1840 United States presidential election in Kentucky

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United States presidential election in Kentucky, 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 15 0
Popular vote 58,488 32,616
Percentage 64.20% 35.80%

President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

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

Kentucky voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Kentucky by a margin of 28.4%.

With 64.20% of the popular vote, Kentucky would prove to be Harrison's strongest state in the 1840 election.[1]

Results[edit]

United States presidential election in Kentucky, 1840[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 58,488 64.20% 15 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 32,616 35.80% 0 0.00%
Total 91,104 100.00% 15 100.00%

References[edit]

  1. ^ "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1840 Presidential General Election Results - Kentucky". U.S. Election Atlas. Retrieved 23 December 2013.