Norah has $50,000 to invest. She is considering two investment options. Option A pays 1.5% simple interest. Option B pays 1.4% interest compounded annually. Drag dollar amounts to the table to show the value of each investment option after 5 years, 10 years, and 20 years rounded to the nearest dollar.

Respuesta :

Answer:

Part 1) Option A

a) [tex]A=\$53,750[/tex]

b) [tex]A=\$57,500[/tex]

c) [tex]A=\$65,000[/tex]

Part 2) Option B

a) [tex]A=\$53,864[/tex]  

b) [tex]A=\$58,027[/tex]  

c) [tex]A=\$67,343[/tex]  

Step-by-step explanation:

Part 1) Option A

Simple Interest

we know that

The simple interest formula is equal to

[tex]A=P(1+rt)[/tex]

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

case a) 5 years

[tex]t=5\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015[/tex]

substitute in the formula above

[tex]A=50,000(1+0.015*5)[/tex]

[tex]A=50,000(1.075)[/tex]

[tex]A=\$53,750[/tex]

case b) 10 years

[tex]t=10\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015[/tex]

substitute in the formula above

[tex]A=50,000(1+0.015*10)[/tex]

[tex]A=50,000(1.15)[/tex]

[tex]A=\$57,500[/tex]

case c) 20 years

[tex]t=20\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015[/tex]

substitute in the formula above

[tex]A=50,000(1+0.015*20)[/tex]

[tex]A=50,000(1.30)[/tex]

[tex]A=\$65,000[/tex]

Part 2) Option B

interest compounded annually

we know that    

The compound interest formula is equal to  

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

case a) 5 years

[tex]t=5\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015\\n=1[/tex]  

substitute in the formula above

[tex]A=50,000(1+\frac{0.015}{1})^{1*5}[/tex]  

[tex]A=50,000(1.015)^{5}[/tex]  

[tex]A=\$53,864[/tex]  

case b) 10 years

[tex]t=10\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015\\n=1[/tex]  

substitute in the formula above

[tex]A=50,000(1+\frac{0.015}{1})^{1*10}[/tex]  

[tex]A=50,000(1.015)^{10}[/tex]  

[tex]A=\$58,027[/tex]  

case c) 20 years

[tex]t=20\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015\\n=1[/tex]  

substitute in the formula above

[tex]A=50,000(1+\frac{0.015}{1})^{1*20}[/tex]  

[tex]A=50,000(1.015)^{20}[/tex]  

[tex]A=\$67,343[/tex]  

Answer:

Option A-

5 years: $53,750

10 years:$57,500

20 years:$65,000

Option B-

5 years:$53,599

10 years:$57,458

20 years:$66,028

Step-by-step explanation: