1900 United States presidential election in Maryland

1900 United States presidential election in Maryland

← 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 8 0 Popular vote 136,185 122,238 Percentage 51.50% 46.23%

County Results McKinley   40-50%   50-60%   60-70% Bryan   40-50%   50-60%

President before election William McKinley Republican Elected President William McKinley Republican

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

Maryland 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 5.27% 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.

As of 2020, this remains the last time Howard County and Baltimore County voted for different candidates.^[1]

In this election, Maryland voted 0.85% more Democratic than the nation at-large.^[2]

Bryan had previous lost Maryland to McKinley four years earlier. In 1908, he would lose the popular vote to William Howard Taft, but would win the electoral vote.