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Ch2Solutions

# Ch2Solutions - Chapter 2 2.1(a(F/P,10,20 = 6.7275(b(A/F,4,8...

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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
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 Therefore, n = 33 years 2.29 P = 20,000(P/A,10%,5) + 2000(P/G,10%,5) = 20,000(3.7908) + 2000(6.8618) = \$89,539.60 2.30 A = 100,000 + 10,000(A/G,10%,5) = 100,000 + 10,000(1.8101) = \$118,101 2.31 P = 0.50(P/A,10%,5) + 0.10(P/G,10%,5) = 0.50(3.7908) + 0.10(6.8618) = \$2.58 2.32 (a) Income = 390,000 – 2(15,000) = \$360,000 (b) A = 390,000 - 15,000(A/G,10%,5) = 390,000 - 15,000(1.8101) = \$362,848.50 2.33 475,000 = 25,000(P/A,10%,8) + G(P/G,10%,8) 475,000 = 25,000(5.3349) + G(16.0287) 16.0287G = 341,627.50 G = \$21,313.49
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Ch2Solutions - Chapter 2 2.1(a(F/P,10,20 = 6.7275(b(A/F,4,8...

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