Logical Reasoning

Games & Tournaments

Hard

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.

Name	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.

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 Zeneca's total score?

✅ Correct Option: 4

Slide 1/11

Understanding the set :-

This is a typical table filling + tournament set,

Here, you have one table in the set and using the clues you just need to fill the table.

Also, in this set the most important thing you need to understand is this line -

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.

It implies that , if you have a bulls eye score in one of the round you will get one bonus round i.e., round 4

And, if you have a bulls eye score in two rounds you will get two bonus rounds i.e., round 4 & 5

And, if you have a bulls eye score in three rounds you will get three bonus rounds i.e., round 4 , 5, & 6.

Now, lets understand the thought process of this set.

	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	5	5	5	1	5	-
Yonita	-	-	3	5	NP	-
Zeneca	-	-	-	5	5	NP

The very first thing we can see over here is;

Tanzi has participated in only one of round 4,5 or 6

=> she has scored a bulls eye or "5" in only one of round 1, 2, or 3.

similiarly, Umeza has scored 5 in 2 rounds from round 1,2 or 3.

W has scored 5 in none of the round.

x has scored 5 in all the three rounds, which we can directly fill to the table.

Y has scored 5 in only one .

Z has scored 5 in two rounds.

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

Clue 4 says,

The number of players hitting bull’s eye in Round 2 was double of that in Round 3.

Now, there are only two cases :

Round 3 has 1 bull's eye and round 2 then will have 2 bull's eye.

Round 3 has 2 bull's eye and round 2 then will have 4 bull's eye.

But, we know, there are 3 people (U, X, Z) who has scored 5 in two rounds.

=> if round 3 has only 1 bull's eye and round 2 has only 2 bull's eye,

then three people can't have 5 in two rounds.

=> second case is true.

=> In round 2, all the leftover places will have 5.

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

Using clue 5;

Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Zeneca score Bull's eye 2 times in round 1 to 3.

If Tanzi scored 1 in round 1, then Zeneca also has to score 1 in round 1,

which means both Tanzi and Zeneca scores in round 3 will be 5, which violates 5.

Hence Tanzi scored 5 in round 1 and Zeneca also scored the same in round 1.

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

Now, we know,

In round 3 only 2 people have bull's eye.

Now, it can't be T (as T only have 1 bull's eye, which is in round 1)

For the same reason it can't be Z, W, Y and X as well.

Therefore, it can only be U.

	Round-1	Round-2	Round-3	Round-4	Round-5	Round-6	Total
Tanzi	5	4	1/4	5	NP	NP	15/18
Umeza	2	5	5	1	2	NP	15
Wangdu	-	4	-	NP	NP	NP
Xyla	5	5	5	1	5	-
Yonita	-	5	3	5	NP	-
Zeneca	5	5	4/1	5	5	NP	24/21

Now, clue 2 says,

Total scores for all players, except one, were in multiples of three.

For T;

we have $5+4+5 = 14$

Therefore, to make it a multiple of 3, the blank place will have either 1 or 4.

For U;

We have $5+5+1+2 = 13$

=> U's score in round 1 will be 2 or 5. but it can't be 5 (as U had 5 in only two rounds)

Therefore, score will be 2

For Z;

We have $5+5+5+5 = 20$

Therefore, Z's score in round 3 can be 1 or 4.

	Round-1	Round-2	Round-3	Round-4	Round-5	Round-6	Total
Tanzi	5	4	1	5	NP	NP	15
Umeza	2	5	5	1	2	NP	15
Wangdu	-	4	-	NP	NP	NP
Xyla	5	5	5	1	5	-
Yonita	-	5	3	5	NP	-	15
Zeneca	5	5	4	5	5	NP	24

Now, clue 1 says,

Tanzi, Umeza and Yonita had the same total score.

We know, U had a score of 15.

=> T and Y will also have a total score of 15.

=> T will have a score of 1 in round 3.

=> Z will have a score of 4 in round 3 (clue 5)

	Round-1	Round-2	Round-3	Round-4	Round-5	Round-6	Total
Tanzi	5	4	1	5	NP	NP	15
Umeza	2	5	5	1	2	NP	15
Wangdu	-	4	-	NP	NP	NP
Xyla	5	5	5	1	5	-
Yonita	1	5	3	5	NP	1	15
Zeneca	5	5	4	5	5	NP	24

For Y,

Z already have $5+3+5 =13$

Therefore, to make it 15, only possibility is 1 in both round 1 and round 6.

	Round-1	Round-2	Round-3	Round-4	Round-5	Round-6	Total
Tanzi	5	4	1	5	NP	NP	15
Umeza	2	5	5	1	2	NP	15
Wangdu	-	4	-	NP	NP	NP	12
Xyla	5	5	5	1	5	-	25
Yonita	1	5	3	5	NP	1	15
Zeneca	5	5	4	5	5	NP	24

Clue 3 says,

The highest total score was one more than double of the lowest total score.

=> highest = 2 (lowest) + 1

now, Highest should be greater then 24 (as Z already have it)

=> only possibility = lowest = 12

and, highest then = $2 * 12 + 1 = 25$

=> W have total of 12

and X have total of 25

	Round-1	Round-2	Round-3	Round-4	Round-5	Round-6	Total
Tanzi	5	4	1	5	NP	NP	15
Umeza	2	5	5	1	2	NP	15
Wangdu	4	4	4	NP	NP	NP	12
Xyla	5	5	5	1	5	4	25
Yonita	1	5	3	5	NP	1	15
Zeneca	5	5	4	5	5	NP	24

Now, for W's total to be 12,

Only possibility = $4+4+4 =12$

And, X will have $25-5-5-5-1-5 = 4$

	Round-1	Round-2	Round-3	Round-4	Round-5	Round-6	Total
Tanzi	5	4	1	5	NP	NP	15
Umeza	2	5	5	1	2	NP	15
Wangdu	4	4	4	NP	NP	NP	12
Xyla	5	5	5	1	5	4	25
Yonita	1	5	3	5	NP	1	15
Zeneca	5	5	4	5	5	NP	24

From the table we can see that,

Zeneca's total score is 24.

Question 21

What was the highest total score?

Question 23

Which of the following statements is true?

Question 24

What was Tanzi's score in Round 3?

What was Zeneca's total score?

Solution

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