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1868 United States presidential election in Maryland

From Wikipedia, the free encyclopedia

United States presidential election in Maryland, 1868

← 1864 November 3, 1868 1872 →
 
Horatio Seymour - Brady-Handysmall.jpg
Ulysses S Grant by Brady c1870-restored (cropped).jpg
Nominee Horatio Seymour Ulysses S. Grant
Party Democratic Republican
Home state New York Illinois
Running mate Francis Preston Blair, Jr. Schuyler Colfax
Electoral vote 7 0
Popular vote 62,357 30,438
Percentage 67.20% 32.80%

President before election

Andrew Johnson
Democratic

Elected President

Ulysses S. Grant
Republican

The 1868 United States presidential election in Maryland took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose 7 representatives, or electors to the Electoral College, who voted for president and vice president.

Maryland voted for the Democratic nominee, Horatio Seymour over the Republican nominee, Ulysses S. Grant. Seymour won the state by a margin of 34.4%.

With 67.2% of the popular vote, Maryland would be Seymour's third strongest victory in terms of percentage in the popular vote after Kentucky and Louisiana.[1]

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Transcription

Results

United States presidential election in Maryland, 1868[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Horatio Seymour of New York Francis Preston Blair, Jr. of Missouri 62,357 67.20% 7 100.00%
Republican Ulysses S. Grant of Illinois Schuyler Colfax of Indiana 30,438 32.80% 0 0.00%
Total 92,795 100.00% 7 100.00%


References

  1. ^ "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1868 Presidential General Election Results - Maryland".


This page was last edited on 31 January 2020, at 20:27
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