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Unformatted text preview: i )n1 + B (1 + i )n2 + + 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)] = $792.28 440 (a) (b) S = $50 ( P/G , 10%, 4) = $50 (4.378) = $218.90 (c) T = $30 ( A/G , 10%, 5) = $30 (1.810) = $54.30 448 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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This note was uploaded on 04/18/2011 for the course ENGINEER 300 taught by Professor Kashayar during the Spring '11 term at California State University Los Angeles .
 Spring '11
 Kashayar

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