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United States presidential election in New York, 1860

From Wikipedia, the free encyclopedia

United States presidential election in New York, 1860
New York (state)
← 1856 November 6, 1860 1864 →
 
Abraham Lincoln by Alexander Helser, 1860-crop.jpg
StephenADouglas.png
Nominee Abraham Lincoln Stephen A. Douglas
Party Republican Democratic
Home state Illinois Illinois
Running mate Hannibal Hamlin Herschel Vespasian Johnson
Electoral vote 35 0
Popular vote 362,646 312,510
Percentage 53.71% 46.29%

President before election

James Buchanan
Democratic

Elected President

Abraham Lincoln
Republican

The 1860 United States presidential election in New York took place on November 6, 1860 throughout all 33 states which was part of the 1860 United States presidential election. Voters chose 35 electors of the Electoral College, who voted for President and Vice President.

New York was won by Republican candidate Abraham Lincoln, who defeated Democratic presidential candidate Stephen A. Douglas by a narrow margin of 7.42%.

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Transcription

Results

United States presidential election in New York, 1860[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Abraham Lincoln of Illinois Hannibal Hamlin of Maine 362,646 53.71% 35 100.00%
Democratic Stephen A. Douglas of Illinois Herschel Vespasian Johnson of Georgia 312,510 46.29% 0 0.00%
Total 675,156 100.00% 35 100.00%

References

This page was last edited on 10 September 2017, at 22:42.
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