1824 United States presidential election in New Jersey
The 1824 United States presidential election in New Jersey took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
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| Elections in New Jersey |
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During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. New Jersey voted for Andrew Jackson over John Quincy Adams, William H. Crawford, and Henry Clay. Jackson won New Jersey by over half of the vote.
Results
| 1824 United States presidential election in New Jersey[1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | Andrew Jackson | 10,332 | 52.08% | 8 | |
| Democratic-Republican | John Quincy Adams | 8,309 | 41.89% | 0 | |
| Democratic-Republican | William H. Crawford | 1,196 | 6.03% | 0 | |
| Totals | 19,837 | 100.0% | 8 | ||
References
- "1824 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 27 February 2013.
State and district results of the 1824 United States presidential election | ||
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