# HW1SOL - 1 7. Problem 1.29 A = \$225,000; n = 3; i = 15% per...

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Homework 1 - Solution 1. Problem 1.14 (a) Payment = 1,600,000(1.10)(1.10) = \$1,936,000 (b) Interest = total amt paid - principal = 1,936,000- 1,600,000 = \$336,000 2. Problem 1.16 F = P(1 + i) 82 , 000 , 000 = x (1 + 0 . 08) x = \$75,925,926 3. Problem 1.20 F = P + P × n × i 850,000 = P + P(4)(0.10) 1.4P = 850,000 P = \$607,143 4. Problem 1.22 (a) F = P + P × n × i 2,800,000 = 2,000,000 + 2,000,000(4)(i) Interest = 10% per year (b) F = P(1 + i) (1 + i) (1 + i) (1 + i) 2,800,000 = 2,000,000(1 + i)4 (1 + i)4 = 1.4000 log(1 + i)4 = log1.400 4log(1 + i) = 0.146 log(1 + i) = 0.0365 (1 + i) = 100.0365 (1 + i) = 1.0877 i = 8.77% 5. Problem 1.23 F = P + P × n × i 3P = P + P(n)(0.20) n = 10 years 6. Problem 1.25 (a) Loan A: interest/year = 500,000(0.10) = \$50,000 Loan B: interest/year = 500,000(0.10) = \$50,000 The same amount of interest will be paid on each loan (b) There was no diﬀerence paid between the two loans

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Unformatted text preview: 1 7. Problem 1.29 A = \$225,000; n = 3; i = 15% per year; F = ? 8. Problem 1.36 (a) F = ?; i = 8%; n = 10; A = \$2000; P = \$10,000 (b) A = ?; i = 12%; n = 30; P = \$16,000; F = 0 (b) P = ?; i = 9%; n = 15; A = \$1000; F = \$700 9. For the compound interest case, ﬁnd the doubling period as a function of interest rate i and the doubling interest rate as a function of number of periods n . F = P (1 + i ) n Here F = 2 P ⇒ 2 P = P (1 + i ) n ⇒ 2 = (1 + i ) n Doubling interest rate: i = 2 1 n-1 Doubling period: n = log 2 log(1+ i ) 10. Problem 1.39 Assuming down is negative: down arrow of \$10,000 in year 0; up arrows in the amount of \$3000 in years 1 thru 5; i = 10% per year; arrow in year 5 identiﬁed as ”F = ?” 2...
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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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HW1SOL - 1 7. Problem 1.29 A = \$225,000; n = 3; i = 15% per...

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