Revision as of 10:40, 18 November 2023 by Admin (Created page with "'''Solution: E''' Let P be the annual interest paid. The present value of John’s payments is <math>Pa_{\overline{X}|0.05}</math>. The present value of Karen’s payments is <math display = "block"> P(1.05)^{-X}\,a_{\underline{\infty}|0.05}=P(1.05)^{-X}\,/\,0.05. </math> Then, <math display = "block"> \begin{array}{l}{{P(1.05)^{-X}/{0.05=1.59P a_{\overline{x}|0.05}}}}\\ {{0.05=1.59-1.05^{-X}}}\\ {{\mathrm{~ln~39=2.59(1.05)^{-X}}}}\\ {{X=10.}}\end{array} </math>...")
Exercise
ABy Admin
Nov 18'23
Answer
Solution: E
Let P be the annual interest paid. The present value of John’s payments is [math]Pa_{\overline{X}|0.05}[/math]. The present value of Karen’s payments is
[[math]]
P(1.05)^{-X}\,a_{\underline{\infty}|0.05}=P(1.05)^{-X}\,/\,0.05.
[[/math]]
Then,
[[math]]
\begin{array}{l}{{P(1.05)^{-X}/{0.05=1.59P a_{\overline{x}|0.05}}}}\\ {{0.05=1.59-1.05^{-X}}}\\ {{\mathrm{~ln~39=2.59(1.05)^{-X}}}}\\ {{X=10.}}\end{array}
[[/math]]