A tractor has a first cost of $40,000 and a salvage value of $15,000 in year...

  

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A tractor has a first cost of $40,000 and a salvage value of $15,000 in year 10. The MARR is 12% per year. An identical tractor can be rented for $32,000 per year. Annual operating cost is same for both options. If x is the minimum number of years the tractor must be used in order to justify its purchase, the relation to find x is represented by: O-40.000(A/P, 12%, 10) - 15,000x + 12,000(A/F, 12%, 10) = -32,000x O-40,000(A/P, 12%. 10) - 15,000x + 12,000(A/F, 12 %, 10) = 32,000x O-32,000(A/P, 10%, 12) - 12,000x + 15,000(A/F, 10%, 12) = 40,000x O-32,000(A/P, 10%, 12) - 12,000x+15,000(A/F, 10%, 12) = -40,000x O-40,000(A/P, 12%. 10 ) + 15,000 (A/F. 12%, 10) = -32,000x O-40,000(A/P, 12%, 10)+15,000(A/F, 12%, 10) = 32,000x O None of them A tractor has a first cost of $40,000 and a salvage value of $15,000 in year 10. The MARR is 12% per year. An identical tractor can be rented for $32,000 per year. Annual operating cost is same for both options. If x is the minimum number of years the tractor must be used in order to justify its purchase, the relation to find x is represented by: O-40.000(A/P, 12%, 10) - 15,000x + 12,000(A/F, 12%, 10) = -32,000x O-40,000(A/P, 12%. 10) - 15,000x + 12,000(A/F, 12 %, 10) = 32,000x O-32,000(A/P, 10%, 12) - 12,000x + 15,000(A/F, 10%, 12) = 40,000x O-32,000(A/P, 10%, 12) - 12,000x+15,000(A/F, 10%, 12) = -40,000x O-40,000(A/P, 12%. 10 ) + 15,000 (A/F. 12%, 10) = -32,000x O-40,000(A/P, 12%, 10)+15,000(A/F, 12%, 10) = 32,000x O None of them

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