Bee Cave, Texas

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Bee Cave is a city in Travis County, Texas, United States. The population was 656 at the 2000 census; in the 2004 census estimate it was 1,700.



Bee Cave is located at 30°18′20″N 97°57′08″W / 30.305430°N 97.952213°W / 30.305430; -97.952213 (30.305430, -97.952213).[3] This is about 12 miles west of Austin in western Travis County.

According to the United States Census Bureau, the city has a total area of 2.6 square miles (6.7 km2), all of it land.


As of the census[1] of 2000, there were 656 people, 207 households, and 172 families residing in the city. The population density was 252.0 people per square mile (97.4/km2). There were 246 housing units at an average density of 94.5/sq mi (36.5/km2). The racial makeup of the city was 94.21% white, 0.15% Native American, 0.15% Asian, 3.51% from other races, and 1.98% from two or more races. Hispanic or Latino of any race were 7.62% of the population.

There were 207 households out of which 50.2% had children under the age of 18 living with them, 79.7% were married couples living together, 2.4% had a female householder with no husband present, and 16.9% were non-families. 9.7% of all households were made up of individuals and 2.4% had someone living alone who was 65 years of age or older. The average household size was 3.17 and the average family size was 3.38.

In the city, the population was spread out with 33.1% under the age of 18, 3.8% from 18 to 24, 34.8% from 25 to 44, 23.2% from 45 to 64, and 5.2% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 108.9 males. For every 100 females age 18 and over, there were 104.2 males.

The median income for a household in the city was $120,871, and the median house value was $130,028. Males had a median income of $100,000 versus $48,125 for females. The per capita income for the city was $45,405. None of the families and 0.5% of the population were living below the poverty line, including no under eighteens and none of those over 64.

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