The correct answer is **B) 7**\n\nHere's a detailed step-by-step solution:\n\n**Step 1: Determine the individual Wins, Draws, and Losses for each team.**\nFor each team, let $W$ be the number of wins, $D$ be the number of draws, and $L$ be the number of losses. Each team played 3 matches, so $W + D + L = 3$. The points ($P$) are calculated as $3W + 1D = P$.\n\n* **Team P (7 points):**\n $3W_P + D_P = 7$\n $W_P + D_P + L_P = 3$\n If $W_P = 0$, $D_P = 7$ (impossible as $D_P \le 3$).\n If $W_P = 1$, $D_P = 4$ (impossible).\n If $W_P = 2$, $D_P = 1$. Then $L_P = 3 - 2 - 1 = 0$. So, P has (2 Wins, 1 Draw, 0 Losses).\n\n* **Team Q (5 points):**\n $3W_Q + D_Q = 5$\n $W_Q + D_Q + L_Q = 3$\n If $W_Q = 0$, $D_Q = 5$ (impossible).\n If $W_Q = 1$, $D_Q = 2$. Then $L_Q = 3 - 1 - 2 = 0$. So, Q has (1 Win, 2 Draws, 0 Losses).\n\n* **Team R (3 points):**\n $3W_R + D_R = 3$\n $W_R + D_R + L_R = 3$\n Possibility 1: If $W_R = 0$, $D_R = 3$. Then $L_R = 3 - 0 - 3 = 0$. R1: (0 Wins, 3 Draws, 0 Losses).\n Possibility 2: If $W_R = 1$, $D_R = 0$. Then $L_R = 3 - 1 - 0 = 2$. R2: (1 Win, 0 Draws, 2 Losses).\n\n* **Team S (1 point):**\n $3W_S + D_S = 1$\n $W_S + D_S + L_S = 3$\n If $W_S = 0$, $D_S = 1$. Then $L_S = 3 - 0 - 1 = 2$. So, S has (0 Wins, 1 Draw, 2 Losses).\n\n**Step 2: Determine total Wins, Draws, and Losses in the tournament.**\nThere are 4 teams, and each plays every other team once, so the total number of matches is $\\binom{4}{2} = \\frac{4 \\times 3}{2} = 6$. \nLet $N_W$ be the number of matches that ended in a win/loss (decided matches), and $N_D$ be the number of matches that ended in a draw (drawn matches).\n\nTotal matches: $N_W + N_D = 6$\nTotal points awarded: In a decided match, 3 points are awarded (3 for winner, 0 for loser). In a drawn match, 2 points are awarded (1 for each team).\nSum of all team points = $7 + 5 + 3 + 1 = 16$.\nSo, $3N_W + 2N_D = 16$.\n\nWe have a system of equations:\n1) $N_W + N_D = 6$\n2) $3N_W + 2N_D = 16$\n\nMultiply equation (1) by 2: $2N_W + 2N_D = 12$.\nSubtract this from equation (2): $(3N_W + 2N_D) - (2N_W + 2N_D) = 16 - 12$\n$N_W = 4$.\nSubstitute $N_W = 4$ into equation (1): $4 + N_D = 6 \\implies N_D = 2$.\n\nSo, there were 4 decided matches and 2 drawn matches in the tournament.\n\n**Step 3: Consolidate team statistics based on global totals.**\nSum of Wins from individual teams must equal $N_W = 4$.\nSum of Draws from individual teams must equal $2 \\times N_D = 2 \\times 2 = 4$.\nSum of Losses from individual teams must equal $N_W = 4$.\n\nLet's check the two possibilities for Team R:\n\n* **Case R1 (0W, 3D, 0L):**\n Total Wins ($W_T$) = $W_P + W_Q + W_R + W_S = 2 + 1 + 0 + 0 = 3$.\n Total Draws ($D_T$) = $D_P + D_Q + D_R + D_S = 1 + 2 + 3 + 1 = 7$.\n Total Losses ($L_T$) = $L_P + L_Q + L_R + L_S = 0 + 0 + 0 + 2 = 2$.\n This case is inconsistent because $W_T \\ne L_T$ (3 wins vs 2 losses) and $D_T$ (7) is odd, meaning an odd number of drawn matches, which is impossible (each draw involves 2 teams, so sum of draws must be even).\n\n* **Case R2 (1W, 0D, 2L):**\n Total Wins ($W_T$) = $W_P + W_Q + W_R + W_S = 2 + 1 + 1 + 0 = 4$.\n Total Draws ($D_T$) = $D_P + D_Q + D_R + D_S = 1 + 2 + 0 + 1 = 4$.\n Total Losses ($L_T$) = $L_P + L_Q + L_R + L_S = 0 + 0 + 2 + 2 = 4$.\n This case is consistent: $W_T = L_T = 4$, and $D_T = 4$ (even). The number of actual drawn matches is $D_T/2 = 4/2 = 2$, which matches $N_D$.\n\nTherefore, the definitive statistics for each team are:\n* **P:** (2 Wins, 1 Draw, 0 Losses)\n* **Q:** (1 Win, 2 Draws, 0 Losses)\n* **R:** (1 Win, 0 Draws, 2 Losses)\n* **S:** (0 Wins, 1 Draw, 2 Losses)\n\n**Step 4: Deduce the outcome of each match.**\nWe know there were 2 draws in total. From the individual draw counts: P (1 draw), Q (2 draws), R (0 draws), S (1 draw).\nSince R had 0 draws, R was not involved in any drawn match. So, the two drawn matches must be between P, Q, and S.\nFor Q to have 2 draws, and P and S to have 1 draw each, the only possible combination for draws is: \n* P drew Q (P-D-Q)\n* Q drew S (Q-D-S)\n\nLet's verify these draws with individual team draw counts: P (1), Q (2), S (1). This matches. \n\nNow, let's assign results for the remaining matches:\n* **Team P:** Has 2 wins and 1 draw. P-D-Q is confirmed. P has no losses ($L_P=0$). So P must have won against R and S.\n * P beat R (P-W-R)\n * P beat S (P-W-S)\n\n* **Team Q:** Has 1 win, 2 draws, and 0 losses ($L_Q=0$). Q-D-P and Q-D-S are confirmed. Q has used up its 2 draws. Q must have won its remaining match against R.\n * Q beat R (Q-W-R)\n\n* **Team R:** Has 1 win, 0 draws, and 2 losses ($D_R=0$). R has lost to P and Q (P-W-R, Q-W-R). R has 2 losses, matching $L_R=2$. Since R has 0 draws, its match against S must have been a win (to achieve $W_R=1$).\n * R beat S (R-W-S)\n\n* **Team S:** Has 0 wins, 1 draw, and 2 losses. S-D-Q is confirmed. S lost to P and R (P-W-S, R-W-S). S has 1 draw and 2 losses, matching $D_S=1, L_S=2$. S has 0 wins, matching $W_S=0$.\n\nAll match outcomes are consistent with the derived statistics.\n\n**Summary of Match Outcomes:**\n1. P drew Q\n2. P beat R\n3. P beat S\n4. Q drew S\n5. Q beat R\n6. R beat S\n\n**Step 5: Answer the specific question.**\nThe question asks: \"If the match between Team Q and Team S had resulted in a win for Team Q instead of a draw