assignment: applications 2 - 6a\nquestions gefa82h.ap2 - 6.01\nnote: please make sure to properly format…
assignment: applications 2 - 6a\nquestions gefa82h.ap2 - 6.01\nnote: please make sure to properly format your answers. all dollar figures in the answers need to include the dollar sign and any amount over 1,000 should include the comma ($2,354.67). all percentage values in the answers need to include a percentage sign (%). for all items without specific rounding instructions, round your answers to two decimal places, show both decimal places (5.06).\na bank representative studies compound interest so she can better serve customers. she analyses what happens when $2,000 earns interest several different ways at a rate of 2% for 3 years.\na. find the interest if it is computed using simple interest.\nb. find the interest if it is compounded annually.\nc. find the interest if it is compounded semi - annually.\nd. find the interest if it is compounded quarterly.\ne. find the interest if it is compounded monthly.\nf. find the interest if it is compounded daily.\ng. find the interest if it is compounded hourly.\nh. find the interest if it is compounded every minute.\ni. find the interest if it is compounded continuously.\nj. what is the difference in total interest if computed using simple interest or if compounded continuously?
Answer
Explanation:
Step1: Recall simple - interest formula
The simple - interest formula is $I = Prt$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), and $t$ is the time in years. Given $P=$2000$, $r = 0.02$, and $t = 3$. $I=Prt=2000\times0.02\times3$ $I = 120$
Step2: Recall compound - interest formula
The compound - interest formula is $A=P(1 + \frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after $n$ years, including interest, $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. The interest $I=A - P$.
a. Simple interest
$I = 120.00$
b. Compounded annually ($n = 1$)
$A=2000(1+\frac{0.02}{1})^{1\times3}=2000(1.02)^{3}=2000\times1.061208 = 2122.42$ $I=A - P=2122.42-2000=$122.42$
c. Compounded semi - annually ($n = 2$)
$A=2000(1+\frac{0.02}{2})^{2\times3}=2000(1.01)^{6}=2000\times1.061520150601\approx2123.04$ $I=A - P=2123.04 - 2000=$123.04$
d. Compounded quarterly ($n = 4$)
$A=2000(1+\frac{0.02}{4})^{4\times3}=2000(1.005)^{12}\approx2000\times1.0616778169 = 2123.36$ $I=A - P=2123.36-2000=$123.36$
e. Compounded monthly ($n = 12$)
$A=2000(1+\frac{0.02}{12})^{12\times3}=2000(1+\frac{0.02}{12})^{36}\approx2000\times1.061797724 = 2123.59$ $I=A - P=2123.59 - 2000=$123.59$
f. Compounded daily ($n = 365$)
$A=2000(1+\frac{0.02}{365})^{365\times3}=2000(1+\frac{0.02}{365})^{1095}\approx2000\times1.06183131\approx2123.66$ $I=A - P=2123.66-2000=$123.66$
g. Compounded hourly ($n = 365\times24$)
$A=2000(1+\frac{0.02}{365\times24})^{365\times24\times3}=2000(1+\frac{0.02}{8760})^{26280}\approx2000\times1.06183694\approx2123.67$ $I=A - P=2123.67-2000=$123.67$
h. Compounded every minute ($n = 365\times24\times60$)
$A=2000(1+\frac{0.02}{365\times24\times60})^{365\times24\times60\times3}=2000(1+\frac{0.02}{525600})^{525600\times3}\approx2000\times1.06183719\approx2123.67$ $I=A - P=2123.67-2000=$123.67$
i. Compounded continuously
The formula for continuous compounding is $A = Pe^{rt}$, where $e\approx2.71828$. $A=2000e^{0.02\times3}=2000e^{0.06}\approx2000\times1.061836547\approx2123.67$ $I=A - P=2123.67-2000=$123.67$
j. Difference between simple and continuous compounding
$123.67 - 120.00=$3.67$
Answer:
a. $120.00 b. $122.42 c. $123.04 d. $123.36 e. $123.59 f. $123.66 g. $123.67 h. $123.67 i. $123.67 j. $3.67