A person wants to establish an annuity for retirement purpos
Solution
36 (A). The annuity payment formula is P = Ar/[ 1-(1+r)-n] where P is the periodic payment, r is the rate of interest per period , n is the number of eriods and A is the present value. Hete P = 5000, n = 10*4 = 40 and r = (7.32/100)*1/4 = 0.00183 . Then we have 5000 = A*0.00183/[ 1- (1.00183)-40] = 0.00183 A/( 1- 1/1.07587373) = 0.00183 A/(1- 0.929477105) = 0.00183 A/0.070522895 so that A = 5000*(0.070522895)/ 0.00183 = 352.614475/ 0.00183 = $ 192685.51 ( on rounding off to the nearest cent). Thus, there will have to be $ 192685.51 in the account by the time the person retires.
(B) The formula for the future value of annuity is F = P[{(1+r)n -1}/r] *(1+r) , where P, n, r are as above. Here F = $ 192685.51, n = 20*4 = 80, r = 0.00183. Then 192685.51 = P [ {(1.00183)80-1}/0.00183]* 1.00183 = P[ (1.157504283-1)/0.00183]* 1.00183 = P [(0.157504283)/0.00183]*1.00183 = ( 86.06791421)*(1.00183)P = 86.22541849 P. Therefore P = 192685.51/86.22541849 = $2234.67 ( on rounding off to the nearest cent). Thus, $ 2234.67 has to be deposited every quarter for 20 years in order to have $ 192685.51 in the account at the end of 20 years.
( C ). The person has deposited $ 2234.67 * 80 = $ 178773.60 in his account and he receives $ 5000*40 = 200000. Therefore, the interest earned during the 30 year period is $ 200000- $ 178773.60 = $ 21226.40

