Practice Questions: Team Formation- 2 | Logical Reasoning (LR) and Data Interpretation (DI) - CAT PDF Download

Direction:
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.

Q1: The two teams that defeated the leader of Stage-I are:
(a) B & F
(b) E & F
(c) B & D
(d) E & D
(e) F & D
Ans: (b)
Sol:
There are a total of ⁶C₂ 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.

Q2: The only team(s) that won both matches in Stage-II is (are):
(a) B
(b) E & F
(c) A, E & F
(d) B, E & F
(e) B & F
Ans: (d)
Sol:
There are a total of ⁶C₂ 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.

Q3: 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
Ans: (e)  
Sol:
There are a total of ⁶C₂ 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.

Q4: The team(s) with the most wins in the event is (are):
(a) A
(b) A & C
(c) F
(d) E
(e) B & E
Ans: (e)  
Sol:
There are a total of ⁶C₂ 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.

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

Q5: In how many ways a team can be constituted so that the team includes N?
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6
Ans: (e)  
Sol:
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

Q6: Who can be a member of a team of size 5?
(a) K
(b) L
(c) M
(d) P
(e) R
Ans: (c)  
Sol:
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.

Q7: What could be the size of a team that includes K?
(a) 2 or 3
(b) 2 or 4
(c) 3 or 4
(d) Only 2
(e) Only 4
Ans: (e)
Sol:
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 can be 4.

Q8: What would be the size of the largest possible team?
(a) 8
(b) 7
(c) 6
(d) 5
(e) cannot be determined
Ans: (d)
 Sol:
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.

Q9: What could be the size of a team that includes K?
(a) 2 or 3
(b) 2 or 4
(c) 3 or 4
(d) Only 2
(e) Only 4
Ans: (e)
Sol:
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.

Q10: Who cannot be a member of a team of size 3?
(a) L
(b) M
(c) N
(d) P
(e) Q
Ans: (a)  
Sol:
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.

Q11: What would be the size of the largest possible team?
(a) 8
(b) 7
(c) 6
(d) 5
(e) Cannot be determined
Ans: (d)  
Sol:
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.

Q12: Who cannot be a member of a team of size 3? [CAT 2006]
(a) L
(b) M
(c) N
(d) P
(e) Q
Ans: (a)  
Sol:
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.

Q3: Who can be a member of a team of size 5?
(a) K
(b) L
(c) M
(d) P
(e) R
Ans: (c)  
Sol:
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.

Q14: In how many ways a team can be constituted so that the team includes N?
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6
Ans: (e)  
Sol:
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