I will use the fact that she deposits $M for the entire period plus and addition $M for the last 15 years, (making it 2M for the last 15 years)
i = .06/12 = .005
M(1.005)^480 - 1)/.005 + M(1.005)^180 - 1)/.005 = 1000000
M( 1.005^480 -1 + 1.005^180 - 1)/.005 = 1000000
I get M = 438.15
check my calculations
or
amount of the first 300 payments of M at the end of 40 years
= M(1.005^300 - 1)/.005 (1.005)^180
amount of the last 180 payments of 2M
= 2M(1.005^180 - 1)/.005
M(1.005^300 - 1)/.005 (1.005)^180 + 2M(1.005^180 - 1)/.005 = 1000000
M(1700.672022 + 2(290.8187124) = 1000000
M(2282.309447) = 1000000
M = 438.15
clearly my first method is the easier of the two.
Felicity has just commenced work and is investigating superannuation funds.
She calculates that she will need $1 000 000 in her fund when she retires in 40
years. She finds a fund guaranteeing to pay interest on each deposit at a rate of
6% per annum compounded monthly.
She intends to deposit $M into the fund at the beginning of each month for the
first 25 years (300 months). For the remaining 15 years (180 months), she will
deposit $2M into the fund.
Find the value of M.
1 answer