1884 United States presidential election in Indiana

Main article: 1884 United States presidential election

1884 United States presidential election in Indiana

←  1880 November 4, 1884 1888  →

Nominee Grover Cleveland James G. Blaine Party Democratic Republican Home state New York Maine Running mate Thomas A. Hendricks John A. Logan Electoral vote 15 0 Popular vote 245,005 238,489 Percentage 49.46% 48.15%

County Results Cleveland   40–50%   50–60%   60–70%   70–80% Blaine   40–50%   50–60%   60–70%

President before election Chester A. Arthur Republican Elected President Grover Cleveland Democratic

A presidential election was held in Indiana on November 4, 1884, as part of the 1884 United States presidential election. Voters chose 15 representatives, or electors to the Electoral College, who voted for president and vice president.

Indiana voted for the Democratic nominee, Grover Cleveland over the Republican nominee, James G. Blaine. Cleveland won the state by a narrow margin of 1.31%.

Results

1884 United States presidential election in Indiana [1]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Democratic	Grover Cleveland of New York	Thomas Andrews Hendricks of Indiana	245,005	49.46%	15	100.00%
	Republican	James Gillespie Blaine of Maine	John Alexander Logan of Illinois	238,489	48.15%	0	0.00%
	Greenback	Benjamin Franklin Butler of Massachusetts	Absolom Madden West of Mississippi	8,810	1.78%	0	0.00%
	Prohibition	John Pierce St. John of Kansas	William Daniel of Maryland	3,028	0.61%	0	0.00%
Total				495,332	100.00%	15	100.00%

See also

• United States presidential elections in Indiana

References

1. "1884 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved December 23, 2013.