Revision as of 21:59, 26 November 2023 by Admin (Created page with "'''Solution: C''' Note that 6 years <math>=6 \times 12</math> months <math>=72</math> months; 4 years <math>=48</math> months; 2 years <math>=24</math> months. After 6 years, the first two years of payments have grown to <math>50 \ddot{s}_{\overline{24} \mid j}(1+i)^4</math>. After 6 years, the next two years of payments have grown to <math>100 \ddot{s}_{\overline{24} \mid j}(1+i)^2</math>. After 6 years, the last two years of payments have grown to <math>150 \ddot{s}_...")
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Exercise


Nov 26'23

Answer

Solution: C

Note that 6 years [math]=6 \times 12[/math] months [math]=72[/math] months; 4 years [math]=48[/math] months; 2 years [math]=24[/math] months.

After 6 years, the first two years of payments have grown to [math]50 \ddot{s}_{\overline{24} \mid j}(1+i)^4[/math]. After 6 years, the next two years of payments have grown to [math]100 \ddot{s}_{\overline{24} \mid j}(1+i)^2[/math]. After 6 years, the last two years of payments have grown to [math]150 \ddot{s}_{\overline{24} \mid j}(1+i)^0[/math]. Let [math]T[/math] be the total after 6 years. Then the total after 7 years is [math]T(1+i)[/math]. Thus

[[math]] 10000=(1+i)\left[50 \ddot{s}_{\overline{24} \mid j}(1+i)^4+100 \ddot{s}_{\overline{24} \mid j}(1+i)^2+150 \ddot{s}_{\overline{24} \mid j}(1+i)^0\right] . [[/math]]

Divide by 50 and factor to get

[[math]] 200=\ddot{s}_{\overline{24} \mid j}(1+i)\left[(1+i)^4+2(1+i)^2+3\right] . [[/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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