Revision as of 03:01, 20 January 2024 by Admin (Created page with "'''Answer: E''' In the final year: <math>\left({ }_{24} V+P\right)(1+\mathrm{i})=b_{25}\left(q_{68}\right)+1\left(p_{68}\right)</math> Since <math>b_{25}=1</math>, this reduces to <math>\left({ }_{24} V+P\right)(1+i)=1 \Rightarrow(0.6+P)(1.04)=1 \Rightarrow P=0.36154</math> Looking back to the <math>12^{\text {th }}</math> year: <math>\left({ }_{11} V+P\right)(1+i)=b_{12}\left(q_{55}\right)+{ }_{12} V\left(p_{55}\right)</math> <math>\Rightarrow(5.36154)(1.04)=14(0.15...")
Exercise
ABy Admin
Jan 20'24
Answer
Answer: E
In the final year: [math]\left({ }_{24} V+P\right)(1+\mathrm{i})=b_{25}\left(q_{68}\right)+1\left(p_{68}\right)[/math]
Since [math]b_{25}=1[/math], this reduces to [math]\left({ }_{24} V+P\right)(1+i)=1 \Rightarrow(0.6+P)(1.04)=1 \Rightarrow P=0.36154[/math]
Looking back to the [math]12^{\text {th }}[/math] year: [math]\left({ }_{11} V+P\right)(1+i)=b_{12}\left(q_{55}\right)+{ }_{12} V\left(p_{55}\right)[/math]
[math]\Rightarrow(5.36154)(1.04)=14(0.15)+{ }_{12} V(0.85) \Rightarrow{ }_{12} V=4.089[/math]