The most and least expensive American cities for dining out

Photo via Getty Images/Lauri Patterson

The most and least expensive American cities for dining out

Travel

The most and least expensive American cities for dining out

The next time your friends ask you where you’d like to go for dinner, you might want to suggest Little Rock. According to a recent study, restaurants in the Arkansas capital and those in Cranston, Rhode Island have the least expensive menus in the United States, with average meal prices that are 46% lower than the ones being served in the priciest city (and that one might not be where you think it is).

The math wizards at CreditLoan crunched a lot of numbers on the Foursquare Menu API, studying thousands of menus in hundreds of cities to determine which destinations, regions and even cuisines are the most and least affordable. (And, sorry West Virginia and Vermont: there weren’t enough menu items to get accurate pricing information for your restaurants).

After reading countless descriptions of everything from chicken fingers to filet mignon, CreditLoan determined that Little Rock and Cranston are the most affordable cities, with an average menu price of $8.76. Those cities were followed by Rochester, NY ($8.78), Tucson, AZ ($8.81) and Syracuse, NY ($8.82).

On the other side of the spectrum, the most expensive city was the Vegas suburb of Paradise, NV ($16.02), followed by three reasons why you should let your grandparents pay when you visit them in Florida: Miami Beach ($15.33), Manhattan ($14.37; THERE YOU ARE, NYC), Boca Raton, FL ($14.10), and Miami ($13.98).

Among CreditLoan’s other interesting findings, Michigan is overall the most affordable state for eating out, while Nevada is the priciest.  The East South Central region – which includes Tennessee, Mississippi, and Kentucky – has the cheapest tacos, and the price of a burger in New England might make you slightly less hungry.

Dinner tomorrow? Arkansas it is!

Latest

More Eat Sip Trip
Home