Question
If Derek plans to deposit $11,118.00 into his retirement account on each birthday beginning with his 26th and the account earns 10.00%, how long will it take him to accumulate $2,076,331.00?
Answers
GPT-4o mini
To determine how long it will take for Derek to accumulate $2,076,331.00 by depositing $11,118.00 annually into his retirement account at an annual interest rate of 10%, we can use the future value of an annuity formula. The formula for the future value of an annuity is:
\[
FV = P \times \frac{(1 + r)^n - 1}{r}
\]
Where:
- \( FV \) = future value of the annuity (the amount he wants to accumulate, $2,076,331.00)
- \( P \) = annual deposit (the amount he deposits each year, $11,118.00)
- \( r \) = annual interest rate (10% or 0.10)
- \( n \) = number of years (the variable we want to solve for)
Plugging in the values we have:
\[
2,076,331.00 = 11,118.00 \times \frac{(1 + 0.10)^n - 1}{0.10}
\]
Now, let's isolate \((1 + 0.10)^n - 1\):
\[
\frac{2,076,331.00}{11,118.00} = \frac{(1.10)^n - 1}{0.10}
\]
Calculating the left side:
\[
\frac{2,076,331.00}{11,118.00} \approx 186.1
\]
Now, we rewrite the equation:
\[
186.1 = \frac{(1.10)^n - 1}{0.10}
\]
Multiplying both sides by 0.10 gives:
\[
18.61 = (1.10)^n - 1
\]
Adding 1 to both sides:
\[
19.61 = (1.10)^n
\]
Now, to solve for \( n \), we take the logarithm of both sides:
\[
n \cdot \log(1.10) = \log(19.61)
\]
Thus,
\[
n = \frac{\log(19.61)}{\log(1.10)}
\]
Calculating each logarithm:
1. \(\log(19.61) \approx 1.292\)
2. \(\log(1.10) \approx 0.0414\)
Now:
\[
n = \frac{1.292}{0.0414} \approx 31.23
\]
Since \( n \) must be a whole number (as he makes deposits at the end of each year), we round up to the next whole number, giving \( n = 32 \).
Thus, it will take Derek approximately **32 years** to accumulate $2,076,331.00 by making annual deposits of $11,118.00 at an interest rate of 10%.
\[
FV = P \times \frac{(1 + r)^n - 1}{r}
\]
Where:
- \( FV \) = future value of the annuity (the amount he wants to accumulate, $2,076,331.00)
- \( P \) = annual deposit (the amount he deposits each year, $11,118.00)
- \( r \) = annual interest rate (10% or 0.10)
- \( n \) = number of years (the variable we want to solve for)
Plugging in the values we have:
\[
2,076,331.00 = 11,118.00 \times \frac{(1 + 0.10)^n - 1}{0.10}
\]
Now, let's isolate \((1 + 0.10)^n - 1\):
\[
\frac{2,076,331.00}{11,118.00} = \frac{(1.10)^n - 1}{0.10}
\]
Calculating the left side:
\[
\frac{2,076,331.00}{11,118.00} \approx 186.1
\]
Now, we rewrite the equation:
\[
186.1 = \frac{(1.10)^n - 1}{0.10}
\]
Multiplying both sides by 0.10 gives:
\[
18.61 = (1.10)^n - 1
\]
Adding 1 to both sides:
\[
19.61 = (1.10)^n
\]
Now, to solve for \( n \), we take the logarithm of both sides:
\[
n \cdot \log(1.10) = \log(19.61)
\]
Thus,
\[
n = \frac{\log(19.61)}{\log(1.10)}
\]
Calculating each logarithm:
1. \(\log(19.61) \approx 1.292\)
2. \(\log(1.10) \approx 0.0414\)
Now:
\[
n = \frac{1.292}{0.0414} \approx 31.23
\]
Since \( n \) must be a whole number (as he makes deposits at the end of each year), we round up to the next whole number, giving \( n = 32 \).
Thus, it will take Derek approximately **32 years** to accumulate $2,076,331.00 by making annual deposits of $11,118.00 at an interest rate of 10%.