Question

4.

The Weidmans want to save $20,000 in 4 years for a down payment on a house. If they make monthly deposits in an account paying 12%, compounded monthly, what is the size of the payments that are required to meet their goal? 

5.

 (a) Patty Stacey deposits $2200 at the end of each of 5 years in an IRA. If she leaves the money that has accumulated in the IRA account for 25 additional years, how much is in her account at the end of the 30-year period? Assume an interest rate of 6%, compounded annually. (Round your answer to the nearest cent.)

 (b) Suppose that Patty's husband delays starting an IRA for the first 10 years he works but then makes $2200 deposits at the end of each of the next 15 years. If the interest rate is 6%. compounded annualy, and if he leaves the money in his account for S additional years, how much will be an has account at the end of the 30 year period? (Round your anawer to the nearest cent.)

 (c) Does Patty or her husband have more IRA money?

 Patty

 Patty's husband

Answer 1

4.

Formula for FV of annuity can be used to compute monthly deposits as:

FV = C x [(1+i) ⁿ -1/i]

C = FV / [(1+i) ⁿ -1/i]

FV = Future value of annuity = $ 20,000

C = Periodic cash deposit

i = Periodic rate =0.12/12 = 0.01 p.m.

n = Number of periods = 4 years x 12 months = 48 periods

C = $ 20,000 / [(1+0.01)⁴⁸ -1/0.01]

    = $ 20,000 / [(1.61222607768247-1)/0.01]

    = $ 20,000 / (0.61222607768247/0.01)

   = $ 20,000 /61.2226077682465

   = $ 326.676708638555 or $ 326.68

Monthly payment of $ 326.68 is needed to achieve the goal.

5.

(a)

Computation of FV of fund size at the end of year 5:

FV = C x [(1+i) ⁿ -1/i]

C = $ 2,200

i = 6 % or 0.06 p.a.

n = 5

FV = $ 2,200 x [(1+0.06) ⁵ -1/0.06]

      = $ 2,200 x [(1.06) ⁵ -1/0.06]

      = $ 2,200 x [(1.3382255776 -1)/0.06]

      = $ 2,200 x [(0.3382255776/0.06)

     = $ 2,200 x 5.63709296

      = $ 12,401.604512

Computation of future value of $ 12,401.604512 after 25 additional years:

FV = PV x (1+r) ⁿ

      = $ 12,401.604512 x (1+0.06)²⁵

      = $ 12,401.604512 x (1.06)²⁵

     = $ 12,401.604512 x 4.29187071974349

     = $ 53,226.0832828916 or $ 53,226.08

Patty Stacey will have $ 53,226.08 in her account after 30 years.

(b)

Computation of FV of fund size at the end of year 15:

FV = C x [(1+i) ⁿ -1/i]

C = $ 2,200

i = 6 % or 0.06 p.a.

n = 15

FV = $ 2,200 x [(1+0.06) ¹⁵ -1/0.06]

      = $ 2,200 x [(1.06) ¹⁵ -1/0.06]

      = $ 2,200 x [(2.3965581931-1)/0.06]

      = $ 2,200 x [(1.3965581931/0.06)

     = $ 2,200 x 23.2759698849949

      = $ 51,207.1337469887

Computation of future value of $ 51,207.1337469887 after 5 additional years:

FV = PV x (1+r) ⁿ

      = $ 51,207.1337469887 x (1+0.06)⁵

      = $ 51,207.1337469887 x (1.06)⁵

     = $ 51,207.1337469887 x 1.3382255776

     = $ 68,526.6961358045 or $ 68,526.70

Patty’s husband will have $ 68,526.70 in her account after 30 years.

(c)

Patty’s husband will have more IRA money.