Nolan is going to invest 96,000 and leave it in an account for 6 years. Assuming the interest is compounded monthly, what interest rate, to the nearest hundredth of a percent, would be required in order for Nolan to end up with142,000?
The Correct Answer and Explanation is:
To determine the interest rate that would allow Nolan to grow his investment from $96,000 to $142,000 in 6 years with monthly compounding, we can use the compound interest formula:
[
A = P \left(1 + \frac{r}{n}\right)^{nt}
]
where:
- ( A ) is the amount of money after interest (future value), which is $142,000.
- ( P ) is the initial principal (starting amount), which is $96,000.
- ( r ) is the annual interest rate (as a decimal).
- ( n ) is the number of compounding periods per year (12 for monthly compounding).
- ( t ) is the number of years the money is invested, which is 6.
We need to rearrange this formula to solve for ( r ), as we’re trying to find the annual interest rate.
Step 1: Set Up the Equation
Substitute the known values into the formula:
[
142,000 = 96,000 \left(1 + \frac{r}{12}\right)^{12 \times 6}
]
Step 2: Isolate (\left(1 + \frac{r}{12}\right))
Divide both sides by 96,000:
[
\frac{142,000}{96,000} = \left(1 + \frac{r}{12}\right)^{72}
]
Calculating the left side:
[
\frac{142,000}{96,000} \approx 1.4792
]
So, we have:
[
1.4792 = \left(1 + \frac{r}{12}\right)^{72}
]
Step 3: Take the 72nd Root
To solve for ( r ), take the 72nd root of both sides:
[
1 + \frac{r}{12} = 1.4792^{\frac{1}{72}}
]
Calculating (1.4792^{\frac{1}{72}}), we get:
[
1 + \frac{r}{12} \approx 1.0055
]
Step 4: Solve for ( r )
Subtract 1 from both sides:
[
\frac{r}{12} \approx 0.0055
]
Multiply by 12 to solve for ( r ):
[
r \approx 0.066 \text{ or } 6.6\%
]
Answer
The required annual interest rate, to the nearest hundredth of a percent, is 6.6%.
Explanation
In this problem, we applied the compound interest formula to determine the necessary interest rate. Since compounding occurs monthly, the interest compounds 12 times a year. We set up the formula with $142,000 as the target balance after 6 years. After isolating ( r ), we simplified using roots to find the rate that achieves this growth. Compound interest grows money faster than simple interest because each period’s interest is added to the balance and compounded again, leading to exponential growth. This explains why only a 6.6% rate is needed for Nolan’s investment to grow by $46,000 over 6 years.