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Engineering Fundamentals An Introduction to Engineering (SI Edition) 7th Edition By Saeed Moaveni (Solutions Manual)

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Solutions Manual for Engineering Fundamentals An Introduction to Engineering (SI Edition) 7th Edition By Saeed Moaveni (All Chapters, 100% Original Verified, A+ Grade) Solutions Manual for Engineering Fundamentals An Introduction to Engineering (SI Edition) 7th Edition By Saeed Moaveni (All Chapters, 100% Original Verified, A+ Grade) Solutions Manual for Engineering Fundamentals An Introduction to Engineering (SI Edition) 7th Edition By Saeed Moaveni (All Chapters, 100% Original Verified, A+ Grade) Solutions Manual for Engineering Fundamentals An Introduction to Engineering (SI Edition) 7th Edition By Saeed Moaveni (All Chapters, 100% Original Verified, A+ Grade)

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Uploaded on
January 6, 2025
Number of pages
508
Written in
2024/2025
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Solutions Manual for Engineering
Fundamentals An Introduction
to Engineering (SI Edition) 7th
Edition By Saeed Moaveni (All
Chapters 1-21, 100% Original
Verified, A+ Grade)
All Chapters Arranged Reverse:
21-1
This is the Original Solutions Manual
for 7th Edition, All Other Files in the
Market are Wrong/Old Questions.

, Solution and Answer Guide: Moaveni, Engineering Fundamentals: An Introduction to Engineering 7e, SI Edition, 2024,
9780357684429; Chapter 21: Engineering Economics



Solution and Answer Guide
Moaveni, Engineering Fundamentals: An Introduction to Engineering 7e, SI Edition, 2024,
9780357684429; Chapter 21: Engineering Economics


End of Section Problem Solutions

Note: Some solutions are not provided due to the nature of the problems.



21.1 Compute the future value of the following deposits made today:

(a) $10,000 at 6.75% compounding annually for 10 years

(b) $10,000 at 6.75% compounding quarterly for 10 years

(c) $10,000 at 6.75% compounding monthly for 10 years



SOLUTION


(a)
F = P (1 + i ) n = 10000(1 + .0675)10 = $19216.70


(b)
i nm 0.0675 (10 )( 4 ) 0.0675 40
F = P (1 + ) = 10000(1 + ) = 10000(1 + ) = $19530.02
m 4 4


(c)
i nm 0.0675 (10 )(12 ) 0.0675 120
F = P (1 + ) = 10000(1 + ) = 10000(1 + ) = $19603.21
m 12 12




© 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible 21-1
website, in whole or in part.

, Solution and Answer Guide: Moaveni, Engineering Fundamentals: An Introduction to Engineering 7e, SI Edition, 2024,
9780357684429; Chapter 21: Engineering Economics

21.2 Compute the interest earned on the deposits made in Problem 21.1.



SOLUTION


(a) Interest = $19216.70 − $10000 = $9216.70


(b) Interest = $19530.02 − $10000 = $9530.02


(c) Interest = $19603.21 − $10000 = $9603.21


21.3 How much money do you need to deposit in a bank today if you are planning to
have $5000 in four years by the time you get out of college? The bank offers a 6.75%
interest rate that compounds monthly.


SOLUTION


F 5000
P= nm
= ( 4 )(12 )
= $3819.78
 i   0.0675 
1 +  1 + 
 m  12 



21.4 How much money do you need to deposit in a bank each month if you are planning
to have $5000 in four years by the time you get out of college? The bank offers a
6.75% interest rate that compounds monthly.


SOLUTION


 i   0.0675 
   
A=F  m  = 5000  12  = $91.02
nm ( 4 )(12 )
 i     0.0675  
 1 +  − 1   1 +  − 1
 m    12  


© 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible 21-2
website, in whole or in part.

, Solution and Answer Guide: Moaveni, Engineering Fundamentals: An Introduction to Engineering 7e, SI Edition, 2024,
9780357684429; Chapter 21: Engineering Economics

21.5 Determine the effective rate corresponding to the following nominal rates:

(a) 6.25% compounding monthly

(b) 9.25% compounding monthly

(c) 16.9% compounding monthly



SOLUTION
12
i  0.0625 
(a) ieff = (1 + ) m − 1 = 1 +  − 1 = 0.0643 or 6.43%
m  12 
12
i  0.0925 
(b) ieff = (1 + ) m − 1 = 1 +  − 1 = 0.0965 or 9.65%
m  12 
12
i  0.169 
(c) ieff = (1 + ) m − 1 = 1 +  − 1 = 0.1827 or 18.27%
m  12 


21.6 Using Excel or a spreadsheet of your choice, create interest-time factor tables,
similar to Table 21.9, for i = 6.5% and i = 6.75%.


SOLUTION


The Interest-Time Factors for i = 6.5%

n (F/P, i, n) (P/F, i, n) (P/A, i, n) (A/P, i, n) (F/A, i, n) (A/F, i, n)

1 1.06500000 0.93896714 0.93896714 1.06500000 1.00000000 1.00000000

2 1.13422500 0.88165928 1.82062642 0.54926150 2.06500000 0.48426150

3 1.20794963 0.82784909 2.64847551 0.37757570 3.19922500 0.31257570

4 1.28646635 0.77732309 3.42579860 0.29190274 4.40717462 0.22690274

5 1.37008666 0.72988084 4.15567944 0.24063454 5.69364098 0.17563454

6 1.45914230 0.68533412 4.84101356 0.20656831 7.06372764 0.14156831

7 1.55398655 0.64350621 5.48451977 0.18233137 8.52286994 0.11733137

8 1.65499567 0.60423119 6.08875096 0.16423730 10.07685648 0.09923730

9 1.76257039 0.56735323 6.65610419 0.15023803 11.73185215 0.08523803



© 2024 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible 21-3
website, in whole or in part.

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