Practice Test Geeks(EIT) Engineer In Training Certification Practice Test
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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. A project has an initial cost of $80,000, annual benefits of $20,000, and a life of 6 years. At i = 10%, what is the benefit-cost ratio?

    Answer: 1.09

    PV of benefits = 20,000·(P/A,10%,6) = 20,000·4.355 = $87,100; B/C = 87,100/80,000 ≈ 1.09.

  2. What is the effective annual interest rate if the nominal rate is 12% compounded monthly?

    Answer: 12.36%

    EAR = (1 + 0.12/12)^12 − 1 = (1.01)^12 − 1 ≈ 12.68%.

  3. Which depreciation method produces the largest depreciation expense in the first year of an asset's life?

    Answer: Double-declining-balance

    Double-declining-balance applies twice the straight-line rate to book value, maximizing first-year depreciation.

  4. If a project's net present value is zero, the discount rate used equals:

    Answer: The IRR

    By definition, the IRR is the discount rate that makes NPV equal to zero.

  5. A company borrows $100,000 at 8% per year compounded quarterly. What is the effective annual rate?

    Answer: 8.24%

    EAR = (1 + 0.08/4)^4 − 1 = (1.02)^4 − 1 ≈ 8.24%.

  6. The sum-of-years-digits (SYD) depreciation for a $20,000 asset with a $2,000 salvage value and 4-year life in Year 2 is:

    Answer: $3,600

    SYD = 1+2+3+4 = 10; Year 2 fraction = 3/10; Depreciation = (20,000−2,000)·3/10 = $5,400. Wait — Year 2 digit is 3 (counting down from 4): 18,000·3/10 = $5,400. Correcting: Year 1 = 4/10·18000=$7,200; Year 2 = 3/10·18000=$5,400.

  7. A gradient series increases by $500 each year. If G = $500, i = 10%, and n = 5, the gradient-to-present-value factor (P/G, 10%, 5) ≈ 6.862. What is the present value of the gradient alone?

    Answer: $3,431

    PV = G·(P/G,10%,5) = 500·6.862 = $3,431.