1895 Rhode Island gubernatorial election

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1895 Rhode Island gubernatorial election
Flag of Rhode Island (1882-1897).svg
  1894 April 3, 1895 1896  
  CharlesWLippitt.jpg
Dem
PRO
Nominee Charles W. Lippitt George L. Littlefield Smith Quimby
Party Republican Democratic Prohibition
Popular vote25,09814,2892,624
Percentage56.89%32.39%5.95%

1895 Rhode Island gubernatorial election results map by county.svg
County results
Lippitt:      50–60%     60–70%

Governor before election

Daniel Russell Brown
Republican

Elected Governor

Charles W. Lippitt
Republican

The 1895 Rhode Island gubernatorial election was held on April 3, 1895. Republican nominee Charles W. Lippitt defeated Democratic nominee George L. Littlefield with 56.89% of the vote.

Contents

General election

Candidates

Major party candidates

Other candidates

Results

1895 Rhode Island gubernatorial election [1]
PartyCandidateVotes%±%
Republican Charles W. Lippitt 25,098 56.89%
Democratic George L. Littlefield14,28932.39%
Prohibition Smith Quimby2,6245.95%
Socialist Labor George Boomer 1,7303.92%
Populist William Foster Jr.3790.86%
Majority10,809
Turnout
Republican hold Swing

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References

  1. Moore, John Leo, ed. (1994). Congressional Quarterly's Guide to U.S. elections. CQ Press. ISBN   9780871879967 . Retrieved July 18, 2020.