# Concepts in Engineering Economics

Please refer attached file for complete details.

1. The value of P (The present worth) is most nearly

$ 367

$ 371

$ 377

$ 386

2. The value of P (the present worth) is most nearly:

$80

$83

$85

$89

3. The value of Q is most nearly

$193

$243

$258

$272

4. The present value, P is most nearly:

$265

$275

$290

$305

5. (Be careful with the time line.) The value of P is most nearly

$ 1640

$ 1670

$ 1690

$ 1710

6. Consider the following Cash Flow table for questions 6 - 8...

Year Cash Flow

0 - 30,000

1 + 10,000

2 + 10,000

3 + 20,000

4 + 20,000

5 - 5,000

The interest rate is 10 %

The Present Worth of costs is most nearly

$ 29,100

$ 31,400

$ 33,100

$ 34,400

7. The Present Worth of Benefits is most nearly

$44,200

$44,600

$45,100

$46,000

8. The Net Present Worth is most nearly

$ 12,300

$ 12,600

$ 12,900

$ 13,300

9. The correct Analysis Period is:

10 years

15 years

30 years

150 years

10. The Present Worth (PW) of the cost (- Installed cost and operating cost + Salvage value) of Westinghome is most nearly

- $ 87,400

- $ 92,700

- $ 93,600

- $ 95,100

11. The Present Worth (PW) of the cost ( -Installed cost and operating cost + Salvage Value) of Itis is most nearly:

- $ 92,500

- $ 92,700

- $ 94,600

- $ 95,100

12. The recommendation should be to accept the bid of (Points : 1)

Westinghome

Itis

13. The annual income from a rental house is $ 14,000. The annual expenses are $ 4,000. The house is bought for $ 90,000. At the end of eight years the house is sold for $ 140,000. For an interest rate of 8 %, the Net Present Worth of this investment is most nearly

$ 42,400

$ 42,900

$ 43,100

$ 45,400

14. A machine is purchased for $ 10,000. Its life is six years and there is no salvage value. Annual maintenance is paid at the end of each year. The first year maintenance is $ 2,000 and maintenance increases by 15 % each year. The interest rate is 8 %. The Present Value of the cost is most nearly

$ 21,900

$ 22,400

$ 23,100

$ 23,800

15. The present worth of the cost for alternative B is most nearly

- $ 1,105

- $ 1,215

- $ 1,295

- $ 1,305

16. Which course of action should be selected?

Alternative A

Alternative B

17. The present worth of costs is most nearly:

$ 6,000

$ 6,025

$ 6,150

$ 6,200

18. The present worth of benefits is most nearly:

$ 6,000

$ 6,025

$ 6,100

$ 6,150

19. The appropriate recommendation should be:

Invest in the equipment

Don't invest in the equipment

20. The NPW of alternative B is most nearly

$3.50

$6.80

$7.70

$9.20

21. The NPW of alternative A is most nearly

$3.50

$6.80

$7.70

$9.20

22. The NPW of alternative C is most nearly

$3.50

$6.80

$7.70

$12.40

23. The NPW of alternative D is most nearly

$3.50

$7.70

$9.20

$32.40

24. The NPW of alternative E is most nearly:

$7.70

$9.20

$12.40

$32.40

25. The NPW of alternative F is most nearly:

$3.50

$7.70

$12.40

$32.40

26. Which alternative should be selected?

A

B

C

D

E

F

27. The amount of money deposited 50 years ago is most nearly:

$2,600

$5,300

$7,100

$20,400

28. The capitalized cost of the dam, including maintenance is most nearly

$ 2.2 million

$ 2.3 million

$ 2.4 million

$ 2.5 million

29. The maximum equal annual withdrawal is most nearly

$ 1150

$ 1180

$ 1270

$ 1390

https://brainmass.com/economics/macroeconomics/concepts-engineering-economics-504966

#### Solution Preview

1.

P=50/(1+10%)+100/(1+10%)^2+150/(1+10%)^3+200/(1+10%)^4=$377.399

Option C i.e. $377 is correct.

2.

P=30/(1+15%)^0+20/(1+15%)^1+20/(1+15%)^2+30/(1+15%)^3=82.2397

Option B i.e. $83 is correct.

3.

Periodic Cash inflows=C=$50

Number of periods=n=5

PV of cash inflows=Co+ C/r*(1-1/(1+r)^n)=50+50/12%*(1-1/(1+12%)^5)=230.2388

PV of cash outflows=Q/(1+12%)^1

So, Q/1.12=230.2388

Q=257.8675

Option C i.e. $258 is correct

4.

P=50/(1+10%)^4+120/(1+10^)^5+50/(1+10%)^6+120/(1+10%)^7+50/(1+10%)^8+120/(1+10%)^9=$272.68

Correct option is $275

5.

(P/G,I,n)=((1+i)^n-(i*n)-1)/(i^2*(1+i)^n)=((1+6%)^7-(6%*7)-1)/(6%^2*(1+6%)^7)= 15.44969

PV of cash inflows=100*15.44969=1544.969

PV of cash outflows=P/(1+6%)^1

P/(1+6%)^1=1544.969

P=1544.969*1.06=1637.667

Option A i.e. $1640 is correct.

6.

The interest rate is 10 %

PV of costs=-30000/(1+10%)^0-5000/(1+10%)^5=-33104.61

Option C i.e. $33100 is correct

7.

PV of ...

#### Solution Summary

There are 29 basic problems related to time value of money concepts. Solutions to given problems illustrate the methodology to find out the present value and NPV of cash flows in different cases. Solutions are derived with the help of suitable formulas.