Revision as of 17:42, 26 November 2023 by Admin (Created page with "'''Solution: D''' Let <math>j=</math> semi annual interest. <math>2600=1000(1+j)^4+1500(1+j)^2</math> This is a quadratic in <math>x=(1+j)^2</math>, which simplifies to <math>10 x^2+15 x-26=0</math> so <math display = "block">x=\frac{-15 \pm \sqrt{15^2-4(10)(-26)}}{2(10)}=1.028342</math> Thus <math>(1+j)^2=1.028342</math> so <math>j=1.028342^5-1=0.01407199=\frac{i^{(2)}}{2}</math>. Finally <math>i^{(2)}=2(.01407199)=.02814=2.81 \%</math>. '''References''' {{cite web...")
Exercise
ABy Admin
Nov 26'23
Answer
Solution: D
Let [math]j=[/math] semi annual interest. [math]2600=1000(1+j)^4+1500(1+j)^2[/math] This is a quadratic in [math]x=(1+j)^2[/math], which simplifies to [math]10 x^2+15 x-26=0[/math] so
[[math]]x=\frac{-15 \pm \sqrt{15^2-4(10)(-26)}}{2(10)}=1.028342[[/math]]
Thus [math](1+j)^2=1.028342[/math] so [math]j=1.028342^5-1=0.01407199=\frac{i^{(2)}}{2}[/math]. Finally [math]i^{(2)}=2(.01407199)=.02814=2.81 \%[/math].
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.