Chapter 8.7, Problem 47PPS

(a)

To determine

To find: The money would in the account of $5$ year.

Answer to Problem 47PPS

The total amount after $5$ year is $1001.86$ .

Explanation of Solution

Given information:

The given deposited $5$ and interest $4.5\%$ .

Calculation:

Now

  $A=P{e}^n$

  $p=$ Principal amount

  $r=$ Annual interest rate

  $t=$ Number of year

  $A=$ Amount

Calculate the solution

  $\begin{array}{c}A=800e^{0.045*5}\\ =1001.86\end{array}$

The total amount of $5$ year .

  

Therefore the total amount after $5$ year $1001.86$ .

(b)

To determine

To find: The time to double your money .

Answer to Problem 47PPS

The after $15.4$ year the deposited money will be doubled.

Explanation of Solution

Given information:

The given deposited $@800$ and interest $4.5\%$ .

Calculation:

Calculate solution.

  $\begin{array}{c}p=800\\ A=1600\\ r=0.045\\ 800e^{0.045t}=1600\\ e^{0.045t=2}\\ natural\log arithmofbothsides\\ \ln e^{0.045t}=\ln 2\\ 0.045t=\ln 2\\ t=\frac{1}{0.045}\ln 2\\ =15.4\end{array}$

  

Therefore after $15.4$ year the deposited money will be doubled.

(c)

To determine

To find: The rate to double your money in $9$ year .

Answer to Problem 47PPS

The required interest rate $7.7\%$ .

Explanation of Solution

Given information:

The given deposited $@800$ and interest $4.5\%$ .

Calculation:

Calculate the equation.

  $\begin{array}{l}P{e}^{9\frac{r}{100}}=2P\\ e^{9\frac{r}{100}}=2\end{array}$

The natural logarithm of both sides

  $\begin{array}{l}\ln e^{9\frac{r}{100}}=\ln 2\\ \frac{9r}{100}=\ln 2\\ r=\frac{100}{9}\ln 2\\ =7.7\end{array}$

Use $TI83$ to calculate to computer value

  

  $=7.7\%$

  

Therefore the required interest rate to double money is $7.7\%$ .

(d)

To determine

To find: The deposit to increase amount.

Answer to Problem 47PPS

The deposit is $@5655.25$ need to be deposited in that account.

Explanation of Solution

Given information:

The given deposited $\$10000$ and interest $4.75\%$

Calculation:

Calculate the equation .

  $\begin{array}{l}P{e}^{0.0475*12}=10000\\ P=10000e^{-0.0475*12}\\ =5655.25\end{array}$

Use $TI83$ to calculate to computer value

  

Therefore the $@5655.25$ need to be deposited in that account.