A BMW costs $35 000, its salvage value declines by declining-balance depreciation of 10% per year, and its maintenance costs are $100 in the first year and go up by $500 a year. Mary and Tom both like to drive BMWs. Mary trades her car in for a new model every year, whereas Tom keeps his until it reaches its economic life, then trades it in. Mary and Tom both have MARRs of 5%. How much more does Mary pay for her car per year, on average, than Tom?
A)They each pay the same.
B)$569
C)$615
D)$769
E)$869

Question

Quizplus · Accepted Answer

The correct answer is C.The present worth of Mary's costs is:PW = -35,000 + 100(P/A,5%,1) + 100(P/G,5%,1)(P/F,5%,1) + 100(P/G,5%,1)^2(P/F,5%,2) + ...= -35,000 + 100(12.4622) + 100(11.6523)(0.9524) + 100(11.6523)^2(0.9070) + ...= -35,000 + 1246.22 + 1091.47 + 992.36 + ...= -35,000 + 15,792.31= $15,792.31The present worth of Tom's costs is:PW = -35,000 + 100(P/A,5%,n) + 100(P/G,5%,n)(P/F,5%,n)= -35,000 + 100(12.4622) + 100(11.6523)(0.9524)^n= -35,000 + 1246.22 + 1091.47(0.9524)^nFor n = 1, PW = $15,792.31For n = 2, PW = $14,923.78For n = 3, PW = $14,055.26For n = 4, PW = $13,186.74For n = 5, PW = $12,318.22For n = 6, PW = $11,449.70For n = 7, PW = $10,581.18For n = 8, PW = $9,712.66For n = 9, PW = $8,844.14For n = 10, PW = $7,975.62The economic life of the car is 10 years.The average annual cost for Mary is:AAC = PW(A/P,5%,10)= 15,792.31(0.1358)= $2,139.92The average annual cost for Tom is:AAC = PW(A/P,5%,10)= 7,975.62(0.1358)= $1,084.96The difference in average annual cost is:AAC = 2,139.92 - 1,084.96= $1,054.96Therefore, Mary pays $1,054.96 more for her car per year, on average, than Tom.

A BMW Costs $35 000, Its Salvage Value Declines by Declining-Balance