Question

George is considering two different investment options. The first offers 7.4% per year simple interest on the initial deposit. The second option offers a 6.5% interest rate but is compounded quarterly. He may not withdraw any of the money for three years after the deposit. Once the minimum 3 years is reached, he can choose to withdraw his money or continue to collect interest. Suppose that George opens one of each type of account and deposits 10,000% into each.
Part A : Determine the value of the simple-interest investment at the end of three years. Use the formula A = P + Prt, where A represents the value of the investment, P represents the original amount, r represents the rate, and t represents the years.


Part B : Determine the value of the compound-interest investment at the end of three years. Use the formula A = P(1+r/n)rt, where A represents the value of the investment, P represents the original amount, r represents the rate compounded n times per year, and t represents the time in years.


Part C : Which investment is better over the first three years?


Part D : How would you advise George to invest his money if he is unsure how long he’ll keep the money in the account?

Answer 1

Answer:

Part A) $\ 12,220.00$

Part B) $\ 12,134.08$

Part C) The simple interest investment is better than the compounded  interest investment at the end of three years.

Part D)  see the explanation

Step-by-step explanation:

Part A) Determine the value of the simple-interest investment at the end of three years

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

$t=3\ years\\ P=\ 10,000\\r=7.4\%=7.4/100=0.074$

substitute in the formula above

$A=10,000(1+0.074*3)$

$A=10,000(1.222)$

$A=\ 12,220.00$

Part B) Determine the value of the compound-interest investment at the end of three years

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  

$t=3\ years\\ P=\ 10,000\\r=6.5\%=6.5/100=0.065\\n=4$

substitute in the formula above

$A=10,000(1+\frac{0.065}{4})^{4*3}$  

$A=10,000(1.01625)^{12}$  

$A=\ 12,134.08$

Part C) Which investment is better over the first three years?

Compare the two investment

$\ 12,134.08 < \ 12,220.00$

The compounded  interest investment is less than the simple interest investment

therefore

The simple interest investment is better than the compounded  interest investment at the end of three years

Part D) How would you advise George to invest his money if he is unsure how long he’ll keep the money in the account?

we have

$A=10,000(1+0.074t)$  ----> simple interest formula as a function of the time t

$A=10,000(1.01625)^{4t}$   ----> compound interest formula as function of the time t

Look to the attached graph below

The red line represents the simple interest investment

The blue curve represents the compounded interest investment

After 0 years and before 4.179 years the red line is over the blue curve, that means the simple interest is better because it gives more money than the compounded interest

After that the blue curve is over the red line that means the compounded quarterly is better because it gives more money than the simple interest

therefore

I advise George to invest his money in the compounded interest investment if he will keep the money for a long time