1840 United States presidential election in Maryland

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United States presidential election in Maryland, 1840

← 1836 October 30 - December 2, 1840 1844 →
  Harrison crop.jpg Van Buren.jpg
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 10 0
Popular vote 33,528 28,752
Percentage 53.83% 46.17%

President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

The 1840 United States presidential election in Maryland took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose ten representatives, or electors to the Electoral College, who voted for President and Vice President.

Maryland voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Maryland by a margin of 7.66%.

Results[edit]

United States presidential election in Maryland, 1840[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 33,528 53.83% 10 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 28,752 46.17% 0 0.00%
Total 62,280 100.00% 10 100.00%

References[edit]

  1. ^ "1840 Presidential General Election Results - Maryland". U.S. Election Atlas. Retrieved 23 December 2013.