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1890 Rhode Island gubernatorial election

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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%

County results
Davis:      50–60%
Ladd:      40–50%      50–60%

Governor before election

Herbert W. Ladd
Republican

Elected Governor

John W. Davis
Democratic

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.

General election

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Candidates

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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

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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

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  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.