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Engineering Economics Flashcards

7 cards from real EIT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Engineering Economics flashcards as text
  1. An investment of $10,000 grows to $14,802 in 5 years with continuous compounding. What is the nominal annual interest rate?

    Answer: 7.0%

    Using FV = PV·e^(rn): 14802 = 10000·e^(5r), so r = ln(1.4802)/5 ≈ 0.07 = 7%.

  2. A perpetuity pays $500 per year forever. If the interest rate is 8%, what is the present value?

    Answer: $6,250

    PV of a perpetuity = A/i = 500/0.08 = $6,250.

  3. A machine costs $50,000 and has a salvage value of $5,000 after 10 years. Using straight-line depreciation, what is the annual depreciation?

    Answer: $4,500

    Annual depreciation = (Cost − Salvage)/Life = (50,000 − 5,000)/10 = $4,500.

  4. Two mutually exclusive projects have IRRs of 12% and 15%. The MARR is 10%. Which project should be selected?

    Answer: The one with the higher NPV at MARR

    For mutually exclusive projects, select by NPV or incremental IRR analysis, not simply highest IRR.

  5. What does the capital recovery factor (A/P, i, n) calculate?

    Answer: The annual payment to repay a present loan

    The capital recovery factor converts a present amount P into an equivalent uniform annual series A.

  6. A bond with a face value of $1,000 pays 6% annual coupons and matures in 5 years. If the market interest rate is 8%, what is the bond's present value (approximately)?

    Answer: $921

    PV = 60·(P/A,8%,5) + 1000·(P/F,8%,5) = 60·3.993 + 1000·0.681 ≈ $921.

  7. Using the MACRS 5-year class, what percentage of the asset cost is depreciated in Year 1?

    Answer: 20%

    MACRS 5-year class uses the half-year convention; Year 1 depreciation rate is 20%.