Question

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.

Answer 1

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

Explanation:

Answer 2

Part 1) Option A

a)
`A=\$53,750`

b)
`A=\$57,500`

c)
`A=\$65,000`

Part 2) Option B

a)
`A=\$53,864`

b)
`A=\$58,027`

c)
`A=\$67,343`

Explanation:

Part 1) Option A

Simple Interest

we know that

The simple interest formula is equal to

`A=P(1+rt)`

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

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

substitute in the formula above

`A=50,000(1+0.015*5)`

`A=50,000(1.075)`

`A=\$53,750`

case b) 10 years

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

substitute in the formula above

`A=50,000(1+0.015*10)`

`A=50,000(1.15)`

`A=\$57,500`

case c) 20 years

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

substitute in the formula above

`A=50,000(1+0.015*20)`

`A=50,000(1.30)`

`A=\$65,000`

Part 2) Option B

interest compounded annually

we know that

The compound interest formula is equal to

`A=P(1+(r)/(n))^(nt)`

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

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

substitute in the formula above

`A=50,000(1+(0.015)/(1))^(1*5)`

`A=50,000(1.015)^(5)`

`A=\$53,864`

case b) 10 years

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

substitute in the formula above

`A=50,000(1+(0.015)/(1))^(1*10)`

`A=50,000(1.015)^(10)`

`A=\$58,027`

case c) 20 years

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

substitute in the formula above

`A=50,000(1+(0.015)/(1))^(1*20)`

`A=50,000(1.015)^(20)`

`A=\$67,343`