1884 United States presidential election in New Jersey

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United States presidential election in New Jersey, 1884

← 1880 November 4, 1884 1888 →
  StephenGroverCleveland.png JamesGBlaine.png
Nominee Grover Cleveland James G. Blaine
Party Democratic Republican
Home state New York Maine
Running mate Thomas A. Hendricks John A. Logan
Electoral vote 9 0
Popular vote 127,798 123,440
Percentage 48.98% 47.31%

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

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

New Jersey voted for the Democratic nominee, Grover Cleveland, over the Republican nominee, James G. Blaine. Cleveland won his birth state by a very narrow margin of 1.67%.

Results[edit]

United States presidential election in New Jersey, 1884[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Grover Cleveland of New York Thomas Andrews Hendricks of Indiana 127,798 48.98% 9 100.00%
Republican James Gillespie Blaine of Maine John Alexander Logan of Illinois 123,440 47.31% 0 0.00%
Prohibition John Pierce St. John of Kansas William Daniel of Maryland 6,159 2.36% 0 0.00%
Greenback Benjamin Franklin Butler of Massachusetts Absolom Madden West of Mississippi 3,496 1.34% 0 0.00%
N/A Others Others 28 0.01% 0 0.00%
Total 260,921 100.00% 9 100.00%

References[edit]

  1. ^ "1884 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.