FM-09-05sol

# FM-09-05sol - EXAM 2/FM SAMPLE QUESTIONS SOLUTIONS SOCIETY...

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EXAM 2/FM SAMPLE QUESTIONS SOLUTIONS 11/03/04 1 SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM FM FINANCIAL MATHEMATICS EXAM FM SAMPLE SOLUTIONS Copyright 2005 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study note are taken from past SOA/CAS examinations. FM-09-05 PRINTED IN U.S.A.

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EXAM 2/FM SAMPLE QUESTIONS SOLUTIONS 11/03/04 2 The following model solutions are presented for educational purposes. Alternate methods of solution are, of course, acceptable. 1. Solution: C Given the same principal invested for the same period of time yields the same accumulated value, the two measures of interest i (2) and δ must be equivalent, which means: δ e i = + 2 ) 2 ( ) 2 1 ( over one interest measurement period (a year in this case). Thus, e = + 2 ) 2 04 . 1 ( or e = + 2 ) 02 . 1 ( and 0396 . ) 02 . 1 ln( 2 ) 02 . 1 ln( 2 = = = or 3.96%. ---------------------------- 2. Solution: E Accumulated value end of 40 years = 100 [(1+i) 4 + (1+i) 8 + ……. .(1+i) 40 ]= 100 ((1+i) 4 )[1-((1+i) 4 ) 10 ]/[1 - (1+i) 4 ] (“Sum of finite geometric progression = 1 st term times [1 – (common ratio) raised to the number of terms] divided by [1 –common ratio]”) and accumulated value end of 20 years = 100 [(1+i) 4 + (1+i) 8 + ……. .(1+i) 20 ]=100 ((1+i) 4 )[1-((1+i) 4 ) 5 ]/[1 - (1+i) 4 ] But accumulated value end of 40 years = 5 times accumulated value end of 20 years Thus, 100 ((1+i) 4 )[1-((1+i) 4 ) 10 ]/[1 - (1+i) 4 ] = 5 {100 ((1+i) 4 )[1-((1+i) 4 ) 5 ]/[1 - (1+i) 4 ]} Or, for i > 0, 1-((1+i) 40 = 5 [1-((1+i) 20 ] or [1-((1+i) 40 ]/[1-((1+i) 20 ] = 5 But x 2 - y 2 = [x-y] [x+y], so [1-((1+i) 40 ]/[1-((1+i) 20 ]= [1+((1+i) 20 ] Thus, [1+((1+i) 20 ] = 5 or (1+i) 20 = 4. So X = Accumulated value at end of 40 years = 100 ((1+i) 4 )[1-((1+i) 4 ) 10 ]/[1 - (1+i) 4 ] =100 (4 1/5 )[1-((4 1/5 ) 10 ]/[1 – 4 1/5 ] = 6194.72 Alternate solution using annuity symbols : End of year 40, accumulated value = ) / ( 100 | 4 | 40 a s , and end of year 20 accumulated value = ) / ( 100 | 4 | 20 a s . Given the ratio of the values equals 5, then 5 = ] 1 ) 1 [( ] 1 ) 1 /[( ] 1 ) 1 [( ) / ( 20 20 40 | 20 | 40 + + = + + = i i i s s . Thus, (1+i) 20 = 4 and the accumulated value at the end of 40 years is 72 . 6194 ] 4 1 /[ ] 1 16 [ 100 ] ) 1 ( 1 /[ ] 1 ) 1 [( 100 ) / ( 100 5 / 1 4 40 | 4 | 40 = = + + = i i a s
EXAM 2/FM SAMPLE QUESTIONS SOLUTIONS 11/03/04 3 Note : if i = 0 the conditions of the question are not satisfied because then the accumulated value at the end of 40 years = 40 (100) = 4000, and the accumulated value at the end of 20 years = 20 (100) = 2000 and thus accumulated value at the end of 40 years is not 5 times the accumulated value at the end of 20 years.

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EXAM 2/FM SAMPLE QUESTIONS SOLUTIONS 11/03/04 4 3. Solution: C Eric’s interest (compound interest), last 6 months of the 8 th year: ) 2 ( ) 2 1 ( 100 15 i i + Mike’s interest (simple interest), last 6 months of the 8 th year: ) 2 ( 200 i . Thus, ) 2 ( 200 ) 2 ( ) 2 1 ( 100 15 i i i = + or 2 ) 2 1 ( 15 = + i , which means i/2 = .047294 or i = .094588 = 9.46% ------------------------------ 4. Solution: A The payment using the amortization method is 1627.45. The periodic interest is .10(10000) = 1000. Thus, deposits into the sinking fund are 1627.45-1000 = 627.45 Then, the amount in sinking fund at end of 10 years is 627.45 14 . | 10 s Using BA II Plus calculator keystrokes: 2 nd FV (to clear registers) 10 N, 14 I/Y, 627.45 PMT, CPT FV +/- - 10000= yields 2133.18 (Using BA 35 Solar keystrokes are AC/ON (to clear registers) 10 N 14 %i 627.45 PMT CPT FV +/- – 10000 =) ------------------------------- 5. Solution: E Key formulas for estimating dollar-weighted rate of return: Fund January 1 + deposits during year – withdrawals during year + interest = Fund December 31.
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FM-09-05sol - EXAM 2/FM SAMPLE QUESTIONS SOLUTIONS SOCIETY...

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