1884 United States presidential election in Virginia
The 1884 United States presidential election in Virginia took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose 12 representatives, or electors to the Electoral College, who voted for president and vice president.
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
|
Elections in Virginia | ||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|
|
||||||||||||
|
||||||||||||
Virginia voted for the Democratic candidate, New York Governor Grover Cleveland over the Republican candidate, former Secretary of State James G. Blaine. Cleveland won Virginia by a margin of 2.15%.
Results
United States presidential election in Virginia, 1884[1] | |||||
---|---|---|---|---|---|
Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic | Grover Cleveland | 145,491 | 51.05% | 12 | |
Republican | James G. Blaine | 139,356 | 48.90% | 0 | |
Prohibition | John St. John | 130 | 0.05% | 0 | |
Totals | 284,977 | 100.0% | 12 | ||
References
- "1884 Presidential General Election Results - Virginia". U.S. Election Atlas. Retrieved 12 April 2013.
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.