# Engineering Economics

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

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

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

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

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

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

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.

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

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

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

#### Solution Preview

Hello!

Here are your answers.

Question 1 - Answer is C

Since we're using simple (not compounded) interest rate, this can be easily calculated. We want to know how many years it will take to get a 100% return (double the money). Since the interest rate is 10% per year, then it will take 10 years.

Question 2

There appears to be a typo in the choices for this question, as none of them is correct. Time between "one year ago" and "one year from now" is 2 years. Therefore, at a 10% compound rate, $10,000 one year ago should be equal to 10000*(1+0.10)^2 = $12,100 one year from now; that is, we should compound the $10,000 for 2 years at an yearly 10% rate. The correct answer should then be $12,000.

Question 3 - Answer is C

Question 4 - Answer is A

Here we're asked the present value of $20,000 10 years from now, at an interest rate of 12% per year. The formula for the present value would be:

PV = 20000/(1 + 0.12)^10 = $6,439.46

Question 5 - Answer is A

The idea is the same as in question 4, so the formula would be:

PV = 50000/(1 + 0.18)^7 = 15,696.25

Question 6 - Answer is A

Here we must find the present value of the following cashflow:

Year 1 $100,000

Year 2 $90,000

...

Year 5 $60,000

The interest rate is 10% per year. So the formula is:

PV = 100000/1.1 + 90000/1.1^2 + 80000/1.1^3 + 70000/1.1^4 + 60000/1.1^5

The result is $310,460

Question 7 - Answer is B

The idea is the same as in the previous question, but now the cash flow is always the same: $100,000, and the interest rate is 18%. So we get:

PV = 100000/1.18 + 100000/1.18^2 + ... + 100000/1.18^5 = $312,717

Question 8 - Answer is C

One way to solve this is to find the value of an annual deposit of $1 for the next 5 years. Since the ...