Answer to Question #100906 in Financial Math for Rana

Question

Calculate the compound interest on a term deposit of $30,000 at the rate of 6% annually for 3 years when the investment is compounded: a) annually. b) semi-annually. c) quarterly. d) monthly. e) daily.

Answer 1

Given

P=Principal = $\$30000$

r= rate of interest =$6 \%=0.06$

t =number of time periods elapsed

n = number of times interest applied per time period

(a) compounded annually

n = 1

$$A = P (1 +\frac {r}{n})^{nt}$$

$$A = 30000 (1 +\frac {0.06}{1})^{1\times 3} = 30000 \times (1.06)^3$$

$$A=30000 \times 1.191 =35,730$$

$$Interest = A - P =35730-30000 =\$5730$$

(b) compounded semi-annually

n = 1

$$A = P (1 +\frac {r}{n})^{nt}$$

$$A = 30000 (1 +\frac {0.06}{2})^{2\times 3} = 30000 \times (1.03)^6$$

$$A=30000 \times 1.194 =35,820$$

$$Interest = A - P =35,820-30000 =\$5820$$

(c) compounded quarterly

n= 4

$$A = 30000 (1 +\frac {0.06}{4})^{4\times 3} = 30000 \times (1.015)^{12}$$

$$A=30000 \times 1.1956 =35868$$

$$Interest = A - P =35,868-30000 =\$5868$$

(d) compounded monthly

n=12

$$A = 30000 (1 +\frac {0.06}{12})^{12\times 3} = 30000 \times (1.005)^{36}=35,900$$

$$Interest = A - P =35,900-30000 =\$5900$$

(e) compounded daily

n=365

$$A = 30000 (1 +\frac {0.06}{365})^{365\times 3} = 30000 \times (1.000164385)^{1095}$$

$$A =30000\times 1.1972=\$35916$$

$$Interest = A - P =35,916-30000 =\$5916$$