Hardy, Iowa

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Hardy is a city in Humboldt County, Iowa, United States. The population was 57 at the 2000 census.


Hardy is located at 42°48′37″N 94°3′6″W / 42.81028°N 94.05167°W / 42.81028; -94.05167 (42.810207, -94.051713)[1].

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


As of the census[2] of 2000, there were 57 people, 25 households, and 20 families residing in the city. The population density was 128.9 people per square mile (50.0/km²). There were 30 housing units at an average density of 67.9/sq mi (26.3/km²). The racial makeup of the city was 96.49% White, and 3.51% from two or more races.

There were 25 households out of which 36.0% had children under the age of 18 living with them, 52.0% were married couples living together, 16.0% had a female householder with no husband present, and 20.0% were non-families. 16.0% of all households were made up of individuals and 8.0% had someone living alone who was 65 years of age or older. The average household size was 2.28 and the average family size was 2.35.

In the city the population was spread out with 21.1% under the age of 18, 7.0% from 18 to 24, 35.1% from 25 to 44, 17.5% from 45 to 64, and 19.3% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 90.0 males. For every 100 females age 18 and over, there were 95.7 males.

The median income for a household in the city was $28,929, and the median income for a family was $26,250. Males had a median income of $28,750 versus $16,875 for females. The per capita income for the city was $12,910. There were 5.6% of families and 15.4% of the population living below the poverty line, including 22.2% of under eighteens and none of those over 64.


Bode | Bradgate | Dakota City | Gilmore City‡ | Hardy | Humboldt | Livermore | Lu Verne‡ | Ottosen | Pioneer | Renwick | Rutland | Thor

Avery | Beaver | Corinth | Delana | Grove | Humboldt | Lake | Norway | Rutland | Vernon | Wacousta | Weaver

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