5. The Smiths bought a new house 4 years ago for $200,000. The house is now worth $275,000. Assuming a ste a dy rate of growth, what was the yearly rate of appreciation?
5. The Smiths bought a new house 4 years ago for $200,000. The house is now worth $275,000. Assuming a ste a dy rate of growth, what was the yearly rate of appreciation?
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Real Estate Appreciation Problem
**Problem Statement:**
The Smiths bought a new house 4 years ago for $200,000. The house is now worth $275,000. Assuming a steady rate of growth, what was the yearly rate of appreciation?
### Explanation
To calculate the yearly rate of appreciation, we can use the formula for compound interest growth, which is applicable here to calculate the appreciation:
\[ A = P (1 + r)^n \]
Where:
- \( A \) is the amount after n years (\$275,000 in this case)
- \( P \) is the principal amount or the initial value of the house (\$200,000)
- \( r \) is the annual appreciation rate
- \( n \) is the number of years (4 years)
Rearranging the formula to solve for \( r \):
\[ \left( \frac{A}{P} \right) = (1 + r)^n \]
\[ \left( \frac{275,000}{200,000} \right) = (1 + r)^4 \]
\[ 1.375 = (1 + r)^4 \]
Taking the fourth root of both sides:
\[ 1 + r = \left( 1.375 \right)^{\frac{1}{4}} \]
\[ 1 + r \approx 1.0825 \]
\[ r \approx 0.0825 \]
Thus, the annual appreciation rate is approximately \( 0.0825 \) or \( 8.25\% \).
### Conclusion
The Smiths' house appreciated at an estimated yearly rate of **8.25%** over the span of 4 years. This steady rate of growth resulted in their house's value increasing from \$200,000 to \$275,000.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F80b1b221-3b74-460d-9079-f2de5ffa7b08%2F11f208f6-e85d-41d9-9d74-2e5c24809034%2Fzosf05s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Real Estate Appreciation Problem
**Problem Statement:**
The Smiths bought a new house 4 years ago for $200,000. The house is now worth $275,000. Assuming a steady rate of growth, what was the yearly rate of appreciation?
### Explanation
To calculate the yearly rate of appreciation, we can use the formula for compound interest growth, which is applicable here to calculate the appreciation:
\[ A = P (1 + r)^n \]
Where:
- \( A \) is the amount after n years (\$275,000 in this case)
- \( P \) is the principal amount or the initial value of the house (\$200,000)
- \( r \) is the annual appreciation rate
- \( n \) is the number of years (4 years)
Rearranging the formula to solve for \( r \):
\[ \left( \frac{A}{P} \right) = (1 + r)^n \]
\[ \left( \frac{275,000}{200,000} \right) = (1 + r)^4 \]
\[ 1.375 = (1 + r)^4 \]
Taking the fourth root of both sides:
\[ 1 + r = \left( 1.375 \right)^{\frac{1}{4}} \]
\[ 1 + r \approx 1.0825 \]
\[ r \approx 0.0825 \]
Thus, the annual appreciation rate is approximately \( 0.0825 \) or \( 8.25\% \).
### Conclusion
The Smiths' house appreciated at an estimated yearly rate of **8.25%** over the span of 4 years. This steady rate of growth resulted in their house's value increasing from \$200,000 to \$275,000.
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