HW-Chap-6-1 - the computations. 2. Quarterly interest...

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Engineering Economy GE 272 Chapter 6 6-1 C = $15 + $15 (A/G, 10%, 4) = $15 + $15 (1.381) = $35.72 6-8 A = F[(er −1)/(em − 1)] = $5 × 106 [(e0.15 − 1)/(e(0.15)(40) − 1)] = $5 × 106 [0.161834/402.42879] = $2,011 6-16 Equivalent total taxes if all were paid on April 1st: = $425 + $425 (F/P, ¾%, 4) = $425 + $425 (1.030) = $862.75 Equivalent uniform monthly payment: = $862.75 (A/P, ¾%, 12) = $862.75 (0.0800) = $69.02 Therefore the monthly deposit is $69.02. Amount to deposit September 1: = Future worth of 5 months deposits (May − Sept) = $69.02 (F/A, ¾%, 5) = $69.02 (5.075) = $350.28 Notes: 1. The fact that the tax payments are for the fiscal year, July 1 through June 30, does not affect
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Unformatted text preview: the computations. 2. Quarterly interest payments to the savings account could have an impact on the solution, but they do not in this problem. 3. The solution may be verified by computing the amount in the savings account on Dec. 1 just before making the payment (about $560.03) and the amount on April 1 after making that payment ($0). 6-26 Alternative A EUAC = A = [$2,000 + $500 (P/F, 12%, 1)] (A/P, 12%, 5) = [$2,000 + $500 (0.8929)] (0.2774) = $678.65 Alternative B EUAC = A = $3,000 (F/P, 12%, 1) (A/F, 12%, 5) = $3,000 (1.120) (0.1574) = $528.86 To minimize EUAC, select B....
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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 .

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