Romanian general election, 1928

General elections were held in Romania in December 1928.[1] The Chamber of Deputies was elected on 12 December, whilst the Senate was elected in three stages on 15, 17 and 19 December.[1] The result was a victory for the governing National Peasants' Party-led alliance, which won 348 of the 387 seats in the Chamber of Deputies and 105 of the 110 seats in the Senate elected through universal male vote.[2] Of the 348 Chamber seats won by the alliance, the National Peasants' Party won 326, the Social Democratic Party won 9, the German Party won 8, the Hungarian People's Party won 2 and three were won by Jewish candidates.[2] It is generally regarded as the freest election ever held in Romania until the 1990s.[3]


Party Chamber Senate
Votes % Seats +/– Votes % Seats +/–
National Peasants' Party-led alliance2,208,92279.2348+294105+88
National Liberal Party185,9396.713–3050–92
Magyar Party172,0996.216New3New
National PartyPeople's Party70,7902.55+500
Peasants' Party–Lupu70,5062.55New0New
Peasant Workers' Bloc38,8511.40New0New
National-Christian Defense League32,2731.20New0New
Other parties8,0510.303
Jewish Party02New
Invalid/blank votes53,249
Registered voters/turnout3,671,35277.4
Source: Nohlen & Stöver


  1. 1 2 Dieter Nohlen & Philip Stöver (2010) Elections in Europe: A data handbook, p1591 ISBN 978-3-8329-5609-7
  2. 1 2 Nohlen & Stöver, pp1610-1611
  3. Romania from Encyclopædia Britannica.
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