Fall 2002 - Course 3 Solutions
Fall 2002 - Course 3 Solutions
Fall 2002 - Course 3 Solutions
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Question #17<br />
Answer: B<br />
10 30:<br />
40 10 30 10 40<br />
10<br />
10<br />
10 10<br />
d 10 30 id 10 40 ib1<br />
g<br />
10<br />
b10E30gb 10 E40gb1<br />
ig<br />
E = p p v = p v p v + i<br />
= +<br />
= b054733 . gb053667 . gb179085<br />
. g<br />
= 052604 .<br />
10<br />
The above is only one of many possible ways to evaluate 10 p 30 10 p 40 v , all of which should<br />
give 0.52604<br />
a = a − E a<br />
30: 4010 : 30: 40 10 30: 40 30+ 10:<br />
40+<br />
10<br />
= ba&& 30: 40 −1g −b0. 52604gba&&<br />
40:<br />
50 −1g<br />
b g b gb g<br />
= 13. 2068 − 0. 52604 114784 .<br />
= 71687 .<br />
Question #18<br />
Answer: A<br />
Equivalence Principle, where π is annual benefit premium, gives<br />
d<br />
b g<br />
1000 35 35<br />
π =<br />
d<br />
A + IA × π = a&&<br />
x π<br />
i<br />
1000A35<br />
1000·<br />
042898 .<br />
=<br />
a&&<br />
- IA ( 1199143 . - 616761 . )<br />
b g<br />
35 35<br />
We obtained a&&<br />
35 from<br />
a&&<br />
35<br />
i<br />
42898 .<br />
=<br />
582382 .<br />
= 7366 .<br />
1 A35<br />
1 0.<br />
42898<br />
= − = − = 11.<br />
99143<br />
d 0.<br />
047619
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