Question

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

Answer 1

Answer:

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

Step-by-step explanation:

Part 1)

we know that    

The compound interest formula is equal to  

$A=P(1+\frac{r}{n})^{nt}$  

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  

$A=4,500(1+\frac{0.055}{4})^{4*7}$  

$A=4,500(1.01375)^{28}$

$A=\ 6,595.94$    

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  

$A=P(1+\frac{r}{n})^{nt}$  

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  

$40,000=P(1+\frac{0.0475}{12})^{12*18}$  

$40,000=P(\frac{12.0475}{12})^{216}$  

$P=40,000/[(\frac{12.0475}{12})^{216}]$  

$P=\ 17,040.06$  

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  

$A=P(1+\frac{r}{n})^{nt}$  

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  

$A=11,500(1+\frac{0.056}{2})^{2*8}$  

$A=11,500(1.028)^{16}$

$A=\ 17,889.07$  

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