Football at the 1960 Summer Olympics – Group 4 explained
Group 4 of the 1960 Summer Olympics football tournament took place from 26 August to 1 September 1960.[1] [2] The group consisted of Hungary, Peru, India and France. The top team, Hungary, advanced to the semi-finals.
Teams
| Team | Region | Method of
qualification | Date of
qualification | Finals
appearance | Last
appearance | Previous best
performance |
|---|
| style=white-space:nowrap | | Europe | Europe Group 7 winners | 6 April 1960 | 5th | | data-sort-value="8" | Gold medal (1952) |
| style=white-space:nowrap | | Americas | Americas second round runners-up | 24 April 1960 | 2nd | | data-sort-value="4" | Quarter-finals (1936) |
| style=white-space:nowrap | | Asia | Asia second round winner | 30 April 1960 | 4th | | data-sort-value="5" | Fourth place (1956) |
| style=white-space:nowrap | | Europe | Europe Group 6 winners | 1 May 1960 | 8th | | data-sort-value="7" | Silver medal (1900) | |
Standings
In the semi-finals, the winners of Group 4, Hungary, advanced to play the winner of Group 3, Denmark.
Matches
All times listed are local, CET (UTC+1).
Hungary vs India
France vs Peru
France vs India
Hungary vs Peru
Hungary vs France
Peru vs India
See also
Notes and References
- Web site: Olympic Games 1960 » Group 4. worldfootball.net. 1 August 2024.
- Web site: XVII. Olympiad Rome 1960 Football Tournament. Reyes. Macario. RSSSF. 20 June 2019.
