Revision as of 21:40, 26 November 2023 by Admin (Created page with "Two annuities have the same present value. The first annuity is a 12-year annuity-immediate paying $K per year. The second annuity is an 4-year annuity-immediate paying $2K per year. Both annuities are based on an annual effective interest rate of i, i > 0. Determine i. <ul class="mw-excansopts"> <li>0.1185</li> <li>0.1278 </li> <li>0.1312</li> <li> 0.136</li> <li> 0.142</li> </ul> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.htm...")
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ABy Admin
Nov 26'23

Exercise

Two annuities have the same present value. The first annuity is a 12-year annuity-immediate paying $K per year. The second annuity is an 4-year annuity-immediate paying $2K per year. Both annuities are based on an annual effective interest rate of i, i > 0.

Determine i.

  • 0.1185
  • 0.1278
  • 0.1312
  • 0.136
  • 0.142

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

ABy Admin
Nov 26'23

Solution: B

[math]K a_{\overline{12 \mid}}=2 K a_{\overline{4} \mid}[/math] so [math]a_{\overline{12 \mid}}=2 a_{\overline{4} \mid}[/math] and [math]\frac{1-(1+i)^{-12}}{i}=2 \frac{1-(1+i)^{-4}}{i}[/math]. Let [math]x=(1+i)^{-4}[/math]. Then [math]1-x^3=2(1-x)[/math] so [math]x^3-2 x+1=0[/math]. One root is clearly [math]x=1[/math] so [math](x-1)[/math] is a factor.

Thus [math](x-1)\left(x^2+x-1\right)=0[/math]. By the quadratic formula, [math]x=\frac{-1 \pm \sqrt{1+4}}{2}[/math]. But [math]x=(1+i)^{-4}[/math] lies between 0 and 1 . So [math](1+i)^{-4}=x=\frac{-1+\sqrt{5}}{2}=.618034[/math]. Thus [math]i=(.618034)^{-1 / 4}-1=.12784=12.784 \%[/math].

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

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