HW2SOL - Homework 2 - Solution 1. Problem 2.5 The value of...

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Homework 2 - Solution 1. Problem 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. Problem 2.9 P = 95,000,000(P/F,12%,3) = 95,000,000(0.7118) = $67,621,000 3. Problem 2.16 A = 140,000(4000)(A/P,8%,3) = 560,000,000(0.38803) = $217,296,800 4. Problem 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 5. Problem 2.25 P = 225,000(P/A,15%,3) = 225,000(2.2832) = $513,720 1
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6. Problem 2.29 P = 20,000(P/A,10%,5) + 2000(P/G,10%,5) = 20,000(3.7908) + 2000(6.8618) = $89,539.60 7. Problem 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 8. Problem 2.40 First find Pg and then convert to F Pg = 8000[10/(1 + 0.10)] = $72,727 F = 72,727(F/P,10%,10) = 72,727(2.5937) = $188,632 9. Problem 2.43 First find Pg and then convert to F: Pg = 2000[1 - (1.15/1.10)7]/(0.10 - 0.15)
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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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HW2SOL - Homework 2 - Solution 1. Problem 2.5 The value of...

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