United States presidential election in Virginia, 1892
Main article: United States presidential election, 1892
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The 1892 United States presidential election in Virginia took place on November 8, 1892, as part of the 1892 United States presidential election. Voters chose twelve representatives, or electors to the Electoral College, who voted for President and Vice President.
Virginia voted for the Democratic candidate, former President Grover Cleveland over the Republican candidate, incumbent President Benjamin Harrison. Cleveland won the state by a margin of 17.46%.
Results
| United States presidential election in Virginia, 1892[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Grover Cleveland | 164,136 | 56.17% | 12 | |
| Republican | Benjamin Harrison (inc.) | 113,098 | 38.70% | 0 | |
| Populist | James B. Weaver | 12,275 | 4.20% | 0 | |
| Prohibition | John Bidwell | 2,729 | 0.93% | 0 | |
| Totals | 292,238 | 100.0% | 12 | ||
References
- ↑ "1892 Presidential General Election Results - Virginia". U.S. Election Atlas. Retrieved 12 April 2013.
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