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?
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
Section: Chapter Questions
Problem 1PS
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8311114-a7cb-49e3-b44e-1e05de4bc29a%2F2218b546-912a-4cba-91df-d9b6ebb92dc3%2F6beaz4.png&w=3840&q=75)
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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