Practice Questions: Team Formation- 2

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 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)
Explanation:
There are a total of ⁶C₂ matches => 15 matches. Stage-I contains 9 matches and Stage-II contains the remaining 6 matches.
From Stage-I clues, we can list definite matches and results: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won).
Only one team won all three Stage-I matches. Since B, C, D, E and F each lost at least one match in Stage-I, A must be the team that won all three Stage-I matches. From the matches already deduced, A still needs two opponents, B needs two opponents and C and F need one opponent each. C and F both lost all their Stage-I matches, so they could not have played each other. Also, F did not play the Stage-I leader A. The remaining Stage-I fixtures consistent with these constraints are: A-B (A won), A-C (A won) and B-F (B won).
Thus, Stage-I matches (with results) are: 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).
The six Stage-II fixtures are then the pairs not yet played: D-B, D-E, E-A, F-A, B-C and C-F. The leader of Stage-I (A) lost both Stage-II matches, so the two teams that beat A are E and F. Hence, the correct option is (b).

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 the 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)  
Explanation:
From the reconstructed results, the total wins for each team are: A - 3 wins; B - 4 wins; C - 0 wins; D - 2 wins; E - 4 wins; F - 2 wins.
The highest number of wins is 4, achieved jointly by B and E. Hence, option (e) is correct.

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 can a team be constituted so that the team includes N?
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6
Ans: (e)  
Explanation:
If N is included, then L and K cannot be in the team (condition 5 and the K↔L dependency).
Choose one of M or Q: 2 choices.
Consider the choice among P, R and S (exactly one required):
- If S is chosen then, by condition 4, U and W must also be included. That gives one team composition for each choice of (M or Q).
- If S is not chosen, we must choose exactly one from {P, R}, which gives 2 choices for the P/R slot, again for each choice of (M or Q).
Total = 2 (M/Q) × (1 (S,U,W case) + 2 (P or R cases)) = 2 × 3 = 6 ways. Hence option (e).

Q6: Who can be a member of a team of size 5?
(a) K
(b) L
(c) M
(d) P
(e) R
Ans: (c)  
Explanation:
To obtain the largest team size, include S which forces U and W to be included (so S,U,W are in). Also include N (possible only when L and K are excluded). Include one of M or Q (but not both). That gives a 5-member team: S, U, W, N and (M or Q).
Therefore, M (or Q) can be a member of a size-5 team; among the options given, only M (option c) is correct.

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)
Explanation:
If K is included, then L must also be included (condition 3). If L is in the team, then N and U cannot be members (conditions 5 and 6). Because U is excluded, S and W (which require U when S is chosen) cannot be present. The team must include exactly one of P, R or S - since S is excluded, that leaves one choice from {P, R}. Also include exactly one of M or Q. So the smallest feasible composition including K is {K, L, (P or R), (M or Q)}, which has 4 members. No 2- or 3-member team including K is possible under these constraints. Hence, the team size must be 4 (option e).

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)
 Explanation:
To maximise team size, include S, which forces U and W to be included (S,U,W). Do not include L or K (since including L would exclude N and U, reducing size). Include N (possible when L and K are excluded). Include one of M or Q. Thus the largest feasible team is S, U, W, N and (M or Q) - five members. Therefore, the largest possible team size is 5 (option d).

Q9: Who cannot be a member of a team of size 3?
(a) L
(b) M
(c) N
(d) P
(e) Q
Ans: (a)  
Explanation:
If L is in a team, then K must also be in it (condition 3). Additionally, a team must include exactly one among P,R,S and exactly one among M,Q. Thus any team containing L must contain at least K, L, one from {P,R,S} and one from {M,Q} - that is, at least 4 members. Therefore, L cannot be a member of any 3-member team. Hence, option (a) is correct.