1900 United States presidential election in Pennsylvania

1900 United States presidential election in Pennsylvania

← 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 32 0
Popular vote 712,665 424,232
Percentage 60.74% 36.16%

County Results

President before election

William McKinley
Republican

Elected President

William McKinley
Republican

The 1900 United States presidential election in Pennsylvania took place on November 6, 1900 as part of the 1900 United States presidential election. Voters chose 32 representatives, or electors to the Electoral College, who voted for president and vice president.

Pennsylvania overwhelmingly voted for the Republican nominee, President William McKinley, over the Democratic nominee, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan. McKinley won Pennsylvania by a landslide margin of 24.31% in this rematch of the 1896 United States presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley to score a decisive victory.

With 60.74% of the popular vote, Pennsylvania would be McKinley's fourth strongest victory in terms of percentage in the popular vote after Vermont, North Dakota and Maine.[1]

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

  1. ^ "1900 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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