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1884 United States presidential election in Maryland

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1884 United States presidential election in Maryland

← 1880 November 4, 1884 1888 →
 
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

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

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.

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

Notes

References

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