1. Chapters 1 & 2 Practice Problems for Engineering Economy Class

2. Example 1.1 If $10,000 is borrowed at an interest rate of 10% per year, determine the amount of money that is owed at the end of 3 years if the interest is (a) simple, and (b) compound Solution: (a)Simple interest is charged on the principal. Therefore, the amount of money owed after 3 years is: Amt owed year 3 = principal + interest = 10,000 + 10,000 (0.10) (3) = $13,000 (b) Compound interest is charged on the principal and the accrued interest. Thus, $10,000(F/P, 10%, 3) = $13,310

3. American Airlines estimates that unpainted airplanes use less fuel (because of less weight) to the extent of $20,000 per year per plane. How much could the company afford to spend now to strip the paint from an airplane if it expects to recover its investment in five years at an interest rate of 20% per year? (A) $ 53,626 (B) $ 59,812 (C) $ 64,701 (D) $ 66,892 P = 20,000 (P/A, 20%, 5) = 20,000 (2.9906) = $59,812 Answer is (B)

4. Wendy's International has received a bid of $9,000 to re-coat a parking lot with a water sealer and repaint the parking-space lines. What is the equivalent annual cost of the job if the company expects to recover its investment in four years at an interest rate of 20% per year? (A) $ 2,991 (B) $ 3,102 (C) $ 3,477 (D) $ 4,316 A = 9,000 (A/P, 20%, 4) = 9,000 (0.38629) = $3476.61 Answer is (C)

5. A large bank that was trying to attract new customers started a program wherein a person starting a checking account could borrow up to $10,000 at an interest rate of 6% per year simple interest for up to 3 years. A person who borrows the maximum amount for the maximum time would owe how much money at the end of the 3 year period? (A) $10,600 (B) $11,200 (C) $11,800 (D) $11,910 Amt owed = Principal + interest = 10,000 + 10,000 (0.06) (3) = $11,800 Answer is (C)

6. An engineer invested his bonus check of $7,000 in a mutual fund two years ago. If the value of his investment now is $10,000, the rate of return he made on his investment was closest to: (A) Less than 10%/yr. (B) 14.5%/yr. (C) 19.5%/yr. (D) Over 20%/yr. 7000 (1 + i) (1 + i) = 10,000 (1 + i) 2 = 1.4286 1 + i = 1.1952 i = 19.52% Answer C or $7000(F/P, i,2) = $10000 i = 10000/7000 factor (1.4286) on interest table for F/P at n=2 i = 19.5% between 18% & 20% tables Answer: C

7. According to the rule of 72, the length of time it would take for $1000 to accumulate to $4,000 at an interest rate of 4% per year would be closest to: (A) 36 years (B) 24 years (C) 18 years (D) less than 15 years Time to double = 72 / 4 = 18 yrs Time to quadruple = 2 * 18 = 36 years Answer = A

8. A sum of $50,000 now would be equivalent to how much money two years from now at an interest rate of 20% per year? (A) Less than $65,000 (B) $70,000 (C) $72,000 (D) Over $73,000 Solution: 50000 (F/P, 20%, 2) = 50000 (1.4400) = $72,000

9. A sum of $30,000 one year from now would be equivalent to how much money now, at an interest rate of 25% per year? (A) Less than $24,000 (B) $24,000 (C) $27,500 (D) $37,500 Present worth = 30,000 / ( 1 + 0.25 ) = $24,000 Answer is (B) or 30000 (P/F, 25%, 1) = 30000 (0.8000) = $24,000 Answer (B)

10. How much money could a start-up software company afford to borrow now if it promises to repay the loan with five equal year- end payments of $10,000 if the interest rate is 10% per year? (A) $37,910 (B) $32,640 (C) $41,620 (D) $50,000 Solution: The amount that could be borrowed is the present worth of the $10,000 series: P = 10,000 (P/A, 10%, 5) = 10,000 (3.7908) = $37,908 Answer is (A)

11. Pebble Beach Country Club installed a new computer-controlled irrigation system that uses reclaimed sewage for watering the fairways. The cost of the pumps, piping, and controls was $1,100,000. If the club expects to recover its investment in 10 years using an interest rate of 10% per year, the annual savings in water cost must be nearest to: (A) $ 179,025 (B) $ 186,193 (C) $ 201,250 (D) $ 127,362 A = 1,100,000 (A/P, 10%, 10) = 1,100,000 (0.16275) = $179,025

12. An elastomeric roofing material can be installed on a parts warehouse for $10,000. If the company expects to recover its investment in seven years through reduced energy costs, the required annual savings at an interest rate of 20% per year is closest to: (A) $ 2,067 (B) $ 2,774 (C) $ 2359 (D) $ 3,067 A = 10,000 (A/P, 20%, 7) = 10,000 (0.27742) = $2774.20 Answer is (B)

13. How much should a company which manufactures corrugated pipe be willing to pay a contractor who claims he has a device which will reduce the company's energy bill by at least $4,000 per year. Assume the company wants to recover its investment in five years at an interest rate of 15% per year. (A) $ 19,362 (B) $ 11,156 (C) $ 15,329 (D) $ 13,409 P = 4,000 (P/A, 15%, 5) = 4,000 (3.3522) = $13,408.80