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Engineering economics: board exam questions with solutions

On the ECE exam under General Engineering and Applied Sciences (GEAS) and the REE exam under Engineering Sciences and Allied Subjects (ESAS). Try each question before you open its solution, then practise with fresh numbers below: 4 computed question types on this topic.

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Questions

1. How much simple interest does ₱50,000 earn at 8% per year for 3 years?

  1. ₱12,985.60
  2. ₱4,000
  3. ₱62,000
  4. ₱12,000
Show the answer and solution

Answer: D. ₱12,000

I = Pin = 50 000 × 0.08 × 3 = ₱12,000. ₱12,985.60 is the compound interest.

2. ₱100,000 is deposited at 6% compounded quarterly. What is it worth after 5 years?

  1. ₱134,685.50
  2. ₱133,822.56
  3. ₱130,000.00
  4. ₱134,885.02
Show the answer and solution

Answer: A. ₱134,685.50

F = P(1 + i)ⁿ = 100 000(1.015)²⁰ = ₱134,685.50. ₱133,822.56 is the annually compounded value.

3. What is the effective annual rate of 12% compounded monthly?

  1. 12.00%
  2. 12.36%
  3. 12.55%
  4. 12.68%
Show the answer and solution

Answer: D. 12.68%

i_eff = (1 + 0.12/12)¹² − 1 = 1.01¹² − 1 = 0.1268 = 12.68%. 12.36% is the semi-annual figure and 12.55% the quarterly one.

4. What is the present worth of ₱20,000 received at the end of each year for 10 years at 8%?

  1. ₱200,000.00
  2. ₱134,201.63
  3. ₱289,731.25
  4. ₱92,638.70
Show the answer and solution

Answer: B. ₱134,201.63

P = A[1 − (1.08)⁻¹⁰]/0.08 = 20 000 × 6.7101 = ₱134,201.63. ₱289,731.25 is the future worth.

5. Equipment costs ₱500,000 and has a ₱50,000 salvage value after 10 years. Using straight-line depreciation, what is its book value at the end of year 4?

  1. ₱300,000
  2. ₱180,000
  3. ₱320,000
  4. ₱350,000
Show the answer and solution

Answer: C. ₱320,000

Annual depreciation = (500 000 − 50 000)/10 = ₱45,000. BV₄ = 500 000 − 4 × 45 000 = ₱320,000.

6. A product sells for ₱50 per unit with a variable cost of ₱30 per unit, and fixed costs are ₱200,000. What is the break-even quantity?

  1. 10,000 units
  2. 4,000 units
  3. 6,667 units
  4. 2,500 units
Show the answer and solution

Answer: A. 10,000 units

Q = F/(p − v) = 200 000/(50 − 30) = 10,000 units.

7. What equal year-end deposit into a fund earning 10% builds ₱1,000,000 in 5 years?

  1. ₱200,000.00
  2. ₱263,797.48
  3. ₱163,797.48
  4. ₱124,184.26
Show the answer and solution

Answer: C. ₱163,797.48

A = F·i/[(1 + i)ⁿ − 1] = 1 000 000 × 0.10/(1.61051 − 1) = ₱163,797.48. ₱263,797.48 is the capital recovery factor applied to a present amount.

8. A facility costs ₱2,000,000 and needs ₱50,000 of maintenance every year forever. At 5% interest, what is its capitalized cost?

  1. ₱2,050,000
  2. ₱3,000,000
  3. ₱2,500,000
  4. ₱1,000,000
Show the answer and solution

Answer: B. ₱3,000,000

Capitalized cost = first cost + A/i = 2 000 000 + 50 000/0.05 = 2 000 000 + 1 000 000 = ₱3,000,000.

Worked examples with computed numbers

These numbers are fixed on this page; in practice they change every time.

9. ₱67,000 is deposited at 10% compounded annually. How much is in the account after 3 years?

  1. ₱89,415
  2. ₱89,177
  3. ₱81,070
  4. ₱87,100
Show the answer and solution

Answer: B. ₱89,177

F = P(1 + i)ⁿ = ₱67,000 × 1.1^3 = ₱89,177.

10. What is the present worth of ₱21,000 paid at the end of each year for 13 years at 6% per year?

  1. ₱127,993
  2. ₱396,525
  3. ₱273,000
  4. ₱185,906
Show the answer and solution

Answer: D. ₱185,906

P = A[1 − (1 + i)^−n]/i = 21000 × [1 − 1.06^−13]/0.06 = ₱185,906.

11. What is the effective annual rate of 17% compounded quarterly?

  1. 4.25%
  2. 17%
  3. 18.11%
  4. 21.25%
Show the answer and solution

Answer: C. 18.11%

i = (1 + j/m)^m − 1 = (1 + 17%/4)^4 − 1 = 18.11%.

12. Equipment costs ₱277,000 with a salvage value of ₱42,000 after 14 years. By straight-line depreciation, what is its book value after 3 years?

  1. ₱92,357
  2. ₱217,643
  3. ₱226,643
  4. ₱50,357
Show the answer and solution

Answer: C. ₱226,643

Annual depreciation (C − S)/n = ₱16,785.71; book value after 3 years = ₱277,000 − 3 × ₱16,785.71 = ₱226,643.

Other topics in the same subjects

Physics: mechanics · Physics: waves, electricity and magnetism · General chemistry · Thermodynamics · Engineering materials · Engineering management · Computer fundamentals and programming · RA 9292 and ethics (electronics) · Fluid mechanics · Strength of materials · RA 7920 and ethics (electrical) · PEC Part 1: general requirements

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