1895 Rhode Island gubernatorial election

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.

1895 Rhode Island gubernatorial election

← 1894 April 3, 1895 1896 →
 
Dem
PRO
Nominee Charles W. Lippitt George L. Littlefield Smith Quimby
Party Republican Democratic Prohibition
Popular vote 25,098 14,289 2,624
Percentage 56.89% 32.39% 5.95%

Governor before election

Daniel Russell Brown
Republican

Elected Governor

Charles W. Lippitt
Republican

General election

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Candidates

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Major party candidates

  • Charles W. Lippitt, Republican
  • George L. Littlefield, Democratic

Other candidates

  • Smith Quimby, Prohibition
  • George Boomer, Socialist Labor
  • William Foster Jr., People's

Results

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1895 Rhode Island gubernatorial election[1]
Party Candidate Votes % ±%
Republican Charles W. Lippitt 25,098 56.89%
Democratic George L. Littlefield 14,289 32.39%
Prohibition Smith Quimby 2,624 5.95%
Socialist Labor George Boomer 1,730 3.92%
Populist William Foster Jr. 379 0.86%
Majority 10,809
Turnout
Republican hold Swing

References

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