1890 Rhode Island gubernatorial election

The 1890 Rhode Island gubernatorial election was held on April 2, 1890. Democratic nominee John W. Davis defeated incumbent Republican Herbert W. Ladd with 48.76% of the vote.

1890 Rhode Island gubernatorial election

← 1889 April 2, 1890 1891 →
 
Nominee John W. Davis Herbert W. Ladd
Party Democratic Republican
Popular vote 20,548 18,988
Percentage 48.76% 45.06%

Governor before election

Herbert W. Ladd
Republican

Elected Governor

John W. Davis
Democratic

General election edit

Candidates edit

Major party candidates

  • John W. Davis, Democratic
  • Herbert W. Ladd, Republican

Other candidates

  • John H. Larry, Prohibition
  • Arnold B. Chace, Union[1][2][3]

Results edit

1890 Rhode Island gubernatorial election[4]
Party Candidate Votes % ±%
Democratic John W. Davis 20,548 48.76%
Republican Herbert W. Ladd (incumbent) 18,988 45.06%
Prohibition John H. Larry 1,820 4.32%
Independent Arnold B. Chace 752 1.78%
Majority 1,560
Turnout
Democratic gain from Republican Swing

References edit

  1. ^ "A Quaker for Governor". The sun. New York, N.Y. March 2, 1890. p. 18. Retrieved January 9, 2021.
  2. ^ "For the Suppression of Rum". Pittsburg dispatch. Pittsburg, Pa. February 26, 1890. p. 4. Retrieved January 9, 2021.
  3. ^ "Personal and Political". The herald. Milbank, S.D. March 4, 1890. p. 2. Retrieved January 9, 2021.
  4. ^ Moore, John Leo, ed. (1994). Congressional Quarterly's Guide to U.S. elections. CQ Press. ISBN 9780871879967. Retrieved July 27, 2020.