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Solutions to Extra Problems 2a

# Solutions to Extra Problems 2a - a 1 05 b& 1 0495 2 ± 2...

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ACTSC 231 Solutions to Problem Set 1a 1. 1000 e R 2 0 s ds +500 e R 2 : 5 s ds = 1000 e R 2 0 3+2 s 50 ds +500 e R 2 : 5 3+2 s 50 ds = 1000 e 3 s + s 2 50 2 0 + 500 e 3 s + s 2 50 2 : 5 = 1000 e 3(2)+2 2 3(0) & 0 2 50 + 500 e 3(2)+2 2 3( : 5) ( : 5) 2 50 = 1000 e 1 = 5 + 500 e 33 = 200 = 1811 : 099 3 2. Find t such that 300 1 + : 06 2 ± 2( t 1) = 3 n 200 1 + : 04 4 ± 4 t o = ) (1 : 03) 2( t 1) = 2 (1 : 01) 4 t = ) 2 ( t 1) ln 1 : 03 = ln 2 + 4 t ln 1 : 01 = ) t [2 ln 1 : 03 4 ln 1 : 01] = ln 2 + 2 ln 1 : 03 = ) t = ln 2+2 ln 1 : 03 2 ln 1 : 03 4 ln 1 : 01 = 38 : 944 597 years or 38 years and 345 days. 3. The annual accumulation factors are:
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Unformatted text preview: a. 1 : 05 b. & 1 + : 0495 2 ± 2 = 1 : 050112563 c. (1 & : 0475) & 1 = 1 : 049868766 d. e : 0485 = 1 : 049695372 e. & 1 & : 048 12 ± & 12 = 1 : 04927165 Borrowers prefer the smallest accumulation factor. best: e,d,c,a,b worst 1...
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