1884 United States presidential election in Maryland

The 1884 United States presidential election in Maryland took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for president and vice president.

1884 United States presidential election in Maryland
Flag of the United States (1877–1890).svg
← 1880 November 4, 1884 1888 →
  StephenGroverCleveland.png Unsuccessful 1884.jpg
Nominee Grover Cleveland James G. Blaine
Party Democratic Republican
Home state New York Maryland
Running mate Thomas A. Hendricks John A. Logan
Electoral vote 8 0
Popular vote 96,866 85,748
Percentage 52.07% 46.10%

Maryland Presidential Election Results 1884.svg
County Results

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

Maryland voted for the Democratic nominee, Grover Cleveland, over the Republican nominee, James G. Blaine. by a margin of 5.98%.[1]

ResultsEdit

1884 United States presidential election in Maryland[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Grover Cleveland of New York Thomas Andrews Hendricks of Indiana 96,866 52.07% 8 100.00%
Republican James Gillespie Blaine of Maryland John Alexander Logan of Illinois 85,748 46.10% 0 0.00%
Prohibition John Pierce St. John of Kansas William Daniel of Maryland 2,827 1.52% 0 0.00%
Greenback Benjamin Franklin Butler of Massachusetts Absolom Madden West of Mississippi 578 0.31% 0 0.00%
Total 186,019 100.00% 8 100.00%

See alsoEdit

NotesEdit

ReferencesEdit

  1. ^ Humanities, National Endowment for the (1884-11-19). "Evening capital. (Annapolis, Md.) 1884-1910, November 19, 1884, Image 1". ISSN 2643-8593. Retrieved 2021-11-29.
  2. ^ "1884 Presidential General Election Results – Maryland".