1900 United States presidential election in Vermont

1900 United States presidential election in Vermont

← 1896 November 6, 1900 1904 →

Nominee William McKinley William Jennings Bryan Party Republican Democratic Home state Ohio Nebraska Running mate Theodore Roosevelt Adlai E. Stevenson Electoral vote 4 0 Popular vote 42,569 12,849 Percentage 75.73% 22.86%

County results Municipality results McKinley   50-60%   60-70%   70-80%   80-90%   90-100% Bryan   40-50%   50-60%   70-80% Tie   40-50%

President before election William McKinley Republican Elected President William McKinley Republican

Elections in Vermont
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Ballot measures 2022 Proposal 5
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v t e

The 1900 United States presidential election in Vermont took place on November 6, 1900 as part of the 1900 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Vermont overwhelmingly voted for the Republican nominees, incumbent President William McKinley of Ohio and his running mate Theodore Roosevelt of New York. They defeated the Democratic nominees, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan and his running mate, former Vice President Adlai Stevenson I. McKinley won Vermont by a landslide margin of 52.87% in this rematch of the 1896 presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley to score a decisive victory.

Four years earlier, McKinley had won Vermont with 80.08% of the popular vote, making it his strongest victory in the 1896 presidential election in terms of percentage in the popular vote as well as the best performance of any presidential candidate in Vermont to date. Vermont would once again be McKinley's strongest state in popular vote percentage, though with a slightly reduced margin of 75.73%.^[1]

Bryan had previously lost Vermont to McKinley four years earlier and would later lose the state again in 1908 to William Howard Taft.