1884 United States presidential election in New Jersey
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| Elections in New Jersey |
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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 9 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
| 1884 United States presidential election in New Jersey[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
- ^ "1884 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.
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