Brownsville named the poorest city in America

Poorest cities: Brownsville, Texas: A man stands out of the wind during a downpour in downtown Brownsville. Poorest cities: Brownsville, Texas: A man stands out of the wind during a downpour in downtown Brownsville. Photo: Dave Einsel, Getty Images Photo: Dave Einsel, Getty Images Image 1 of / 12 Caption Close Brownsville named the poorest city in America 1 / 12 Back to Gallery

According to U.S. Census Bureau data and economic website 24/7 Wall St., two of the poorest cities in the nation are in Texas.

Data shows that the Brownsville-Harlingen area has overtaken McAllen as the poorest city in the country, putting them in the first and second spots.

Rounding out the poorest top five cities are Lake Havasu City, Ariz., Dalton, Ga., and Gadsden, Ala.

The richest city in the country is California's San Jose-Sunnyvale-Santa Clara area. Other cities the richest top five include Boston, San Francisco, Washington, D.C., and Bridgeport, Conn.

Of the Brownsville-Harlingen area's 415,557 residents, 36 percent live below the poverty level, compared to just 10.8 percent in San Jose. Texas' overall poverty rate is 17 percent. In Brownsville, 22 percent of the population has only a high school diploma or other equivalent certificate. 37 percent do not have a high school diploma.