Four football teams — A, B, C and D — played a single round-robin tournament: each team played every other team exactly once, for a total of six matches. A win earned 3 points, a draw 1 point to each side and a loss 0 points.
The final table below shows the points of each team and the goals scored (GF) by each team over the tournament. Goals conceded (GA) are shown only for A and D.
| Team | Points | GF | GA |
|---|---|---|---|
| A | 7 | 6 | 2 |
| B | 4 | 4 | – |
| C | 4 | 3 | – |
| D | 1 | 2 | 6 |
Additional information:
- No match had more than 3 goals in total.
- A won more than one match, and A's margin of victory against D was strictly greater than its margin of victory in any other match.
The final table below shows the points of each team and the goals scored (GF) by each team over the tournament. Goals conceded (GA) are shown only for A and D.
| Team | Points | GF | GA |
|---|---|---|---|
| A | 7 | 6 | 2 |
| B | 4 | 4 | – |
| C | 4 | 3 | – |
| D | 1 | 2 | 6 |
Additional information:
- No match had more than 3 goals in total.
- A won more than one match, and A's margin of victory against D was strictly greater than its margin of victory in any other match.
Question 1
Single correct option
What was the result of the match between B and C?
- A. B won 2–0
- B. B won 1–0
- C. C won 2–0
- D. The match was drawn 1–1
Show answer and solution
Answer: (A) B won 2–0
Deduction. Points force the records: A = 7 → 2W 1D; B = C = 4 → 1W 1D 1L each; D = 1 → 1D 2L. So exactly 2 matches were drawn.
A's matches total 6–2 with at most 3 goals per match. If A's draw were 0–0, its two wins would have to total 6–2, impossible with ≤ 3 goals each (the only 3-goal wins are 3–0 and 2–1). So the draw is 1–1 and the wins total 5–1: they must be 3–0 and 2–1. A's bigger win, 3–0, is against D.
D's other matches (vs B and C) total 2–3: one draw and one loss. A 0–0 draw would need a 2–3 loss (5 goals), so D drew 1–1 and lost 1–2.
Cases. If A beat C 2–1 and drew with B, then C (having lost to A) must draw with D and beat B, while B beats D 2–1; B's goals then force B–C to be 1–1 — a contradiction. Hence A beat B 2–1 and drew 1–1 with C; B drew 1–1 with D; C beat D 2–1; and B's goal total (4) gives B 2–0 C.
The unique set of results:
| Match | Score |
|---|---|
| A vs B | 2–1 |
| A vs C | 1–1 |
| A vs D | 3–0 |
| B vs C | 2–0 |
| B vs D | 1–1 |
| C vs D | 2–1 |
Hence B beat C 2–0.
A's matches total 6–2 with at most 3 goals per match. If A's draw were 0–0, its two wins would have to total 6–2, impossible with ≤ 3 goals each (the only 3-goal wins are 3–0 and 2–1). So the draw is 1–1 and the wins total 5–1: they must be 3–0 and 2–1. A's bigger win, 3–0, is against D.
D's other matches (vs B and C) total 2–3: one draw and one loss. A 0–0 draw would need a 2–3 loss (5 goals), so D drew 1–1 and lost 1–2.
Cases. If A beat C 2–1 and drew with B, then C (having lost to A) must draw with D and beat B, while B beats D 2–1; B's goals then force B–C to be 1–1 — a contradiction. Hence A beat B 2–1 and drew 1–1 with C; B drew 1–1 with D; C beat D 2–1; and B's goal total (4) gives B 2–0 C.
The unique set of results:
| Match | Score |
|---|---|
| A vs B | 2–1 |
| A vs C | 1–1 |
| A vs D | 3–0 |
| B vs C | 2–0 |
| B vs D | 1–1 |
| C vs D | 2–1 |
Hence B beat C 2–0.
Question 2
Numerical answer
How many matches in the tournament ended in a draw?
Show answer and solution
Answer: 2
No detailed grid is needed. A (7 points) must have 2 wins and 1 draw; B and C (4 points each) must have 1 win, 1 draw and 1 loss; D (1 point) has 1 draw and 2 losses.
Each drawn match gives two 'draw appearances'. Total draw appearances = 1 + 1 + 1 + 1 = 4, so the number of drawn matches = 4/2 = 2 (A–C 1–1 and B–D 1–1).
Check: total points = 3 × (decisive matches) + 2 × (draws) = 3 × 4 + 2 × 2 = 16 = 7 + 4 + 4 + 1. ✓
Each drawn match gives two 'draw appearances'. Total draw appearances = 1 + 1 + 1 + 1 = 4, so the number of drawn matches = 4/2 = 2 (A–C 1–1 and B–D 1–1).
Check: total points = 3 × (decisive matches) + 2 × (draws) = 3 × 4 + 2 × 2 = 16 = 7 + 4 + 4 + 1. ✓
Question 3
Numerical answer
How many goals did C concede in the tournament?
Show answer and solution
Answer: 4
The unique set of results:
| Match | Score |
|---|---|
| A vs B | 2–1 |
| A vs C | 1–1 |
| A vs D | 3–0 |
| B vs C | 2–0 |
| B vs D | 1–1 |
| C vs D | 2–1 |
C's matches: drew 1–1 with A (conceded 1), lost 0–2 to B (conceded 2), beat D 2–1 (conceded 1).
GA(C) = 1 + 2 + 1 = 4. Consistency check: GA(B) = 2 (vs A) + 0 (vs C) + 1 (vs D) = 3, and GA(B) + GA(C) = 7 = 15 − 2 − 6. ✓
| Match | Score |
|---|---|
| A vs B | 2–1 |
| A vs C | 1–1 |
| A vs D | 3–0 |
| B vs C | 2–0 |
| B vs D | 1–1 |
| C vs D | 2–1 |
C's matches: drew 1–1 with A (conceded 1), lost 0–2 to B (conceded 2), beat D 2–1 (conceded 1).
GA(C) = 1 + 2 + 1 = 4. Consistency check: GA(B) = 2 (vs A) + 0 (vs C) + 1 (vs D) = 3, and GA(B) + GA(C) = 7 = 15 − 2 − 6. ✓
Question 4
Single correct option
Against which team(s) did D score its goals?
- A. B and C only
- B. A and C only
- C. B only
- D. C only
Show answer and solution
Answer: (A) B and C only
The unique set of results:
| Match | Score |
|---|---|
| A vs B | 2–1 |
| A vs C | 1–1 |
| A vs D | 3–0 |
| B vs C | 2–0 |
| B vs D | 1–1 |
| C vs D | 2–1 |
D lost 0–3 to A, drew 1–1 with B and lost 1–2 to C. So D's two goals came one each against B and C.
This follows directly from the deduction that A's 3–0 win over D is forced: D scored nothing against A, and D's two remaining matches (one draw, one loss with ≤ 3 goals each and 3 goals conceded between them) must be 1–1 and 1–2.
| Match | Score |
|---|---|
| A vs B | 2–1 |
| A vs C | 1–1 |
| A vs D | 3–0 |
| B vs C | 2–0 |
| B vs D | 1–1 |
| C vs D | 2–1 |
D lost 0–3 to A, drew 1–1 with B and lost 1–2 to C. So D's two goals came one each against B and C.
This follows directly from the deduction that A's 3–0 win over D is forced: D scored nothing against A, and D's two remaining matches (one draw, one loss with ≤ 3 goals each and 3 goals conceded between them) must be 1–1 and 1–2.
Question 5
Single correct option
B and C finished level on points. If ties on points are broken by goal difference (GF − GA), which team finished second?
- A. B
- B. C
- C. D
- D. Cannot be determined
Show answer and solution
Answer: (A) B
The unique set of results:
| Match | Score |
|---|---|
| A vs B | 2–1 |
| A vs C | 1–1 |
| A vs D | 3–0 |
| B vs C | 2–0 |
| B vs D | 1–1 |
| C vs D | 2–1 |
Goal difference: B = 4 − 3 = +1 (conceded 2 to A, 0 to C, 1 to D); C = 3 − 4 = −1.
So B finishes second, ahead of C. It cannot be 'cannot be determined', because the clues pin down every score uniquely; D, on 1 point, is clearly last.
| Match | Score |
|---|---|
| A vs B | 2–1 |
| A vs C | 1–1 |
| A vs D | 3–0 |
| B vs C | 2–0 |
| B vs D | 1–1 |
| C vs D | 2–1 |
Goal difference: B = 4 − 3 = +1 (conceded 2 to A, 0 to C, 1 to D); C = 3 − 4 = −1.
So B finishes second, ahead of C. It cannot be 'cannot be determined', because the clues pin down every score uniquely; D, on 1 point, is clearly last.
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