Broward County, FL vs Sandoval County, NM
Property Tax Rate Comparison 2025
Quick Answer
Broward County, FL: 0.88% effective rate · $3,539/yr median tax · median home $402,176
Sandoval County, NM: 0.84% effective rate · $1,304/yr median tax · median home $155,264
Side-by-Side Comparison
| Metric | Broward County, FL | Sandoval County, NM | National Avg |
|---|---|---|---|
| Effective Tax Rate | 0.88% | 0.84% | 1.06% |
| Median Annual Tax | $3,539 | $1,304 | $2,778 |
| Median Home Value | $402,176 | $155,264 | $268,728 |
| Population | 1,944,375 | 148,834 | — |
Broward County, FL
- Effective Rate
- 0.88%
- Median Annual Tax
- $3,539
- Median Home Value
- $402,176
- Population
- 1,944,375
Sandoval County, NM
- Effective Rate
- 0.84%
- Median Annual Tax
- $1,304
- Median Home Value
- $155,264
- Population
- 148,834
Frequently Asked Questions
What is the property tax difference between Broward County and Sandoval County?
Broward County, FL has an effective property tax rate of 0.88% with a median annual tax of $3,539. Sandoval County, NM has a rate of 0.84% with a median annual tax of $1,304. The difference is 0.04 percentage points.
Which county has higher property taxes, Broward County or Sandoval County?
Broward County, FL has the higher effective property tax rate at 0.88% compared to 0.84%.
How do Broward County and Sandoval County compare to the national average?
The national average effective property tax rate is 1.06%. Broward County is below average at 0.88%, and Sandoval County is below average at 0.84%.