Question

1. You are saving to buy a new house in 7 years. If you invest $4,500 now at 5.5% interest compounded

quarterly, how much money will you have to use for your down payment?

2. You have your heart set on buying a new car in 2 years. You invest $3,200 at 3.75% interest
compounded continuously. How much money will have to use for your down payment on your new
car?

3. You are a proud new parent of a baby girl. In eighteen years, you will want to help her pay for college.
How much do you need to invest now at 4.75% interest compounded monthly so you can help her pay
for the $40,000 expense of college?

4. After building your new home, you decide you would like to install an in-ground pool in 7 years. How
much do you need to invest now at 5.25% interest compounded continuously to have the $30,000 you
will need to build the pool?

5. You recently received an inheritance of $11,500 from your grandparents. Which option is the best way
to invest your money and how much better is the best investment? Option A: 5.6% interest
compounded semi-annually for 8 years or Option B: 3.45% interest compounded continuously for 5
years.
Part A)
Part B)
Part C)

Answer 1

Part 1)
`\$6,595.94`

Part 2)
`\$3,449.23`

Part 3)
`\$17,040.06`

Part 4)
`\$20,773.90`

Part 5) The Option A is the best way to invest the money by $4,223.94 than Option B

Explanation:

Part 1)

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

`t=7\ years\\ P=\$4,500\\ r=5.5\%=5.5/100=0.055\\n=4`

substitute in the formula above

`A=4,500(1+(0.055)/(4))^(4*7)`

`A=4,500(1.01375)^(28)`

`A=\$6,595.94`

Part 2)

we know that

The formula to calculate continuously compounded interest is equal to

`A=P(e)^(rt)`

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

e is the mathematical constant number

we have

`t=2\ years\\ P=\$3,200\\ r=3.75\%=3.75/100=0.0375`

substitute in the formula above

`A=3,200(e)^(0.0375*2)`

`A=\$3,449.23`

Part 3)

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

`t=18\ years\\ A=\$40,000\\ r=4.75\%=4.75/100=0.0475\\n=12`

substitute in the formula above

`40,000=P(1+(0.0475)/(12))^(12*18)`

`40,000=P((12.0475)/(12))^(216)`

`P=40,000/[((12.0475)/(12))^(216)]`

`P=\$17,040.06`

Part 4)

we know that

The formula to calculate continuously compounded interest is equal to

`A=P(e)^(rt)`

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

e is the mathematical constant number

we have

`t=7\ years\\ A=\$30,000\\ r=5.25\%=5.25/100=0.0525`

substitute in the formula above

`30,000=P(e)^(0.0525*7)`

`30,000=P(e)^(0.3675)`

`P=30,000/(e)^(0.3675)`

`P=\$20,773.90`

Part 5)

Option A

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

`t=8\ years\\ P=\$11,500\\ r=5.6\%=5.6/100=0.056\\n=2`

substitute in the formula above

`A=11,500(1+(0.056)/(2))^(2*8)`

`A=11,500(1.028)^(16)`

`A=\$17,889.07`

Option B

we know that

The formula to calculate continuously compounded interest is equal to

`A=P(e)^(rt)`

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

e is the mathematical constant number

we have

`t=5\ years\\ P=\$11,500\\ r=3.45\%=3.45/100=0.0345`

substitute in the formula above

`A=11,500(e)^(0.0345*5)`

`A=11,500(e)^(0.1725)`

`A=\$13,665.13`

Compare the options

Option A ------>
`\$17,889.07`

Option B ----->
`\$13,665.13`

so

Option A > Option B

Find out the difference

`\$17,889.07-$13,665.13=$4,223.94`

therefore

The Option A is the best way to invest the money by $4,223.94 than Option B