Fruithurst, Alabama

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Fruithurst is a city in Cleburne County, Alabama, United States. At the 2000 census the population was 270.


Fruithurst is located at 33°43'51.971" North, 85°25'54.332" West (33.731103, -85.431759)[1].

According to the U.S. Census Bureau, the city has a total area of 1.0 square miles (2.6 km²), all of it land.


As of the census[2] of 2000, there were 270 people, 107 households, and 74 families residing in the city. The population density was 264.8 people per square mile (102.2/km²). There were 111 housing units at an average density of 108.9/sq mi (42.0/km²). The racial makeup of the city was 99.63% White and 0.37% Black or African American. 0.74% of the population were Hispanic or Latino of any race.

There were 107 households out of which 30.8% had children under the age of 18 living with them, 53.3% were married couples living together, 10.3% had a female householder with no husband present, and 30.8% were non-families. 28.0% of all households were made up of individuals and 13.1% had someone living alone who was 65 years of age or older. The average household size was 2.41 and the average family size was 2.93.

In the city the population was spread out with 20.4% under the age of 18, 12.6% from 18 to 24, 24.4% from 25 to 44, 27.8% from 45 to 64, and 14.8% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 98.5 males. For every 100 females age 18 and over, there were 87.0 males.

The median income for a household in the city was $27,813, and the median income for a family was $31,667. Males had a median income of $22,188 versus $16,250 for females. The per capita income for the city was $18,130. About 17.0% of families and 17.9% of the population were below the poverty line, including 19.5% of those under the age of eighteen and 2.6% of those sixty five or over.


Coordinates: 33°43′52″N 85°25′54″W / 33.731103°N 85.431759°W / 33.731103; -85.431759

Fruithurst | Heflin

Edwardsville | Ranburne

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