Palma wants to establish a fund for her granddaughter's college education. What lump sum must she deposit in an account that pays an annual interest rate of 6%, compounded monthly, if she wants to have $10,000 in 10 years?

Essentials Of Investments
11th Edition
ISBN:9781260013924
Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.
Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.
Chapter1: Investments: Background And Issues
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**Problem Statement:**

Palma wants to establish a fund for her granddaughter's college education. What lump sum must she deposit in an account that pays an annual interest rate of 6%, compounded monthly, if she wants to have $10,000 in 10 years?

**Solution Details:**

To determine the required lump sum deposit, we can use the formula for compound interest:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]

Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested for, in years.

Given:
- \( A = 10,000 \)
- \( r = 0.06 \)
- \( n = 12 \) (monthly compounding)
- \( t = 10 \)

We need to solve for \( P \) (the principal amount).

Rearranging the formula to solve for \( P \):

\[ P = \frac{A} {\left(1 + \frac{r}{n}\right)^{nt}} \]

Plugging in the given values:

\[ P = \frac{10,000}{\left(1 + \frac{0.06}{12}\right)^{12 \times 10}} \]

\[ P = \frac{10,000}{\left(1 + 0.005\right)^{120}} \]

\[ P = \frac{10,000}{\left(1.005\right)^{120}} \]

By calculating the compound factor:

\[ (1.005)^{120} \approx 1.8194 \]

So:

\[ P = \frac{10,000}{1.8194} \]

\[ P \approx 5496.86 \]

Therefore, Palma needs to deposit approximately $5,496.86 in the account today to have $10,000 in 10 years, given the conditions mentioned.
Transcribed Image Text:**Problem Statement:** Palma wants to establish a fund for her granddaughter's college education. What lump sum must she deposit in an account that pays an annual interest rate of 6%, compounded monthly, if she wants to have $10,000 in 10 years? **Solution Details:** To determine the required lump sum deposit, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for, in years. Given: - \( A = 10,000 \) - \( r = 0.06 \) - \( n = 12 \) (monthly compounding) - \( t = 10 \) We need to solve for \( P \) (the principal amount). Rearranging the formula to solve for \( P \): \[ P = \frac{A} {\left(1 + \frac{r}{n}\right)^{nt}} \] Plugging in the given values: \[ P = \frac{10,000}{\left(1 + \frac{0.06}{12}\right)^{12 \times 10}} \] \[ P = \frac{10,000}{\left(1 + 0.005\right)^{120}} \] \[ P = \frac{10,000}{\left(1.005\right)^{120}} \] By calculating the compound factor: \[ (1.005)^{120} \approx 1.8194 \] So: \[ P = \frac{10,000}{1.8194} \] \[ P \approx 5496.86 \] Therefore, Palma needs to deposit approximately $5,496.86 in the account today to have $10,000 in 10 years, given the conditions mentioned.
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