SOLUTION: TIME MACHINE PROBLEM

SO = $4750

N = 10 YEARS

R = 4 %

TABLE 11 - I (10, 4): 1.48

1 1.48 x $4750 = $7030

2 1.48 x $7030 = $10,404

3 1.48 x $10,404 = $15,397

4 1.48 x $15,397 = $22,788

5 1.48 x $22,788 = $33,727

6 1.48 x $33,727 = $49,916

7 1.48 x $49,916 = $73,875

The inventor must make 6 trips into the past (10 years), deposit his money (initially $4750), return to the future (today), withdraw the new principal, and return to the past….