1. A gold mine in Mount Province has an annual output of 25,000 tons of ore. It is expected that the ore body will be exhausted within a period of 20 years. The management cost annually is P750,000 , and the operating cost of the mine and the smelter plants is P120 per ton. The processed ore produce an income of P450 per ton. If the annual dividend rate is 12% payable annually, and the sinking fund rate is 9% annually, determine the valuation of the property now.

GIVEN: n = 20 r = 12% i = 9% REQ’D: P = ?SOLUTION:

Annual gross income = (25000)(P450) = P11, 250, 000 Annual management and operating costs = P750, 000 + 25000(120) = P3, 750, 000

R = P11, 250, 000 - P3, 750, 000 = P7, 500 000

P= Rr+( A /F ,9% , n )

= 75000000.12+0.01954648

= P53, 745, 535. 31 ans.

2. Timber land in Palawan was purchased for P8,000,000 and earned an average annual profit of P1,400,000 for 14 years, at the end of which time the land was sold for P200,000. Assuming that a sinking fund earning 7% was established to provide for depletion, determine the investment rate.

GIVEN:n = 14 years i = 7% R = P1, 400, 000

REQ’D: r = ?SOLUTION:

P=8,000,000−200,000¿ = P7, 922, 436.55

P= Rr+¿¿

r=RP

−( A/ F ,i% , n)

r= 14000007922436.

−0.07¿¿

r = 13.2%3. The San Fernando Manufacturing Company owns four different production machine with data

tabulated below :

Machine Number Number Owned First Cost Salvage Value Expected Life 1 8 P40, 000 P12, 000 12 2 6 32, 000 8, 000 10 3 4 18, 000 4, 000 10 4 4 24, 000 6, 000 8

One-half of the machines of the kind will be replaced after 8 years and the rest will be sold after 12 years. Compute the total annual straight-line depreciation charges by (a) the group depreciation method, and (b) the composite depreciation method.

SOLUTION:a.) Group Depreciation Method

AVERAGE DEPRECIATION=12+10+10+84

=10 years

ANNUALGROUP DEPRECIATION=TOTAL DEPRECIATION AMOUNTAVERAGE LIFE

¿8 (40000 )−12000¿+6 (32000−8000 )+4 (18000−4000 )+4(24000−6000) ¿10

=49600010

=P49 ,600 per year

ANNUAL DEPRECIATION /MACHINE= 490008+6+4+4

=P2254.55 /MACHINE

DEPRECIATION FOR EACH YEAR IS SHOWN IN THE TABLE BELOW.

YEAR NUMBER IN SERVICE ANNUAL DEPRECIATION ACCOMULATED 1 2 3 4 DEPRECIATION

1 – 8 8 6 4 4 P49, 600 P396, 4009 – 12 4 3 2 2 P24, 000.05 P198, 400.40

P595, 200.05 DEPRECIATION FOR YEARS 9 – 12 = P2254.55(4 + 3 + 2 + 2) = P24, 800.05

b.) Composite Depreciation Method:The annual depreciation amounts for each machine are:

D1=8 (40000−12000 )

12=P18 ,666.67 / year

D2=6(32000−8 ,000)

10=P14 ,400/ year

D3=4 (18000−4000)

10=P5 ,600 / year

D4=4 (24 ,000−6000)

8=P9 ,000/ year

Composite Depreciation amounts:

Years 1 – 8:

P18, 600.67 + P14, 400 + P5, 600 + P9, 000 = P47, 666.67

Accumulated Depreciation after 8 years = 8(47, 666.67) = P381, 333.36

Years 9 – 10:

18, 666.67 + P14, 400 + P5, 600 = P38, 666.67

Accumulated Depreciation after 10 years = P381, 333.36 + 2(38, 666.67) = P458, 666.70

Years 11 – 12: (Machine 1 only)

Accumulated Depreciation after 12 years = P450, 666.70 + 2(P18, 666.67) = P496, 000.00

The total accumulated depreciation by either method after 12 years = P496, 000.00

4. A firm plans to market a new minicomputer that will sell for $10,000. This financing scheme is being considered by the company: Payments of $872 each year for 20 years would be made by the purchaser after an initial down payment is made. If their interest is charge is 6% compounded monthly, what down deposit should the company request?

GIVEN:Cost = $10, 000 A =$872 n = 20 years i = 6% compounded monthly

REQ’D:DOWN DEPOSIT = ?

SOLUTION:

ER=¿

COST=DOWN DEPOSIT + A ( P/ A , i , n )

1000=DOWN DEPOSIT+872¿

DOWN DEPOSIT=10000−9866.95

DOWN DEPOSIT=$133.05

5. A man deposit $2,000 an a savings account when his son was born. The nominal interest rate was 8% per year, compounded continuously. On the son’s birthday, the accumulated sum is withdrawn from the account. How much would this accumulated amount be?

GIVEN:P = $2000Nominal rate of interest = 8% per year compound continuouslyN = 18

REQ’D:

F = ?

SOLUTION:

Effective Rate of Interest=e0.08– 1=0.08328708

F=P ¿

F=$2000¿

F = $8, 441

6. Maintenance costs for a new bridge with an expected 50-year life are estimated to be $1,000 each year for the first 5 years, followed by a $10,000 expenditure in the 15 th year and a $10,000 expenditure in year 30. If I = 10% per year, what is the equivalent uniform annual cost over the entire 50-year period?

GIVEN:

i = 10%n = 50 yearsA1 = $ 1,000 maintenance cost each year for the first 5 years

C1=1,000[ (1+0.10 )5−10.10 ]=$6,105.1 money worth after 5 years

C2 = $ 10,000 expenditure during the 15th yearC3 = $ 10,000 expenditure during the 30th year

Required:

P = ? & A = ?

SOLUTION:

F=6,105.1 (1+0.10 )45+10,000 (1+0.10 )35+10,000 (1+0.10 )20

F=$793,303.0599

Since,

F=P (1+ i )n

P=793,303.0599(1+0.10 )50

P=$6,757.79

Therefore;

A= Fi(1+i )n−1

=793,303.0599(0.10)

(1+0.10 )50−1=$681.59

7. Uncle Wilbur’s trout ranch is now for sale for $40,000. Annual property taxes, maintenance, supplies, and so on, are estimated to continue to be $3,000 per year. revenues from the ranch are expected to be $10,000 next year and then decline by $500 per year thereafter through the tenth year. if you bought the ranch, you would plan to keep it only 5 years and at the time sell it for the value of the land, which is $15,000. If your desired annual rate of return is 12% should you become a trout rancher?

SOLUTION: TOTAL EXPENSES = 15, 000+ 40, 000 = $55, 000 for 5 years An expenses every year = 11, 000 TOTAL REVENGE = 45, 000 + 15, 000 = $60, 000 for 5 years An expenses every year = 12, 000 FPROFIT EVERY YEAR = 12, 000 – 11, 000 = 1, 000

RATE OF RETURN = 100012000

x100=9.09%<12% (do not buy theranch)

8. The heat loss through the exterior walls of a certain poultry processing plant is estimated to cost the owner $3,00 next year. a salesman from super fier Insulation, Inc. has told you, the plant engineer, that he can reduce the heat loss by 80% with the insallation of $15,000 worth of Super fiber now. If the cost of heat loss rises by $200 per year (gradient) after next year and the ownder plans to keep the present building for 10 more years, what would yoy recommended if the money is 12% per year?

Illustration:

1 2 3 4 5 6 7 8 9 10

Without Super Fiber Insulation

1 2 3 4 5 6 7 8 9 10

With Super Fiber Insulation

SOLUTION:

When i = 12%

Without Super Fiber Insulation

F=3000[ (1+0.12 )10−10.12 ]+ 2000.12 [ (1+0.12 )10−1

0.12−10]=$65 ,227.43033

P=65,227.43033(1+0.12 )10

=$21,001.49

With Super Fiber Insulation

F=600[ (1+0.12 )10−10.12 ]+ 2000.12 [ (1+0.12 )10−1

0.12−10]+15000 (1+0.12 )10=$69 ,698.18929

P=69 ,698.18929

(1+0.12 )10=$22,440.95

P = ?

3000

F = ?

3200

36003400

3800

4000

4200

4400

4600

P = ?

F = ? 600

800

12001000

1400

1600

1800

2000

2200

2400

15000

4800

Therefore:

The owner must not accept the offer of the Salesman.

9. Solve for the value of F below so that the left-hand cash flow diagram is equivalent to the one on the right. Let I = 8% per year.

10. A corporations floats P200,000 worth of ten-year callable bonds is P1,000 denominations. The bond rate is 7% compounded annually. Prepare an amortization.

GIVEN:

F = P200, 000 n = 10 i = 7%

SOLUTION:

no .of bonds=2000001000

=P200bonds

A ¿¿

28475.50 – 14000 = P14, 475.50no .of bonds=14475.50

1000=15bonds

YEAR PRINCIPALINTEREST AT

7%NO. OF BONDS

RETIRED

AMOUNT OF PRINCIPAL

REPAID

YEAR END PAYMENT

1 P200, 000 P14, 000 14 P14, 000 P28, 0002 186, 000 13, 020 15 15, 000 28, 0203 171, 000 11, 970 17 17, 000 28, 9704 154, 000 10, 780 18 18, 000 28, 7805 136, 000 9, 520 19 19, 000 28, 5206 117, 000 8, 190 20 20, 000 28, 1907 97, 000 6, 790 22 22, 000 28, 7908 75, 000 5, 250 23 23, 000 28, 2509 52, 000 3, 640 25 25, 000 28, 640

10 27, 000 1, 890 27 27, 000 28, 890TOTALS P1, 215, 000 P85, 120 200 P200, 000 P285, 120

11. A man borrowed P150, 000 from a bank for home improvement, to be repaid by month-end payment for 60 months. The current rate of the interest charge by banks is 19% compounded monthly. Based on this rate, prepare an amortization schedule.

GIVEN:

P = 150, 000.00

n = 60 months (period)

I = 19% compounded monthly = 19%12

=1.58333333%

SOLUTON:

A = [ i

1−i¿−n ¿]=150000¿

PERIOD

PRINCIPAL AT THE BEGINNING

OF EACH 6 M0NTHS

INTEREST AT 4% PER PERIOD

PAYMENT AT END OF EACH

PERIOD

PERIODIC PAYMENT TO

PRINCIPAL

1 150000 2374.999995 3891.08 1516.080005

2 148483.92 2350.995395 3891.08 1540.084605

3 146943.8354 2326.610722 3891.08 1564.469278

4 145379.3661 2301.839959 3891.08 1589.240041

5 143790.1261 2276.676991 3891.08 1614.403009

6 142175.7231 2251.11561 3891.08 1639.96439

7 140535.7587 2225.149508 3891.08 1665.930492

8 138869.8282 2198.772275 3891.08 1692.307725

9 137177.5205 2171.977403 3891.08 1719.102597

10 135458.4179 2144.758278 3891.08 1746.321722

11 133712.0961 2117.108184 3891.08 1773.971816

12 131938.1243 2089.020297 3891.08 1802.059703

13 130136.0646 2060.487685 3891.08 1830.592315

14 128305.4723 2031.503307 3891.08 1859.576693

15 126445.8956 2002.06001 3891.08 1889.01999

16 124556.8756 1972.150526 3891.08 1918.929474

17 122637.9461 1941.767477 3891.08 1949.312523

18 120688.6336 1910.903362 3891.08 1980.176638

19 118708.457 1879.550565 3891.08 2011.529435

20 116696.9275 1847.701349 3891.08 2043.378651

21 114653.5489 1815.347854 3891.08 2075.732146

22 112577.8168 1782.482095 3891.08 2108.597905

23 110469.2188 1749.095961 3891.08 2141.984039

24 108327.2348 1715.181214 3891.08 2175.898786

25 106151.336 1680.729483 3891.08 2210.350517

26 103940.9855 1645.732267 3891.08 2245.347733

27 101695.6378 1610.180928 3891.08 2280.899072

28 99414.7387 1574.066693 3891.08 2317.013307

29 97097.72539 1537.380649 3891.08 2353.699351

30 94744.02604 1500.113743 3891.08 2390.966257

31 92353.05979 1462.256777 3891.08 2428.823223

32 89924.23656 1423.800409 3891.08 2467.279591

33 87456.95697 1384.735149 3891.08 2506.344851

34 84950.61212 1345.051356 3891.08 2546.028644

35 82404.58348 1304.739236 3891.08 2586.340764

36 79818.24271 1263.78884 3891.08 2627.29116

37 77190.95155 1222.190064 3891.08 2668.889936

38 74522.06162 1179.93264 3891.08 2711.14736

39 71810.91426 1137.00614 3891.08 2754.07386

40 69056.8404 1093.399971 3891.08 2797.680029

41 66259.16037 1049.10337 3891.08 2841.97663

42 63417.18374 1004.105407 3891.08 2886.974593

43 60530.20914 958.3949761 3891.08 2932.685024

44 57597.52412 911.9607966 3891.08 2979.119203

45 54618.40492 864.7914094 3891.08 3026.288591

46 51592.11633 816.8751734 3891.08 3074.204827

47 48517.9115 768.2002638 3891.08 3122.879736

48 45395.03176 718.7546681 3891.08 3172.325332

49 42222.70643 668.5261837 3891.08 3222.553816

50 39000.15261 617.5024151 3891.08 3273.577585

51 35726.57503 565.6707701 3891.08 3325.40923

52 32401.1658 513.0184574 3891.08 3378.061543

5329023.10426 459.5324831 3891.08 3431.547517

54 25591.55674 405.1996475 3891.08 3485.880352

56 18564.60293 293.9395458 3891.08 3597.140454

57 14967.46248 236.984822 3891.08 3654.095178

58 11313.3673 179.1283152 3891.08 3711.951685

59 7601.415613 120.3557469 3891.08 3770.724253

60 3830.69136 60.65261307 3891.08 3830.427387

TOTAL 5271477.736 83465.06397 233464.8 150000

12. (M.E. Board, November 1983) On January 1, 1978 the purchasing manager of a cement company bought a new machine costing P140,000. Depreciation has been computed by the straight-line method, based on an estimated useful life of 5 years and residual scrap value) 12, 800.

On January 2, 1981 extraordinary repairs (almost equivalent to a rebuilding of the machinery) were performed at a cost of P30,400. Because of the thorough going nature of these repairs, the normal life of machinery was extended materially; the revised estimate of useful life was 4 years in 1981.

Determine the annual provision for depreciation for the years 1978 to 1980 and the adjusted provision for the depreciation of December 31, 1981. Assume payment in cash for the machine and the repairs.

13. (ECE Board, August 1975) A broadcasting corporation purchased equipment worth P53,000 and paid P1, 500 for freight and deliviery charges to the site. The equipment has a normal life of 10 years with a trade-in value of P5,000 agianst the purchase of new equipment at the end of life.

(a) Determine the annual depreciation cost by the straight-line method.(b) Determine the annual depreciation cost by the sinking fund method. Assume

interest is 6% compounded annually.

GIVEN: CO=53000+1500=P54 ,500

Delivery Change = P1, 800N = 10 years Cn=P5 ,000

SOLUTION:a.) Straight – line Formula

d=Co−Cn

n=54500−5000

10=P4 ,980

b.) Sinking Fund Formulad= (Co−Cn )¿

d= (54500−5000 ) [ 0.06

(1.06 )10−1 ]d=P3 ,755.46

14. (M.E. Board, June 1990) A machinery supplier is offering a certain machine on a 10% down payment and the balance payable in equal year-end payments without interest for 2 years. Under this arrangement the price is pegged at {150,000. However, for cash purchase the

machine would only cost P195, 000. What is the equivalent interest rate that is charged on the two-year payment plan of interest is compounded quarterly?

GIVEN:C1 = P 250,000 Price of the machine for 2-year paymentC0 = P 195,000 Original cost of machine when price purchasedn = 2 yearsd = down payment = .10(250,000) = 25,000C1-new = 250,000 – 25,000 = 225,000 Remaining balanceCA = 225,000/2 = 112,500 Cash to be paid every year-endC0 = 195,000 – 25,000 = 170,000 Remaining amount to be paid and to be compounded

Therefore:

F=112,500 [ (1+i )2−1i ] Expected total compounded amount of money after 2 years

F=170,000 (1+i )2 Total compounded money after 2 years

SOLUTION:

170,000 (1+i )2=112,500 [ (1+i )2−1i ]

170,000112,500

(1+i )2i=(1+i )2−1

6845

(1+i )2 i=(1+i )2−1

68 (1+i )2i=45 (1+ i)2−45

68 (1+i )2i−45 (1+ i)2=−45

(1+i )2 (68 i−45 )+45=0

By trial and error:i=0.2091effective rate compounded annually

(1+.2091 )2 (68(.2091)−45 )+45=0.0002616≅ 0

To get for the equivalent effective rate compounded quarterly:

0.2091=(1+ i4 )

4

−1

1.2091=(1+ i4 )

4

4√1.2091=∜ (1+ i4 )

4

1.08614=1+ i4

i=0.194456≅ 0.1945THEREFORE:

15. A debt of P10, 000 with interest at the rate law of 8% payable semi-annually is to be amortized by equal payments at the end of each 6 months for 4 years. Find the semi-annual payment and contract and amortization schedule.

GIVEN:

i=19.45% compounded quarterly

P = 10, 000 n = 4(2) = 8 quarters i=8%2

=4%

SOLUTION:

A = P (A /P , i , n¿=10000¿

PERIOD

PRINCIPAL AT THE BEGINNING

OF EACH 6 M0NTHS

INTEREST AT 4% PER PERIOD

PAYMENT AT END OF EACH

PERIOD

PERIODIC PAYMENT TO

PRINCIPAL

1 P10, 000.00 P400.00 P1,485.28 P1, 085.28

2 8, 917.72 356.59 P1,485.28 1,228.69

3 8, 786.03 311.44 P1,485.28 1, 173.84

4 6, 612.19 264.49 P1,485.28 1, 220.79

5 5, 391.70 451.66 P1,485.28 1, 269.62

6 4, 121.78 164.87 P1,485.28 1, 320.41

7 2, 801.37 112.05 P1,485.28 1, 373.23

8 1, 428.14 57. 13 P1,485.28 1, 428.15

TOTALS P47, 055.63 P1882.23 P11, 882.24 P10, 000.00