Geneva, Iowa

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Geneva is a city in Franklin County, Iowa, United States. The population was 171 at the 2000 census.

[edit] Geography

Geneva is located at 42°40′33″N, 93°7′53″W (42.675905, -93.131396)^GR1.

According to the United States Census Bureau, the city has a total area of 1.1 km² (0.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 171 people, 69 households, and 42 families residing in the city. The population density was 153.5/km² (400.4/mi²). There were 77 housing units at an average density of 69.1/km² (180.3/mi²). The racial makeup of the city was 100.00% White. Hispanic or Latino of any race were 0.58% of the population.

There were 69 households out of which 29.0% had children under the age of 18 living with them, 55.1% were married couples living together, 4.3% had a female householder with no husband present, and 37.7% were non-families. 34.8% of all households were made up of individuals and 23.2% had someone living alone who was 65 years of age or older. The average household size was 2.48 and the average family size was 3.28.

In the city the population was spread out with 29.8% under the age of 18, 5.8% from 18 to 24, 22.2% from 25 to 44, 24.0% from 45 to 64, and 18.1% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 78.1 males. For every 100 females age 18 and over, there were 81.8 males.

The median income for a household in the city was $33,542, and the median income for a family was $40,625. Males had a median income of $33,750 versus $20,000 for females. The per capita income for the city was $15,112. None of the families and 5.9% of the population were living below the poverty line, including no under eighteens and 8.5% of those over 64.