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# HW-Chap-4-2 - i)n−1 B(1 i)n−2 … B Subtract equation(2...

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Engineering Economy ENGR 300 Chapter 4 4-35 \$3,575 = \$375 + \$93.41 (P/A, i %, 45) ( P/A , i %, 45) = (\$3,575 − \$375)/\$93.41 = 34.258 From compound interest tables, i = 1.25% per month. For an \$800 down payment, unpaid balance is \$2775. P = \$2,775, n = 45 months, i = 1.25%, A = ? A = \$2,775 (A/P, 1.25%, 45)* = \$2,775 (0.0292) = \$81.03 Effective interest rate = (1 + i)12 − 1 = (1.0125)12 – 1 = 0.161 = 16.1% per year * Note that no interpolation is required as ( A/P, 1.25%, 45) = 1/( P/A , i %, 45) = 1/34.258 = 0.0292 4-38 The solution may follow the general approach of the end-of-year derivation in the book. (1) F = B (1 + i )n +.… + B (1 + i )1 Divide equation (1) by (1 + i): (2) F (1 + i )−1 = B (1 +
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Unformatted text preview: i )n−1 + B (1 + i )n−2 + … + B Subtract equation (2) from equation (1): (1) − (2) F − F (1 + i )−1 = B [(1 + i )n − 1] Multiply both sides by (1 + i): F (1 + i ) − F = B [(1 + i )n+1 − (1 + i)] So the equation is: F = B[(1 + i )n+1 − (1 + i)]/i Applied to the numerical values: F = 100/0.08 [(1 + 0.08)7 − (1.08)] = \$792.28 4-40 (a) (b) S = \$50 ( P/G , 10%, 4) = \$50 (4.378) = \$218.90 (c) T = \$30 ( A/G , 10%, 5) = \$30 (1.810) = \$54.30 4-48 P = \$100 (P/A, 10%, 4) + \$100 (P/G, 10%, 4) = \$100 (3.170 + 4.378) = \$754.80 Also: P = 4B (P/A, 10%, 4) − B (P/G, 10%, 4) Thus, 4B (3.170) − B (4.378) = \$754.80 B = \$754.80/8.30 = \$90.94...
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