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A tractor has a first cost of $40,000, a monthly op- erating cost of $1500, and a salvage value of $12,000 in 10 years
A tractor has a first cost of $40,000, a monthly op- erating cost of $1500, and a salvage value of $12,000 in 10 years. The MARR is 12% per year. An identical tractor can be rented for $3200 per month (operating cost not included). If n is the minimum number of months per year the tractor must be used in order to justify its purchase, the relation to find n is represented by: (a) -40,000(A/P,1%,10) - 1500n + 12,000(A/F,1% 10) = 3200n (b) -40,000(A/P,12%,10) - 1500n + 12,000(A/F,12%,10) = 3200n (c) -40,000(A/P,1%,120) - 1500n + 12,000(A/F,1%,120) = 3200n (d) -40,000(A/P,11.4%,10) - 1500n + 12,000(A/F,11.4%,10) = 3200n
3) (3pts) 13.58: No change (Determine the number of month for breakeven) *Note: Any answer choice shown in the textbook (multiple choices of 13.58) is NOT correct.
Expert Solution
Solution:
Present worth of operating cost= 1,500*A
=1,500*5.65
Present worth of operating cost =8,475
where,
A=(1/1.12+(1/1.12)^2+...+(1.12)^10)
Present worth of salvage value = 12,000*(1/1.12)^10
=12,000*0.32
Present worth of salvage value =3,840
Actual Present worth = 8,470+3,840+40,000
=52,310
Actual Present worth must be equal to the income of rent over a period of n months
Therefore,
52,310 = 3,200 n
n = 52310 / 3200
= 16.6months or
= 17 months to justify the purchase
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