1884 United States presidential election in Maryland

The 1884 United States presidential election in Maryland took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for president and vice president.

1884 United States presidential election in Maryland

November 4, 1884

Nominee Grover Cleveland James G. Blaine Party Democratic Republican Home state New York Maryland Running mate Thomas A. Hendricks John A. Logan Electoral vote 8 0 Popular vote 96,866 85,748 Percentage 52.07% 46.10%

County Results Cleveland   40-50%   50-60%   60-70% Blaine   40-50%   50-60%

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

Maryland voted for the Democratic nominee, Grover Cleveland, over the Republican nominee, James G. Blaine. by a margin of 5.98%.

Results

1884 United States presidential election in Maryland[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Democratic	Grover Cleveland of New York	Thomas Andrews Hendricks of Indiana	96,866	52.07%	8	100.00%
	Republican	James Gillespie Blaine of Maryland	John Alexander Logan of Illinois	85,748	46.10%	0	0.00%
	Prohibition	John Pierce St. John of Kansas	William Daniel of Maryland	2,827	1.52%	0	0.00%
	Greenback	Benjamin Franklin Butler of Massachusetts	Absolom Madden West of Mississippi	578	0.31%	0	0.00%
Total				186,019	100.00%	8	100.00%

See also

• United States presidential elections in Maryland

Notes

References

1. "1884 Presidential General Election Results – Maryland".