Chapter 9A, Problem 10P

To determine

(a)

The value of account after $10$ years.

Answer to Problem 10P

The value of account after $10$ year is $\$75805$.

Explanation of Solution

Given:

Salary is $\$60000$.

Interest rate for stock fund is $7.3\%$.

Calculation:

Make the table for cash flows.

Here, the salary is increasing every year by $3\%$.

Write the expression to calculate the salary for each year.

$Salary={\left(\$60000\times 1.03\right)}^n$         ...... (I)

Here, the number of year is $n$.

Write the expression to calculate the saving of the employee.

$S_{EE}=\left(Salar{y}_n\times 0.04\right)$         ...... (II)

Here, the saving by employee is $S_{EE}$ and salary for the current year is $Salar{y}_n$.

Write the expression to calculate the future worth.

$F=S_{EE}{\left(1+0.073\right)}^{10-n}$         ...... (III)

Here the future worth in account is $F$.

Write the expression to calculate the saving of the employer.

$S_E=\left(Salar{y}_n\times 0.04\right)$         ...... (IV)

Here, the saving by employee is $S_{EE}$ and salary for the current year is $Salar{y}_n$.

Write the expression to calculate the total savings.

$TS=F+S_E$

Using the table, the future worth at the end of $10$ year is $\$75805$.

Conclusion:

Thus, the value of account after $10$ years is $\$75805$.

To determine

(b)

The value of employer’s matching fund.

Answer to Problem 10P

The value of the employer’s matching fund at the end of $10$ year is $\$37902$.

Explanation of Solution

Given:

Salary is $\$60000$.

Interest rate for stock fund is $7.3\%$.

Calculation:

Make the table for cash flows.

Here, the salary is increasing every year by $3\%$.

Write the expression to calculate the salary for each year.

$Salary={\left(\$60000\times 1.03\right)}^n$         ...... (I)

Here, the number of years is $n$.

Write the expression to calculate the saving of the employee.

$S_{EE}=\left(Salar{y}_n\times 0.04\right)$         ...... (II)

Here, the saving by employee is $S_{EE}$ and salary for the current year is $Salar{y}_n$.

Write the expression to calculate the future worth.

$F=S_{EE}{\left(1+0.073\right)}^{10-n}$         ...... (III)

Here the future worth in account is $F$.

Write the expression to calculate the saving of the employer.

$S_E=\left(Salar{y}_n\times 0.04\right)$         ...... (IV)

Here, the saving by employee is $S_{EE}$ and salary for the current year is $Salar{y}_n$.

Write the expression to calculate the total saving.

$TS=F+S_E$

Using the table, the values of employer matching fund at the end of $10$ year is $\$37902$.

Conclusion:

Thus, the value of employer matching fund at the end of $10$ years is $\$37902$.

To determine

(c)

The value of account after $40$ years.

Answer to Problem 10P

The value of the account at the end of $40$ year is $\$1505521$.

Explanation of Solution

Given:

Salary is $\$60000$.

Interest rate for stock fund is $7.3\%$.

Calculation:

Make the table for cash flows.

Here, the salary is increasing every year by $3\%$.

Write the expression to calculate the salary for each year.

$Salary={\left(\$60000\times 1.03\right)}^n$         ...... (I)

Here, the number of years is $n$.

Write the expression to calculate the saving of the employee.

$S_{EE}=\left(Salar{y}_n\times 0.04\right)$         ...... (II)

Here, the saving by employee is $S_{EE}$ and salary for the current year is $Salar{y}_n$.

Write the expression to calculate the future worth.

$F=S_{EE}{\left(1+0.073\right)}^{40-n}$         ...... (III)

Here the future worth in account is $F$.

Write the expression to calculate the saving of the employer.

$S_E=\left(Salar{y}_n\times 0.04\right)$         ...... (IV)

Here, the saving by employee is $S_{EE}$ and salary for the current year is $Salar{y}_n$.

Write the expression to calculate the total saving.

$TS=F+S_E$

Using the table, the value of account at the end of $40$ years is $\$1505521$.

Conclusion:

Thus, the value of account at the end of $40$ years is $\$1505521$.