Joe won a lottery jackpot that will pay him $12,000 each year for the next ten years. If the market interest r

Question

Joe won a lottery jackpot that will pay him $12,000 each year for the next ten years. If the market interest rates are currently 12%, how much does the lottery have to invest today to pay out this prize to Joe over the next ten years?

Answer 1

Approximately $67,802 dollars. The thing to remember is that you can slowly draw down the principal and ultimately end up with no principal.

Here's how it would work:
Year 0 -- Lottery invests $67,802
Year 1 -- Joe gets $12000 (8136 interest, 3864 principal), remaining principal is 63938
Year 2 -- Joe gets $12000 (7673 interest, 4327 principal), remaining principal is 59611
Year 3 -- Joe gets $12000 (7153 interest, 4847 principal), remaining principal is 54764
Year 4 -- Joe gets $12000 (6572 interest, 5428 principal), remaining principal is 49336
Year 5 -- Joe gets $12000 (5920 interest, 6080 principal), remaining principal is 43256
Year 6 -- Joe gets $12000 (5191 interest, 6806 principal), remaining principal is 36447
Year 7 -- Joe gets $12000 (4374 interest, 7626 principal), remaining principal is 28821
Year 8 -- Joe gets $12000 (3459 interest, 8541 principal), remaining principal is 20280
Year 9 -- Joe gets $12000 (2434 interest, 9566 principal), remaining principal is 10714
Year 10 -- Joe gets $12000 (1286 interest, 10714 principal), remaining principal is 0

So each year Joe gets $12,000 (partially from interest and partially from principal). In the end, the lottery's initial $67,802 results in $120,000 in principal and interest paid to Joe.

Answer 2

It depends on how interest is compounded.

If it's compounded continuously, then you can use the formula A=Pe^(rt), where P is the initial amount invested, r is the interest rate (in this case 0.12), t is the amount of time for which the money is invested, in years, and A is the total balance after t years, including interest.

Answer 3

You need to learn how to do your own homework honey. Good luck.