415583543-EE1 - econ PDF

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100 SolvedProblems in(EngineeringEconomy) Interest (p. 1) Equation of Values (p. 2) Compounded continuously(p. 2) Discount (p. 2-3) Inflation (p. 3) Annuity (p. 3 - 4) Capitalized Cost (CASE 1) (p. 5) Comparing Alternatives (p. 5 - 6) Capitalized Cost (CASE2 & 3) (p. 6) Depreciatio...

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100 Solved Problems in (Engineering Economy)  Interest (p. 1)  Equation of Values (p. 2)  Compounded continuously (p. 2)  Discount (p. 2-3)  Inflation (p. 3)  Annuity (p. 3 - 4)  Capitalized Cost (CASE 1) (p. 5)  Comparing Alternatives (p. 5 - 6)  Capitalized Cost (CASE2 & 3) (p. 6)  Depreciation(p. 6- 7)  Uniform Arithmetic Gradient(p. 7)  Geometric Gradient(p. 7)  Bonds(p. 8)  Present Economy(p. 8-9)  Methods for Making Economic Study(p. 9-10) CE DEPARTMENT HOLY ANGEL UNIVERSITY

Page |1 HOLY ANGEL UNIVERSITY – CE DEPARTMENT ENGINEERING ECONOMY *INTEREST 1.

Determine the ordinary simple interest on ₱700 for 8 months and 15 days if the rate of interest is 15%. a. P74.38 b. P84.38 c. P94.38 d. P104.38

2.

Determine the exact simple interest on ₱500 for the period from January 10 to October 28, 1996 at 16% interest. a.

3.

P73.83

b. ₱22400

c. ₱33400

d. ₱44400

b. 12.68%

c. 15.68%

d. 18.68%

Find the amount at the end of two years and seven months if P1000 is invested of 8% compounded quarterly using simple interest for anytime less than a year interest period. a. P4226.336

6.

d. 20

Find the nominal rate which if converted quarterly could be used instead of 12% compounded monthly. What is the corresponding effective rate? a.21.68%

5.

c. P63.83

What will be the future worth of money after 14 months, if a sum of ₱10,000 is invested today at a simple interest rate of 12% per year? a. ₱11400

4.

b. 0

b. P3226.336

c. P2226.336

d. P1226.336

A P2000 loan was originally made at 8% simple interest for 4 years. At the end of this period the loan was extended 3 years, without the interest for 4years. At the end of this interest rate was made 10% compounded semi-annually. How much should the borrower pay at the first 4 years? a. P2905

7.

b. P2500

c. P2640

d. P2000

How much should the borrower pay at the end of 7 years?

a. P3005.11

b. P3537.86

c. P2640.51

d. P3800.02

SITUATION 1: What payment amount 15 years from now is equivalent to a payment of Php 3,500 five years from now, if interest is 11% and is compound?

8.

[a] Compounded semi-annually a. P 10,212.15

9.

b. P 21,212.15

c. P 11,111.11

d. P 9,876.54

b. P 12,462.02

c. P 10,462.02

d. P 9,462.02

[b] Compounded monthly a. P 11,462.02

10. If you have to invest, which is better: 12% compounded semi-annually or 12% compounded monthly? a. compounded monthly

b. compounded semi-annually

c. both a & b

d. none

ENGINEERING ECONOMY

Page |2 *EQUATION OF VALUES 11. A man bought a lot worth P1,000,000 if paid in cash. On installment basis, he paid a down payment of P200,000; P300,000 at the end of one year. P400,000 at the end of three years and a final payment at the end of five years. What was the final payment if interest was 20%?

. a. P792, 576

b. P877, 522

c. P999, 999

d. P671, 123

*COMPOUNDED CONTIUOUSLY SITUATION 2: How many years are required for your money to triple if it is invested at 9% compounded?

12. [a] annually

a. 12.48yrs

b. 12.34yrs

c. 12.75yrs

d. 12.21yrs

b. 12.34yrs

c. 12.75yrs

d. 12.21yrs

b. 12.34yrs

c. 12.75yrs

d. 12.21yrs

b. 12.34yrs

c. 12.75yrs

d. 12.21yrs

13. [b] semi-annually a. 12.48yrs 14. [c] quarterly, a. 12.48yrs 15. [d] continuously? a. 12.48yrs

16. What is the accumulated amount after 3 years of P1, 000 invested at the rate of 8% per year compounded continuously? a. P1, 777.25

b. P1, 271.25

c. P1, 721.25

d. P1, 127.25

*DISCOUNT

SITUATION 3: Mr. J. Dela Cruz borrowed money from a bank. He received from the bank P1, 342 and promises to pay P1, 500 at the end of 9 months. Determine the following:

17. a) Rate of discount (Banker’s Discount). a.10.53%

b. 15.70%

c. 17.65%

d. 11.77%

b. 15.70%

c. 17.65%

d. 11.77%

18. b) Rate of interest. a.10.53%

19. c) Rate of interest for one year

ENGINEERING ECONOMY

Page |3 a.10.53%

b. 15.70%

c. 17.65%

d. 11.77%

20. A man borrows P10, 000 from a loan firm. The rate of simple interest is 15%, but the interest is to be deducted from the loan at the time the money is borrowed. At the end of one year, he has to pay back P10, 000. What is the actual rate of interest? a.10.53%

b. 15.70%

c. 17.65%

d. 11.77%

*INFLATION 21. A certain product presently costs P500.If inflation rate is at the rate of 6% per year, what will be the cost of this product in 4years? a. P 631.24

b. P 731.24

c. P 531.24

d. P 831.24

22. In year zero, you invest P10, 000 in a 15% security for 5 years. During that time, the average annual inflation is 6%. How much, in terms of year zero pesos will be in the account at maturity? a. P12, 000.03

b. P13, 123.50

c. P21, 130.03

d. P15, 030.03

23. An economy is experiencing inflation of an annual rate of 8% if this continuous what will be P1000 be worth two years from now in term of today pesos? a. P757.34

b. P857.34

c. P657.34

d. P957.34

*ANNUITY SITUATION 4: What are the present worth and future worth of P600 deposited at the end of every month for 4 years if the interest rate is 12% compounded quarterly? 24. What is the present worth? a. P22, 835.43

b. P21, 538.15

c. P23, 225.78

d. P36, 641.32

c. P23, 225.78

d. P36, 641.32

25. What is the future worth? a. P22, 835.43

b. P21, 538.15

26. The buyer of a certain machine may pay either P2, 000 cash down payment and P2, 000 annually for the next 6 years, or pay P3, 500 cash and P2, 000 annually for the next 5 years. If money is worth 12% compounded annually, which method of payment is better for the buyer and by how much? a. 1ST CHOICE, P486.74 b. 2ND CHOICE, P991.23 c. 1ST CHOICE, P 684.74 d. 2ND CHOICE, P563.23 27. How much money would you have to deposit for five consecutive years starting one year from now if you want to be able to withdraw P50, 000 ten years from now? Assume the interest is 14% compounded annually. a. P 3,289.25 28.   

b. P 3,892.82

c. P 3,928.60

d. P 3,982.75

A corporation will make the following disbursements: P50,000 on Dec. 31, 1991 P100,000 on Dec. 31, 1992 P200,000 on Dec. 31, 1993

To accumulate these sums, a sinking fund is established by making equal year-end deposits starting Dec. 31, 1986 up to the end of 1993. If the fund earns 9% interest compounded annually, what is the required amount of the annual deposit? a. P33, 404.89

b. P33, 555.99

c. P33, 786.98

d. P33, 123.66

ENGINEERING ECONOMY

Page |4 SITUATION 5: A man bought a property and he paid Php 100 000 cash and agreed to pay Php 20 000 at the end of each 6 months for 5 years. He failed to pay the first 5 payments. At the end of 3 years, he is required to pay by the seller the entire debt consisting of his accumulated and future liabilities, otherwise the farm would be foreclosed by the seller. 29. What is the Present Worth? a. P274, 102.447

b. P247, 201.741

c. P244, 012.14

d. P248, 221.74

30. What must he pay if money is worth 12% compounded semi-annually? a. P288, 708.43

b. P218.556.32

c. P207, 153.62

d. 208,807.43

31. A lathe for a machine shop costs P60, 000 if paid in cash. On the installment plan, a purchaser should pay P20, 000 down payment and 10 quarterly installments, the first due at the end of the first year after purchase. If money is worth 15% compounded quarterly, determine the quarterly installment. a. P5999.12

b. P5693.64

c. P5439.18

d.P5878.56

32. A man invests P10, 000 now for the college education of his 2-year old son. If the fund earns 14% effective, how much will the son get each year starting from his 18th to the 22nd birthday? a. P20, 791.64

b. P21.556.32

c. P17, 153.62

d. 28,807.43

33. A person buys a piece of property for P100, 000 down payment and ten deferred semi-annual payments of P8, 000 each starting three years from now. What is the present value of the investment if the rate of interest is 12% compounded semi-annually? a. P134, 898.26

b. P143, 999.08

c. P143, 019.15

d. P134, 190.51

34. A farmer bought a tractor costing P25000, payable in 10 semiannual payments, each instalment payable at the beginning of each period. If the rate or interest is 26% compounded semiannually, determine the amount of each instalment. a. P5, 707.21

b. P4, 077.20

c. P3, 770.22

35. A certain manufacturing plant is being sold and was submitted by interested buyers. The first bid offered to payment is being made at the beginning of each year. The first year, P180,000 the second year, and P270,00 each being made at the beginning of each year. If money is should the owner of the plant accept? a. P241, 282.32

b. P242, 828.32

d. P5, 007.23

submitted for bidding. Two birds were pay P200, 000 each year for 5 years, each second bidder offered to pay P120,000 the year for the next 3 years, all payments worth 12% compounded annually, which bid

c. P240, 288.32

d. P241, 882.32

SITUATION 6: If money is worth 8% compounded quarterly, determine the present value of the following:

36. An annuity of P1, 000 payable quarterly for 50 yrs. a. ₱50, 074.35

b. ₱49, 981.85

c. ₱49, 047.35

d. ₱50,000.00

37. An annuity of P1, 000 payable quarterly for 100 yrs. a. ₱50, 074.35

b. ₱49, 981.85

c. ₱49, 047.35

d. ₱50,000.00

c. ₱49, 047.35

d. ₱50,000.00

38. A perpetuity of P1, 000 payable quarterly? a. ₱50, 074.35

b. ₱49, 981.85

39. What amount of money invested today at 15% interest can provide the following scholarships: P30, 000 at the end of each year for 6 years; P40000 for the next 6 years and P50, 000 thereafter? a. ₱241, 282.32

b. ₱242, 828.32

c. ₱241, 822.32

d. ₱242, 228.32

ENGINEERING ECONOMY

Page |5 *CAPITALIZED COST

40. Determine the capitalized cost of a structure that requires an initial investment of P1, 500,000 and an annual maintenance of P150, 000. Interest is 15%. a. ₱2050.00

b. ₱5200.00

c. ₱2500.00

d. ₱2005.00

SITUATION 7: A car costs 1,500,000 and has an annual maintenance of 30,000. if the interest is 15% compounded monthly:

41. Solve for the annual interest. a. 16.08%

b. 28.06%

c. 32.12%

d. 18.18%

c. P1,787,619.95

d. P1,878,619.95

42. Solve for the capitalized cost. a. P1,686,619.95

b. P1,868,619.95

43. A manufacturing plant installed a new broiler at a total cost of P150, 000 and is estimated to have a useful life of 10 years. It is estimated to have a scrap value at the end of its useful life of P5, 000. If interest is 12% compounded annually, determine its capitalized cost. a. P219, 952.86

b. P217, 588.86

c. P216, 935.86

d. P21, 855.86

*COMPARING ALTERNATIVES

SITUATION 8: An industrial plant has an engine costing P7200 which is now 5 years. Old its working lie is 15 years and salvage value of 500. The average operating cost per year thus far has been P4200. A new engine costing P12, 000 with an estimated life of 15 years and a salvage value of P800 is guaranteed to have an operating cost of P 3500 per year. The new engine is considered as a replacement of the old. Will replacement be justified if P4500 can be obtained from the sale of the old engine? 44. Solve for the depreciation OLD engine. Use straight line depreciation with 6% interest a. P400.00 45. Solve

for

a. P790.12

b. P450.00 the

depreciation

c. P500.00 NEW

engine.

b. P746.67

Use

straight

d. P480.00 line

depreciation

c. P820.57

d. P680.97

c. P4625.00

d. P4200

c. P4625.00

d. P4200

c. 4.17%

d. 2.30%

with

6%

interest

46. Solve for the total cost of the Old engine. a. P4600

b. P4246.67

47. Solve for the total cost of the New Engine. a. P4600

b. P4246.67

48. Solve for the rate of return a.15.30%

b. 8.20%

ENGINEERING ECONOMY

Page |6 SITUATION 9: is estimated same output value of P6,

A plastic extrusion machine purchased 4 years ago has annual operating costs of P12, 000 and to have a service life of 6 more years with salvage value of P4, 500. A new machine with the will cost P42, 000, with annual operating costs of 7,200, a life of 12 years and salvage 000. For this class of asset, the manufacturer expects an 18% rate of return.

49. What are the depreciations? Use sinking fund method. a.0.1059X – 476.55, P1030.60

b.0.2859X + P11, 523.45, P1030.60

c.0.2859X + P11, 523.45, P7560

d. NONE

50. Determine the minimum trade in value that the manufacturer should receive and still purchase the new machine. a. P15, 925.32

b. P14, 925.32

c. P16, 925.32

d. P17, 925.32

CAPITALIZED COST: CASE 2 51. What is the capitalized cost if an iShop Computer Services purchased a computer unit at a total cost of P12, 000 with a useful life of 7 years? Scrap value at the end of its useful life is P9, 000and the interest is 9% compounded annually. a. P16251.08 b. P15623.01 c. P12653.63 d. P16253.36 52. A new engine was installed by a textile plant at a cost of P300, 000 and projected to have a useful life of 15 years. At the end of its useful life, it is estimated to have a salvage value of P30, 000. Determine its capitalized cost if interest is 18% compounded annually. a. P213, 653.26 b. P231, 524.09 c. P321, 657.35 d. P320, 604.17 CAPITALIZED COST: CASE 3 53. Determine the capitalized cost of a research laboratory which requires P5,000,000 for original construction; P100,000 at the end of every year for the first 6 years and then P120,000 each year thereafter for operating expenses, and P500,000 every 5 years for replacement of equipment with interest at 12% per annum? a. P6 573 645.73 b. P6 563 523.68 c. P6 562 321.36 d. P6 685 325.51 SITUATION 10: A financial analysis of two types of bridges based on capitalized cost and on the following data is to be made: BRIDGE A INITIAL COST P 200,000 COST OF RENEWAL P 200,000 SALVAGE VALUE 0 ANNUAL MAINTENANCE P 1,000 REPAIRS EVERY 5 YRS P 10,000 LIFE YEARS 30 If the rate of interest is 8% compounded annually.

BRIDGE B P 240,000 P 240,000 P 20,000 NONE P 5,000 40

54. Determine the capitalized cost for Bridge A. a. ₱ 235,635.66

b.₱ 252,315.38

c. ₱255,875.64

d. ₱225,325.37

c. ₱261,268.97

d. ₱216,628.79

55. Determine the capitalized cost for Bridge B. a. ₱263,261.36

b. ₱236,216.63

56. Among the two types of bridges, what would you must choose? How much you’ll save? a. BRIDGE A: P 5393.33 b. BRIDGE B: P 9560.26 c. BRIDGE A: P 9560.26 d. BRIDGE B: P 5393.33 DEPRECIATION (STRAIGHT LINE METHOD): SITUATION 11: An electronic balance costs P90, 000 and has an estimated salvage value of P8, 000 at the end of its 10 years life time. Use the straight line method. 57. What is the depreciation in three years? a. 25, 300

b. 26, 400

c. 23, 500

d. P24, 600

58. What is the book value in three years?

ENGINEERING ECONOMY

Page |7 a. P63, 400

b. P65, 400

c. P64, 500

d. P64, 300

SITUATION 12: A tax and duty free importation of a 30HP sand mill (for paint manufacturing) cost P360, 000, CIF Manila Bank charges brokerage cost P5, 000. Foundation and installation costs were P25, 000. Other incidental expenses amounted to P20, 000. Salvage value of the mill is estimated to be P60, 000 after 20 years. 59. Find the appraisal value of the mill, using straight line depreciation, at the end of 10 years. a. P235, 000 b. P253, 000 c. P325, 000 d. P352, 000 60. Find the appraisal value of the mill, using straight line depreciation, at the end of 15 years. a. P174, 560

b. P147, 560

c. P156, 740

d. P165, 470

61. A certain company makes it the policy that for any new piece of equipment the annual depreciation cost should not exceed 10% of the original cost at any time with no salvage or scrap value. Determine the length of service life necessary if the depreciation method used is straight line formula. a. 11.5 years

b. 12 years

c. 10 years

d. 10.5 years

DEPRECIATION (SINKING FUND METHOD): SITUATION 13: A broadcasting corporation purchased equipment worth P53, 000 and paid P1, 500 for freight and delivery charges to the site. The equipment has a normal life of ten years with a trade-in value of P5, 000 against the purchase of new equipment at the end of life. The interest is 6%. 62. What is the annual depreciation cost? a. P3375.63 b. P1536.35 c. P3755.46 63. What is the total depreciation cost in 7 years? a. P33 564.78

b. P31 522.75

d. P3575.58

c. P32 512.34

d. P30 658.14

c. P27 572.50

d. P25 775.15

64. What is the book value in 7 years? a. P22 977.25

b. P29 277.75

DEPRECIATION (DOUBLE DECLINING BALANCE METHOD): SITUATION 14: Determine the rate of depreciation, the total depreciation up to the end of the 8th year and the book value at the end of 8 years for an asset that costs P15, 000 new and has an estimated scrap value of P2, 000 at the end of 10 years by: 65. Using the declining balance method. a. P12008.42

b. P12007.44

c. P12017.14

d. P12177.11

c. P12817.14

d. P12483.42

66. Using the double declining balance method. a. P12158.42 UNIFORM ARITHMETIC GRADIENT

b. P12381.44

SITUATION 15: A loan was to be amortized by a group of four end-of-year payments forming an ascending arithmetic progression. The initial payment was to be P5, 000 and the difference between successive payments was to be P400. But the loan was negotiated to provide for the payment of equal rather than uniformly varying sums. The interest rate of the loan was 15%. 67. Determine the present worth value. a. P 14,279.89

b. P 17,699.18

c. P 14,189.91

d. P 13,299.85

68...