1900 United States presidential election in Delaware

1900 United States presidential election in Delaware

← 1896 November 6, 1900 1904 →
 
Nominee William McKinley William Jennings 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%

County Results
McKinley
  40-50%
  50-60%


President before election

William McKinley
Republican

Elected President

William McKinley
Republican

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. State 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. 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 the state by a margin of 8.77% in this rematch of the 1896 presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley score a decisive victory.

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


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