Question

B) Option 2 : Deposit regular monthly payments into an investment account. Formula given = A=Px [(1+r/12)^12t -1]/ (r/12)

     i) John doesn't have enough to make a one time deposit, but he can make monthly payments into a retirement account. Suppose John makes $125 monthly payments into an account with a rate of return of 9.8%. How much would he have at age of 50? (Hit: John is 18 years old.)

   ii) If John were to increase his monthly deposit amount to $200, How old (to the nearest year) would John be when the balance reaches $1,000,000? (Hit: John is 18 years old.)

iii) What monthly deposit amount would be required for John to get $1,000,000 by age 50? (Hit: John is 18 years old.)

1. How much money would John actually have invested under these circumstances?
2. How much interest would have been earned in that time?

iv) What if John decided to wait until age 60 to retire?   In other words, if he made the same payments found in part iii above, how much would be in his account at that time? Is that total more than, less than, or equal to what you would expect? Give an explanation as to why you believe that is the case. (Hit: John is 18 years old.)

Note: please turn in all the exercises the following (step by step).

Answer 1

Part i)-

p=$125; r=9.8%per annum; t=(50-18) = 32years

Amount = P*[(1+r/12)^12t-1]/(r/12)

= 125*[(1+9.8%/12)^(12*32) -1]/(9.8%/12)

= 125*[(1+0.008167)^384 -1]/0.008167

= 125*[(1.008167)^384 -1]/0.008167

= 125*(22.7204-1)/0.008167

= 125*(21.7204/0.008167)

= 125*2659.6429

= $332,455.37

Part ii)-

p=$200; r=9.8%per annum; Amount = $1,000,000; t=?

Amount = P*[(1+r/12)^12t-1]/(r/12)

1000000 = 200*[(1+9.8%/12)^(12*t) -1]/(9.8%/12)

1000000 = 200*[(1+0.008167)^12t -1]/0.008167

8167 = 200*[(1.008167)^12t -1]

40.835 = (1.008167)^12t -1

41.835 = (1.008167)^12t

(1.008167)^459 = (1.008167)^12t

12t = 459; t = 38.25 years

Age to reach $1,000,000 is 38.25years + 18 years = 56.25 years or 56 years 3 months

Part iiia)-

P=?; r=9.8%per annum; Amount = $1,000,000; t=(50-18) = 32years

Amount = P*[(1+r/12)^12t-1]/(r/12)

1000000 = P*[(1+9.8%/12)^(12*32) -1]/(9.8%/12)

1000000 = P*[(1+0.008167)^384 -1]/0.008167

8167 = P*[(1.008167)^384 -1]

8167 = P*(22.7204-1)

8167 = P*21.7204

P = 8167/21.7204

P = $376 ; Monthly deposit is $376

Part iiib)-

Interest earned = Total amount - investment

= $1,000,000 - ($376*32years*12months)

= $1,000,000 - $144,384

= $855,616

Part iv)-

p=$376; r=9.8%per annum; t=(60-18) = 42years

Amount = P*[(1+r/12)^12t-1]/(r/12)

= 376*[(1+9.8%/12)^(12*42) -1]/(9.8%/12)

= 376*[(1+0.008167)^504 -1]/0.008167

= 376*[(1.008167)^504 -1]/0.008167

= 376*(60.2971-1)/0.008167

= 376*(59.2971/0.008167)

= 376*7260.5726

= $2,729,975.30