Chapter 7, Problem 25PT

(a)

To determine

To find: The explanation of the numbers in this expression.

Answer to Problem 25PT

The expressions are explained.

Explanation of Solution

Given information:

Given that money invested in a certificate of deposit (CD). Suppose $\$4000$ money invested in a two year (CD).And the interest rate is $5\%$ and the expression is $4000{\left(1+0.05\right)}^2$ .

Calculation:

Consider the expression,

  $I=4000{\left(1+0.05\right)}^2$

Total amount of money $t$ years is,

  $I=P{\left(1+r\right)}^t$

Where, $P$ is the principal, $t$ is the time and $r$ is the rate.

So, $4000$ is the principal, $2$ year is the time and $0.05$ is the rate.

Therefore, the expressions are explained.

(b)

To determine

To find: The amount at the end of two years.

Answer to Problem 25PT

The amount at the end of two years is $\$4410.0$ .

Explanation of Solution

Given information:

Given that money invested in a certificate of deposit (CD). Suppose $\$4000$ money invested in a two year (CD). And the interest rate is $5\%$ and the expression is $4000{\left(1+0.05\right)}^2$ .

Calculation:

Consider the equation

  $I=4000{\left(1+0.05\right)}^2$

Using identity of square of a sum

  ${\left(a+b\right)}^2=a^2+2ab+b^2$

Substitute the values

  $I=4000\left(1^2+2\left(1\right)(0.05)+{0.05}^2\right)$

  $\begin{array}{l}I=4000\left(1+0.1+0.0025\right)\\ I=4000\left(1.1025\right)\\ I=4.4100\times {10}^3\\ I=4410.0\end{array}$

Therefore, the amount at the end of two years is $\$4410.0$

(c)

To determine

To find: The total amount of money after $4$ years.

Answer to Problem 25PT

The total amount of money after $4$ years is $\$12744.29321$ .

Explanation of Solution

Given information:

Given that money invested $\$10,000$ in a (CD) for $4$ years at an annual rate of $6.25\%$ .

Calculation:

  $P=10,000$ ,

  $\begin{array}{c}r=6.25\%\\ =0.0625\end{array}$

  $t=4$

Total amount of money $t$ years is,

  $I=P{\left(1+r\right)}^t$

Substitute the values,

  $\left(10,000\right){\left(1+0.0625\right)}^4={\left[\left(10,000\right){\left(1+0.0625\right)}^2\right]}^2$

  $\begin{array}{l}I={\left[\left(10,000\right){\left(1+0.0625\right)}^2\right]}^2\\ I=\left(10,000\right){\left[1^2+2\times 1\left(0.0625\right)+{0.0625}^2\right]}^2\\ I=\left(10,000\right){\left[1+0.1250+0.00390625\right]}^2\\ I=\left(10,000\right){\left[1+0.12890625\right]}^2\\ I=\left(10,000\right)\left[1^2+2\times 1\times \left(0.12890625\right)+{\left(0.12890625\right)}^2\right]\\ I=\left(10,000\right)\left[1.274429321\right]\\ I=12744.29321\end{array}$

Therefore, the total amount of money after $4$ years is $\$12744.29321$ .