Election to the United States House of Representatives in 1990

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A total of 435 seats
  • Dem . : 267
  • GOP : 167
  • Otherwise: 1

On November 6, 1990, the United States ' House of Representatives was elected. The election was part of the general election for the 102nd United States Congress that year, which also elected a third of US Senators . Since the elections, about the middle of the term of Republican President George HW Bush took place ( Midterm Election ) , they were regarded as a vote on the current policy of the President.

The number of MPs to be elected was 435. The distribution of seats was based on the 1980 census .

In the elections there was again a slight win for the Democrats, who took 7 more seats from the Republicans than in the 1988 elections. This increased their absolute majority to 267 mandates. Because Bernie Sanders (later Democrat) was also elected as an independent candidate in Vermont , the Republicans lost 8 seats. The shifts in favor of the Democrats show the dissatisfaction of some electoral groups with the previous policy of the Federal Government and President George Bush Sen. Since the Democrats were able to maintain their majority in the Senate, the President had to grapple with an opposition Congress in the second half of his term of office .

Election result

Total: 435 (435)

The results of the last election two years earlier are in brackets. Changes during the legislative period that do not affect the elections themselves are not included in these figures, but are noted in the article on the 102nd Congress in the section on the members of the House of Representatives under the appropriate names of the representatives. As a result, the sources sometimes contain different information, as changes during the legislative period were sometimes incorporated into the figures and sometimes not. The election result shown is based on the source below (Party Divisions).

See also

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