1896 United States presidential election in Maryland

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United States presidential election in Maryland, 1896

← 1892 November 3, 1896 1900 →
  William McKinley by Courtney Art Studio, 1896.jpg WilliamJBryan1902.png
Nominee William McKinley William J. Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Garret Hobart Arthur Sewall
Electoral vote 8 0
Popular vote 136,959 104,150
Percentage 54.73% 41.62%

President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

The 1896 United States presidential election in Maryland took place on November 3, 1896. All contemporary 45 states were part of the 1896 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, former Ohio Governor William McKinley and his running mate Garret Hobart of New Jersey.

This was the first time that a Republican won Maryland since 1864 and the Democrats would not win the state's popular vote until 1912.

Results[edit]

United States presidential election in Maryland, 1896[1]
Party Candidate Votes Percentage Electoral votes
Republican William McKinley 136,959 54.73% 8
Democratic William Jennings Bryan 104,150 41.62% 0
Prohibition Joshua Levering 5,918 2.36% 0
National Democratic John M. Palmer 2,499 1.00% 0
Socialist Labor Charles Matchett 587 0.23% 0
National Prohibition Charles Bentley 136 0.05% 0
Totals 250,249 100.00% 8
Voter turnout

References[edit]

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

Notes[edit]