1940 United States presidential election in Iowa

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United States presidential election in Iowa, 1940

← 1936 November 5, 1940[1] 1944 →

All 11 Iowa votes to the Electoral College
  WendellWillkie.jpg FDRoosevelt1938.png
Nominee Wendell Willkie Franklin D. Roosevelt
Party Republican Democratic
Home state Indiana New York
Running mate Charles L. McNary Henry A. Wallace
Electoral vote 11 0
Popular vote 632,370 578,800
Percentage 52.0% 47.6%

President before election

Franklin D. Roosevelt
Democratic

Elected President

Franklin D. Roosevelt
Democratic

The 1940 United States presidential election in Iowa took place on November 5, 1940, as part of the 1940 United States presidential election. Iowa voters chose eleven[2] representatives, or electors, to the Electoral College, who voted for president and vice president.

Iowa was won by Wendell Willkie (RIndiana), running with Minority Leader Charles L. McNary, with 52.03% of the popular vote, against incumbent President Franklin D. Roosevelt (DNew York), running with Secretary Henry A. Wallace, with 47.62% of the popular vote.[3][4]

Results[edit]

United States presidential election in Iowa, 1940
Party Candidate Votes %
Republican Wendell Willkie 632,370 52.03%
Democratic Franklin D. Roosevelt (inc.) 578,800 47.62%
Write-in 4,260 0.35%
Total votes 1,215,430 100%

References[edit]

  1. ^ "United States Presidential election of 1940 - Encyclopædia Britannica". Retrieved August 19, 2018.
  2. ^ "1940 Election for the Thirty-ninth Term (1941-45)". Retrieved August 19, 2018.
  3. ^ "1940 Presidential General Election Results - Iowa". Retrieved August 19, 2018.
  4. ^ "The American Presidency Project - Election of 1940". Retrieved August 19, 2018.