The correct answer is **A) 0-1** **Step-by-step derivation:** **1. Determine Wins, Draws, and Losses for each team based on Points:** Each team played 3 matches. Possible point combinations (Wins, Draws, Losses) are unique for certain point totals: * **Team P (4 points):** Must have (1 Win, 1 Draw, 1 Loss). ($1 \times 3 + 1 \times 1 + 1 \times 0 = 4$). So, $W_P=1, D_P=1, L_P=1$. * **Team R (4 points):** Must have (1 Win, 1 Draw, 1 Loss). ($1 \times 3 + 1 \times 1 + 1 \times 0 = 4$). So, $W_R=1, D_R=1, L_R=1$. * **Team S (2 points):** Must have (0 Wins, 2 Draws, 1 Loss). ($0 \times 3 + 2 \times 1 + 1 \times 0 = 2$). So, $W_S=0, D_S=2, L_S=1$. **2. Deduce specific match outcomes (Win/Draw/Loss):** There are a total of $\frac{4 \times 3}{2} = 6$ matches in a round-robin tournament. Let's use the given conditions and the (W,D,L) records: * **Condition 3:** "Team R's only loss was against Team P." This means P beat R. ($P \succ R$) * This accounts for P's 1 Win and R's 1 Loss. * **Condition 2:** "The match between P and S ended in a draw." ($P \leftrightarrow S$) * This accounts for P's 1 Draw and one of S's 2 Draws. Now, P's record (1W, 1D, 1L) is accounted for: P beat R (1W), P drew with S (1D). Therefore, P's only loss must be against Q. ($Q \succ P$) * This accounts for P's 1 Loss and one of Q's Wins. Remaining (W,D,L) counts after P's matches: * P: (1W, 1D, 1L) - All matches determined (beat R, drew S, lost to Q). * R: (1W, 1D, 1L) - 1 Loss to P. Needs 1 Win and 1 Draw from matches against Q and S. * S: (0W, 2D, 1L) - 1 Draw with P. Needs 1 Draw and 1 Loss from matches against Q and R. * Q: ($W_Q$, $D_Q$, $L_Q$) - Has 1 Win against P. Needs 2 more matches to determine (against R and S). * **Condition 1:** "Team Q lost its match against Team R." ($R \succ Q$) * This accounts for one of R's Wins and one of Q's Losses. Now R's record (1W, 1D, 1L) is accounted for: R lost to P (1L), R beat Q (1W). Therefore, R's remaining match against S must be a Draw. ($R \leftrightarrow S$) * This accounts for R's 1 Draw and S's second Draw. Now S's record (0W, 2D, 1L) is accounted for: S drew with P (1D), S drew with R (1D). Therefore, S's remaining match against Q must be a Loss. ($Q \succ S$) * This accounts for S's 1 Loss and another of Q's Wins. **3. Final W/D/L records and Points for Q:** * Q beat P ($1W_Q$) * Q lost to R ($1L_Q$) * Q beat S ($1W_Q$) So, Team Q's final record is (2 Wins, 0 Draws, 1 Loss). $W_Q=2, D_Q=0, L_Q=1$. Points for Q ($P_Q$) = $2 \times 3 + 0 \times 1 + 1 \times 0 = 6$ points. **Summary of Match Outcomes (Win/Draw/Loss):** * P vs Q: Q wins * P vs R: P wins * P vs S: Draw * Q vs R: R wins * Q vs S: Q wins * R vs S: Draw **4. Determine Goal Scores:** Let $S_{XY}$ be the goals scored by team X against team Y. * **Condition 4:** "Team S scored a total of 1 goal in the tournament." ($GF_S = S_{SP} + S_{SQ} + S_{SR} = 1$). This implies S scored 1 goal in one match and 0 goals in the other two. * **Condition 5:** "The sum of goals scored by all teams is 6." (Total $GF = 6$). This means the sum of goals in all 6 matches is 6. Let $TG_{XY}$ be total goals in match X-Y. $TG_{PQ} + TG_{PR} + TG_{PS} + TG_{QR} + TG_{QS} + TG_{RS} = 6$. From the W/D/L outcomes: * P vs Q (Q wins): $TG_{PQ} \ge 1$ (e.g., 0-1, 1-2) * P vs R (P wins): $TG_{PR} \ge 1$ (e.g., 1-0, 2-1) * P vs S (Draw): $TG_{PS} \ge 0$ (e.g., 0-0, 1-1) * Q vs R (R wins): $TG_{QR} \ge 1$ (e.g., 0-1, 1-2) * Q vs S (Q wins): $TG_{QS} \ge 1$ (e.g., 1-0, 2-1) * R vs S (Draw): $TG_{RS} \ge 0$ (e.g., 0-0, 1-1) Let's analyze the implications of $GF_S = 1$ in three cases: **Case 1: S scored its goal against P ($S_{SP}=1$).** * P vs S: Draw, so $S_{PS}=S_{SP}=1$. Score is 1-1. ($TG_{PS}=2$) * S scored 0 goals in other matches: $S_{SQ}=0$ and $S_{SR}=0$. * Q vs S: Q wins, S scored 0 goals. So Q must score at least 1 goal. Score is $x-0$, where $x \ge 1$. ($TG_{QS}=x$) * R vs S: Draw, S scored 0 goals. So R must score 0 goals. Score is 0-0. ($TG_{RS}=0$) Substitute these into the total goals equation: $TG_{PQ} + TG_{PR} + TG_{QR} + TG_{PS} + TG_{QS} + TG_{RS} = 6$ $(S_{PQ}+S_{QP}) + (S_{PR}+S_{RP}) + (S_{QR}+S_{RQ}) + 2 + x + 0 = 6$ Let $T_D = (S_{PQ}+S_{QP}) + (S_{PR}+S_{RP}) + (S_{QR}+S_{RQ})$ be the sum of goals in the three decisive matches P-Q, P-R, Q-R. $T_D + x + 2 = 6 \implies T_D + x = 4$. Since $T_D \ge 1+1+1 = 3$ (each decisive match has at least 1 goal) and $x \ge 1$, the only possible solution is $T_D=3$ and $x=1$. This means: * $T_D=3$: This forces each of the decisive matches (P-Q, P-R, Q-R) to have a total of 1 goal. The only way this is possible with a winner/loser is (0-1) or (1-0). * P vs Q (Q wins): Score must be 0-1. ($S_{PQ}=0, S_{QP}=1$) * P vs R (P wins): Score must be 1-0. ($S_{PR}=1, S_{RP}=0$) * Q vs R (R wins): Score must be 0-1. ($S_{QR}=0, S_{RQ}=1$) * $x=1$: Q vs S score must be 1-0. So, in Case 1, the unique scores are: * P vs Q: 0-1 * P vs R: 1-0 * P vs S: 1-1 * Q vs R: 0-1 * Q vs S: 1-0 * R vs S: 0-0 Total GF = $GF_P+GF_Q+GF_R+GF_S = (0+1+1) + (1+0+1) + (0+1+0) + (1+0+0) = 2+2+1+1=6$. (Consistent) **Case 2: S scored its goal against Q ($S_{SQ}=1$).** * Q vs S: Q wins. S scored 1 goal. So Q must have scored $k$ goals, where $k > 1$. Score is $k-1$. ($TG_{QS}=k+1$) * S scored 0 goals in other matches: $S_{SP}=0$ and $S_{SR}=0$. * P vs S: Draw, S scored 0 goals. So P must score 0 goals. Score is 0-0. ($TG_{PS}=0$) * R vs S: Draw, S scored 0 goals. So R must score 0 goals. Score is 0-0. ($TG_{RS}=0$) Substitute these into the total goals equation: $T_D + 0 + (k+1) + 0 = 6 \implies T_D + k + 1 = 6 \implies T_D + k = 5$. Since $T_D \ge 3$ and $k > 1 \implies k \ge 2$, the only possible solution is $T_D=3$ and $k=2$. This means: * $T_D=3$: Forces P-Q: 0-1, P-R: 1-0, Q-R: 0-1. * $k=2$: Q vs S score must be 2-1. So, in Case 2, the unique scores are: * P vs Q: 0-1 * P vs R: 1-0 * P vs S: 0-0 * Q vs R: 0-1 * Q vs S: 2-1 * R vs S: 0-0 Total GF = $GF_P+GF_Q+GF_R+GF_S = (0+1+0) + (1+0+2) + (0+1+0) + (0+1+0) = 1+3+1+1=6$. (Consistent) **Case 3: S scored its goal against R ($S_{SR}=1$).** * R vs S: Draw, so $S_{RS}=S_{SR}=1$. Score is 1-1. ($TG_{RS}=2$) * S scored 0 goals in other matches: $S_{SP}=0$ and $S_{SQ}=0$. * P vs S: Draw, S scored 0 goals. So P must score 0 goals. Score is 0-0. ($TG_{PS}=0$) * Q vs S: Q wins, S scored 0 goals. So Q must score at least 1 goal. Score is $x-0$, where $x \ge 1$. ($TG_{QS}=x$) Substitute these into the total goals equation: $T_D + 0 + x + 2 = 6 \implies T_D + x = 4$. Since $T_D \ge 3$ and $x \ge 1$, the only possible solution is $T_D=3$ and $x=1$. This means: * $T_D=3$: Forces P-Q: 0-1, P-R: 1-0, Q-R: 0-1. * $x=1$: Q vs S score must be 1-0. So, in Case 3, the unique scores are: * P vs Q: 0-1 * P vs R: 1-0 * P vs S: 0-0 * Q vs R: 0-1 * Q vs S: 1-0 * R vs S: 1-1 Total GF = $GF_P+GF_Q+GF_R+GF_S = (0+1+0) + (1+0+1) + (0+1+1) + (0+0+1) = 1+2+2+1=6$. (Consistent) **Conclusion:** In all three possible scenarios (each assuming S scored its only goal in a different match), the scores for the matches P-Q, P-R, and Q-R are uniquely determined as 0-1, 1-0, and 0-1, respectively. Specifically, the match between Team Q and Team R always resulted in a score of 0-1, with Team R winning. The final answer is $\boxed{\text{0-1}}$