Arenzville

Arenzville is a village in Cass County, Illinois, United States. The population was 409 at the 2010 census.

Arenzville is located at 39°52′39″N 90°22′15″W  /  39.87750°N 90.37083°W  / 39.87750; -90.37083 (39.877519, -90.370955).[2]

According to the 2010 census, Arenzville has a total area of 0.78 square miles (2.02 km2), all land.[3]

As of the census[5] of 2000, there were 419 people, 172 households, and 120 families residing in the village. The population density was 534.0 people per square mile (207.4/km2). There were 183 housing units at an average density of 233.2 per square mile (90.6/km2). The racial makeup of the village was 99.76% White, and 0.24% from two or more races.

There were 172 households, out of which 32.0% had children under the age of 18 living with them, 62.2% were married couples living together, 6.4% had a female householder with no husband present, and 30.2% were non-families. 26.7% of all households were made up of individuals, and 16.3% had someone living alone who was 65 years of age or older. The average household size was 2.44 and the average family size was 2.95.

In the village, the population was spread out, with 26.3% under the age of 18, 6.0% from 18 to 24, 25.8% from 25 to 44, 22.0% from 45 to 64, and 20.0% who were 65 years of age or older. The median age was 40 years. For every 100 females, there were 102.4 males. For every 100 females age 18 and over, there were 86.1 males.

The median income for a household in the village was $37,500, and the median income for a family was $46,023. Males had a median income of $35,208 versus $17,250 for females. The per capita income for the village was $19,730. None of the families and 2.1% of the population were living below the poverty line, including no under eighteens and 4.5% of those over 64.