have 0 5’s . Yonita will have one 5. Zeneca will have 2 5’s.
  2. (Clue 4) implies No. of 5’s in Round 2 is 2 times no. of 5’s in round 3. So, no. of 5’s in Round 2 can be 2 or 4 corresponding to 1 or 2 respectively in Round 3. Let’s assume there are 4 5’s in Round 2(this assumption fills R2 column entirely). Even if you try to fill R3 column with 1 5 and R2 column with 2 5’s, it wont be possible to satisfy both the constraints simultaneously.
  3. (Clue 1+2) implies Tanzi, Yonita & Umeza will have same scores in multiples of 3, as if even one of them is not a multiple of 3, all 3 won’t be a multiple of 3, which violates condition 2. Tanzi’s total score except for the blank cell is 5+4+5+x=14+x. To make it a multiple of 3, x can be either 1 or 4. 4 is eliminated as 4 will make Tanzi’s total=18, which isn’t possible for Yonita without having 2 5’s in 1st 3 rounds. So only option left for Tanz is 1 in the blank cell making the total=15. For Umeza, the value in the blank would be(15-13)=2. For Yonita, 15-13=2.
  4. (Clue 5) implies scores in scores in 1st round is same for both Tanzi & Zeneca(5). Then Zeneca’s 3rd column will be filled with 4(different from Tanzi’s 3rd column which is 1). Making total score of Zeneca 24.
  5. (Clue 3) implies highest scorer should have a value which is an odd number. Zeneca’s score is 24, which is even so Zeneca can’t be the highest scorer which leaves Xyka. Except for R6 value Xyka’s score is 21, and it needs to be greater than 24. Maximum value it can have is 5 which makes the value 21+5=16. But that isn’t possible as the highest score should be an odd number. The only possiblity is 25 or 21+4.Fill the R6 column of Xyka as 4.
  6. Highest score is 25, so the lowest should be 12. Wangdu has the R2 filled as 4, R1 & R3 will be filled as 4 making the sum as 12. Note that none of the values can be 5 for Wangdu as he didn’t get to participate in any of the Bonus rounds, so only possibility is 3 fours.

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