1940 United States presidential election in Iowa

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.

Main article: 1940 United States presidential election

1940 United States presidential election in Iowa

November 5, 1940[1]
All 11 Iowa votes to the Electoral College
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%
County Results Willkie   40-50%   50-60%   60-70% Roosevelt   40-50%   50-60%
President before election Franklin D. Roosevelt Democratic Elected President Franklin D. Roosevelt Democratic

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

Results

1940 United States presidential election in Iowa

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

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.