ngineer deposits $8,000 in year 4, $8,500 in year 5, and amounts increasing by $500 per year through year 10. At an interest rate of 1% per year, the equation to find the present worth in year 0 is: (a) 18000(P/A i, 7) + 500(P/G i, 7)](P/F i, 3) (b) [8000(P/A i, 7) + 500(P/G i, 7)] (P/F i, 4) (c) [8000(P/A i, 6) + 500(P/G i, 6)](P/F i, 3) (d) [8000(P/A i, 6) + 500(P/G i, 6)] (P/F i, 4)
ngineer deposits $8,000 in year 4, $8,500 in year 5, and amounts increasing by $500 per year through year 10. At an interest rate of 1% per year, the equation to find the present worth in year 0 is: (a) 18000(P/A i, 7) + 500(P/G i, 7)](P/F i, 3) (b) [8000(P/A i, 7) + 500(P/G i, 7)] (P/F i, 4) (c) [8000(P/A i, 6) + 500(P/G i, 6)](P/F i, 3) (d) [8000(P/A i, 6) + 500(P/G i, 6)] (P/F i, 4)
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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An engineer deposits $8,000 in year 4, $8,500 in year 5, and amounts increasing by $500 per year through year 10. At an interest rate of 1% per year, the equation to find the present worth in year 0 is:
(a) 18000(P/A i, 7) + 500(P/G i, 7)](P/F i, 3)
(b) [8000(P/A i, 7) + 500(P/G i, 7)] (P/F i, 4)
(c) [8000(P/A i, 6) + 500(P/G i, 6)](P/F i, 3)
(d) [8000(P/A i, 6) + 500(P/G i, 6)] (P/F i, 4)
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