1828 United States presidential election in New Jersey
The 1828 United States presidential election in New Jersey took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
| ||||||||||||||||||||||||||
|
Elections in New Jersey |
---|
New Jersey voted for the National Republican candidate, John Quincy Adams, over the Democratic candidate, Andrew Jackson. Adams won New Jersey by a narrow margin of 4.26%.
Results
1828 United States presidential election in New Jersey[1] | |||||
---|---|---|---|---|---|
Party | Candidate | Votes | Percentage | Electoral votes | |
National Republican | John Quincy Adams | 23,753 | 52.12% | 8 | |
Democratic | Andrew Jackson | 21,809 | 47.86% | 0 | |
N/A | Other | 8 | 0.02% | 0 | |
Totals | 45,570 | 100.0% | 8 | ||
References
- "1828 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 28 February 2013.
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.