# DI - Games & Tournaments - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic DI - Games & Tournaments. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**Answer the next 4 questions based on the information given **

The game of Chango is a game where two people play against each other; one of them wins and the other loses, i.e., there are no drawn Chango games. 12 players participated in a Chango championship. They were divided into four groups: Group A consisted of Aruna, Azul, and Arif; Group B consisted of Brinda, Brij, and Biju; Group C consisted of Chitra, Chetan, and Chhavi; and Group D consisted of Dipen, Donna, and Deb.

Players within each group had a distinct rank going into the championship. The players have NOT been listed necessarily according to their ranks. In the group stage of the game, the second and third ranked players play against each other, and the winner of that game plays against the first ranked player of the group. The winner of this second game is considered as

the winner of the group and enters a semi-final.

The winners from Groups A and B play against each other in one semi-final, while the winners from Groups C and D play against each other in the other semi-final. The winners of the two semi-finals play against each other in the final to decide the winner of the championship.

It is known that:

- Chitra did not win the championship.
- Aruna did not play against Arif. Brij did not play against Brinda.
- Aruna, Biju, Chitra, and Dipen played three games each, Azul and Chetan played two games each, and the remaining players played one game each.

**CAT 2021 LRDI Slot 2 | DI - Games & Tournaments**

Who among the following was DEFINITELY NOT ranked first in his/her group?

- A.
Brij

- B.
Chitra

- C.
Dipen

- D.
Aruna

Answer: Option C

**Explanation** :

According to the information given we can draw the following table.

The winner of any group will have to play at least 2 games (at least one game in the group stage and 1 semi-final)

**Let us first consider group B:**

The winner of group B cannot be Brinda or Brij hence it has to be Biju.

Biju must have played one game each with Brinda and Brij in any order. Biju must have won both his group matches and qualified for the semi-finals.

Now, in the semi-final Biju must have played against the winner of group A. Biju would have lost this match since Biju only plays 3 matches.

∴ The winner of semi-final between group A and B is the winner of group B.

**Let us now consider group D:**

This case is same as that of group B.

Here, Dipen must have played one game each with Donna and Deb in any order. Dipen must have won both his group matches and qualified for the semi-finals.

Now, in the semi-final Dipen must have played against the winner of group C. Dipen would have lost this match since Dipen only plays 3 matches.

∴ The winner of semi-final between group C and D is the winner of group C.

∴ Final will be played between winners of group A and C.

Now winners of group A and group C must have played at least 3 matches (at least one match in group stage, 1 semi-final and 1 final)

∴ Winner of group A can only be Aruna and winner of group C can only be Chitra.

Since Chitra cannot be the champion hence Aruna will win the final and become the Champion.

As per the rules of the game, one of 2nd or 3rd ranked player of any group will play 2 matches in the group stage and the 1st ranked player will play only 1 match in group stage.

In group D, Dipen is definitely playing 2 matches during group stage, hence Dipen is definitely not the 1st ranked player in group D.

Hence, option (c).

Workspace:

**CAT 2021 LRDI Slot 2 | DI - Games & Tournaments**

Which of the following pairs must have played against each other in the championship?

- A.
Azul, Brij

- B.
Chitra, Dipen

- C.
Deb, Donna

- D.
Donna, Chetan

Answer: Option B

**Explanation** :

Consider the solution to first question of this set.

Chitra and Dipen play against each other in 2nd semi-final.

Hence, option (b).

Workspace:

**CAT 2021 LRDI Slot 2 | DI - Games & Tournaments**

Who won the championship?

- A.
Brij

- B.
Chitra

- C.
Aruna

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

Aruna won the championship.

Hence, option (c).

Workspace:

**CAT 2021 LRDI Slot 2 | DI - Games & Tournaments**

Who among the following did NOT play against Chitra in the championship?

- A.
Aruna

- B.
Biju

- C.
Chetan

- D.
Dipen

Answer: Option B

**Explanation** :

Consider the solution to first question of this set.

Biju definitely did not play against Chitra in the championship.

Hence, option (b).

Workspace:

**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.

**CAT 2019 LRDI Slot 1 | DI - Games & Tournaments**

What was the highest total score?

- A.
23

- B.
21

- C.
24

- D.
25

Answer: Option D

**Explanation** :

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 4.

Workspace:

**CAT 2019 LRDI Slot 1 | DI - Games & Tournaments**

What was Zeneca's total score?

- A.
23

- B.
22

- C.
21

- D.
24

Answer: Option D

**Explanation** :

Consider the solution to first question of this set.

Zeneca’s total score was 24.

Hence, option 4.

Workspace:

**CAT 2019 LRDI Slot 1 | DI - Games & Tournaments**

Which of the following statements is true?

- A.
Xyla was the highest scorer.

- B.
Zeneca’s score was 23.

- C.
Zeneca was the highest scorer.

- D.
Xyla’s score was 23.

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

Xyla was the highest scorer.

Hence, option 1.

Workspace:

**CAT 2019 LRDI Slot 1 | DI - Games & Tournaments**

What was Tanzi's score in Round 3?

- A.
1

- B.
5

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

Tanzi’s score in Round 3 was 1.

Hence, option 1.

Workspace:

**Answer the following questions based on the information given below.**

Ten players, as listed in the table below, participated in a rifle shooting competition comprising of 10 rounds. Each round had 6 participants. Players numbered 1 through 6 participated in Round 1, players 2 through 7 in Round 2,..., players 5 through 10 in Round 5, players 6 through 10 and 1 in Round 6, players 7 through 10, 1 and 2 in Round 7 and so on. The top three performances in each round were awarded 7, 3 and 1 points respectively. There were no ties in any of the 10 rounds. The table below gives the total number of points obtained by the 10 players after Round 6 and Round 10.

The following information is known about Rounds 1 through 6:

- Gordon did not score consecutively in any two rounds.
- Eric and Fatima both scored in a round.

The following information is known about Rounds 7 through 10:

- Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds.
- Joshin scored in Round 7, while Amita scored in Round 10.
- No player scored in all the four rounds.

**CAT 2019 LRDI Slot 2 | DI - Games & Tournaments**

What were the scores of Chen, David, and Eric respectively after Round 3?

- A.
3, 0, 3

- B.
3, 3, 0

- C.
3, 3, 3

- D.
3, 6, 3

Answer: Option C

**Explanation** :

As the top three performances in each round were awarded 7, 3 and 1 respectively, remaining all performances were awarded 0 points.

Amita was participant in round 1 and in rounds 6 through 10.

So, in all she scored 8(= 1 + 7) points in round 1 and round 6. Also she scored 10 points in rounds 7 through 10 together.

Bala was participant in rounds 1 and 2 and also in rounds 7 through 10.

As he scored 2 points in first two rounds, he must have scored 1 point in each of the two rounds. He scored 3(= 3 or 1+1+1) points in rounds 7 through 10.

Chen scored 3 points in first three rounds and as he scored in three consecutive rounds among 7 through 10, he must have scored 1 point each in round 8, 9 and 10.

David scored 6 points in first four rounds together and 0 (i.e., 6 – 6) each in 9th and 10th rounds.

Eric scored 3(= 3 or 1+1+1) points in five rounds together and hence 10 – 3 = 7 points in the 10th round.

Gordon scored 17 points in rounds 2 through 6 together and hence, 17 – 17 = 0 points in the 7th round.

As he did not scored consecutively in any two rounds, he must have scored 0 in 3rd and 5th round and 7, 7, and 3 points in 2nd, 4th and 6th rounds, not necessarily in the given sequence.

Hansa scored 1 point in one of the rounds 3-6. So, she must have scored 4 – 1 = 3 points in either 7th or 8th rounds.

Ikea scored 2 points in 4th, 5th and 6th rounds together. Therefore, remaining 15 points must have been scored in 7th, 8th and 9th round in some order.

As Chen scored 1 point in the 8th and 9th round, Ikea must have scored 1 point in 7th round and 7 points each in the 8th and 9th round.

Joshin scored 14 points in 5th and 6th rounds. So he scored 7 points in each of 5th and 6th rounds.

Now we need to decide about remaining 3 points in the last four rounds.

As Joshin scored in round 7, and Chen has scored 1 point in each of the last three rounds, Joshin must have scored all 3 points in the 7th round and 0 points in the last three rounds.

Therefore, we can tabulate this information as

As Joshin scored 3 points in the 7th round, Hansa must have scored 3 in 8th round and 0 in the 7th round. Now consider 7th round. Only Amita must have scored 7 in the 7th round.

As she has scored in round 10, the remaining 3 points she must have scored in round 10. Thus, we can now conclude that Bala scored 3 points in the 9th round.

As Joshin scored 7 points in the 6th round Amita must have scored 1 point in this round and hence7 points in the first round.

Gordon scored 7, 7 and 3 points in 2nd, 4th and 6th rounds. Joshin scored 7 points in the 6th round so, Gordon scored 7, 7 and 3 points in 2nd, 4th and 6th rounds respectively.

So we can fill remaining values as 0 in the table for the 6th round.

Thus, Ikea scored 1 point each in 4th and 5th rounds. And hence, Hansa scored 1 point in the 3rd round.

So far we have entered values 1 and 7 for all the rounds. Now we have to enter 3 points of each of Bala, Chen, David and Fatima and 7 points for Fatima.

Eric and Fatima both scored in a round. This is possible only when Fatima scored 7 points and Eric scored 3 points. This must be round 3.

No one other than Fatima scored 3 points in round 5. Similarly, David scored 3 points in the 4th round. Chen and David scored 3 points in 1st and 2nd round in some order.

Thus we have

After round 3, the scores of Chen, David and Eric were 3, 3, 3.

Hence, option (c).

Workspace:

**CAT 2019 LRDI Slot 2 | DI - Games & Tournaments**

Which three players were in the last three positions after Round 4?

- A.
Bala, Ikea, Joshin

- B.
Bala, Hansa, Ikea

- C.
Bala, Chen, Gordon

- D.
Hansa, Ikea, Joshin

Answer: Option D

**Explanation** :

Consider the solution to first question of this set.

The three players in the last position after Round 4 were Hansa, Ikea and Joshin.

Hence, option (d).

Workspace:

**CAT 2019 LRDI Slot 2 | DI - Games & Tournaments**

Which player scored points in maximum number of rounds?

- A.
Ikea

- B.
Chen

- C.
Amita

- D.
Joshin

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

Ikea scored in maximum number of rounds i.e., in 5 rounds.

Hence, option (a).

Workspace:

**CAT 2019 LRDI Slot 2 | DI - Games & Tournaments**

Which players scored points in the last round?

- A.
Amita, Chen, David

- B.
Amita, Bala, Chen

- C.
Amita, Chen, Eric

- D.
Amita, Eric, Joshin

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

Amita, Chen and Eric scored in the last round.

Hence, option (c).

Workspace:

**Answer the following questions based on the information given below:**

In a sports event, six teams (A, B, C, D, E and F) are competing against each other. Matches are scheduled in two stages. Each team plays three matches in Stage-I and two matches in Stage-II. No team plays against the same team more than once in the event. No ties are permitted in any of the matches. The observations after the completion of Stage-I and Stage-II are as given below.

Stage-I:

- One team won all the three matches.
- Two teams lost all the matches.
- D lost to A but won against C and F.
- E lost to B but won against C and F.
- B lost at least one match.
- F did not play against the top team of Stage-I.

Stage-II:

- The leader of Stage-I lost the next two matches.
- Of the two teams at the bottom after Stage-I, one team won both matches, while the other lost both matches.
- One more team lost both matches in Stage-II.

**CAT 2008 LRDI | DI - Games & Tournaments**

The team(s) with the most wins in the event is (are):

- A.
A

- B.
A & C

- C.
F

- D.
E

- E.
B & E

Answer: Option E

**Explanation** :

Let the bold letters denote the teams that have lost.

From condition 3 of stage I,

D lost to A.

D won against C.

D won against F.

These can be represented as:

**D** -- A

D -- **C**

D -- **F**

Similarly, condition 4 of stage I can be represented as:

**E** -- B

E -- **C**

E -- **F**

Since D and E have participated in three matches in stage I, they would not be involved in any other match in stage I.

From the above representations it is clear that all other teams except A have lost at least one match.

∴ From condition 1, of stage I, only A has won all the three matches in stage I.

Also, A will participate in 2 more matches as every team participates in 3 matches in stage I.

∴ A will win in 2 of the remaining 3 matches.

Also A is the top team as it wins all matches in stage I.

From condition 6 of stage I,

F did not play against A.

∴ A won against B and C which can be represented as:

**B** -- A

A --** C**

The only 2 teams which have not won even a single match so far is C and F.

From statement 6 of stage I, F loses in the remaining match against B, which can be

represented as:

**F** -- B

Stage I can be represented as:

**D** -- A **B** -- A

D --** C** A -- **C**

D -- **F ** **F** -- B

**E **-- B

E -- **C**

E -- **F**

From condition 1 of stage II,

A lost both matches in stage II.

Also, since no team plays against the same team more than once in the event, A plays matches against E and F.

**A** -- E

**A **-- F

Since one of the two teams at the bottom after stage I won both matches in stage II, F is the team which has won both the matches in stage II.

Also C lost both matches in stage II.

F --** C**

B -- **C**

The last condition states that one more team lost both matches in stage II.

∴ D lost both matches in stage II.

**D** -- B

**D** -- E

Stage II can be represented as:

**A** -- E

**A** -- F

F -- **C**

B -- **C**

**D** -- B

**D **-- E

Now, we can calculate the number of times each team has won.

It can be observed from the above table that B and E have most wins in the event.

Hence, option 5.

Workspace:

**CAT 2008 LRDI | DI - Games & Tournaments**

The two teams that defeated the leader of Stage-I are:

- A.
F & D

- B.
E & F

- C.
B & D

- D.
E & D

- E.
F & D

Answer: Option B

**Explanation** :

E and F defeated A.

Hence, option 2.

Workspace:

**CAT 2008 LRDI | DI - Games & Tournaments**

The only team(s) that won both the matches in Stage-II is (are):

- A.
B

- B.
E & F

- C.
A, E & F

- D.
B, E & F

- E.
B & F

Answer: Option D

**Explanation** :

B, E and F are the three teams that won both matches in stage II.

Hence, option 4.

Workspace:

**CAT 2008 LRDI | DI - Games & Tournaments**

The teams that won exactly two matches in the event are:

- A.
A, D & F

- B.
D & E

- C.
E & F

- D.
D, E & F

- E.
D & F

Answer: Option E

**Explanation** :

From the table it is clear that the team that won excatly two matches in the event are D and F.

Hence, option 5.

Workspace:

**Answer the following question based on the information given below.**

In the table below is the listing of players, seeded from highest (#1) to lowest (#32), who are due to play in an Association of Tennis Players (ATP) tournament for women. This tournament has four knockout rounds before the final, i.e., first round, second round, quarterfinals, and semi-finals. In the first round, the highest seeded player plays the lowest seeded player (seed # 32) which is designated match No. 1 of first round; the 2nd seeded player plays the 31st seeded player which is designated match No. 2 of the first round, and so on. Thus, for instance, match No. 16 of first round is to be played between 16th seeded player and the 17th seeded player. In the second round, the winner of match No. 1 of first round plays the winner of match No. 16 of first round and is designated match No. 1 of second round. Similarly, the winner of match No. 2 of first round plays the winner of match No. 15 of first round, and is designated match No. 2 of second round. Thus, for instance, match No. 8 of the second round is to be played between the winner of match No. 8 of first round and the winner of match No. 9 of first round. The same pattern is followed for later rounds as well.

**CAT 2005 LRDI | DI - Games & Tournaments**

If there are no upsets (a lower seeded player beating a higher seeded player) in the first round, and only match Nos. 6, 7, and 8 of the second round result in upsets, then who would meet Lindsay Davenport in quarter finals, in case Davenport reaches quarter finals?

- A.
Justine Henin

- B.
Nadia Petrova

- C.
Patty Schnyder

- D.
Venus Williams

Answer: Option D

**Explanation** :

The table shows the match nos. and the seed numbers of players playing those matches in Round 1 and 2.

As there are no upsets in the first round, players seeded 1 to 16 reach round 2.

There are upsets only in matches 6, 7 and 8 in round 2. So, seed numbers 1, 2, 3, 4, 5, 11, 10 and 9 reach the quarter finals. Then Davenport who is seed no. 2 plays seed no. 10, who is Venus Williams.

Hence, option 4.

Workspace:

**CAT 2005 LRDI | DI - Games & Tournaments**

If Elena Dementieva and Serena Williams lose in the second round, while Justine Henin and Nadia Petrova make it to the semifinals, then who would play Maria Sharapova in the quarterfinals, in the event Sharapova reaches quarterfinals?

- A.
Dinara Safina

- B.
Justine Henin

- C.
Nadia Petrova

- D.
Patty Schnyder

Answer: Option C

**Explanation** :

Seed numbers 6 and 8 lose in the second round and seed numbers 7 and 9 reach the semi-finals.

Seed number 9 plays matches 9, 8 and 1 in rounds 1, 2 and the quarterfinals.

Sharapova, who is seed number 1, plays match no. 1 in every round. Thus, Sharapova plays seed number 9, Nadia Petrova, in the quarterfinals.

Hence, option 3.

Workspace:

**CAT 2005 LRDI | DI - Games & Tournaments**

If, in the first round, all even numbered matches (and none of the odd numbered ones) result in upsets, and there are no upsets in the second round, then who could be the lowest seeded player facing Maria Sharapova in semi-finals?

- A.
Anastasia Myskina

- B.
Flavia Pennetta

- C.
Nadia Petrova

- D.
Svetlana Kuznetsova

Answer: Option A

**Explanation** :

The matches in rounds 1 and 2, quarterfinals and semi-finals are as shown in the table.

Sharapova is seeded 1. The lowest seed that could face her in the semi-finals could be seed no. 13, which is Anastasia Myskina.

Hence, option 1.

Workspace:

**CAT 2005 LRDI | DI - Games & Tournaments**

If the top eight seeds make it to the quarterfinals, then who, amongst the players listed below, would definitely not play against Maria Sharapova in the final, in case Sharapova reaches the final?

- A.
Amelie Mauresmo

- B.
Elena Dementieva

- C.
Kim Clijsters

- D.
Lindsay Davenport

Answer: Option C

**Explanation** :

The top 8 seeds make it to the quarterfinals. Thus matches 1 to 4 in quarter finals are between 1 and 8, 2 and 7, 3 and 6, and 4 and 5.

Sharapova is seeded 1. If she reaches the finals, she definitely beats seed number 8 in the quaterfinals and one of seed numbers 4 or 5 in the semi-finals. So, she can play seed numbers 2, 3, 6, or 7 in the finals. Kim Clijsters is seeded 4. Thus, she will definitely not play against Sharapova in the final.

Hence, option 3.

Workspace:

**Answer the following question based on the information given below.**

The year was 2006. All six teams in Pool A of World Cup hockey, play each other exactly once. Each win earns a team three points, a draw earns one point and a loss earns zero points. The two teams with the highest points qualify for the semi-finals. In case of a tie, the team with the highest goal difference (Goal For – Goals Against) qualifies.

In the opening match, Spain lost to Germany. After the second round (after each team played two matches), the pool table looked as shown below.

In the third round, Spain played Pakistan, Argentina played Germany, and New Zealand played South Africa. All the third round matches were drawn. The following are some results from the fourth and fifth round matches.

a. Spain won both the fourth and fifth round matches.

b. Both Argentina and Germany won their fifth round matches by 3 goals to 0.

c. Pakistan won both the fourth and fifth round matches by 1 goal to 0.

**CAT 2004 LRDI | DI - Games & Tournaments**

Which one of the following statements is true about matches played in the first two rounds?

- A.
Pakistan beat South Africa by 2 goals to 1.

- B.
Argentina beat Pakistan by 1 goal to 0.

- C.
Germany beat Pakistan by 2 goals to 1.

- D.
Germany beat Spain by 2 goals to 1.

Answer: Option B

**Explanation** :

Since Pakistan’s ‘Goals For’ (GF) and ‘Goals Against’ (GA) is 2 and 1 respectively, and it won 1 game and lost the other, the only possible combination of scores for Pakistan in its two matches will be 2:0 and 0:1.

Similarly, Germany’s scores in its two matches will be 2:1 and 1:0; while Argentina’s scores in its two matches will be 1:0 and 1:0.

Hence, we have the following table:

(G stands for Germany, A for Argentina, S for Spain, P for Pakistan, NZ for New Zealand and SA for South Africa)

Since Spain lost to Germany in the opening round, one of Spain’s score could be either 0:1 or 1:2. Since Spain’s total GF = 5 and GA = 2 for the first 2 rounds, we have the following possible scores for Spain:

Since the GF for both South Africa and New Zealand is 1, for each of these teams the GF for one match will be 1 while the GF of the second match will be 0. Hence, the different possibilities are as follows:

Combining the above possibilities for Spain, SA and NZ with the table showing the scores of Germany, Argentina and Pakistan, we see that the 1st possibility for SA is not possible since no possibility for any other team has the score 3:1. Similarly, the 2nd and 3rd possibility for NZ can also be eliminated. Hence, we have,

Looking at the GF and GA values, we can guess which teams played against others, for certain cases. (For example, SA has a score of 1:2 and Germany is the only team to have 2:1.) These are filled in the first column of the above table.

Also, since Germany beats SA with a score of 2:1, it beats Spain with a score of 1:0. This eliminates the 2nd possibility for Spain. This implies that Spain beats NZ with 5:1, eliminating NZ’s 1st possibility. Hence, we have,

Compiling the above information, we have the following:

Argentina beat Pakistan by 1 goal to 0.

Hence, option 2.

Workspace:

**CAT 2004 LRDI | DI - Games & Tournaments**

Which one of the following statements is true about matches played in the first two rounds?

- A.
Germany beat New Zealand by 1 goal to 0.

- B.
Spain beat New Zealand by 4 goals to 0.

- C.
Spain beat South Africa by 2 goals to 0.

- D.
Germany beat South Africa by 2 goals to 1.

Answer: Option D

**Explanation** :

Germany beat SA by 2 goals to 1.

Hence, option 4.

Workspace:

**CAT 2004 LRDI | DI - Games & Tournaments**

If Pakistan qualified as one of the two teams from Pool A, which was the other team that qualified?

- A.
Argentina

- B.
Germany

- C.
Spain

- D.
Cannot be determined

Answer: Option D

**Explanation** :

The following table gives the distribution of games within each round:

(Note: There is an inconsistency in the question. In Round 4, only 4 teams have played while in Round 5, two teams (NZ and SA) have played 2 matches. This, as we will see later, causes an inconsistency in the question as well.)

The points table will be as follows:

Since, 4 teams are tied with the same number of points, the qualifying teams will be chosen based on their Goal Difference (GD).

The GFs and GAs for the first two rounds is already given. In the third round, the match of Germany versus Argentina was drawn. Hence, both their GFs and GAs will be equal (say, X). In the fourth round, Spain won the match against Argentina. Hence, if Spain’s score is (say) A:B, where A > B, then Argentina’s will be B:A.

Following along similar lines, we have the following table:

Hence,

GD for Pakistan = (2 + Y + 1 + 1) – (1 + Y + 0 + 0) = 3

GD for Germany = (3 + X + 0 + 3) – (1 + X + 1 + 0) = 4

GD for Argentina = (2 + X + B + 3) – (0 + X + A + 0) = 5 + (B – A)

Since B < A, hence (B – A) < 0

∴ GD for Argentina < 5

GD for Spain = (5 + Y + A + C) – (2 + Y + B + D) = 3 + (A – B) + (C – D)

Since A > B and C > D, hence (A − B) ≥ 1 and (C – D) ≥ 1

∴ GD for Spain ≥ 5

Hence, Pakistan cannot qualify for the finals! The closest option that can be marked is 'Cannot be determined'.

Hence, option 4.

Workspace:

**CAT 2004 LRDI | DI - Games & Tournaments**

Which team finished at the top of the pool after five rounds of matches?

- A.
Argentina

- B.
Germany

- C.
Spain

- D.
Cannot be determined

Answer: Option C

**Explanation** :

The following table gives the distribution of games within each round:

(Note: There is an inconsistency in the question. In Round 4, only 4 teams have played while in Round 5, two teams (NZ and SA) have played 2 matches. This, as we will see later, causes an inconsistency in the question as well.)

The points table will be as follows:

Since, 4 teams are tied with the same number of points, the qualifying teams will be chosen based on their Goal Difference (GD).

The GFs and GAs for the first two rounds is already given. In the third round, the match of Germany versus Argentina was drawn. Hence, both their GFs and GAs will be equal (say, X). In the fourth round, Spain won the match against Argentina. Hence, if Spain’s score is (say) A:B, where A > B, then Argentina’s will be B:A.

Following along similar lines, we have the following table:

Hence,

GD for Pakistan = (2 + Y + 1 + 1) – (1 + Y + 0 + 0) = 3

GD for Germany = (3 + X + 0 + 3) – (1 + X + 1 + 0) = 4

GD for Argentina = (2 + X + B + 3) – (0 + X + A + 0) = 5 + (B – A)

Since B < A, hence (B – A) < 0

∴ GD for Argentina < 5

GD for Spain = (5 + Y + A + C) – (2 + Y + B + D) = 3 + (A – B) + (C – D)

Since A > B and C > D, hence (A − B) ≥ 1 and (C – D) ≥ 1

∴ GD for Spain ≥ 5

it is clear that Germany, Argentina, Spain and Pakistan are tied after 5 rounds, but only Spain’s GD ≥ 5.

Hence, option 3.

Workspace:

**CAT 1998 LRDI | DI - Games & Tournaments**

A, B, C, D, ..., X, Y, Z are the players who participated in a tournament. Everyone played with every other player exactly once. A win scores 2 points, a draw scores 1 point and a loss scores 0 point. None of the matches ended in a draw. No two players scored the same score. At the end of the tournament, by ranking list is published which is in accordance with the alphabetical order. Then

- A.
M wins over N

- B.
N wins over M

- C.
M does not play with N

- D.
None of these

Answer: Option A

**Explanation** :

It can be seen that each of the 26 players played 25 matches.

Since none of the matches ended in a draw, the scores for each of the players has to be even (since a win gives 2 points). So the highest score possible for a player would be 50 and the lowest would be 0.

Since all 26 of them had different scores varying between 0 and 50, the scores should indeed be all the even numbers between 0 and 50. And since the ranks obtained by players are in alphabetical order, it can be concluded that A scored 50, B scored 48, C scored 46 and so on and Z scored 0.

Now the only way A can score 50 is, if he wins all his matches, i.e. he defeats all other players. Now B has scored 48. So he has lost only one of his matches, which incidentally is against A. He must have defeated all other players.

Similarly, C has scored 46 in 25 matches. So he must have lost two matches, (i.e. to A and B) and defeated all other players. So we conclude that a player whose name appears alphabetically higher up in the order has defeated all the players whose name appear alphabetically lower down.

Hence, M should win over N.

Workspace:

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