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
April 3, 1895
 
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

Elections in Rhode Island

General election

Candidates

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
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%
People's William Foster Jr. 379 0.86%
Majority 10,809
Turnout
Republican hold Swing

References
  1. "Congressional Quarterly's Guide to U.S. elections". Books.google.com. Retrieved July 18, 2020.
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