United States presidential election in New Jersey, 1828
Main article: United States presidential election, 1828
| ||||||||||||||||||||||||||||||||
| ||||||||||||||||||||||||||||||||
|
Elections 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.
New Jersey voted for the National Republican candidate, John Quincy Adams, over the Democratic candidate, Andrew Jackson. Adams won New Jersey by a margin of 4.26%.
Results
United States presidential election in New Jersey, 1828[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 - version of the Monday, August 10, 2015. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.