1974 Wimbledon Championships – Girls' singles
Appearance
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Girls' singles | ||||||||||||
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1974 Wimbledon Championships | ||||||||||||
Final | ||||||||||||
Champion | Socialist Federal Republic of Yugoslavia Mima Jaušovec | |||||||||||
Runner-up | Romania Mariana Simionescu | |||||||||||
Score | 7–5, 6–4 | |||||||||||
Details | ||||||||||||
Draw | 25 | |||||||||||
Events | ||||||||||||
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1974 tennis event results
Main article: 1974 Wimbledon Championships
Mima Jaušovec defeated Mariana Simionescu in the final, 7–5, 6–4 to win the girls' singles tennis title at the 1974 Wimbledon Championships.[1]
Draw
[edit ]Key
[edit ]- Q = Qualifier
- WC = Wild card
- LL = Lucky loser
- Alt = Alternate
- SE = Special exempt
- PR = Protected ranking
- ITF = ITF entry
- JE = Junior exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
- SR = Special ranking
Finals
[edit ] Semifinals
Final
Top half
[edit ] First round
Second round
Quarterfinals
Semifinals
Austria E Celigoj
0
4
Brazil Léa Rios de Almeida
4
6
4
West Germany I Pohmann
6
6
West Germany Inge Pohmann
6
3
6
Bottom half
[edit ] First round
Second round
Quarterfinals
Semifinals
Canada DA Mushlian
6
3
2
Netherlands B Sluyter
0
4
United Arab Emirates Hoda Baligh
0
0
France Christine Gimmig
3
3
Denmark Helle Sparre
6
6
References
[edit ]- ^ "Girls' Singles Finals 1947-2017". wimbledon.com. Wimbledon Championships . Retrieved 13 August 2017.
External links
[edit ]- Source for the draw at Wimbledon.com