Chapter 4, Problem 11CRT

(a)

To determine

To find: The amount in the account be after 3 years.

Answer to Problem 11CRT

  $\$29,396.15$

Explanation of Solution

Given: A sum of $25,000 is deposited into an account paying 5.4% interest per year, compounded daily.

Formula Used: $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Sum, $P=\$25,000$

Rate of interest, $r=5.4\%PA=0.054$

Time, $t=3\text{ years}$

Number of term, $n=365$

Put above value sin formula.

  $\begin{array}{l}A=25000{\left(1+\frac{0.054}{365}\right)}^{365\times 3}\\ A\approx 29396.15\end{array}$

Hence, the amount in the account after 3 years would be $\$29,396.15$

(b)

To determine

To find: The time when amount in account becomes $35,000.

Answer to Problem 11CRT

6 years

Explanation of Solution

Given: A sum of $25,000 is deposited into an account paying 5.4% interest per year, compounded daily.

Formula Used: $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Sum, $P=\$25,000$

Rate of interest, $r=5.4\%PA=0.054$

Amount, $A=\$35,000$

Number of term, $n=365$

Put above value into formula.

  $\begin{array}{c}35000=25000{\left(1+\frac{0.054}{365}\right)}^{365\times t}\\ {\left(1+\frac{0.054}{365}\right)}^{365\times t}=\frac{25}{35}\\ 365t=\frac{\ln \left(1.4\right)}{\ln \left(1.0001\right)}\\ t\approx 6\end{array}$

Hence, after 6 years amount reach to $35,000

(c)

To determine

To find: The time when initial value double.

Answer to Problem 11CRT

4 years 2 months

Explanation of Solution

Given: A sum of $25,000 is deposited into an account paying 5.4% interest per year, compounded daily.

Formula Used: $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Sum, $P=\$25,000$

Rate of interest, $r=5.4\%PA=0.054$

Amount, $A=\$50,000$

Number of term, $n=365$

Put above value into formula.

  $\begin{array}{c}50000=25000{\left(1+\frac{0.054}{365}\right)}^{365\times t}\\ {\left(1+\frac{0.054}{365}\right)}^{365\times t}=\frac{50}{25}\\ 365t=\frac{\ln \left(2\right)}{\ln \left(1.0001\right)}\\ t\approx 13\end{array}$

Hence, after 13 years amount will be double.