Revision as of 00:31, 27 November 2023 by Admin (Created page with "'''Solution: D''' SOLUTION 1: Interest rate is .04 per six months so <math>(1+j)^6=1.04</math>. Thus <math>j=.006558</math>. Then <math>220000=X a_{\overline{300} \mid j}</math> so <math>X=\frac{220000}{a_{\overline{300} \mid j}}=\frac{220000}{131.0276}=1679.04</math>. SOLUTION 2: Interest rate is <math>i=.04</math> for a period of 6 months. There are <math>25(2)=50</math> periods. <math>a_{\overline{50} |}^{(6)}</math> means <math>\mathrm{PV}</math> of payments of...")
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Exercise


ABy Admin
Nov 27'23

Answer

Solution: D

SOLUTION 1:

Interest rate is .04 per six months so [math](1+j)^6=1.04[/math]. Thus [math]j=.006558[/math]. Then [math]220000=X a_{\overline{300} \mid j}[/math] so [math]X=\frac{220000}{a_{\overline{300} \mid j}}=\frac{220000}{131.0276}=1679.04[/math].

SOLUTION 2:

Interest rate is [math]i=.04[/math] for a period of 6 months. There are [math]25(2)=50[/math] periods. [math]a_{\overline{50} |}^{(6)}[/math] means [math]\mathrm{PV}[/math] of payments of [math]1 / 6[/math] per month. [math]6 a_{\overline{50}|}^{(6)}[/math] means PV of payments of 1 per month. [math]X a_{\overline{50}|}^{(6)}[/math] means PV of payments of [math]X[/math] per month. Then [math]220000=6 X a_{\overline{50} \mid}^{(6)}[/math] so [math]X=220000 / 6 a_{\overline{50}|}^{(6)}[/math]. Now

[[math]]6 a_{\overline{50} \mid}^{(6)}=6 \frac{1-v^{50}}{i^{(6)}}=6 \frac{1-1.04^{-50}}{6\left[1.04^{1 / 6}-1\right]}=.859287385 / .0006558197=131.0249[[/math]]

so [math]X=220000 / 131.0249=[/math] 1679.07.

References

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

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