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1884 United States presidential election in Delaware

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1884 United States presidential election in Delaware

← 1880 November 4, 1884 1888 →
 
Nominee Grover Cleveland James G. Blaine
Party Democratic Republican
Home state New York Maine
Running mate Thomas A. Hendricks John A. Logan
Electoral vote 3 0
Popular vote 16,957 12,953
Percentage 56.55% 43.20%

County Results
Cleveland
  50-60%
  60-70%


President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

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

Delaware voted for the Democratic nominee, Grover Cleveland, over the Republican nominee, James G. Blaine. Cleveland won the state by a margin of 13.35%.

Results

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1884 United States presidential election in Delaware[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Grover Cleveland of New York Thomas Andrews Hendricks of Indiana 16,957 56.55% 3 100.00%
Republican James Gillespie Blaine of Maine John Alexander Logan of Illinois 12,953 43.20% 0 0.00%
Prohibition John Pierce St. John of Kansas William Daniel of Maryland 64 0.21% 0 0.00%
Greenback Benjamin Franklin Butler of Massachusetts Absolom Madden West of Mississippi 10 0.03% 0 0.00%
Total 29,984 100.00% 3 100.00%

See also

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References

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