MATHEMATICS ENGINEERING ECONOMY CE LICENSURE EXAMINATION PROBLEMS ENGINEERING ECONOMY
6.
Find the effective rate on a nominal rate of 8.5% compounded continuously. (M01 M 8) a. 9.98 % c. 8.57 % b. 8.87 % d. 10.12 %
7.
Which of the following has the greatest effective rate? (M02 M 2) a. 12.31 % compounded quarterly b. 12.20 % compounded monthly c. 12.35 % compounded annually d. 12.32 % compounded semi-annually
8.
A nominal annual rate of interest of 7%, compounded continuously, has an effective annual interest rate of: (N03 M 22) a. 7.25 % c. 7.47 % b. 7.53 % d. 7.14 %
SIMPLE INTEREST 1.
A time deposit of P110,000 for 31 days earns P890.39 on maturity date after deducting the 20% withholding tax on interest income. Find the rate of interest per annum. (M97 M 26) a. 12.5 % c. 11.75 % b. 12.25 % d. 11.95 %
2.
The tag price of a certain commodity is for 100 days. If paid in 31 days, there is a 3% discount. What is the simple interest paid? (M99 M 9) a. 12.15 % c. 22.32 % b. 6.25 % d. 16.14 %
3.
Terms of sales: P60,000 payable in 60 days 15 days. Find the equivalent annual rate of in 60 days. (M03 M 19) a. 22.5 % c. b. 34.8 % d.
PRESENT-FUTURE VALUATION RELATIONSHIPS or P57,500 payable in simple interest if paid
9.
Accumulate P5000 for 10 years at 8% compounded quarterly. Find the compounded interest at the end of time. (M94 M 1) a. P7040.60 c. P5080.20 b. P6040.20 d. P4080.20
10.
Accumulate P5000 for 10 years at 8% compounded quarterly. Find the compounded interest at the end of time. (M94 M 4) a. P7040.60 c. P6040.20 b. P4080.20 d. P5080.60
11.
P500,000 was deposited 20.15 years ago at an interest rate of 7% compounded semi-annually. How much is the sum now? (N94 M 22) a. P1.4 million c. P1.8 million b. P1.6 million d. P2.0 million
12.
How long (years) will it take money to quadruple if it earns 7% compounded semi-annually? (M95 M 13) a. 26.30 c. 33.15 b. 40.30 d. 20.15
24.7 % 32.9 %
EQUIVALENT RATES 4.
5.
What rate (%) compounded quarterly is equivalent to 6% compounded semi-annually? (M98 M 6) a. 5.93 c. 5.96 b. 5.99 d. 5.90 Which of the following has the least effective annual interest rate? (N99 M 3) a. 12 % compounded quarterly b. 11.7 % compounded semi-annually c. 11.5 % compounded monthly d. 12.2 % compounded annually
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MATHEMATICS ENGINEERING ECONOMY 13.
P200,000 was deposited at an interest of 24% compounded semiannually. After how many years will the sum be P621,170? (N95 M 15) a. 4 c. 6 b. 5 d. 7
14.
15.
19.
P200,000 was deposited on Jan. 1, 1988 at an interest rate of 24% compounded semi-annually. How much would the sum on Jan. 1, 1993? (M96 M 12) a. P421,170 c. P401,170 b. P521,170 d. P621,170
A house and lot costing P2 million was bought at a down payment of P500,000 and P1 million after one year. The remaining balance will be paid at the end of the third year. If the interest rate is 24% compounded semi-annually, what is the required payment? (N01 M 11) a. P1,432,876.00 c. P1,274,324.00 b. P1,763,985.00 d. P1,387,215.00
20.
At what interest rate, compounded quarterly, will an investment double in five years? (N02 M 22) a. 16.6 % c. 14.1 % b. 15.1 % d. 13.7 %
If P500,000 is deposited at a rate of 11.25% compounded monthly, determine the compounded interest after 7 years and 9 months. (N96 M 25) a. P660,592 c. P670,258 b. P680,686 d. P690,849
ANNUITIES
16.
One hundred thousand pesos was placed in a time deposit which earned 9% compounded quarterly, tax free. After how many years would it be able to earn a total interest of fifty thousand pesos? (N98 M 24) a. 4.56 c. 4.32 b. 4.47 d. 4.63
17.
How long will it take for money to quadruple itself if invested at 20% compounded quarterly? (M00 M 1) a. 10.7 years c. 9.5 years b. 6.3 years d. 7.1 years
18.
P1,000,000 was invested to an account earning 8% compounded continuously. What is the amount after 20 years? (N00 M 14) a. P4,452,796.32 c. P5,365,147.25 b. P4,953,032.42 d. P3,456,254.14
21.
A man inherited a regular endowment of P100,000 every end of 3 months for x years. However, he may choose to get a single lump sum of P3,702,939.80 at the end of 4 years. If the rate of interest was 14% compounded quarterly, what is the value of x? (N94 M 24) a. 13 c. 12 b. 10 d. 11
22.
A man paid a 10% down payment of P200,000 for a house and lot and agreed to pay the balance of monthly installments for 5 years at an interest rate of 15% compounded monthly. What was the monthly installment in pesos? (M95 M 14) a. P44,528.34 c. P43,625.92 b. P42,821.87 d. P45,825.62
23.
A service car whose cash price was P540,000 was bought with a down payment of P162,000 and monthly of P10,874.29 for 5 years. What was the rate of interest if compounded monthly? (M96 M 29) a. 20 % c. 24 % b. 26 % d. 22 %
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MATHEMATICS ENGINEERING ECONOMY 24.
A man inherited a regular endowment of P100,000 every end of 3 months for 10 years. However, he may choose to get a single lump sum payment at the end of 4 years. How much is this lump sum if the cost of money is 14% compounded quarterly? (N96 M 24) a. P3,502,546 c. P3,802,862 b. P3,702,939 d. P3,602,431
25.
The present value of an annuity of R pesos payable annually for 8 years, with the first payment at the end of 10 years, is P187,481.25. Find the value of R if money is worth 5%. (M98 M 12) a. P45,000 c. P42,000 b. P44,000 d. P43,000
26.
27.
28.
A debt of x pesos, with interest rate of 7% compounded annually will be retired at the end of 10 years through the accumulation of deposit in the sinking fund invested at 6% compounded semiannually. The deposit in the sinking fund every end of six months is P21,962.68. What is the value of x? (N98 M 14) a. P300,000 c. P350,000 b. P250,000 d. P400,000 A man paid 10% down payment of P200,000 for a house and lot and agreed to pay the balance on monthly installments for 60 months at an interest rate of 15% compounded monthly. Determine the required monthly payment. (M99 M 10) a. P4,282 c. P58,477 b. P42,822 d. P5,848 To maintain its newly acquired equipment, the company needs P40,000 per year for the first five years and P60,000 per year for the next five years. In addition, an amount of P140,000 would also be needed at the end of the fifth and the eighth years. At 6%, what is the present worth of these costs? (N99 M 4) a. P689,214 c. P549,812 b. P512,453 d. P586,425
29.
Five thousand pesos is deposited at the end of each year for 15 years into an account earning 7.5% compounded continuously. Find the amount after 15 years. (M00 M 5) a. P133,541.3 c. P148,365.9 b. P152,754.2 d. P112,854.1
30.
A man made a year-end payment of P100,000 to an account earning 8% annually for 10 years. How much is in the account after 20 years? (N00 M 15) a. P3,127,540.18 c. P3,327,452.36 b. P4,075,458.99 d. P3,247,111.25
31.
Determine the equal payment series future worth factor of an annuity of P15,000 per year for 25 years at 18% interest annually. (M01 M 14) a. 0.786 c. 342.6 b. 1.389 d. 167.21
32.
A man deposits P6,000.00 every end of three months for his retirement. If the interest rate is 10% annually compounded quarterly, what lump sum value can he expect after 20 years? (M02 M 19) a. P528,000.00 c. P1,245,278.32 b. P785,500.00 d. P1,490,296.28
33.
What is the difference between the sums of an annuity due and an ordinary annuity for the following data: (N02 M 21) Periodic payment : P14,000 Term : 16 years Interest rate : 10% compounded quarterly Payment interval : 3 months a. P58,598 c. P61,763 b. P53,992 d. P67,042
34.
Find the present value, in pesos, of an annuity of P20,000 payable annually for 8 years, with the first payment at the end of 10 years, if money is worth 5%. (M03 M 20) a. P79,354 c. P83,325 b. P77,451 d. P81,424
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MATHEMATICS ENGINEERING ECONOMY 35.
What is the difference between the sums of an annuity due and an ordinary annuity for the following data: (N03 M 23) Periodic payment : P100,000 Term : 20 years Payment interval : 1 year Interest rate : 12% compounded annually a. P965,732 c. P864,629 b. P657,432 d. P753,458
39.
A machine costing P45,000 is estimated to have a salvage value of P4,350 when retired at the end of 6 years. Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation in %? (M96 M 5) a. 33.25 c. 35.25 b. 32.25 d. 34.25
40.
An engineer bought an equipment for P500,000. Other expenses, including installation, amounted to P30,000. At the end of its estimated useful life of 10 years, the salvage value will be 10% of the first cost. Using the straight-line method of depreciation, what is the book value after 5 years? (N97 M 10) a. P291,500 c. P271,500 b. P281,500 d. P301,500
41.
A machine having a first cost of P60,000.00 will be retired at the end of 8 years. Depreciation cost is computed using a constant percentage of the declining book value. What is the total cost of depreciation, in pesos, up to the time the machine is retired if annual rate of depreciation is 28.72%? (N98 M 25) a. P56,000 c. P58,000 b. P57,000 d. P59,000
42.
A machine costing P480,000 has a life expectancy of 12 years with a salvage value of 10% of the first cost. What is the book value after five years using the declining balance method? (M00 M 11) a. P183,892 c. P196,432 b. P152,758 d. P214,785
43.
A certain copier machine cost P150,000 with a trade-in value of P15,000 after making 800,000 copies. Using the declining balance method, what is the book value when the machine had made 300,000 copies? (M00 M 28) a. P68,111 c. P62,531 b. P64,896 d. P63,254
PERPETUITY 36.
37.
Find the present value in pesos, of a perpetuity of P15,000 payable semi-annually if money is worth 8%, compounded quarterly. (N95 M 13) a. P371,287 c. P392,422 b. P386,227 d. P358,477 What is the present worth of a perpetuity of P10,000 annually if money is worth 10% compounded monthly? (N01 M 6) a. P95,499.07 c. P100,000.00 b. P96,783.98 d. P105,342.12
DEPRECIATION 38.
An engineer bought equipment for P500,000. He spent an additional amount of P30,000 for installation and other expenses. The estimated useful life of the equipment is 10 years. The salvage value is x% of the first cost. Using the straight-line method of depreciation, the book value at the end of 5 years will be P291,500. What is the value of x? (N94 M 23) a. 10 c. 20 b. 5 d. 15
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MATHEMATICS ENGINEERING ECONOMY 44.
An equipment having a first cost of P450,000 has a life expectancy of 10 years with a final salvage value of P80,000. Using the double-declining balance method, what is its book value after 6 years? (M01 M 7) a. P117,964.80 c. P184,320.00 b. P147,456.00 d. P230,400.00
45.
A machine having a total first cost of P100,000 has a salvage cost of P10,000 after 10 years. Determine the book value after five years using the double-declining balance method. (M02 M 15) a. P32,768.00 c. P40,960.00 b. P26,214.40 d. P34,452.00
46.
Cost of machine = P140,000; useful life = 8 years; salvage value = P10,000. Determine the 4th year depreciation using the doubledeclining balance method, in pesos. (M03 M 18) a. P14,765 c. P16,182 b. P17,312 d. P15,634
47.
An equipment is bought at P420,000 with an economic life of 6 years and a salvage value of P50,000. The first year depreciation is P105,714. The cost of money is 12% per year. What method of depreciation was used? (N03 M 24) a. straight-line method b. sinking-fund method c. SOYD method d. declining balance method
49.
DEPLETION 50.
Machine cost = $15,000; life = 8 years; salvage value = $3,000. What minimum cash return would the investor demand annually from the operation of this machine if he desires interest annually at the rate of 8% on his investment and accumulates a capital replacement fund by investing annual deposits at 5%? (M97 M 24) a. $2,456.66 c. $2,576.60 b. $2,435.38 d. $2,365.30
An investor pays P1,100,000 for a mine which will yield a net income of P200,000 at the end of each year for 10 years and then will become useless. He accumulates a replacement fund to recover his capital by annual investments at 4.5%. At what rate (%) does he receive interest on his investment at the end of each year? (M95 M 15) a. 10.04 % c. 11.5 % b. 8.5 % d. 17.5 %
CAPITALIZED COST 51.
At 6%, find the capitalized cost of a bridge whose cost is P200M and life is 20 years, if the bridge must be partially rebuilt at a cost of P100M at the end of each 20 years. (N96 M 23) a. P245.3 M c. P215 M b. P210 M d. P220 M
52.
A company uses a type of truck which costs P2M, with life of 3 years and a final salvage value of P320,000. How much could the company afford to pay for another type of truck for the same purpose, whose life is 4 years with a final salvage value of P400,000, if money is worth 4%? (M97 M 23) a. P2,805,120 c. P2,585,870 b. P2,718,480 d. P2,642,320
ANNUAL COST 48.
A contractor can buy trucks for P800,000 each, or rent them for P1,200 per truck per day. The truck has a salvage value of P100,000 at the end of its useful life of 5 years. The annual maintenance cost is P20,000 per truck. Using the annual-cost method and 14% interest rate, determine the number of days per year that each truck must be used to warrant its purchase. Use sinking-fund method of depreciation. (N02 M 20) a. 187 c. 155 b. 177 d. 199
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MATHEMATICS ENGINEERING ECONOMY 53.
Determine the capitalized cost of a project that cost P324,000 with an anticipated salvage value of P50,000 at the end of four years, if money is worth 8%. (M99 M 5) a. P1,342,846 c. P1,084,079 b. P1,480,792 d. P1,120,304
57.
BREAK-EVEN ANALYSIS 54.
Determine the break-even point in terms of number of units produced per month using the following data: (the costs are in pesos per unit) (M98 M 7) Selling price per unit = P600.00 Total monthly overhead expenses = P428,000.00 Labor cost = P115.00 Cost of materials = P76.00 Other variable cost = P2.32 a. 1036 c. 1053 b. 1044 d. 1025
55.
The cost of producing an item is as follows: Labor cost per unit = P315.00 Material cost per unit = P100.00 Variable cost per unit = P3.00 Monthly overhead cost = P461,600.00 If each unit is sold at P995.00, how many units must be produced each month to break-even? (M99 M 11) a. 800 c. 850 b. 750 d. 950
56.
The cost of producing a certain commodity consists of P45.00 per unit for labor and material cost and P15.00 per unit for other variable cost. The fixed cost per month amounts to P450,000. If the commodity is sold at P250.00 each, what is the break-even quantity? (N00 M 16) a. 2014 c. 2589 b. 2178 d. 2367
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The cost of producing a commodity consists of P65.00 per unit for labor and material cost and P25.00 per unit for other variable cost. The fixed cost per month amounts to P700,000. If the commodity is sold at P290.00 each, what is the break-even quantity? (N01 M 17) a. 3200 c. 3600 b. 3500 d. 3400
