Exercise


Nov 26'23

Answer

Solution: B

We work with rate [math]i=.05[/math] for 6 months. Let [math]\mathrm{X}[/math] be the initial deposit. The Jan 1, 2005 deposit is [math]X(1.1025)^{10}[/math]. Note that [math]1.05^2=1.1025[/math]. The Jan 1 deposits accumulate, measuring from 2005 down, to

[[math]] A 1=X(1.1025)^{10}(1.05)^2+\cdots+X(1.1025)^0(1.05)^{22}=11 X(1.1025)^{11} [[/math]]

The July 1 deposits accumulate, measuring from 2005 down, to

[[math]] A 2=X(1.1025)^{11}(1.05)^1+X(1.1025)^{10}(1.05)^3+\cdots+X(1.1025)^1(1.05)^{21}=11 X(1.1025)^{11.5} [[/math]]

Thus [math]11000=11 X\left(1.1025^{11}+1.1025^{11.5}\right)[/math] so [math]X=1000 /\left(1.1025^{11}+1.1025^{11.5}\right)=[/math] 166.7560 .

References

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

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