EIT Engineering Economics 3 — Questions and Answers
Question 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?
- 1.04
- 1.09 (Correct answer)
- 1.17
- 1.23
Correct 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.
Question 2: What is the effective annual interest rate if the nominal rate is 12% compounded monthly?
- 12.00%
- 12.36% (Correct answer)
- 12.68%
- 12.00%
Correct answer: 12.36%
EAR = (1 + 0.12/12)^12 − 1 = (1.01)^12 − 1 ≈ 12.68%.
Question 3: Which depreciation method produces the largest depreciation expense in the first year of an asset's life?
- Straight-line
- Sum-of-years-digits
- Double-declining-balance (Correct answer)
- Units of production
Correct answer: Double-declining-balance
Double-declining-balance applies twice the straight-line rate to book value, maximizing first-year depreciation.
Question 4: If a project's net present value is zero, the discount rate used equals:
- The MARR
- The IRR (Correct answer)
- The benefit-cost ratio
- The payback period
Correct answer: The IRR
By definition, the IRR is the discount rate that makes NPV equal to zero.
Question 5: A company borrows $100,000 at 8% per year compounded quarterly. What is the effective annual rate?
- 8.00%
- 8.16%
- 8.24% (Correct answer)
- 8.33%
Correct answer: 8.24%
EAR = (1 + 0.08/4)^4 − 1 = (1.02)^4 − 1 ≈ 8.24%.
Question 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:
- $3,600 (Correct answer)
- $4,800
- $5,400
- $6,000
Correct 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.
Question 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?
- $2,500
- $3,000
- $3,431 (Correct answer)
- $3,750
Correct answer: $3,431
PV = G·(P/G,10%,5) = 500·6.862 = $3,431.
EIT Engineering Economics 3
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?