1900 United States presidential election in Delaware

The 1900 United States presidential election in Delaware took place on November 6, 1900. All contemporary 45 states were part of the 1900 United States presidential election. Delaware voters chose three electors to the Electoral College, which selected the president and vice president.

Delaware was won by the Republican nominees, incumbent President William McKinley of Ohio and his running mate Theodore Roosevelt of New York.

United States presidential election in Delaware, 1900

November 6, 1900
  William McKinley by Courtney Art Studio, 1896 WilliamJBryan1902
Nominee William McKinley William J. Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Theodore Roosevelt Adlai Stevenson I
Electoral vote 3 0
Popular vote 22,535 18,852
Percentage 53.67% 44.90%

President before election

William McKinley

Elected President

William McKinley


United States presidential election in Delaware, 1900[1]
Party Candidate Votes Percentage Electoral votes
Republican William McKinley 22,535 53.67% 3
Democratic William Jennings Bryan 18,852 44.90% 0
Prohibition John Woolley 546 1.30% 0
Social Democratic Eugene Debs 56 0.13% 0
Totals 41,989 100.00% 3
Voter turnout


  1. ^ Dave Leip’s U.S. Election Atlas; Presidential General Election Results – Delaware
Other 1900

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