Ch2Solutions

Ch2Solutions - Chapter 2 2.1 (a) (F/P,10%,20) = 6.7275 (b)...

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Chapter 2 2.1 (a) (F/P,10%,20) = 6.7275 (b) (A/F,4%,8) = 0.10853 (c) (P/A,8%,20) = 9.8181 (d) (A/P,20%,28) = 0.20122 (e) (F/A,30%,15) = 167.2863 2.2 F = 180,000(F/P,10%,3) = 180,000(1.3310) = $239,580 2.3 F = 2,700,000(F/P,20%,3) = 2,700,000(1.7280) = $4,665,600 2.4 F = 20(649)(F/P,8%,2) = 12,980(1.1664) = $15,139.87 2.5 The value of the system is the interest saved on $20 million for 2 years. F = 20,000,000(F/P,15%,2) = 20,000,000(1.3225) = $26,450,000 Interest = 26,450,000 - 20,000,000 = $6,450,000 2.6 P = 2,100,000(P/F,15%,2) = 2,100,000(0.7561) = $1,587,810 2.7 P = 40,000(P/F,12%,4) = 40,000(0.6355) = $25,420 2.8 P = 85,000(P/F,18%,5) = 85,000(0.4371) = $37,154 2.9 P = 95,000,000(P/F,12%,3) = 95,000,000(0.7118) = $67,621,000 2- 1
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2.10 F = 175,000(F/P,10%,6) = 175,000(1.7716) = $310,030 2.11 F = 150,000(F/P,8%,8) = 150,000(1.8509) = $277,635 2.12 P = 7000(P/F,10%,2) + 9000(P/F,10%,3) + 5000(P/F,10%,5) = 7000(0.8264) + 9000(0.7513) + 5000(0.6209) = $15,651 2.13 P = 600,000(0.10)(P/F,10%,2) + 1,350,000(0.10)(P/F,10%,5) = 60,000(0.8264) + 135,000(0.6209) = $133,406 2.14 P = 8,000,000(P/A,10%,5) = 8,000,000(3.7908) = $30,326,400 2.15 A = 10,000,000(A/P,10%,10) = 10,000,000(0.16275) = $1,627,500 2.16 A = 140,000(4000)(A/P,8%,3) = 560,000,000(0.38803) = $217,296,800 2.17 P = 1,500,000(P/A,8%,4) = 1,500,000(3.3121) = $4,968,150 2.18 A = 2,550,000(A/P,14%,6) = 2,550,000(0.25716) = $655,758 2.19 P = 280,000(P/A,18%,8) = 280,000(4.0776) = $1,141,728 2.20 A = 3,500,000(A/P,20%,5) = 3,500,000(0.33438) = $1,170,330 2- 2
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2.21 A = 5000(7)(A/P,10%,10) = 35,000(0.16275) = $5696.25 2.22 F = 70,000(F/P,10%,5) + 90,000(F/P,10%,3) = 70,000(1.6105) + 90,000(1.3310) = $232,525 2.23 F = (458-360)(0.90)(20,000)(F/A,10%,5) = 1,764,000(6.1051) = $10,769,396 2.24 100,000(F/P,9%,3) + 75,000(F/P,9%,2) + x(F/P,9%,1) = 290,000 100,000(1.2950) + 75,000(1.1881) + x(1.0900) = 290,000 1.09x = 71.392.50 x = $65,498 2.25 P = 225,000(P/A,15%,3) = 225,000(2.2832) = $513,720 2.26 F = P(F/P,12%,n) 4P = P(F/P,12%,n) (F/P,12%,n) = 4.000 From 12% interest tables, n is between 12 and 13 years Therefore, n = 13 years 2.27 1,200,000 = 400,000(F/P,10%,n) + 50,000(F/A,10%,n) Solve for n by trial and error: Try n = 5: 1,200,000 = 400,000(F/P,10%,5) + 50,000(F/A,10%,5) 1,200,000 = 400,000(1.6105) + 50,000(6.1051) 1,200,000 = 949,455 n too low Try n = 8: 1,200,000 = 400,000(2.1436) + 50,000(11.4359) 1,200,000 = 1,429,235 n too high By interpolation, n is between 6 and 7 Therefore, n = 7 years 2- 3
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2.28 2,000,000 (F/P,7%,n) = 158,000(F/A,7%,n) (thousands) Solve for n by trial and error: Try n = 30: 2,000,000(F/P,7%,30) = 158,000(F/A,7%,30) 2,000,000(7.6123) = 158,000(94.4608) 15,224,600 = 14,924,806 n too low Try n = 32: 2,000,000(8.7153) = 158,000(110.2182) 17,430,600 = 17,414,476 n too low Try n = 33: 2,000,000(9.3253) = 158,000(118.9334) 18,650,600 = 18,791,447 n too high By interpolation, n is between 32 and 33
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This note was uploaded on 03/24/2011 for the course ISEN 302 taught by Professor Ko during the Spring '08 term at Texas A&M.

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Ch2Solutions - Chapter 2 2.1 (a) (F/P,10%,20) = 6.7275 (b)...

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