Chapter 11.7, Problem 20E

(a)

To determine

To find:

The interest rate with an annual compounding given data.

Answer to Problem 20E

  $r=3.034\%$

Explanation of Solution

Given:

The balance after $x$ year is

  $P{\left(1+r\right)}^x$

Concept used:

Formula:

The equation is

  $A=P{\left(1+r\right)}^x$

Calculation:

The table is

Date	Jan $1$	Mar. $31$	June $30$	Sept. $30$	Dec. $31$
Balance	$\$1000$	$1007.50$	$1015.06$	$1022.67$	$1030.34$

  $A=P{\left(1+r\right)}^x\mathrm{....................}\left(1\right)$

Since, $A=1034.34,P=1000,x=1$

From equation (1)

  $\begin{array}{l}A=P{\left(1+r\right)}^x\\ 1034.34=1000{\left(1+r\right)}^1\\ \frac{1034.34}{1000}=\left(1+r\right)\\ 1.03034=r+1\\ 0.03034=r\\ r=3.034\%\end{array}$

(b)

To determine

To find:

The function that models the amount given data.

Answer to Problem 20E

  $1.03034={\left(1+r\right)}^x$

Explanation of Solution

Given:

The balance after $x$ year is

  $P{\left(1+r\right)}^x$

Concept used:

Formula:

The equation is

  $A=P{\left(1+r\right)}^x$

Calculation:

The table is

Date	Jan $1$	Mar. $31$	June $30$	Sept. $30$	Dec. $31$
Balance	$\$1000$	$1007.50$	$1015.06$	$1022.67$	$1030.34$

  $A=P{\left(1+r\right)}^x\mathrm{....................}\left(1\right)$

Since, $A=1034.34,P=1000$

From equation (1)

  $\begin{array}{l}A=P{\left(1+r\right)}^x\\ 1034.34=1000{\left(1+r\right)}^x\\ \frac{1034.34}{1000}={\left(1+r\right)}^x\\ 1.03034={\left(1+r\right)}^x\end{array}$

(c)

To determine

To find:

The equation in terms of base $e$ .

Answer to Problem 20E

  $\frac{1034.34}{1000}=e^r$

Explanation of Solution

Given:

The balance after $x$ year is

  $P{\left(1+r\right)}^x$

Concept used:

Formula:

The equation is

  $P=P_0{e}^{kt}$

Calculation:

The table is

Date	Jan $1$	Mar. $31$	June $30$	Sept. $30$	Dec. $31$
Balance	$\$1000$	$1007.50$	$1015.06$	$1022.67$	$1030.34$

  $P=P_0{e}^{kt}\mathrm{....................}\left(1\right)$

From equation (1)

  $\begin{array}{l}P=P_0{e}^{kt}\\ 1034.34=1000e^r\\ \frac{1034.34}{1000}=e^r\end{array}$

(d)

To determine

To find:

The balances for the rate with continuous compounding.

Answer to Problem 20E

  $r=2.99\%$

Explanation of Solution

Given:

The balance after $x$ year is

  $P{\left(1+r\right)}^x$

Concept used:

Formula:

The equation is

  $A=P{e}^{rt}$

Calculation:

The table is

Date	Jan $1$	Mar. $31$	June $30$	Sept. $30$	Dec. $31$
Balance	$\$1000$	$1007.50$	$1015.06$	$1022.67$	$1030.34$

Since, $A=1034.34,P=1000,t=1$

  $P=P_0{e}^{kt}\mathrm{....................}\left(1\right)$

From equation (1)

  $\begin{array}{l}P=P_0{e}^{kt}\\ 1034.34=1000e^r\\ \frac{1034.34}{1000}=e^r\\ 1.03034=e^r\\ \ln \left(1.03034\right)=r\\ 0.0290\approx r\\ r=2.99\%\end{array}$