On sam's 40th birthday, he decided to start investing $100 at the end of every month, until his retirement on his 60th birthday. After his 60th birthday, he withdraws half of the money in his account and spends it. The other half of the money is reinvested as a lump sum at 3% per annum compounded monthly. How much will he have when he is 75 years old?
For the purposes of the question, I have assumed a monthly compounded interest rate of 3% per annum of the $100 monthly savings, as the question does not specify. Apply the formula where:
F=the future amount: unknown
x = the monthly payment ($100)
i= interest: 3% p a compounded monthly = `0.03/12 =` 0.0025
n=number of months: 20 years x 12 months per year=240. We multiply by the 12to get the months because t is compounded monthly. If it was compounded annualy, we would not convert the 20 years into months.
`therefore F = (x[(1+i)^n-1])/i` becomes `F=(100[(1+0.0025)^240 - 1])/ 0.0025`
`therefore F=(100[(1.0025)^240 - 1])/0.0025`
Care to ensure that the operations are performed in the correct order and do not round off until you get to a final answer:
`therefore F= 82.07549953/0.0025`
Now we know he spends half and invests the remainder. Therefore we have $32830.20 divided by 2= $16415.10. Use the formula to establish
the amount (A): unknown
after 15 years x 12 months: therefore n= 180
from the current investment (P) which is $16415.10
where i = 3%: i= 0.0025 ( as it is compounded monthly)
`therefore A=P(1+i)^n` becomes
`A = 16415,10(1+0.0025)^180`
`therefore A= $25 729.55`
Care to do the operations in the correct order and do not round off too soon.
Ans: When Sam is 75 he will have $25 729.55.