One of Florida’s most popular travel destinations was recently named the world’s most expensive city for a family vacation, according to research by Compare the Market.
Orlando, home to a handful of family-friendly theme parks and tourist sites, will cost a family of four on a seven-night vacation an average of $7,350, the report showed. The family in the study consisted of two adults, a 10-year-old and a 5-year-old.
That breaks down to $4,138 in lodging during peak season, $3,148 in activities and around $64 per day in other costs.
The Australian travel insurance comparison site ranked the cities based on the average combined cost of activities, hotels and daily expenses.
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Not far behind in second place was Rovaniemi, Finland, with the average vacation amount coming in at $7,082. Cost breakdown in this city, known as the “official home of Santa Claus,” was as follows: $4,138 for lodging during peak season, $476 in activities and $52 per day in other expenses.
The Gold Coast in Australia took the third spot with an average vacation cost of $6,620. Lodging during peak season in this city would cost around $5,600, activities would be $972 and other daily costs would be around $48.
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Rounding out the rest of the rankings were:
- Vancouver, Canada: $4,836
- London, United Kingdom: $4,139
- Barcelona, Spain: $3,460
- Queenstown, New Zealand: $3,400
- Vienna, Austria: $2,776
- Billund, Denmark: $2,711
- Bali, Indonesia: $2,636
- Osaka, Japan: $2,313
The least expensive city included in the study for a family of four was Nairobi, Kenya. For a seven-night stay, it would cost an average of $1,974 – $1,853 for lodging, $93 for activities, and $28 for other daily costs.
Compare the Market said prices were converted from local currencies to American dollars. The costs portrayed above do not include airfare. The “daily costs” are considered taxis and other forms of public transportation, cheaper priced food and coffee.
Overall prices were determined based on hotels and “family friendly” activities listed on TripAdvisor.
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