# Team Formation Questions for CAT

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## Team Formation Questions for CAT

Download important CAT Team Formation Questions with Solutions PDF based on previously asked questions in CAT exam. Practice Team Formation Questions with Solutions for CAT exam.

Instructions

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 â€“ I and two matches in Stage â€“ 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 â€“ I and Stage â€“ II are as given below.

Stage-I:

â€¢ One team won all the three matches.

â€¢ Two teams lost all the matches.

â€¢ D lost to A but won against C and F.

â€¢ E lost to B but won against C and F.

â€¢ B lost at least one match.

â€¢ F did not play against the top team of Stage-I.

Stage-II:

â€¢ The leader of Stage-I lost the next two matches

â€¢ Of the two teams at the bottom after Stage-I, one team won both matches, while the other lost both matches.

â€¢ One more team lost both matches in Stage-II.

Question 1:Â The two teams that defeated the leader of Stage-I are:

a)Â B & F

b)Â E & F

c)Â B & D

d)Â E & D

e)Â F & D

Question 2:Â The only team(s) that won both matches in Stage-II is (are):

a)Â B

b)Â E & F

c)Â A, E & F

d)Â B, E & F

e)Â B & F

Question 3:Â The teams that won exactly two matches in the event are:

a)Â A, D & F

b)Â D & E

c)Â E & F

d)Â D, E & F

e)Â D & F

Question 4:Â The team(s) with the most wins in the event is (are):

a)Â A

b)Â A & C

c)Â F

d)Â E

e)Â B & E

Instructions

K, L, M, N, P, Q, R, S, U and W are the only ten members in a department. There is a proposal to form a team from within the members of the department, subject to the following conditions:

1. Â A team must include exactly one among P,R and S.
2. Â A team must include either M or Q, but not both.
3. Â If a team includes K, then it must also include L, and vice versa.
4. Â If a team includes one among S, U and W, then it should also include the other two.
5. Â L and N cannot be members of the same team.
6. Â L and U cannot be members of the same team.

The size of a team is defined as the number of members in the team.

Question 5:Â What could be the size of a team that includes K?

a)Â 2 or 3

b)Â 2 or 4

c)Â 3 or 4

d)Â Only 2

e)Â Only 4

Question 6:Â In how many ways a team can be constituted so that the team includes N?

a)Â 2

b)Â 3

c)Â 4

d)Â 5

e)Â 6

Question 7:Â What would be the size of the largest possible team?

a)Â 8

b)Â 7

c)Â 6

d)Â 5

e)Â cannot be determined

Question 8:Â Who can be a member of a team of size 5?

a)Â K

b)Â L

c)Â M

d)Â P

e)Â R

Question 9:Â Who cannot be a member of a team of size 3?

a)Â L

b)Â M

c)Â N

d)Â P

e)Â Q

Question 10:Â What would be the size of the largest possible team?

[CAT 2006]

a)Â 8

b)Â 7

c)Â 6

d)Â 5

e)Â Cannot be determined

Question 11:Â Who can be a member of a team of size 5?

[CAT 2006]

a)Â K

b)Â L

c)Â M

d)Â P

e)Â R

Question 12:Â Who cannot be a member of a team of size 3?

[CAT 2006]

a)Â L

b)Â M

c)Â N

d)Â P

e)Â Q

Question 13:Â In how many ways a team can be constituted so that the team includes N?

[CAT 2006]

a)Â 2

b)Â 3

c)Â 4

d)Â 5

e)Â 6

Question 14:Â What could be the size of a team that includes K?

[CAT 2006]

a)Â 2 or 3

b)Â 2 or 4

c)Â 3 or 4

d)Â Only 2

e)Â Only 4

There are a total of $^6 C_2$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches – stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the two teams that won against stage 1 leader A are E and F.

There are a total of $^6 C_2$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches – stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the teams that won both of their stage 2 matches are B, E and F.

There are a total of $^6 C_2$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches – stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the wins by each team are A (3), B(4), C(0), D(2), E(4), F(2). Hence, D and F won exactly 2 matches.

There are a total of $^6 C_2$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches – stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the wins by each team are A(3), B(4), C(0), D(2), E(4), F(2). Hence, most wins are by B and E.

A team which has K should have L also.

Since L is there in the team, N and U should not be there in the team. Since U is not there in the team, S and W should not be there in the team.

So, the team will have K, L, one of P and R and one of M or Q.

So, the team size will be 4.

Since N is in the team, L and K cannot be in the team.
The team can have one of M and Q. So, 2 ways of selection.
If the team has S, then it should have U and W as well.
If the team has no S, then it should have one of P or R.
So, the number of ways of forming the team is 2*(1+2) = 6 ways

Out of P, R and S only 1 can be in the team. If S is there, U and W will also be there. So, P and R should not be in the team for its size to be maximum.

Out of M and Q, only 1 can be there.

If L is there in the team, N and U cannot be in the team.

If L is not there in the team, then K is also not there in the team but N and U can be in the team.

So, the maximum team size is 5 consisting of S, U, W, (M or Q), N.

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Out of P, R and S only 1 can be in the team. If S is there, U and W will also be there. So, P and R should not be in the team for its size to be maximum.

Out of M and Q, only 1 can be there.

If L is there in the team, N and U cannot be in the team.

If L is not there in the team, then K is also not there in the team but N and U can be in the team.

So, the maximum team size is 5 consisting of S, U, W, (M or Q), N.

So, M can be a member of team size 5.

If L is in the team, the team should include K also. The team should have one among P, R and S and one among M and Q.

So, the team size cannot be 3 if L is in the team.

Out of P, R and S only 1 can be in the team. If S is there, U and W will also be there. So, P and R should not be in the team for its size to be maximum.

Out of M and Q, only 1 can be there.

If L is there in the team, N and U cannot be in the team.

If L is not there in the team, then K is also not there in the team but N and U can be in the team.

So, the maximum team size is 5 consisting of S, U, W, (M or Q), N.

Out of P, R and S only 1 can be in the team. If S is there, U and W will also be there. So, P and R should not be in the team for its size to be maximum.

Out of M and Q, only 1 can be there.

If L is there in the team, N and U cannot be in the team.

If L is not there in the team, then K is also not there in the team but N and U can be in the team.

So, the maximum team size is 5 consisting of S, U, W, (M or Q), N.

So, M can be a member of team size 5.

If L is in the team, the team should include K also. The team should have one among P, R and S and one among M and Q.

So, the team size cannot be 3 if L is in the team.

Since N is in the team, L and K cannot be in the team.

The team can have one of M and Q. So, 2 ways of selection.

If the team has S, then it should have U and W as well.

If the team has no S, then it should have one of P or R.

So, the number of ways of forming the team is 2*(1+2) = 6 ways