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# Engineering Economics

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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

https://brainmass.com/economics/the-time-value-of-money/engineering-economics-81451

#### Solution Preview

Hello!

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 ...

\$2.19

## Engineering Economics Questions

See the attached file.

Question 2

Yakima is retiring this year with his savings in an investment fund worth \$750,000. The fund has an average annual return of 9.00% (EAR).

a How much can Yakima withdraw at the end of each month (12 months per year) to have the fund last 30 years and still have \$100,000 in the fund at the end of the 30 years (just to be safe)?

b How much can Yakima withdraw up front to invest in his home and still be able to withdraw \$5,000 monthly for 30 years, with a zero balance at the end?

Initial Investment \$750,000
EAR 9%
APR

Month Interest Withdraw Balance
0 0 0 \$750,000
1 \$67,500
2
3
4
5
6
7
8
9
10
11
12

Question 3 Score 0

The Smothers are looking at retiring and would like to be debt free when this starts. They presently owe \$225,000 on their home mortgage that has a rate of 4.25% APR with monthly compounding.

a If they make payments of \$3,000 monthly, how many years until they can retire (when the house loan is paid off).

b If in the upcoming 5 years, they make \$2,000 monthly payments, and then sell their house for \$500,000, how much will they have to buy a home in a retirement development? (All taxes and fees should be ignored)

A
Payments 3000
Rate 4.25%
Period 12
Time
PV \$35,184.77

Monthly Rate 0.3542%

B
Presently Owe \$225,000
Rate 0.3542%
Payments \$2,000
Period 60
Amount Paid \$120,000
\$0.00

Question 4

Save-Your-Day loans advertises no-interest loans of \$100 on which only a \$1 fee per week is charged. Collateral is usually a car title. The borrower will not pay anything until the loan is paid off. Then the amount to be paid back is \$1.00 for each week per \$100 of loan value. For a loan of \$200 (two loans of \$100) for 3 weeks, the pay back amount would be \$200 + \$1.00 * 2 loans * 3 weeks = \$200 + \$6= only \$206.
Louie borrows \$500 and pays it back in 4 weeks. What effective annual rate will Save-Your-Day earn on the loan to Louie?

Question 5

Joan Dale inherited \$500,000 that she invested at 12% APR compounded monthly and will not make any further payments. In 15 years when she retires, she will reinvest it in a more conservative fund that earns 5% APR with monthly compounding.

If she wants the retirement funds to last 25 years, how much will she receive monthly in her retirement years.

Question 6

Veronica borrowed \$5,000 from her Uncle and the agreed upon interest rate is 4% annually (EAR)?

a What would be the result at the end of five years If she pays her uncle \$100 a month?

b If each month she pays the interest (only) on the loan, how much would this be?

Question 7
Your client is considering two investments and has asked you to evaluate these alternatives. Provide financial and risk advice for your client regarding which to purchase.

a 500 shares of a stock that can be purchased for \$80 a share. Forecasts are that it can be sold in 5 years for \$135 a share. It also is forecasted to pay quarterly dividends of \$2.00 per share in the upcoming 5 years.

b A bond with a face value of \$50,000 that matures in 5 years. It can be purchased for \$40,000 today, and it has a coupon rate of 5.00% that is paid semi-annually.

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