Answers of Chapter 1 2 and 3 - 3.30 15,000 = 2000 +...

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0 1 2 3 4 5 6 7 8 F= ? $9000 $3000 $10,000 i = 10% Chapter 1 Problems: 20, 29,39, 46 1.20 F = 240,000 + 240,000(0.10)(3) = $312,000 1.29 (a) F = ?; i = 7%; n = 10; A = $2000; P = $9000 (b) A = ?; i = 11%; n = 20; P = $14,000; F = 0 (c) P = ?; i = 8%; n = 15; A = $1000; F = $800 1.39 Chapter 2 Problems: 1, 9, 22, 30, 40 2.1 1. (F/P,8%25) = 6.8485; 2. (P/A,3%,8) = 7.0197; 3. (P/G,9%,20) = 61.7770; 4. (F/A,15%,18) = 75.8364; 5. (A/P,30%,15) = 0.30598
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2.9 2. F = 1,700,000(F/P,18%,1) = 1,700,000(1.18) = $2,006,000 2.22 3. P = 2000(P/A,8%,35) = 2000(11.6546) = $23,309 2.30 300,000 = A + 10,000(A/G,10%,5) 300,000 = A + 10,000(1.8101) A = $281,899 2.40 32. For g = i, P = 60,000(0.1)[15/(1 + 0.04)] = $86,538 Chapter 3 Problems: 20, 30, 36, 46, 48 3.20 F = 9000(F/P,8%,11) + 600(F/A,8%,11) + 100(F/A,8%,5) = 9000(2.3316) + 600(16.6455) + 100(5.8666) = $31,558
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Unformatted text preview: 3.30 15,000 = 2000 + 2000(P/A,15%,3) + 1000(P/A,15%,3)(P/F,15%,3) + x(P/F,15%,7) 15,000 = 2000 + 2000(2.2832) + 1000(2.2832)(0.6575) + x(0.3759) x = $18,442 3.36 P for maintenance = [11,500(F/A,10%,2) + 11,500(P/A,10%,8) + 1000(P/G,10%,8)](P/F,10%,2) = [11,500(2.10) + 11,500(5.3349) + 1000(16.0287)](0.8264) = $83,904 P for accidents = 250,000(P/A,10%,10) = 250,000(6.1446) = $1,536,150 Total savings = 83,904 + 1,536,150 = $1,620,054 Build overpass 3.46 Find P in year –1 and then move to year 0. P (yr –1) = 15,000{[1 – (1 + 0.10) 5 /(1 + 0.16) 5 ]/(0.16 – 0.10)} = 15,000(3.8869) = $58,304 P = 58,304(F/P,16%,1) = 58,304(1.16) = $67,632 3.48 P = [2000(P/A,12%,6) – 200(P/G,12%,6)](F/P,12%,1) = [2000(4.1114) – 200(8.9302](1.12) = $7209.1...
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This note was uploaded on 12/06/2009 for the course ELECTRICAL GENG 360 taught by Professor Dr.tarekelmekkawy during the Fall '09 term at Qatar University.

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Answers of Chapter 1 2 and 3 - 3.30 15,000 = 2000 +...

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