Ee4

100 Solved Problems in Engineering Economy                    

Simple Interest (p. 1) Compound Interest (p. 1-2) Effective Rate of interest (p. 2-3) Continuous Compounding (p. 3) Comparing of Alternatives (p. 3-4) Discount (p.4-5) Inflation (p. 5) Geometric Gradient (p.5-6) Service output method (p. 6-7) Straight line method (p.7-8) Sinking Fund Formula (p.8) Declining Balance Method (p. 9-10) Double Declining Balance method (p. 9-10) Sum of the year digits (p. 10-11) Arithmetic Gradient (p.11-15) Annuity (p.11-15) Deferred Annuity (p.11-15) Annuity due (p.11-15) Perpetuity (p.11-15) Capitalize Cost (p.15-16)

CE DEPARTMENT HOLY ANGEL UNIVERSITY

1|Page HOLY ANGEL UNIVERSITY – CE DEPARTMENT

*INSERT SUBJECT HERE “Simple Interest” 1.) Ariel takes a loan of P8,000 to buy a used truck at the rate of 9 % simple Interest. Calculate the annual interest to be paid for the loan amount. a. P 720

c. P 820

b. P 722

d. P 822

2.) A loan of P5,000 is made for a period of 15 months, at a simple interest rate of 15% what future amount is due at the end of the loan period a. P 5,937.50

c. P 5,837.50

b. P 5,637.50

d. P 5,937.50

3.) What is the interest due on a P 1500 note for 4 years and 3 month, if it bears 12% ordinary simple interest? a. P 795

c. P 765

b. P 785

d. P 756

4.) A total of P1,200 is invested at a simple interest rate of 6% for 4 months. How much interest is earned on this investment? a. P 21 b. P 22

c. P 23 d. P 24

5.) Find the simple interest on P10,000 at the rate of 5% for 5 years. Also find the total amount after this time. a. P 12,500 c. P 12,300 b. P 13,500

d. P 13,300

“Compound Interest” 6.) An investment earns 3% compounded monthly. Find the value of an initial investment of P5,000 after 6 years. a. P 5924.74

c. P 5783.74

b. P 5984.74

d. P 5874.74

7.) What is the value of an investment of P3, 500 after 2 years if it earns 1.5% compounded quarterly? a. P 3706.39

c. P 3606.39

b. P 3805.39

d. P 3906.39

2|Page 8.) Mrs. Lalic purchased an antique statue for P450. Ten years later, she sold this statue for P750. If the statue is viewed as an investment, what annual rate did she earn? a. 0.0524

c.

0.0425

b. 0.0542

d.

0.0515

9.) How many years will it take for an investment to double in value if it earns 5% compounded annually? a. 12.2 years

c.

13.2

b. 14.2 years

d.

15.2

10.) Find the compound amount and compound interest on the principal P20,000 borrowed at 6% compounded annually for 3 years. a. P 3812.32 c. P 3820.32 b. P 3281.32

d. P 3720.32

“EFFECTIVE RATE OF INTEREST” 11.)What rate of interest compounded annually is the same as the rate of 8% compounded quarterly a. P b. P

c. P 8.24%

d. P

12.) A man approaches the ABC Loan Agency for P100, 000 to be paid in 24 monthly installments. The agency advertises an interest rate of 1.5% per month. They proceed to calculate the amount of his monthly payment in the following manner.

a. 38.64%

c. 37.64%

b. 39.64%

d. 36.64%

13.) Mr. Smith plans to deposit money in a bank that pays 10% interest per year, compounded daily. What effective rate of interest will he receive yearly? a. 11.615%

c. 10.515%

b. 10.615%

d. 11.515%

14.) What is the nominal interest rate compounded monthly?

if the effective annual interest rate

a.5.5%

c. 6%

b.6.5%

d. 5%

is 6.1678%,

3|Page 15.) What rate of interest compounded annually is the same as the rate of interest of 12% compounded bimonthly? a. 12.60%

c. 12.62%

b. 12.61%

d. 12.63%

“CONTINIOUS COMPOUNDING” 16.) What is the accumulated amount after 3 years of P1000 invested at the rate of 8% per year compounded continuously? a. 1721.25

c. 1712.25

b. 1217.25

d. 1271.25

17.) How many years are required for your money to triple if it is invested at 9% compounded annually? a. 12.21 years

c. 12.48 years

b. 12.75 years

d. 12.34 years

18.) If you invest $1,000 at an annual interest rate of 5% Compounded continously, calculate the final amount you will have in the account after five years. a. 1248.03

c. 1203.48

b.

d. 1230.48

1284.03

19.) If you invest 500 at an annual interest rate of 10% compounded continuously, calculate the final amount you will have in the account after five years. a. 826.36

c. 824.36

b. 825.36

d. 823.36

20.) If you invest 2,000 at an annual interest rate of 13% compounded continuously, calculate the final amount you will have in the account after 20 years. a. 26,927.47

c. 26,972.47

b. 29,627.47

d. 29,972.47

“COMPARING OF ALTERNATIVES” 21.) A company purchases a piece of construction equipment for rental purposes. The expected income is P3100 annually for its useful life of 15 years. Expenses are estimated to be P355 annually. If the purchase price is P25,000 and there is no salvage value, what is the prospective rate of return, neglecting taxes? a. 5.2%

c. P 6.8%

b. 6.4%

d. P 7.0%

4|Page 22-24.) An industrial firm uses an economic analysis to determine which of two different machines to purchase. Each machine is capable of performing the same task in a given amount of time. Assume the minimum attractive rate of return is 8%. Use the following data in this analysis.

Initial cost Estimated life Salvage value Annual maintenance cost

Machine X P6,000 7years none P150

Machine Y P12,000 13years P4,000 P175

22) What is the approximate equivalent uniform annual cost of machine X? a. P 1000

c. P 1190

b. P 1120

d. P 1300

23.) What is the approximate equivalent uniform annual cost of machine Y? a. P 1160

c. P 1490

b. P 1300

d. P 1510

24.) Which, if either, of the two machines should the firm choose, and why? a. machine X because (EUAC)x < (EUAC)y

c. machine Y because (EUAC)x < (EUAC)y

b. machine X because (EUAC)x > (EUAC)y

d. machine Y because (EUAC)x > (EUAC)y

25.) A company produces a gear that is commonly used by several lawn mower manufacturing companies. The base cost of operation (rent, utilities, etc.) is P750,000 per year. The cost of manufacturing is P1.35 per gear. If these gears are sold at P7.35 each, how many must be sold each year to break even? a. P 65,000 per year

c. P 100,000 per year

b. P 90,000 per year

d. P 125,000 per year

“DISCOUNT” 26.) Find the sale price for an item that has a list price of P100 and a discount rate of 25% a. P 85

c. P 75

b. P 72

d. P 82

27.) Find the sale price of an item that has a list price is P24 dollars and a discount rate of 50% a. P 12

c. P 21

b. P 24

d. P 42

28.) Find the sale price of an item that has a list price is P27 and a discount rate of 33.33333% a. P 18.0000002

c. P 17.0000002

b. P 17.0000001

d. P 18.0000001

5|Page 29.) Miss Evilla borrowed money from a bank. She receives from the bank P1,340.00 and promised to pay P1,500.00 at the end of 9 months. Determine the corresponding discount rate or often referred to as the “banker’s discount” a.

12.94%

c. 13.94%

b.

11.94%

d. 14.94%

30.) Mr. J. Dela Cruz borrowed money from a bank. He received from the bank P1,400 and promise to pay P2,000 at the end of 11 months. What is the rate of discount or Banker’s discount a.

30%

c.

20%

b.

40%

d.

25%

“INFLATION” 31.) What is the uninflated present worth of P2000 in 2 years if the average inflation rate is 6% and i is 10%? a. P 1271

c.P 1471

b. P 1472

d.P 1272

32.) What is the future worth value of an amount of P1000 after 10 periods and 4% of inflation rate a. P 500

c. P 665

b. P 755

d. P 555

33.)The bank Pays 5.5% interest compounded annually. Inflation is expected at 2%. What is the Real interest rate ? a. 3.3%

c. 3.5%

b. 3.4%

. 3.6%

34.)A stadium was built in 1955 at P2.4m. At the same year a gift of the same amount was invested at a market rate 8% per year. Inflation is 6% from 1955 to 2010.What is the Real Interest rate ? a. 1.967%

c. 1.525%

b. 1.799%

d. 1.887%

35.)How much the gift will worth in 2010 in terms of 1955 purchasing power? a. P 6,600,440

c. P 6,820,230

b. P 6,710,330

d. P 6,310,410

“GEOMETRIC GRADIENT ” 36.) A mechanical contractor is trying to calculate the present worth of personnel salaries over the next five years. He has four employees whose combined salaries thru the end of this year are ₱150,000. If he expects to give each employee a raise of 5% each year, the present worth of his employees' salaries at an interest rate of 12% per year is nearest to: a. P 591,000

c. P702,900

b. P 816,100

d. P429,300

6|Page 37.) Suppose that the maintenance for a piece of equipment costs ₱300 EOY1 and increases by 15% every year for 5 years. The value of money is 9%. What is the equivalent present cost over the time horizon? a P = ₱2022.71

c. P = ₱1536.22

b. P = ₱1175.16

d. P = ₱1166.90

38.) Determine the present worth of a geometric gradient series with a cash flow of ₱50,000 in year 1 and increases of 6% each year through year 8. The interest rate is 10% per year. a. P 500,546,64

c. P 440,820.64

b. P 315,645.64

d. P 320,572.64

39.) 4. A coal-fired power plant has upgraded an emission control valve. The modification costs only $8000 and is expected to last 6 years with a P200 salvage value. The maintenance cost is expected to be high at $1700 the first year, increasing by 11% per year thereafter. Determine the equivalent present worth of the modification and maintenance cost by hand and by spreadsheet at 8% per year. a. – P17,999

c. -P19,999

b. – P18,999

d. –P20,999

40.) g = 5% i = 7% N = 25 years A= P50,000 what is P? a. P 940,969

c. P 940,696

b. P 940,669

d. P 940,996

“SERVICE OUTPUT METHOD” 41.) Company A purchases a backup generator for $200,000. The estimated serviceable life of the generator is expected to be 5,000 hours. The generator is projected to have a scrap value of $50,000 at the end of its serviceable life. In the current accounting period, the generator ran for approximately 300 hours. determine the depreciation expense a. P8,000

c. P10,000

b. P9,000

d. P11,000

42.) Oil PLC installs a crude oil processing plant costing P12 million with an estimated capacity to process 50 million barrels of crude oil during its entire life. Production during the first year of operation is 2 million barrels. Expected residual value of the processing plant is P2 million. Calculate the Depreciation using service output method. a. P 200,000

c. P 400,000

b. P 300,000

d. P 500,000

43.) A plant costing P110 million was purchased on April 1, 2010. The salvage value was estimated to be P10 million. The expected production was 150 million units. The plant was used to produce 15 million units till the year ended December 31, 2010. Calculate the depreciation on the plant for the year ended December 31, 2011. a. P 10,000,000

c. P 100,000,000

b. P 1,000,000

d. P 100,000

7|Page 44.) A coal mine was purchased by X Corporation for P16 million. It was estimated that the mine has capacity to produce 200,000 tones of coal. The company extracted 46,000 tones during its first year of operation. Calculated the depreciation. a. P 3,980,000

c. P 3,900,000

b. P 3,600,000

d. P 3,680,000

45.) Plastic LTD purchases a steel mould costing $1 million to be used in the production of plastic glasses. The mould could be used in 8 production batches after which it will have a scrap value of $.2 million. During the first year, the company manufactures 2 batches of glasses. Calculated the year Depreciation charge a. P 200,000

c. P 600,000

b. P 400,000

d. P 800,000

“STRAIGHT LINE METHOD” 46.) ABC corporation makes it a policy that for any new equipment purchased; the annual depreciation cost should not exceed 20% of the first cost at any time with no salvage value. Determine the length of service life necessary if the depreciation used is the SL method. a. 4.5 Years

c. 5.5 Years

b. 5 Years

d. 6 Years

47-48.) A tax and duty free importation of a 30HP sand mill (for a paint manufacturing) cost P360,000, CI manila. Bank chargers, arreastre and brokerage cost P5,000. Foundation and installation cost were P25,000. Other incidental expenses ampunted to P20,000. Salvage value of the mill is estimated to be P60,000 after 20 years. Find the appraisal value of the mill, using straight line depreciation, at the end of 47.) 10 years? a. P 225,000

c. P 245,000

b. P 235,000

d. P 255,000

48.) 15 years? a. P 147,000

c. P 149,000

b. P 148,000

d. P 150,000

49.) The Eastern company provides the following information regarding one of its fixed assets that has been purchased on January 1, 2015: Cost of the asset: $35,000, Salvage value: $3,000, Useful life: 10 years. Calculate annual depreciation expense of this asset using straight line method. a. $3,200

c. $3,300

b. $3,100

d. $3,400

8|Page 50.) A fixed asset having a useful life of 3 years is purchased on 1 January 2013. Cost of the asset is $2,000 whereas its residual value is expected to be $500. Calculate depreciation expense for the years ending 30 June 2013 and 30 June 2014. a. $245 and $500

c. $250 and $450

b. $250 and $500

d. $245 and $450

“SINKING FUND FORMULA” 51.)A certain office equipment has a first cost of P20,000 and has a salvage value if P1,000 at the end of 10 years. Determine the depreciation at the end of the 6th year using sinking fund method at 3% interest. a. P10,720

c. P12,420

b. P11,680

d. P9,840

52.) Power to a remote transmitting station is provided by a Diesel-generator unit. The original cost of the unit is P65,000. It costs P2,000 to ship the unit to the job site. An additional cost of P3,000 was incurred for installation. Determine the annual depreciation cost by the sinking fund method, if the unit has an expected life of 10 years. The salvage value of the unit at the end of its life was estimated at P5000. a. P6,500

c. P6,600

b. P6,700

d. P6,400

53.) An industrial plant bought a generator set for 90,000. Other expenses including installation amounted to 10,000. The generator set is to have a life of 17 years with a salvage value at the end of life of 5,000. Determine the depreciation charge during the 13th year by the sinking fund method at 12% . a. 54,741.00

c. 54,414.00

b. 54,471.00

d. 57,471.00

54-55) Power to a remote transmitting station is provided by a Diesel-generator unit. The original cost of the unit is P65,000. It costs P2,000 to ship the unit to the job site. An additional cost of P3,000 was incurred for installation. 54.) Determine the annual depreciation cost by the sinking fund method, if the unit has an expected life of 10 years. The salvage value of the unit at the end of its life was estimated at P5000. a. P 5,500

c. P 6,500

b. P 6,000

d. P 7,000

55.) Determine the annual depreciation cost by the sinking fund method. Assume that the annual charge for depreciation was deposited in a fund drawing compound interest at the rate of 5% a. P 5,167.80

c. P 5,197.80

b. P 6,167.80

d. P 6,190.80

9|Page “DECLINING BALANCE METHOD/CONSTANT PERCENTAGE/MATHESON” & “DOUBLE DECLINING BALANCE METHOD” 56.) An asset costing $20,000 has estimated useful life of 5 years and salvage value of $4,500. Calculate the depreciation for the first year of its life using double declining balance method. a. $8000

c. $8500

b.$7000

d. $7500

57.) An asset costing $20,000 has estimated useful life of 5 years and salvage value of $4,500. Calculate the depreciation of the asset for the second year of its life. a. $4700

c. $4900

b. $4800

d. $4600

58.) An asset costing $20,000 has estimated useful life of 5 years and salvage value of $4,500. Calculate the depreciation of the asset for the 3rd year. a. $2600

c. $2700

b. $2900

d.$2800

59.) The Farma company provides the following information: Cost of the equipment: $500,000, Salvage value: $50,000, Useful life: 5 years. What is the Declining balance rate? a. 35%

c. 30%

b. 45%

d. 40%

60.) The book value of the equipment at the beginning of year 5 is $64,800 whose 40% is $2,592. If this depreciation is used, the book value of the equipment at the end of year 5 will be $38,880 ($64,800 – $25,920) that is less than the salvage value. As stated earlier, the asset is depreciated only to its salvage value under declining balance method. Therefore, the depreciation in year 5 is? a. $14,800

c. $14,600

b. $14,700

d. $14,900

61.) An earth moving equipment that cost P90,000 will have an estimated salvage value of P18,000 at the end of 8 years. Using double-declining balance method, compute the book value and the total depreciation at the end of the 5th year. a. P 10,720

c. P 15,674

b. P 23,183

d. P 21,357

62.) An asset is purchased for P10,000. Its estimated economic life is 10 years after which it will be sold for P1,000. Find the depreciation in the first three years using sum-of-years digit method. a. P 4505

c. P 5120

b. P 5210

d. P 6250

10 | P a g e 63.) Littlefield Company recently purchased a machine at a cost of $12,000. The machine is expected to have a residual value of $2,000 at the end of its useful life in five years. Calculate depreciation expense for 3 years using the double-declining balance (DDB) method. a. year1:$4,800, year2:$2,880, year3:$1,728

c. year1:$4,888, year2:$2,870, year3:$1,708

b. year1:$4,880, year2:$2,800, year3:$1,708

d. year1:$4,808, year2:$2,888, year3:$1,768

64.) An asset has a useful life of 3 years. Cost of the asset is $2,000, residual Value is $500, rate of depreciation is 50%. Depreciation expense for the three years will be as follows: a.

year1: 755, year2: 1225, year3: 1550

c.

year1: 750, year2: 1125, year3: 1500

b

year1: 754, year2: 1325, year3: 1540

d.

year1: 740, year2: 1215, year3: 1400

65.) Company A purchases a backup generator for $200,000. The service life of the generator is expected to be 10 years. The generator is estimated to have a scrap value of $50,000 at the end of its serviceable life. The declining balance depreciation rate would be 20% and in the first year of operation, depreciation expense for this generator would be? a. $41,000

c. $42,000

b. $40,000

d. $43,000

“SUM OF THE YEARS’ DIGIT(SYD) METHOD” 66.) Use sum of the years' digits method of depreciation to prepare a depreciation schedule of the following asset: Cost Salvage Value Useful Life in Years Asset is Depreciated

$45,000 $5,000 4 Yearly

a. $ 35,000

c. $ 40,000

b. $ 30,000

d. $45,000

67.) The initial cost of a piece of construction equipment is Rs.3,000,000 having a useful life of 10 years. The estimated salvage value of the equipment at the end of the useful life is Rs.450,000. Determine the book value (BV5) of the construction equipment at the end of 5th year and depreciation (d5) for 5th year using Sum-of-the-years-digits method? a. BV5= Rs. 1,181,712.19 ;d5= Rs. 278,181.82 278,181.82

c.BV5= Rs. 1,145,454.54 ; d5= Rs.

b. BV5= Rs. 1,181,712.19; d5= Rs. 242,037.44 127,341.13

d. BV5= Rs. 1,725,000.00 ; d5= Rs.

68.) The Monster company purchased a machine on January 1, 2015. The relevant information is given below: Cost of the machine: $250,000, expected useful life of machine: 5 years, Salvage value: $25,000. What is the depreciation expense in 1 year? a. $180,000

c. $185,000

b. $175,000

d. $170,000

11 | P a g e 69.) Following information relates to a fixed asset: Cost $100,000 Residual Value $10,000 Useful Life 3 Years Calculate depreciation expense for year 1 of the asset using the sum of the years' digits method. a. $45,000

c. $47,000

b. $46,000

d. $48,000

70.)A Machine costs of P8,000 and an estimated life of 10 years with a salvage value of P500. What is its book value after 8 years using SYD? a. P 909.9091

c. P 707,7071

b. P 808.8081

d. P 606.6061

“ARITHMETIC GRADIENT” , “ANNUITY” , “DEFERRED ANNUITY” , “ANNUITY DUE” , “PERPETUITY” 71.) A contract has been signed to lease a building at P20,000 per year with an annual increase of P1,500 for 8 years. Payments are to be made at the end of each year, starting one year from now. The prevailing interest rate is 7%. What lump sum paid today would be equivalent to the 8year lease payment plan? a. P147,690.3773

c. P147,069.3773

b. P147,609.7373

d. P147,609.3773

72.) A man wishes to provide a fund for his retirement such that from his 60th to 70th birthdays he will be able to withdraw equal sums of P18, 000 for his yearly expenses. He invests equal amount for his 41st to 59th birthdays in a fund earning 10% compounded annually. How much should each of these amounts be? a. P2,285.00

c. P2,282.00

b. P2,225.00

d. P2,385.00

73-74.) Determine the present worth and the accumulated amount of an annuity consisting of 6 payments of P120, 000 each, the payment are made at the beginning of each year. Money is worth 15% compounded annually. 73.)

What is the present worth?

a. P532,259.00

c.

P542,259.00

b. P522,259.00

d.

P512,259.00

74.) What is the Future worth? a. P 1,280,016.00

c. P 1,208,016.00

b. P 1,028,016.00

d. P 1,208,106.00

75.) Company “Rich” pays P20000 dividends annually and estimates they will pay it indefinitely. How much are investors willing to pay for the dividend with a required rate of return of 5%? a. P 300,000

c. P 400,000

b. P 350,000

d. P 450,000

12 | P a g e 76.) What is the future worth of P600 deposited at the end of every month for 4 years if the interest is 12% compounded quarterly? a. 36,461.00

c. 36,641.00

b. 34,641.00

d. 34,461.00

77.) The buyer of a certain machine may pay either P2000 cash down payment and P2000 annually for the next 6 years or pay P3500 cash and P2000 annually for the next 5 years. If money is worth 12% compounded annually, which method of payment is better? a. Option A

c. Both

b. Option B

d. None

78.) 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 P50000 ten years from now? Assume the interest is 14% compounded annually. a. 3928.60

c. 3892.60

b. 3298.60

d. 3289.60

79.) What is the present worth of a 10 year annuity paying P10,000 at the end of year, with interest at 15% compounded annually a.

50,199

c. 50,166

b.

50,177

d. 50,188

80.) The present worth is 10,000 of a 5 year annuity paying 1,000 at the end of year. What is the interest when compounded semi-annually a. Financer A

c.

Financer C

b. Financer B

d. Financer D

81.) A lathe for a machine shop costs P 60,000 if paid in cash. On the installment plan, a purchase should pay P 20,000 down-payment and 10 quarterly installments, the first due at the end of the first year after purchased. If money is worth 15% compounded quarterly, determine the quarterly installments. a.

5349.18

c.

5349.81

b.

5439.18

d. 5439.81

82.) What is the future worth of P600 deposited at the end of every month for 4 years if the interest is 12% compounded quarterly? A = P600 n = (12) (4) = 48 i = 12% compounded quarterly F =? a. P36,641.13

c. P36,641.31

b. P36,461.13

d. P36,461.31

13 | P a g e 83.) Mr. Reyes borrows P600, 000 at 12% compounded annually, agreeing to repay the loan in 15 equal annual payments. How much of the original principal is still unpaid after he has made the

8th payment? a. 420,042

c. 402,024

b. 402,402

d. 402,042

84-87.) M purchased a small lot in a subdivision, paying P200, 000 down and promising to pay P15, 000 every 3 months for the next 10 years. The seller figured interest at 12% compounded quarterly.

84.) What was the cash price of the lot? a. P546,722 b. P564,722

c. P546,272 d. P547,622

85.) If M missed the first 12 payments, what must he pay at the time the 13th is due to bring him up to date? a. P243,270 b. P234,270

c. P234,720 d. P243,720

14 | P a g e 86.) After making 8 payments, M wished to discharge his remaining indebtedness by a single payment at the time when the 9th regular payment was due, what must he pay in addition to the regular payment then due? a. P300,060

c. P300,006

b. P300,600

d. P300,606

87.) If M missed the first 10 payments, what must he pay when the 11th payment is due to discharge his entire indebtedness? a. P497,984

c. P497,948

b. P479,984

d. P479,948

88-89.) A new office building was constructed 5 years ago by a consulting engineering firm. At that time the firm obtained the bank loan for P 10,000,000 with a 20% annual interest rate, compounded quarterly. The terms of the loan called for equal quarterly payments for a 10-year period with the right of prepayment any time without penalty. Due to internal changes in the firm, it is now proposed to refinance the loan through an insurance company. The new loan is planned for a 20- year term with an interest rate of 24% per annum, compounded quarterly. The insurance company has a onetime service charge 5% of the balance. This new loan also calls for equal quarterly payments. 88.) What is the balance due on the original mortgage (principal) if all payments have been made through a full five years? a. P7,262,747.029

c. P7,222,747.029

b. P7,622,747.029

d. P7,267,247.029

89.) What will be the difference between the equal quarterly payments in the existing arrangement and the revised proposal? a. P121,862

c. P122,862

b. P120,862

d. P123,862

90.) An asphalt road requires no upkeep until the end of 2 years when P60, 000 will be needed for repairs. After this P90, 000 will be needed for repairs at the end of each year for the next 5 years, then P120, 000 at the end of each year for the next 5 years. If money is worth 14% compounded annually, what was the equivalent uniform annual cost for the 12-year period? a. P 77,245.82423

c. P 79,245.82423

b. P 78,245.82423

d. P 76,245.82423

91.) The will of a wealthy philanthropist left P5, 000,000 to establish a perpetual charitable foundation. The foundation trustees decided to spend P1, 200,000 to provide facilities immediately and to provide P100, 000 of capital replacement at the end of each 5 year period. If the invested funds earned 12% per annum, what would be the year end amount available in perpetuity from the endowment for charitable purposes? a. P 441,259

c.P 440,259

b. P 442,259

d.P 443,259

15 | P a g e 92.) Alan wants to retire and receive P12,000 a month. He wants to pass this monthly payment to future generations after his death. He can earn an interest of 8% compounded annually. How much will he need to set aside to achieve his perpetuity goal? a. P 1,700,000

c. P 1,900,000

b. P 1,800,000

d. P 2,000,000

93.) What is the present value of an annuity of P1,500 payable at the end of each 6-month period for 2 years if money is worth 8%, compounded semiannually? a. P5,545.84

c. P4,555.84

b. P4,445.84

d. P5,444.84

94.) Suppose that a court settlement results in a P750,000 award. If this is invested at 9% compounded semiannually, how much will it provide at the beginning of each half-year for a period of 7 years? a. P 70,205.97

c. P 70,520.97

b. P 70,250.97

d. P 70,502.97

95.) A deferred annuity is purchased that will pay P10,000 per quarter for 15 years after being deferred for 5 years. If money is worth 6% compounded quarterly, what is the present value of this annuity? a. P 229,863.85

c. P 229,386.85

b. P 292,863.85

d. P 292,386.85

“CAPITALIZED COST” 96.) A public project has an initial cost of Rs.11000000 and annual operating and maintenance cost of Rs.700000. Further the project will have one time major repair work of Rs.2000000 at the end of 15 year. Find out the capitalized cost of the alternative if interest rate is 12% per year. a.

-Rs.17197833

b. -Rs.17189733

c.

-Rs.17918733

d. -Rs.17198733

97.) A dam will have a first cost of P5,000,000, an annual maintenance cost of P25,000 and mirror reconstruction costs of P100,000 every five years. At an interest rate of 8% per year, the capitalized cost of the dam is nearest to? a. -P5,525,500

c. –P5,312,500

b. –P213,125

d. –P525,625

98.) The first cost of certain piece of equipment is P50,000. It will have an annual operating cost of P20,000 and 5,000 salvage value after its 5 year life. At an interest rate of 10% per year, the capitalized cost of the equipment is closest to? a. –P32,371

c. –P323,710

b. –P122,712

d. –P522.710

16 | P a g e 99.) At 6%, find the capitalized cost of a bridge whose cost is P250M and life is 20 years, if the bridge must be partially rebuilt at a cost of P100M at the end of each 20 years. a. P275.3M

c. P295.3M

b. P265.5M

d.P282.1M

100.) Find the capitalized cost of a Building whose cost 5,000,000 and life of 20 years at rate of 6%, if the building must be partially rebuilt at a cost of 2,500,000 at the end of each 20years. a. P 6312689.874

c. P 6136289.874

b. P 6123689.874

d. P 6132689.874

P a g e | 17 ANSWER KEY 1.)

A

29.) B

57.) B

85.) B

2.)

A

30.) A

58.) C

86.) C

3.)

C

31.) C

59.) D

87.) D

4.)

D

32.) C

60.) A

88.) A

5.)

A

33.) B

61.) D

89.) B

6.)

B

34.) D

62.) C

90.) C

7.)

C

35.) B

63.) A

91.) C

8.)

A

36.) A

64.) C

92. )B

9.)

B

37.) C

65.) B

93. )D

10.) C

38.) D

66.) C

94.) A

11.) B

39.) A

67.) C

95.) D

12.) A

40.) C

68.) B

96.) D

13. )C

41.) B

69.) A

97.) A

14.) C

42.) C

70.) A

98.) C

15.) C

43.) A

71.) D

99.) C

16.) D

44.) D

72.) A

100.) D

17.) A

45.) A

73.) B

18.) B

46.) B

74.)

19.) C

47.) B

75.) C

20.) A

48.) A

76.) C

21.) D

49.) A

77.) A

22.) D

50.)B

78.) A

23.) D

51.) A

79.) D

24.) A

52.) A

80.) A

25.) D

53.) B

81.) B

26.) C

54.) C

82.) C

27.) A

55.) A

83.) D

28.) D

56.) A

84.) A

C

P a g e | 18 “SIMPLE INTEREST” (SOLUTION) 1. 𝐼 = 𝑃𝑖𝑛 P = P8000, i = 9%, n =1

𝐼 = (8000)(.09)(1) 𝐼 = 𝑃720

2. 𝐹 = 𝑃(𝐼 + 𝑖𝑛) F = P5000, i = 15%, n = 15/12

15 𝐹 = 5000(1 + (. 15) ( )) 12 𝐹 = 𝑃5,937.50

3. 𝐼 = 𝑃𝑖𝑛 P = P1500, i = 12%, n = 3/12

𝐼 = 1500(.12)(4 +

3 ) 12

𝐼 = 𝑃765 4. 𝐼 = 𝑃𝑖𝑛 P = P1200, i = 6%, n = 5

𝑛=

4 1 = 12 3

1 𝐼 = (1200)(0.06)( ) 3 𝐼 = 𝑃24 5. 𝐼 = 𝑃𝑖𝑛 P=P10,000, I=0.03/12, n=12(6)

𝐼 = (10,000)(0.05)(5) 𝐼 = 𝑃2,500

“COMPOUND INTEREST” (SOLUTION)

6. 𝐹 = 𝑃(1 + 𝑖)𝑛 P = P5000, I = 0.03/12, n=12(6)

𝐹 = 5000(1 +

0.03 12(6) ) 12

𝐹 = 𝑃5984.74

7. 𝐹 = 𝑃(1 + 𝑖)𝑛 P = P3500, I = 9%, n = 1

𝐹 = 3500(1 +

0.015 4(2) ) 4

𝐹 = 𝑃3606.39

P a g e | 19 𝑟

8. 𝐹 = 𝑃(1 + )𝑚𝑛 𝑚

P = P450, F = 750, m = 1, n = 10

750 = 450(1 +

𝑟 1(10) ) 1

𝑟 = 0.0524 𝑟

9. 𝐹 = 𝑃(1 + )𝑚𝑛 𝑚

P = P, F = 2P, r = 5%, n = 1

2𝑃 = 𝑃(1 +

0.05 1(𝑛) ) 1

2𝑃 = 𝑃(1.05)𝑛 2 = (1.05)𝑛 ln(2) = (𝑛)𝑙𝑛(1.05) 𝑛 = 14.2 𝑦𝑒𝑎𝑟𝑠 10. 𝐹 = 𝑃(1 + 𝑟)𝑛 P = P3500, I = 9%, n = 1

𝐹 = 20000(1 + .06)3 = 23820.32 𝑇ℎ𝑒 𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 = 23820.32 − 20000 𝑇ℎ𝑒 𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 = 3820.32

“EFFECTIVE RATE OF INTEREST”

11. 𝐸𝑅 = ((1 +

0.08 1(4) ) − 1)(100%) 4

𝐸𝑅 = 8.24%

12.

P a g e | 20

13. 𝐸𝑅 = ((1 + 0.10⁄365)1(365) − 1)(100%) 𝐸𝑅 = 10.515%

14. 6.1678% = ((1 +

𝑖 1(12) ) − 1) (100%) 12

𝑖 = 6%

15.

𝐸𝑅 = ((1 +

0.12 1(6) ) − 1) (100%) 6

𝐸𝑅 = 12.62%

“CONTINIOUS COMPOUNDING”

16. 𝐹 = 𝑃𝑒 𝑟𝑛 𝐹 = (1000)𝑒 (0.08)(3) 𝐹 = 1271.24915 17. 𝐿𝑒𝑡 𝑃 = 1000 & 𝐹 = 3000 𝐹 = 𝑃𝑒 𝑟𝑛 3000 = (1000)𝑒 (0.09)(𝑛) n = 12.21% 18. 𝐹 = 𝑃𝑒 𝑟𝑛 𝐹 = (1000)𝑒 (0.05)(5) F = 1284.025417

P a g e | 21

19. 𝐹 = 𝑃𝑒 𝑟𝑛 𝐹 = (500)𝑒 (0.1)(5) F = 824.3606354

20. 𝐹 = 𝑃𝑒 𝑟𝑛 𝐹 = (2000)𝑒 (0.13)(20) F = 26,927.47607

“COMPARING OF ALTERNATIVES” 21.) P 𝑃 P = 0 = −25,000 + (3100) ( , i%, 15) − (355) ( , i%, 15) 𝐴 𝐴 P 𝑃 25,000 = ($2745) ( , i%, 15) 9.1075 = ( , i%, 15) 𝐴 𝐴 𝑃 ( , 6%, 15) = 9.7122 𝐴 P ( , 8%, 15) = 8.5595 𝐴 9.7122 − 9.1075 = 0.5245 9.7122 − 8.5595 i% = 6% + (0.5245)(8% − 6%) i% = 7.049% or (7.0%) 22.) A (EUAC)x = (P6000) ( , 8%, 7) + P150 𝑃 (EUAC)x = (P6000)(0.1921) + P150 (EUAC)x = P1302.60 or (P1300)

P a g e | 22

23.) A 𝐴 (EUAC)x = (P12,000) ( , 8%, 13) + P175 − P4000( , 8%, 13) 𝑃 𝑃 (EUAC)x = (12,000)(0.1921) + P175 − P4000(0.0465) (EUAC)x = P1507 or (P1510)

24. (𝐸𝑈𝐴𝐶)𝑥 < (𝐸𝑈𝐴𝐶)𝑦 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 𝑋 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 (𝐸𝑈𝐴𝐶)𝑥 < (𝐸𝑈𝐴𝐶)𝑦 25. 𝑐𝑜𝑠𝑡𝑠 = $750,000 + (𝑃1.35)(𝑛𝑜. 𝑜𝑓 𝑔𝑒𝑎𝑟𝑠) 𝑟𝑒𝑣𝑒𝑛𝑢𝑒𝑠 = (𝑃7.35)(𝑛𝑜. 𝑜𝑓 𝑔𝑒𝑎𝑟𝑠)𝑃750,000 + (𝑃1.35)(𝑛𝑜. 𝑜𝑓 𝑔𝑒𝑎𝑟𝑠) 𝑟𝑒𝑣𝑒𝑛𝑢𝑒𝑠 = (𝑃7.35)(𝑛𝑜. 𝑜𝑓 𝑔𝑒𝑎𝑟𝑠)𝑃750,000 + (𝑃1.35)𝑛𝑜. 𝑜𝑓 𝑔𝑒𝑎𝑟𝑠 = 𝑃7.35(𝑛𝑜. 𝑜𝑓 𝑔𝑒𝑎𝑟𝑠) 𝑛𝑜. 𝑜𝑓 𝑔𝑒𝑎𝑟𝑠 =

𝑃750,000 = 125,000 𝑃7.35 − 𝑃1.35

“Discount” 26.) 𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 = 𝑃100 × 25% = 𝑃 100 × 0.25 = 𝑃 25 𝑆𝑎𝑙𝑒 𝑝𝑟𝑖𝑐𝑒 = 𝐿𝑖𝑠𝑡 𝑃𝑟𝑖𝑐𝑒 − 𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 𝑃𝑟𝑖𝑐𝑒 = 𝑃 100 − 𝑃25 = 𝑃 75 27.) 𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 = 𝑃24 × 50% = 𝑃24 × 0.50 = 𝑃12 𝑆𝑎𝑙𝑒 𝑃𝑟𝑖𝑐𝑒 = 𝐿𝑖𝑠𝑡 𝑃𝑟𝑖𝑐𝑒 − 𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 = 𝑃 24 − 𝑃 12 = 𝑃12 28.) 𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 = 𝑃 27 × 33.33333% = 𝑃27 × 0.3333333 = 𝑃 8.9999999 𝑆𝑎𝑙𝑒 𝑃𝑟𝑖𝑐𝑒 = 𝐿𝑖𝑠𝑡 𝑃𝑟𝑖𝑐𝑒 − 𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 = 𝑃 27 − 𝑃8.9999999 = 𝑃 18.0000001 29.) 𝐷 = 𝐹𝑊 − 𝑃𝑊 = 1500 − 1340 𝑑=

𝑖=

𝐷 160 = = 0.1066666667 𝐹𝑊 1500

𝑑 0.1066666667 = = 0.1194029851𝑥100 = 11.94% 1 − 𝑑 1 − 0.1066666667

P a g e | 23

30.) 𝐷 = 𝐹𝑊 − 𝑃𝑊 = 2000 − 1400 𝑑=

𝐷 𝐹𝑊

=

600 2000

= 0.3 x 100 = 30%

“INFLATION”

31.) d = i + f + if = 0.06 + 0.10 + (0.06)(0.10) = 0.166 P= (P2000)(1 + d)−𝑛 = (P2000)(1 + 0.166)−2 = P1471 32.) F = 1000 (1 - 0.04)10 F = 665 33.) (1+i)=(1+i’)(1+f) (1+i’)=(1+i)/(1+f) (1+i’)=(1+0.055)/(1.0.02) i’=0.034 or 3.4% 34.)

(1+i)=(1+i’)(1+f) (1+i’)=(1+i)/(1+f) (1+i’)=(1+0.08)/(1.0.06) i’=0.01887 or 1.887%

35.) R Peso base 1955 at 2010= (R Peso base 1955 at 2010) (1 + 𝑖′)𝑛 = P2,400,000(1 + 𝑖′)𝑛 = P2,400,000(1 + 0.01887)55 R Peso base 1955 at 2010=P6,710,330

P a g e | 24

36.) GIVEN: A = 300 g = 0.015 i = 0.15 n = 5 SOLUTION: The cash flow at the end of year 1 is ₱150,000, increasing by g=5% per year. worth is:

𝑃=

(1+0.05)5 ] (1+0.12)5

150000[1−

0.12−0.05

P = 150,000(3.9401) P = ₱591,000 37.) GIVEN: A = 300 g = 0.15 i = 0.15 n = 5

SOLUTION: (1 + 0.15)5 ] (1 + 0.09)5 0.09 − 0.15

300[1 − 𝑃=

P = 300 (5.12) P = ₱1,536.22

Therefore, the present

P a g e | 25

38.) GIVEN: A = ₱50,000 g = 0.06 i = 0.10 n = 8

SOLUTION:

𝑃=

(1+0.06)8 ] (1+0.10)8

50000[1−

0.10−0.06

P = 50000 (5.12) P = ₱320,572.64

39.)

𝑃 = −8000 − 1700 [

1−(

1.11 6 ) 1.08

𝑃

] + 200(𝐹,8%,6)

0.08−0.11

P = – 8000 – 1700 (5.9559) + 126 P = – $17,999

40.) g= 5% i = 7% N = 25 years A = $50,000 𝑃 = 𝑃50,000 [

1 − (1.0525 )(1.07)−25 ] 0.07 − 0.05

𝑃 = 𝑃940,696

P a g e | 26

“SERVICE OUTPUT METHOD” 41.) SO= (P200,000 - P50,000) / 5,000 hours SO= P150,000 / 5,000 hours, or P30 per hour This value is then multiplied by the hours of operation in the current accounting period to determine the depreciation expense: SO= P30 per hour x 300 hours, = P9,000 42.) Depreciation charge for the first year is calculated as follows:

𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝐸𝑥𝑝𝑒𝑛𝑠𝑒 =

(12,000,000 − 2,000,000)(2) 50

𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝐸𝑥𝑝𝑒𝑛𝑠𝑒 = 𝑃400,000 43.) 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝐸𝑥𝑝𝑒𝑛𝑠𝑒 =

(𝑃110,000,000 − 𝑃10,000,000)(15,000,000 𝑢𝑛𝑖𝑡𝑠) 150,000,000 𝑢𝑛𝑖𝑡𝑠

𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝐸𝑥𝑝𝑒𝑛𝑠𝑒 = 𝑃10,000,000

44.) 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 =

(𝑃16,000,000)(40,000 𝑡𝑜𝑛𝑒𝑠) 200,000 𝑡𝑜𝑛𝑒𝑠

𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑃3,680,000

45.) 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝐸𝑥𝑝𝑒𝑛𝑠𝑒 =

(𝑃1,000,000 − 𝑃200,000)(2) 8

𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝐸𝑥𝑝𝑒𝑛𝑠𝑒 = 𝑃200,000

P a g e | 27

“STRAIGHT LINE METHOD” 46.) 𝐶𝐿 = 0 𝑑=

𝐶𝑜 − 𝐶𝐿 𝐿

0.2(𝐶𝑜) =

𝐿=

𝐶𝑜 𝐿

1 = 5 𝑌𝑒𝑎𝑟𝑠 0.2

47.) 𝐶𝑜 = 𝑃360,000 + 𝑃5,000 + 𝑃25,000 + 𝑃20,000 𝐶𝑜 = 𝑃410,000 𝐶𝐿 = 𝑃60,000 𝐿 = 20 𝑦𝑒𝑎𝑟𝑠 𝑑=

(𝑃410,000 − 𝑃60,000) 20 𝑑 = 𝑃17,500 𝐷10 = 10(17,500)

𝐶10 = 410,000 − 175,000 = 235,000 48.) 𝐶15 = 𝑃410,000 − 15(𝑃17,500) 𝐶15 = 𝑃147,500 49.)

=

($35,000 – $3,000) 𝑦𝑒𝑎𝑟𝑠 10 = $3,200

50.)

P a g e | 28

Depreciation expense per annum shall be:



($2000−$500) 3 𝑌𝑒𝑎𝑟𝑠

∶ $500 𝑝. 𝑎.

𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑒𝑥𝑝𝑒𝑛𝑠𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑦𝑒𝑎𝑟 𝑒𝑛𝑑𝑒𝑑 30 𝐽𝑢𝑛𝑒 2013: $500 𝑥

6 = $250 12

As $500 calculated above represents the depreciation cost for 12 months, it has been reduced to 6 months equivalent to reflect the number of months the asset was actually available for use.



𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑒𝑥𝑝𝑒𝑛𝑠𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑦𝑒𝑎𝑟 𝑒𝑛𝑑𝑒𝑑 30 𝐽𝑢𝑛𝑒 2014: $500 𝑥

12 = $500 12

As the asset was available for the whole period, the annual depreciation expense is not apportioned.

“SINKING FUND FORMULA” 51.) 𝐶𝑜 = 𝑃20,000 𝐶𝐿 = 𝑃1,000 𝐿 = 10 = (𝑃20,000 − 𝑃1,000)(

1.036 − 1 ) 1.0310 − 1

= 10,720.61075 52.) The total original cost of the Diesel – generator unit is P65,000 plus the shipment of the unit to the job site and cost for installation.

The annual depreciation cost with an expected unit life of 10 years and a salvage value of P5,000 would be

53.)

P a g e | 29

54.) 𝐶𝑜 = 𝑃65,000 + 𝑃3,000 + 𝑃2,000 = 𝑃70,000 𝑑=

𝑃70,000 − 𝑃5,000 10 𝑑 = 𝑃6,500

55.) 𝐶𝑜 = 𝑃65,000 + 𝑃3,000 + 𝑃2,000 = 𝑃70,000 𝑑=

𝑃70,000 − 𝑃5,000 (1 + 0.05)10 − 1 0.05 𝑑 = 𝑃5,167.80

“DECLINING BALANCE METHOD/CONSTANT PERCENTAGE/MATHESON” 56.) 𝑆𝑡𝑟𝑎𝑖𝑔ℎ𝑡 − 𝑙𝑖𝑛𝑒 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑒 = 1 ÷ 5 = 0.2 = 20% 𝐷𝑒𝑐𝑙𝑖𝑛𝑖𝑛𝑔 𝐵𝑎𝑙𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑒 = 2 × 20% = 40% 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = 40% × $20,000 = $8,000 57.) 𝐷𝑒𝑐𝑙𝑖𝑛𝑖𝑛𝑔 𝐵𝑎𝑙𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑒 = 40% 𝐵𝑜𝑜𝑘 𝑉𝑎𝑙𝑢𝑒 = 𝐶𝑜𝑠𝑡 − 𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = $20,000 − $8,000 = $12,000 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = 40% × $12,000 = $4,800 58.) 𝐷𝑒𝑐𝑙𝑖𝑛𝑖𝑛𝑔 𝐵𝑎𝑙𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑒 = 40% 𝐵𝑜𝑜𝑘 𝑉𝑎𝑙𝑢𝑒 = $20,000 − $8,000 − $4,800 = $7,200 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = 40% × $7,200 = $2,880 𝑇ℎ𝑒 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑎𝑏𝑜𝑣𝑒 𝑤𝑖𝑙𝑙 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑡ℎ𝑒 𝑏𝑜𝑜𝑘 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑠𝑠𝑒𝑡 𝑏𝑒𝑙𝑜𝑤 𝑖𝑡𝑠 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 ($7,200 − $2,880 = $4,320 < $4,500). 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑤𝑜𝑢𝑙𝑑 𝑜𝑛𝑙𝑦 𝑏𝑒 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 𝑢𝑝 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑤ℎ𝑒𝑟𝑒 𝑏𝑜𝑜𝑘 𝑣𝑎𝑙𝑢𝑒 = 𝑠𝑎𝑙𝑣𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒. 𝑇ℎ𝑢𝑠, 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝐴𝑙𝑙𝑜𝑤𝑒𝑑 = $7,200 − $4,500 = $2,700 59.)

𝑆𝑡𝑟𝑎𝑖𝑔ℎ𝑡 − 𝑙𝑖𝑛𝑒 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒:

1 = 0.2 𝑜𝑟 20% 5

𝐷𝑒𝑐𝑙𝑖𝑛𝑖𝑛𝑔 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 𝑟𝑎𝑡𝑒 (𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑒𝑑 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒): 20% × 2 = 40% 60.) = 𝐵𝑜𝑜𝑘 𝑣𝑎𝑙𝑢𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 𝑜𝑓 𝑡ℎ𝑒 𝑦𝑒𝑎𝑟 – 𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 = $64,800 – $50,000

P a g e | 30 = $14,800 “DOUBLE DECLINING BALANCE METHOD” 61.) 2 𝐶5 = 𝑃90000(1 − )5 8 𝐶5 = 𝑃21,357 62.) 𝐶3 = 𝑃9000(1 −

2 3 ) 10

𝐶3 = 𝑃5120 63.) 𝑌𝑒𝑎𝑟 1: (2/5)($12,000) = $4,800 𝑌𝑒𝑎𝑟 2: (2/5)($12,000 − $4,800) = $2,880 𝑌𝑒𝑎𝑟 3: (2/5)($12,000 − $7,680) = $1,728

64.) Depreciation per annum = (Net Book Value - Residual Value) x Rate% R.V

NBV

Rate

Depreciation

Accumalated Depreciation

Year1:

(2000

-

500)

x

50%

=

750

750

Year2:

(1250

-

500)

x

50%

=

375

1125

Year3:

(875

-

500)

x

50%

=

375*

1500

Answer is c 65.) 1st year depreciate expense = 20% 𝑥 $200,000 1st year depreciate expense = 0.2 𝑥 $200,000 = $40,000 “SUM OF THE YEARS’ DIGIT(SYD) METHOD” 66.) 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑌𝑒𝑎𝑟𝑠′ 𝐷𝑖𝑔𝑖𝑡𝑠 = 1 + 2 + 3 + 4 = 4(4 + 1) ÷ 2 = 10 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑏𝑙𝑒 𝐵𝑎𝑠𝑒 = $45,000 − $5,000 = $40,000 Year 1 2 3 4

Depreciable Base $40,000 $40,000 $40,000 $40,000

Depreciation Factor 4/10 3/10 2/10 1/10

Depreciation Expense 4/10 3/10 2/10 1/10

× × × ×

40,000 40,000 40,000 40,000

= 16,000 = 12,000 = 8,000 = 4,000

Accumulated Depreciation $16,000 $28,000 $36,000 $40,000

P a g e | 31

67.) 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡 = 𝑅𝑠. 3000000 − 𝑅𝑠. 450000 = 𝑅𝑠. 2550000 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑦𝑒𝑎𝑟𝑠′ 𝑑𝑖𝑔𝑖𝑡𝑠 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 5𝑡ℎ 𝑦𝑒𝑎𝑟 = (2550000)𝑥

6 = 6 ∗ 46363.63636 = 𝑅𝑠. 278,181.82 55

𝐵𝑜𝑜𝑘 𝑣𝑎𝑙𝑢𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 5𝑡ℎ 𝑦𝑒𝑎𝑟 = (1423636.37 − 278181.82) = 𝑅𝑠. 1,145,454.54 68.)

𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡 (𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑏𝑙𝑒 𝑏𝑎𝑠𝑒): $250,000 – $25,000 = $225,000

𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑒𝑥𝑝𝑒𝑛𝑠𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑦𝑒𝑎𝑟: $225,000 × (

5 ) = $75,000 15

𝐵𝑜𝑜𝑘 𝑣𝑎𝑙𝑢𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑦𝑒𝑎𝑟: $250,000 – $75,000 = $175,000 69) 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑦𝑒𝑎𝑟𝑠′ 𝑑𝑖𝑔𝑖𝑡𝑠 = 3 + 2 + 1 = 6 𝑆𝑡𝑒𝑝 2: 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑏𝑙𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑏𝑙𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 = $100,000 − $10,000 = $90,000 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑢𝑛 − 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑒𝑑 𝑢𝑠𝑒𝑓𝑢𝑙 𝑙𝑖𝑓𝑒

Un- depreciated useful life (years)

Year 1Year 2Year 3 3 2 1

𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑒𝑥𝑝𝑒𝑛𝑠𝑒 𝑌𝑒𝑎𝑟 1: 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑒𝑥𝑝𝑒𝑛𝑠𝑒: 3 = 𝑥 $90,000 6 = $45,000

P a g e | 32

70.) Co=P8000 L=10 CL=500

N=8

𝐷8 =

8[2(10) − 8 + 1] (8,000 − 500) 10(10 + 1) 𝐷8 = 7090.909091

𝐶8 = 8000 − 7090.909091 𝐶8 = 909.0909

“ARITHMETIC GRADIENT” , “ANNUITY” , “DEFERRED ANNUITY” , “ANNUITY DUE” , “PERPETUITY” 71.)

72.)

73.)

P a g e | 33

74.)

75.) P = P20000/5% = P400,000 76.) (1 + 𝑖)12 − 1 = (1 +

0.12 4 ) −1 4

𝑖 = 0.0099 𝑜𝑟 0.99% 𝑛 = 4𝑥12 = 48 (1 + 𝑖)𝑛 − 1 𝐹 = 𝐴( ) 𝑖 𝐹 = 600 (

(1 + 0.0099)48 − 1 ) 0.0099

F = 36,641.00 77.) 𝑃1 = 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑚𝑒𝑡ℎ𝑜𝑑 𝑃 𝑃1 = 𝑃2,000 + 𝑃2,000 ( , 12%, 6) 𝐴 𝑃1 = 𝑃10,222,81 𝑃2 = 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑚𝑒𝑡ℎ𝑜𝑑 𝑃 𝑃2 = 𝑃3,500 + 𝑃2,000 ( , 12%, 5) 𝐴 𝑃2 = 𝑃10,709.55 • 𝐴𝑛𝑠: 𝐹𝑖𝑟𝑠𝑡 𝑀𝑒𝑡ℎ𝑜𝑑 78.) 𝑃 𝑃 𝑃50,000 ( , 14%, 10) = 𝐴 ( , 14%, 5) 𝐹 𝐴 • 𝐴𝑛𝑠: 𝑃3,928.60 79.) 𝑃 𝑃 𝑃 = 𝐴 ( , 𝑖%, 𝑛) = 𝑃10,000 ( , 15%, 10) 𝐴 𝐴 𝑃 = 10,000(

1 − 1.15−10 ) . 15

P=50,188

P a g e | 34 80.) 𝑃 = 𝐴(

(1−𝑖)𝑛 𝑖

10,0000 = 1,000(

)

(1 − 𝑖)−5 ) 𝑖

i =

81.) 3 1 60,000 = 20,000 + 𝑃 ( ) 𝑎 − 𝑎𝑛𝑔𝑙𝑒 − 10(3.75% 1.0375 3 1 (1 − 𝑣 10 ) 40,000 = 𝑃 ( ) , 𝑤ℎ𝑒𝑟𝑒 𝑖 = 0.0375 & 𝑣 = 1/1.0375 1.0375 𝑖

𝑆𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑃 𝑃 = $5,439.1839 82.)

83.)

P a g e | 35

84.)

85.)

86.)

87.)

P a g e | 36 88.)

89.)

90.)

P a g e | 37 91.)

92.) 𝐴 = 𝑃12000⁄0.08/(12) 𝐴 = 𝑃1,800,000

93.) 1 − (1 + 0.04)−4 𝐴 = 𝑃1,500( ) 0.04 𝐴 = 𝑃5444.84

94.) An,due = P750,000, n = 2(7) =14, i = 0.09 / 2 = 0.045 P=? 𝑃750,000 = 𝑃(

1 − (1 + 0.045)−14 )(1 + 0.045) 0.045 𝑃 = 𝑃70,205.97

P a g e | 38 95.) P n k i

= = = =

P10,000 4(15) = 60 4(5) = 20 0.06 / 4 = 0.015 1 − (1 + 0.015)−60 𝐴 = 𝑃10,000( )(1.015)−20 0.015

“CAPITALIZED COST” 96.) 𝑃 700000 𝐶𝑎𝑝𝑖𝑡𝑎𝑙𝑖𝑧𝑒𝑑 𝐶𝑜𝑠𝑡 = −11000000 − 2000000 ( , 𝑖, 𝑛) − 𝐹 𝑖 𝑃 700000 𝐶𝑎𝑝𝑖𝑡𝑎𝑙𝑖𝑧𝑒𝑑 𝐶𝑜𝑠𝑡 = −11000000 − 2000000 ( , 12%, 15) − 𝐹 𝑖 𝐶𝑎𝑝𝑖𝑡𝑎𝑙𝑖𝑧𝑒𝑑 𝐶𝑜𝑠𝑡 = −11,000,000 − 2,000,000𝑥0.1827 − 5,833,333 Capitalized Cost = -Rs.17198733 97.) The P5,000,000 first cost is already a present worth. The P100,000 which occurs every five years can be converted into an infinite A value using the A/F factor for one life cycle. Dividing the A values by i and adding to the P5,000,000 PW will yield the capitalized cost, CC. 𝐶𝐶 = −5,000,000 − 25,000/0.08 − 100,000 (𝐴/𝐹, 8%, 5) / 0.08 𝐶𝐿 = −5,000,000 − 312,500 − 100,000 (0.1705)/0.08 𝐶𝐿 = −𝑃5,525,625

98.) Convert all values into an equivalent A value thru one life cycle and divide by i: A = -50,000 (A/P, 10%, 5) – 20,000 + 5,000 (A/F, 10%, 5) = -50,000 (0.26380) – 20,000 + 5,000 (0.16380) = -32,371 CC = -32,371 / 0.10 = -323,710 99.) 𝐶𝑎𝑝𝑖𝑡𝑎𝑙𝑖𝑧𝑒𝑑 𝑐𝑜𝑠𝑡 = 𝐶𝑜 +

𝐶𝑎𝑝𝑖𝑡𝑎𝑙𝑖𝑧𝑒𝑑 𝑐𝑜𝑠𝑡 = 250 +

𝐶𝑜 − 𝐶𝑛 (1 + 𝑖)𝑛 − 1

100 (1 + 0.06)20 − 1

Capitalized cost = 295.3million

P a g e | 39 100.) 𝐶𝑎𝑝𝑖𝑡𝑎𝑙𝑖𝑧𝑒𝑑 𝑐𝑜𝑠𝑡 = 𝐶𝑜 +

𝐶𝑜 − 𝐶𝑛 (1 + 𝑖)𝑛 − 1

𝐶𝑎𝑝𝑖𝑡𝑎𝑙𝑖𝑧𝑒𝑑 𝑐𝑜𝑠𝑡 = 5,000,000 +

2,500,000 (1 + 0.06)20 − 1

Capitalized cost = 6132689.874