1880 United States presidential election in Minnesota explained

See main article: 1880 United States presidential election.

Election Name:1880 United States presidential election in Minnesota
Country:Minnesota
Type:presidential
Ongoing:no
Previous Election:1876 United States presidential election in Minnesota
Previous Year:1876
Next Election:1884 United States presidential election in Minnesota
Next Year:1884
Election Date:November 2, 1880
Image1:James Abram Garfield, photo portrait seated (cropped).jpg
Nominee1:James A. Garfield
Party1:Republican Party (United States)
Home State1:Ohio
Running Mate1:Chester A. Arthur
Electoral Vote1:5
Popular Vote1:93,902
Percentage1:62.28%
Nominee2:Winfield S. Hancock
Party2:Democratic Party (United States)
Home State2:Pennsylvania
Running Mate2:William H. English
Electoral Vote2:0
Popular Vote2:53,315
Percentage2:35.36%
Map Size:350px
President
Before Election:Rutherford B. Hayes
Before Party:Republican Party (United States)
After Election:James A. Garfield
After Party:Republican Party (United States)
Flag Image:Flag of Minnesota (1893–1957).svg

The 1880 United States presidential election in Minnesota took place on November 2, 1880, as part of the 1880 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.

Minnesota voted for the Republican nominee, James A. Garfield, over the Democratic nominee, Winfield Scott Hancock. Garfield won the state by a margin of 26.92%.

With 62.28% of the popular vote, Minnesota would be Garfield's third strongest victory in terms of percentage in the popular vote after Vermont and Nebraska.[1]

See also

Notes and References

  1. Web site: 1880 Presidential Election Statistics. Dave Leip’s Atlas of U.S. Presidential Elections. 2018-03-05.