United States presidential election in New Jersey, 1900

United States presidential election in New Jersey, 1900
New Jersey
November 6, 1900

 
Nominee William McKinley William Jennings Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Theodore Roosevelt Adlai E. Stevenson
Electoral vote 10 0
Popular vote 221,754 164,879
Percentage 57.27% 41.10%

President before election

William McKinley
Republican

Elected President

William McKinley
Republican

The 1900 United States presidential election in New Jersey took place on November 6, 1900 throughout 45 states. Voters chose 10 representatives, or electors to the Electoral College, who voted for President and Vice President.

New Jersey overwhelmingly voted for the Republican nominee, President William McKinley, over the Democratic nominee, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan. McKinley won New Jersey by a margin of 14.17% in this rematch of the 1896 presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley to score a decisive victory.

Results

United States presidential election in New Jersey, 1900[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican William McKinley of Ohio Theodore Roosevelt of New York 221,754 55.27% 10 100.00%
Democratic William Jennings Bryan of Nebraska Adlai Ewing Stevenson I of Illinois 164,879 41.10% 0 0.00%
Prohibition John Granville Woolley of Illinois Henry Brewer Metcalf of Rhode Island 7,190 1.79% 0 0.00%
Socialist Eugene Victor Debs of Indiana Job Harriman of California 4,611 1.15% 0 0.00%
Socialist Labor Joseph Francis Malloney of Massachusetts Valentine Remmel of Pennsylvania 2,081 0.52% 0 0.00%
People's Wharton Barker of Pennsylvania Ignatius Loyola Donnelly of Minnesota 691 0.17% 0 0.00%
Total 401,206 100.00% 10 100.00%

References

  1. "1900 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.
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