Revision as of 23:53, 18 November 2023 by Admin (Created page with "'''Solution: E''' Let j be the monthly rate and X be the level monthly payment. The principal repaid in the first payment is 1400 = X – 60,000j. The principal repaid in the second payment is 1414 = X –(60,000 – 1400)j. Substituting X = 1400 + 60,000j from the first equation gives 1414 = 1400 + 60,000j – 58,600j or 14 = 1400j and thus j = 0.01 and X = 2000. Let n be the number of payments. Then <math display = "block">60000 = 2000 a_{\overline{n}|0.01}</math> an...")
Exercise
ABy Admin
Nov 18'23
Answer
Solution: E
Let j be the monthly rate and X be the level monthly payment. The principal repaid in the first payment is 1400 = X – 60,000j. The principal repaid in the second payment is 1414 = X –(60,000 – 1400)j. Substituting X = 1400 + 60,000j from the first equation gives 1414 = 1400 + 60,000j – 58,600j or 14 = 1400j and thus j = 0.01 and X = 2000. Let n be the number of payments. Then
[[math]]60000 = 2000 a_{\overline{n}|0.01}[[/math]]
and the calculator (or algebra) gives n = 35.8455. The equation for the drop payment, P, is
[[math]]
60,000=2000a_{\overline{35}|0.01}+P\nu^{36}=58,817.16+0.698925P
[[/math]]
for P = 1962.
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.