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# HW3SOL - (P/A,0.5,n = 48.0000 From 0.5 interest tables n is...

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Homework 3 - Solution 1. Problem 3.4 (a) 2 (b) 6 (c) 1 2. Problem 3.6 (a) r = 4% (b) r = 0.04*2 = 8% (c) r = 0.04*4 = 16% 3. Problem 3.14 (a) Years (b) Semiannual (c) Quarter 4. Problem 3.19 F = 160,000(F/P,4%,14) = 160,000(1.7317) = \$270,072 5. Problem 3.21 F = 2.3(F/P,5%,20) (millions) = 2.3(2.6533) = \$6,102,590 6. Problem 3.29 Monthly difference = 5296 - 3443 = \$1853 F = 1853(F/A,0.5%,480) = 1853(1991.19) = \$3,690,231 7. Problem 3.33 i/yr = (1 + 0.05)2 -1 = 10.25% A = 9,000,000(A/P,10.25%,10) = 9,000,000(0.16450) = \$1,480,476 1

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8. Problem 3.37 Solve for F at the end of year 4 and then convert to A: F = 835,000(F/P,1%,24) + 1,100,000 = 835,000(1.2697) + 1,100,000 = \$2,160,200 A = 2,160,200(A/F,1%,48) = 2,160,200(0.01633) = \$35,276 9. Problem 3.39 840,000 = (30 + 5)(500)(P/A,0.5%,n) 840,000 = 17,500(P/A,0.5%,n)
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Unformatted text preview: (P/A,0.5%,n) = 48.0000 From 0.5% interest tables, n is slightly greater than 55 Therefore, n = 56 months 10. Problem 3.48 Convert all cash ﬂows into present worth and then amortize: i/yr = (1 + 0.0112 -1 = 12.68% Pg = 90,000[1 - (1.03/1.1268)10]/(0.1268 - 0.03) = \$551,073 P = 800,000 + Pg = 800,000 + 551,073 = \$1,351,073 A = 1,351,073(A/P,12.68%,10) = 1,351,073(0.18194) = \$245,813 11. Problem 3.51 Find A per six months and then divide by 6: A/semi = 201,500(A/P,6%,6) = 201,500(0.20336) = \$40,977 A/mo = 40,977/6 = \$6829.51 2 12. Problem 3.54 Move cash ﬂow to end of interest period and then ﬁnd P: Cost/quarter = 100,000(0.019)(3) = \$5700 P = 5700(P/A,3%,12) = 5700(9.9540) = \$56,737 3...
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HW3SOL - (P/A,0.5,n = 48.0000 From 0.5 interest tables n is...

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