test 1 review
Assume you borrow $10,000 today and promise to repay the loan in two payments, one in year 2 and the other in year 4, with the one in year 4 being only half as large as the one in year 2. At an interest rate of 10% per year, the size of the payment in year 4 is closest to: (a) $4280 (b) $3975 (c) $3850 (d) $3690
10,000 = 2x(P/F,10%,2) + x(P/F,10%,4) 10,000 = 2x(0.8264) + x(0.6830) 2.3358x = 10,000 ==>x = $4281
The compound interest rate per year that amounts of $1000 one year ago and $1345.60 one year hence are equivalent to is closest to: (a) 8.5% per year (b) 10.8% per year (c) 20.2% per year (d ) None of the above
1000(1 + i) = 1345.60/(1 + i) ==>(1 + i)^2 = 1345.60/1000 ==>(1 + i) = 1.16 ==>i = 16%
The winner of a multistate mega millions lottery jackpot worth $175 million was given the option of taking payments of $7 million per year for 25 years, beginning 1 year from now, or taking $109.355 million now. The interest rate that renders the two options equivalent to each other is closest to: (a) 4% (b) 5% (c) 6% (d) 7%
109.355 = 7(P/A,i,25) (P/A,i,25) = 15.6221 ==>From tables, i = 4%
You found a report stating that the equivalent annual worth of chemical costs associated with a water treatment process were $125,000 per year for a 5-year period. The report stated that the cost in year 1 was $190,000 and the cost decreased by a uniform amount each year over that 5-year period. However, it did not say how much the decrease was each year. If the interest rate was 20% per year, the amount of the annual decrease, G, is closest to: (a) $27,358 (b) $31,136 (c) $33,093 (d) $39,622
125,000 = 190,000 - G(A/G,20%,5) ==>125,000 = 190,000 - ==>G(1.6405) G = $39,622
The present worth of an increasing geometric gradient is $23,632. The interest rate is 6% per year and the rate of change series is 4% per year. If the cash flow in year 1 is $3,000, the year in which the gradient ends is: (a) 7 (b) 9 (c) 11 (d) 12
23,632 = 3000{1- [(1+0.04)n /(1+0.06)n ]}/(0.06-0.04) [(23,632*0.02)/3000]-1 = (0.98113)n log 0.84245 = nlog 0.98113 n = 9
If you borrow $24,000 now at an interest rate of 10% per year and promise to repay the loan with payments of $3695 per year starting 1 year from now, the number of payments that you will have to make is closest to: (a) 7 (b) 8 (c) 11 (d) 14
24,000 = 3695(P/A,10%,n) (P/A,10%,n) = 6.4952 From 10% tables, n is close to 11
Maintenance costs for a regenerative thermal oxidizer increased according to an arithmetic gradient for 5 years. The cost in year 1 was $7000. If the interest rate is 10% per year and the present worth of the costs for a 5-year period was $28,800, the amount of the yearly increase, G, was closest to: (a) $1670 (b) $945 (c) $620 (d) $330
28,800 = 7000(P/A,10%,5) + G(P/G,10%,5) 28,800 = 7000(3.7908) + G(6.8618) ==>G = $330
A manufacturer of prototyping equipment wants to have $3,000,000 available 10 years from now so that a new product line can be initiated. If the company plans to deposit money each year, starting 1 year from now, the equation that represents how much the company is required to deposit each year at 10% per year interest to have the $3,000,000 immediately after the last deposit is: (a) 3,000,000(A∕F,10%,10) (b) 3,000,000(A∕F,10%,11) (c) 3,000,000 + 3,000,000(A∕F,10%,10) (d) 3,000,000(A∕P,10%,10)
3,000,000(A∕F,10%,10)
Adams Manufacturing spent $30,000 on a new sterilization conveyor belt, which resulted in a cost savings of $4202 per year. The length of time it should take to recover the investment at 8% per year is closest to: (a) Less than 6 years (b) 7 years (c) 9 years (d) 11 years
30,000 = 4202(P/A,8%,n) (P/A,8%,5) = 7.1395 ==> n = 11 years
The cost of tuition at a large public university was $390 per credit hour 5 years ago. The cost today (exactly 5 years later) is $585. The annual rate of increase is closest to: (a) 5% (b) 7% (c) 9% (d) 11%
390 = 585(P/F,i,5) (P/F,i,5) = 0.6667 From tables, i is between 8% and 9%
Aero Serve, Inc. manufactures cleaning nozzles for reverse pulse jet dust collectors. The company spent $40,000 on a production control system that will increase profits by $11,096 per year for 5 years. The rate of return per year on the investment is closest to: (a) 20% (b) 16% (c) 12% (d) Less than 11%
40,000 = 11,096(P/A,i,5) (P/A,i,5) = 3.6049 ==>i = 12 %
You plan to pay $38,000 cash for the new truck you want to buy 5 years from now. You are a very astute investor; all your money earns at 20% per year. If you have already saved $9500, the amount your rich aunt has to give you 2 years from now (as a graduation present) in order for you to have the total amount of $38,000 is closest to: (a) <$7500 (b) $7654 (c) $8,310 (d) $9,880
9,500(F/P,20%,5) + x(F/P,20%,3) = 38,000 ==> 9500(2.4883) + x(1.7280) = 38,000 ==>x = $8311
A company that utilizes carbon fiber 3-D printing wants to have money available 2 years from now to add new equipment. The company currently has $650,000 in a capital account and it plans to deposit $200,000 now and another $200,000 one year from now. The total amount available in 2 years, provided it returns a compounded rate of 15% per year, is closest to: (a) $1,354,100 (b) $1,324,100 (c) $1,125,125 (d) $1,050,000
==>Amount available = total principal in year 0 + interest for 2 years + principal added year 1 + interest for 1 year ==> 850,000(1+0.15)^2 + 200,000 (1+0.15) ==>$1,354,125
The simple interest rate per year that will accumulate the same amount of money in 2 years as a compound interest rate of 20% per year is closest to: (a) 20.5% (b) 21% (c) 22% (d) 23%
==>F in year 2 at 20% compound interest = P(1.20)(1.20) = 1.44P ==>For simple interest, F = P + Pni = P(1 + ni) ==>P(1 + 2i) = 1.44P ==>i=22%
A chip manufacturing company wants to have $10 million available 5 years from now in order to build new warehouse and shipping facilities. If the company can invest money at 10% per year, the amount that it must deposit each year in years 1 through 5 to accumulate the $10 million is closest to: (a) $1,638,000 (b) $2,000,000 (c) $2,638,000 (d) $2,938,000
A = 10,000,000(A/F,10%,5) ==> 10,000,000(0.16380) ==> $1,638,000
Detrich Products is planning to upgrade an aging manufacturing operation 5 years from now at a cost of $100,000. If the company plans to deposit money into an account each year for 4 years beginning 2 years from now (first deposit is in year 2) to pay for the expansion, the amount of the deposit at 10% per year interest is closest to: (a) $30,211 (b) $21,547 (c) $16,380 (d) $14,392
A = 100,000(A/F,10%,4) ==> 100,000(0.21547) ==> $21,547
An engineer who believes in "save now; play later" wants to retire in 30 years with $2.0 million. At 8% per year interest, the amount the engineer will have to invest each year (starting in year 1) to reach the $2 million goal is closest to: (a) $17,660 (b) $28,190 (c) $49,350 (d) $89,680
A = 2,000,000(A/F,8%,30) ==> 2,000,000(0.00883) ==> $17,660
The cost of replacing a high-definition television production line in 6 years is estimated to be $500,000. At an interest rate of 14% per year, compounded semiannually, the uniform amount that must be deposited into a sinking fund every 6 months beginning now is closest to: (a) <$21,000 (b) $21,335 (c) $24,820 (d) $27,950
A = 500,000(A/F,7%,13) ==> 500,000(0.04965) ==> $24,825
At a return rate of 20% per year, the amount of money you must deposit for five consecutive years starting 3 years from now for the account to contain $50,000 fifteen years from now is closest to: (a) $1565 (b) $1759 (c) $1893 (d) $2093
A(F/A,20%,5)(F/P,20%,8) = 50,000 A(7.4416)(4.2998) = 50,000 A = $1563
An upgraded version of a CNC machine has a first cost of $200,000, an annual operating cost of $60,000, and a salvage value of $50,000 after its 8-year life. At an interest rate of 10% per year, the capitalized cost is closest to: (a) $−93,116 (b) $−100,060 (c) $−931,160 (d) $−1,000,600
AW = -200,000(A/P,10%,8) - 60,000 + 50,000(A/F,10%,8) ==> -200,000(0.18744) - 60,000 + 50,000(0.08744) ==> $-93,116 CC = -93,116/0.10 ==> $-931,160
The Gap has some of its jeans stone-washed under a contract with Vietnam Garment Corporation (VGC). If VGC's estimated operating cost per machine is $26,000 for year 1 and it increases by $1500 per year through year 5, the equivalent uniform annual cost per machine over years 1 to 5, at an interest rate of 8% per year, is closest to: (a) $30,850 (b) $28,770 (c) $26,930 (d) $23,670
AW = 26,000 + 1500(A/G,8%,5) ==> $28,770
At a compound interest rate of 10% per year, the amount that $10,000 one year ago is equivalent to now is closest to: (a) $8264 (b) $9091 (c) $11,000 (d) $12,100
Amount now = 10,000 + 10,000(0.10) =$11,000
A small company plans to spend $10,000 in year 2 and $10,000 in year 5. At an interest rate of effective 10% per year, compounded semiannually, the equation that represents the equivalent annual worth in years 1 through 5 is: (a) A = 10,000(P∕F10%,2)(A∕P,10%,5) + 10,000(A∕F,10%,5) (b) A = 10,000(A∕P,10%,4) + 10,000 (A∕F,10%,5) (c) A = 10,000(P∕F,5%,2)(A∕P,5%,10) + 10,000 (A∕F,5%,10) (d) A = [10,000(F∕P,10%,5) + 10,000] (A∕F,10%,5)
Answer is (a)
The present worth of a deposit of $1000 now and $1000 every 6 months for 10 years at an interest rate of 10% per year, compounded semiannually is represented by which of the following equations: (a) P = 1000(P∕A,5%,21)(F∕P,5%,1) (b) P = 1000 (P∕A,5%,20) (c) P = 1000 (P∕A,5%,21) (d) P = 1000 + 1000(P∕A,10.25%,10)
Answer is (a)
A uniform series of payments begins in year 4 and ends in year 11. If you use the P∕A factor with n = 8, the P value you get will be located in year: (a) 0 (b) 3 (c) 4 (d) 5
Answer is (b)
For the cash flows shown, you have been asked to calculate the present worth (in year 0) using i = 10% per year. Which of the following solutions is not correct? Year 0 1 2 3 4 5 Cash Flow, $ 200 200 200 200 300 300 6 300 (a) P = 200 + 200(P∕A,10%,3) + 300(P∕A,10%,3)(P∕F,10%,3) (b) P = 200(P∕A,10%,4) + 300(P∕A,10%,3) (P∕F,10%,3) (c) P = [200(F∕A,10%,7) + 100(F∕A,10%,3)] (P∕F,10%,6) (d) P = [200(P∕A,10%,7) + 100(P∕A,10%,3) (P∕F,10%,4)](F∕P,10%,1)
Answer is (b)
If you deposit $P into a savings account that earns interest at a rate of i% per month for n years, the future worth in year n is represented by all of the following equations, except: (a) F = $P(F∕P, effective i/month, 12n) (b) F = $P(F∕P, effective i/quarter, 3n) (c) F = $P(F∕P, effective i/6-month, 2n) (d) F = $P(F∕P, effective i/year, n)
Answer is (b)
When comparing mutually exclusive alternatives that have different lives by the present worth method, it is necessary to: (a) Always compare them over a period equal to the life of the longer-lived alternative (b) Always compare them over a time period of equal service (c) Always compare them over a period equal to the life of the shorter-lived alternative (d) Find the present worth over one life cycle of each alternative
Answer is (b)
If you make quarterly deposits for 3 years (beginning one quarter from now) into an account that compounds interest at 1% per month, the value of n in the F∕A factor (for determining F at the end of the 3-year period) is: (a) 3 (b) 4 (c) 12 (d) 16
Answer is (c)
If you wish to accumulate $10,000 over a 5-year period by placing $200 a month, starting next month, into a Roth IRA retirement fund that pays 6% per year, compounded quarterly with no interperiod compounding, the NPER function to determine the number of deposits is: (a) = NPER(4.568%,−200,10000) (b) = NPER(1.5%,200,10000) (c) = NPER(1.5%,−600,10000) (d) = NPER(2%,600,−10000)
Answer is (c)
The only time you change the original cash flow diagram in problems involving uniform series cash flows is when the: (a) payment period is longer than the compounding period (b) payment period is equal to the compounding period (c) payment period is shorter than the compounding period (d) stated interest rate is a nominal interest rate
Answer is (c)
The present worth of an alternative that provides infinite service is called its: (a) Net present value (b) Discounted total cost (c) Capitalized cost (d) Perpetual annual cost
Answer is (c)
An arithmetic gradient has cash flows of $1000 in year 4, $1200 in year 5, and amounts increasing by $200 per year through year 10. If you use the factor 200(P∕G,10%,n) to find PG in year 3, the value of n you have to use in the P∕G factor is: (a) 10 (b) 9 (c) 8 (d) 7
Answer is (d)
An engineer analyzed four independent alternatives by the present worth method. On the basis of her results, the alternative(s) she should select are: Alternative A B C Present worth, $ −5000 −2000 −3000 D −1000 (a) Only D (b) Can't tell from this information (c) All of them (d) None of them
Answer is (d)
An interest rate of 2% per month is the same as: (a) 24% per year, compounded monthly (b) a nominal 24% per year, compounded monthly (c) an effective 24% per year, compounded monthly (d) Both (a) and (b)
Answer is (d)
In order to establish a contingency fund to replace equipment after unexpected breakdowns, a manufacturer of thin-wall plastic bottles plans to deposit $100,000 now and $150,000 two years from now into an investment account. Assuming the account grows at 15% per year, the equation that does not represent the future value of the account in year 5 is: (a) F = 100,000(F∕P,15%,5) + 150,000(F∕P,15%,3) (b) F = [100,000(F∕P,15%,2) + 150,000] (F∕P,15%,3) (c) F = [100,000 + 150,000(P∕F,15%,2)] (F∕P,15%,5) (d) F = 100,000(F∕P,15%,5) + 150,000(F∕P,15%,2)
Answer is (d)
The capitalized cost of an initial investment of $200,000 and annual investments of $30,000 forever at an interest rate of 10% per year is closest to: (a) $−230,000 (b) $−300,000 (c) $−500,000 (d) $−2,300,000
CC = -200,000 - 30,000/0.10 ==> $-500,000
The cost of maintaining a permanent monument is expected to be $70,000 now and $70,000 every 10 years forever. At an interest rate of 10% per year, the capitalized cost is nearest: (a) $−11,393 (b) $−58,930 (c) $−84,360 (d) $−113,930
CC = -70,000 - 70,000(A/F,10%,10)/0.10 ==> -70,000[1 + (0.06275)/0.10] ==> $-113,925
An interest rate of 18% per year, compounded continuously, is closest to an effective: (a) 1.51% per quarter (b) 4.5% per quarter (c) 4.6% per quarter (d) 9% per 6 months
Calculate quarterly and semiannual rates to determine correct answer ==>i/quarter = e0.045 -1 = 0.0460 ==>(4.60%) ==>i/6-mths = e0.09-1 = 0.0942 ==>(9.42%)
A single deposit of $25,000 was made by your grandparents on the day you were born 25 years ago. The balance in the account today if it grew at 10% per year is closest to: (a) $201,667 (b) $241,224 (c) $270,870 (d) $296,454
F = 25,000(F/P,10%,25) ==> 25,000(10.8347) ==> $270,868
The time it would take for money to double at a simple interest rate of 5% per year is closest to: (a) 10 years (b) 12 years (c) 15 years (d) 20 years
F = P + Pni 2P = P + P(n)(0.05) n = 20 years
All of the following mean the same as Minimum Attractive Rate of Return except: (a) Hurdle rate (b) Inflation rate (c) Benchmark rate (d) Cutoff rate
Inflation rate
Another name for noneconomic attributes is: A.) Sustainability B.) Intangible factors C.) Equivalence D.)Evaluation Criteria
Intangible factors
Chemical costs associated with a packed-bed flue gas incinerator for odor control have been decreasing uniformly for 5 years because of increases in efficiency. If the cost in year 1 was $100,000 and it decreased by $5,000 per year through year 5, the present worth of the costs at 10% per year is closest to: (a) $344,771 (b) $402,200 (c) $515,400 (d) $590,700
P = 100,000(P/A,10%,5) - 5000(P/G,10%,5) ==> 100,000(3.7908) - 5000(6.8618) ==> $344,771
The maker of a motion-sensing towel dispenser is considering adding new products to enhance offerings in the area of touchless technology. If the company does not expand its product line now, it will definitely do so in 2 years. Assume the interest rate is 10% per year. The amount the company can afford to spend now if the cost 2 years from now is estimated to be $100,000 is closest to: (a) $75,130 (b) $82,640 (c) $91,000 (d) $93,280
P = 100,000(P/F,10%,2) ==> $100,000(0.8264) ==> $82,640
The cost of updating an outdated production process is expected to be $81,000 four years from now. The equivalent present worth of the update at 6% per year interest is closest to: (a) $51,230 (b) $55,160 (c) $60,320 (d) $64,160
P = 81,000(P/F,6%,4) ==> 81,000(0.7921) ==> $64,160
The amount of money the Teachers Credit Union should be willing to loan a developer who will repay the loan in a lump sum amount of $840,000 two years from now at the bank's interest rate of 10% per year is: (a) $694,180 (b) $99,170 (c) $1,106,400 (d) $763,650
P = 840,000(P/F,10%,2) ==> 840,000(0.8264) ==> $694,176
Assume you make monthly deposits of $200 starting 1 month from now into an account that pays 6% per year, compounded semiannually. If you want to know how much you will have after 4 years, the value of i you should use in the F∕A factor, assuming no interperiod interest, is: (a) 0.5% (b) 3.00% (c) 6.0% (d) 12.0%
PP of month < CP of 6 months. Assume no interperiod compounding i/6-mths = 3%
All of the following are examples of noneconomic factors except: A.) Availability of resources B.)Goodwill C.) Customer acceptance D.) Profit
Profit
All of the following are noneconomic attributes except: (a) Sustainability (b) Morale (c) Taxes (d) Environmental acceptability
Taxes
In order to finance a new project costing $30 million, a company borrowed $21 million at 16% per year interest and used retained earnings valued at 12% per year for the remainder of the financing. The company's weighted average cost of capital for the project was closest to: (a) 12.5% (b) 13.6% (c) 14.8% (d) 15.6%
WACC = 0.70(16%) + 0.30(12%) ==> WACC=14.8%
A midcareer engineer hopes to have $2 million available for his use when he retires 20 years from now. He plans to deposit a uniform amount semiannually, beginning now and every 6 months thereafter through the end of year 20. If his investment account has a yearly return of 8% per year, compounded quarterly, the interest rate, i, that must be used in the A∕F equation to determine the size of the uniform deposits is: (a) 2% (b) 8.24% (c) 8.00% (d) 4.04%
i/6-mths = (1 + 0.04/2)2 - 1 = 0.0404 ==> (4.04%)
An interest rate of 12% per year, compounded monthly, is nearest to: (a) 12.08% per year (b) 12.28% per year (c) 12.48% per year (d) 12.68% per year
i/year = (1 + 0.01)12 -1 = 0.1268 ==>(12.68%)
An interest rate of 2% per quarter, compounded continuously, is closest to an effective semiannual rate of: (a) 2.00% per semiannual period (b) 2.02% per semiannual period (c) 4.0% per semiannual period (d) 4.08% per semiannual period
r = 2% per quarter = 4% per 6-mth period ==>i /6-mths = e0.04 -1 = 4.08% per semi
You are planning to make two equal amount deposits, one now and the other 3 years from now in order to accumulate $300,000 ten years in the future. If the interest rate is 14% per year, compounded semiannually, the size of each deposit is: (a) <$46,300 (b) $46,525 (c) $47,835 (d) >$48,200
x(F/P,7%,20) + x(F/P,7%,14) = 300,000 x(3.8697) + x(2.5785) = 300,000 ==> 6.4482x = 300,000 ==> x = $46,525
