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1840 United States presidential election in Vermont

From Wikipedia, the free encyclopedia

1840 United States presidential election in Vermont

← 1836 October 30 - December 2, 1840 1844 →
 
William Henry Harrison (cropped).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 7 0
Popular vote 32,445 18,009
Percentage 63.90% 35.47%

President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

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

Vermont voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Vermont by a margin of 28.43%.

Harrison's 28.43% margin of victory made it his strongest victory in the election while he carried 63.90% of the popular vote made Vermont his second strongest state after Kentucky.[1]

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Transcription

Results

1840 United States presidential election in Vermont[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 32,445 63.90% 7 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 18,009 35.47% 0 0.00%
Liberty James G. Birney of New York Thomas Earle of Pennsylvania 319 0.63% 0 0.00%
Total 50,773 100.00% 7 100.00%

References

  1. ^ "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1840 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved 23 December 2013.
This page was last edited on 1 March 2020, at 21:15
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