QuestionApril 20, 2025

Approximately how much principal would need to be placed into an account earning 3.575% interest compounded quarterly so that it has an accumulated value of 68,000 at the end of 30 years? a. 23,706 b. 23,377 C. 52,069 d. 58,944

Approximately how much principal would need to be placed into an account earning 3.575% interest compounded quarterly so that it has an accumulated value of 68,000 at the end of 30 years? a. 23,706 b. 23,377 C. 52,069 d. 58,944
Approximately how much principal would need to be placed into an account earning 3.575% interest compounded quarterly so that it has an accumulated value of 68,000 at the end of 30 years? a. 23,706 b. 23,377 C. 52,069 d. 58,944

Solution
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Answer

\ 23,377 Explanation 1. Identify the formula Use the compound interest formula: **A = P \left(1 + \frac{r}{n}\right)^{nt}**, where A is the accumulated amount, P is the principal, r is the annual interest rate, n is the number of times interest is compounded per year, and t is the time in years. 2. Substitute known values Given A = 68000, r = 0.03575, n = 4, and t = 30. Substitute these into the formula to find P: \[ 68000 = P \left(1 + \frac{0.03575}{4}\right)^{4 \times 30} \] 3. Simplify and solve for P Calculate \left(1 + \frac{0.03575}{4}\right)^{120} and then solve for P: \[ P = \frac{68000}{\left(1 + \frac{0.03575}{4}\right)^{120}} \] 4. Compute the result Calculate the expression to find P: \[ P \approx \frac{68000}{3.0025} \approx 22648.92 \]

Explanation

1. Identify the formula<br /> Use the compound interest formula: **$A = P \left(1 + \frac{r}{n}\right)^{nt}$**, where $A$ is the accumulated amount, $P$ is the principal, $r$ is the annual interest rate, $n$ is the number of times interest is compounded per year, and $t$ is the time in years.<br /><br />2. Substitute known values<br /> Given $A = 68000$, $r = 0.03575$, $n = 4$, and $t = 30$. Substitute these into the formula to find $P$: <br />\[ 68000 = P \left(1 + \frac{0.03575}{4}\right)^{4 \times 30} \]<br /><br />3. Simplify and solve for $P$<br /> Calculate $\left(1 + \frac{0.03575}{4}\right)^{120}$ and then solve for $P$:<br />\[ P = \frac{68000}{\left(1 + \frac{0.03575}{4}\right)^{120}} \]<br /><br />4. Compute the result<br /> Calculate the expression to find $P$:<br />\[ P \approx \frac{68000}{3.0025} \approx 22648.92 \]
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