Question 1) The time it would take for money to double at a simple interest rate of 10% per year is closest to:
A) 5 Years
B) 7 Years
C) 10 Years
D) 12 Years
Question 2) At a compound interest rate of 10% per year, $10,000 one year ago is equivalent to how much 1 year from now?
A) $8264
B) $9091
C) $11,000
D) $12,000
Question 3) In most engineering economy studies, the best alternative is the one which:
A) Will last the longest time
B) Is easiest to implement
C) Costs the least
D) Is most politically attractive
Question 4) The present worth of an investment of $20,000 in year 10 at an interest rate of 12% per year is closest to:
A) $6440
B) $7560
C) $8190
D) $10,3000
Question 5) The amount of money that could be spent now in lieu of spending $50,000 seven years from now at an interest rate of 18% per year is closest to:
A) $15,700
B) $17,800
C) $19,300
D) $25,100
Question 6) Income from a precious metals mining operation has been decreasing uniformly for 5 years. If income in year 1 was $100,000 and it decreased by $10,000 per year through year 5, the present worth of the income at 10% per year is closest to:
A) $310,500
B) $352,200
C) $379,100
D) $447,700
Question 7) Income from sales of a certain oil additive has been averaging $100,000 per year. At an interest rate of 18% per year, the present worth of the income for 5 years is closest to:
A) $296,100
B) $312,700
C) $328,400
D) $335,100
Question 8) A manufacturing company wants to have $100,000 available in 5 years to replace a production line. The amount of money that would have to be deposited each year at an interest rate of 10 % per year would be closest to:
A) $12,380
B) $13,380
C) $16,380
D) $26,380
Question 9) The present worth of $5000 in year 3, $10,000 in year 5, and $10,000 in year 8 at an interest rate of 12% per year is closest to:
A) $12,100
B) $13,300
C) $14,900
D) $16,200
Question 10) If $10,000 is borrowed now at 10% per year interest, the balance after a $4000 payment is made 5 years from now will be closest to:
A) Less than $13,500
B) $14,300
C) $16,100
D) More than $17,000
Question 11) The present sum needed to provide for an annual withdrawal of $1000 for 25 years beginning 5 years from now at an interest rate of 10% per year is closest to:
A) Less than $6500
B) $7800
C) $8500
D) More than $9000
12) If a company wants to have $100,000 in a contingency fund 10 years from now, the amount the company must deposit each year in years 1 through 5, at an interest rate of 10% per year, is closest to:
A) Less than $8000
B) $8420
C) $9340
D) More than $10,000
Question 13) An interest rate of 1% per month is:
A) A nominal rate
B) A simple rate
C) A effective rate with an unknown compounding period
D) An effective rate with a known compounding period
Question 14) An interest rate of effective 14% per year, compounded weekly, is:
A) An effective rate per year
B) An effective rate per month
C) A nominal rate per year
D) A nominal rate per month
Question 15) An interest rate of 2% per month is the same as:
A) 24% per year
B) A nominal 24% per year, compounded monthly
C) An effective 24% per year, compounded monthly
D) Both (a) and (b)
Question 16) An interest rate of 15% per year, compounded monthly, is nearest to:
A) 1% per month
B) 15.12% per year
C) 16.08% per year
D) 16.92% per year
Question 17) An environmental testing company needs to purchase $40,000 worth of equipment 2 years from now. At an interest rate of 20% per year, compounded quarterly, the present worth of the equipment is closest to:
A) $27,070
B) $27,800
C) $26,450
D) $28,220
Question 18) A steel fabrication company invested $800,000 in a new shearing unit. At an interest rate of 12% per year, compounded monthly, the monthly income required to recover the investment in 3 years is closest to:
A) $221,930
B) $31,240
C) $29,160
D) $26,570
Question 19) Consider the following estimates. The cost of money is 10% per year. The present worth of machine X is closest to?
A) $-68,445
B) $-97,840
C) $-125,015
D) $-223,120
Question 20) Given the following estimates with the cost of money being 10% per year, the capitalized cost of machine Y is closest to:
A) $-30,865
B) $-97,840
C) $-308,650
D) $-684,445
Question 21) 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
Question 22) In comparing alternatives with different lives by the present worth method, it is necessary to:
A) Compare them over a period equal to the life of the longer-lived alternative.
B) Compare them over a time period of equal service.
C) 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.
Question 23) The upgraded version of a machine has a first cost of $20,000, an annual operating cost of $6000, and a salvage value of $5000 after its 8-year life. At an interest rate of 10% per year, the capitalized cost is closest to:
A) $-9,312
B) $-10,006
C) $-93,120
D) $-100,060
Question 24) Find the capitalized cost of a present cost of $30,000, monthly costs of $1000, and periodic costs every 5 years of $5000. Use an interest rate of 12% per year, compounded monthly.
A) $-80,000
B) $-136,100
C) $-195,200
D) $-3,600,000
Question 25) Consider the following estimates, and use an interest rate of 10% per year. The equivalent annual worth of alternative A is closest to:
A) $-25,130
B) $-37,100
C) $-41,500
D) $-42,900