Engineering Economy 3rd Edition Solution Manual by Hipolito

Engineering Economy by Hipolito Sta Maria provides comprehensive solutions for various economic problems encountered in engineering. The third edition includes detailed explanations and calculations for topics such as interest rates, present worth, future worth, and rate of return. This manual is essential for engineering students and professionals looking to enhance their understanding of economic principles in engineering contexts. It covers multiple solved problems that illustrate practical applications of economic analysis in decision-making processes. Ideal for students preparing for exams or professionals seeking to refresh their knowledge in engineering economy.

Key Points

Includes solutions to engineering economy problems related to interest and money-time relationships
Covers calculations for present worth, future worth, and rate of return
Features detailed examples demonstrating economic analysis in engineering decisions
Provides supplementary problems for practice and deeper understanding of concepts

Brian Abadilla

57 pages

Language:English

Type:Book

Ancient History

Brian Abadilla

57 pages

Language:English

Type:Book

Ancient History

247

CHAPTER 2

Interest and Money-Time Relationship

Solved Supplementary Problems

Problem 2.1

What is the annual rate of interest if P265 is earned in four months on an

investment of P15, 000?

Solution:

Let ‘n’ be the number of interest periods. Thus, on the basis of 1 year (12 mo.), the

interest period will be,













Hence, the rate of interest given by the formula, i =





, is computed as

i =



󰇛



󰇜

󰇡





󰇢

= 0.053 or 5.3%

Thus, the annual rate of interest is 5.3%

2-2. A loan of P2, 000 is made for a period of 13 months, from January 1 to January 31 the following

year, at a simple interest of 20%. What is the future amount is due at the end of the loan period?

Solution:

󰇛  󰇜

󰇟  







󰇛



󰇜

󰇠



Answer: 

2-3. If you borrow money from your friend with simple interest of 12%, find the present worth of P20,

000, which is due at the end of nine months.

Given:

Future worth: F = P20, 000

Number of interest period: n =





Simple interest i = 12%

Solution:

󰇛  󰇜



󰇟  







󰇛



󰇜

󰇠





Answer: 

2-4. Determine the exact simple interest on P5, 000 for the period from Jan.15 to Nov.28, 1992, if the

rate of interest is 22%.

Given:

P= P5, 000

i= 22%

Solution:

January 15= 16 (excluding Jan.15)

February = 29

March = 31

April = 30

May = 31

June = 30

July = 31

August = 31

September = 30

October = 31

November 28 = 28 (including Nov.28)

318 days

Exact simple interest = Pin

= 5000×318∕366×0.22

= P955.74

Answer: P955.74

2-5. A man wishes his son to receive P200, 000 ten years from now. What amount should he invest if

it will earn interest of 10% compounded annually during the first 5 years and 12% compounded

quarterly during the next 5 years?

Given:

F= P200, 000;

For compound interest:

i= 10%; n=5

For compound interest

i= 12%∕4= 3%; n= 5×4=20

Solution:

= F (1+ i)

-n

= 200000 (1+0.03)

-20

= P110, 735.15

= P

(1+i)

-n

= 110,735.15 (1+0.10)

-5

= P68, 757.82

Answer: P68, 757.82

2-6. By the condition of a will the sum of P25, 000 is left to be held in trust by her guardian until it

amounts to P45, 000. When will the girl receive the money if the fund is invested at 8%

compounded quarterly?

Given:

P = P25, 000 i = 8%∕4= 2%

F = P45, 000

Solution:

F = P (1+i)

45000= 25000 (1+0.02)

45000∕25000= (1.02)

1.8= (1.02)

In (1.8) = 4nln (1.02)

29.682 = 4n

n = 7.42 years

Answer: 7.42 years

2-7. At a certain interest rate compounded semiannually P5, 000 will amount to P20, 000 after 10

years. What is the amount at the end of 15 years?

2 3 4

200,000

3 4

110 735.15

Overview

Engineering Economy 3rd Edition Solution Manual by Hipolito Sta Maria

/ 57

247

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100 Preposition Examples in Sentences

101 Questions Before Getting Engaged

10th Self-Development and Art Appreciation Marathi

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11 General Orders

FAQs of Engineering Economy 3rd Edition Solution Manual by Hipolito Sta Maria

What are the key concepts covered in Engineering Economy by Hipolito Sta Maria?

Engineering Economy by Hipolito Sta Maria covers essential concepts such as time value of money, interest rates, present worth, future worth, and rate of return. The book emphasizes the application of these concepts in engineering decision-making, providing a framework for analyzing economic feasibility. It also includes various solved problems that illustrate how to apply these principles in real-world scenarios, making it a valuable resource for students and professionals alike.

How does the book approach the topic of interest rates in engineering economy?

The book explains interest rates as a fundamental concept in engineering economy, detailing how they affect investment decisions and financial planning. It introduces various types of interest, including simple and compound interest, and provides formulas for calculating interest over different time periods. By presenting practical examples, the book helps readers understand the implications of interest rates on project evaluations and financial assessments.

What types of problems are included in the solution manual?

The solution manual includes a variety of problems related to engineering economy, such as calculating present worth, future worth, and determining the rate of return on investments. Each problem is accompanied by a step-by-step solution that illustrates the application of economic principles. The manual is designed to enhance the learning experience by providing practical exercises that reinforce theoretical concepts.

Who would benefit from using this solution manual?

This solution manual is beneficial for engineering students, educators, and professionals who need a deeper understanding of economic analysis in engineering contexts. It serves as a study aid for students preparing for exams and as a reference for professionals involved in project evaluations and financial decision-making. The comprehensive nature of the solutions provided makes it a valuable tool for anyone looking to master the principles of engineering economy.

What is the significance of the rate of return in engineering economy?

The rate of return is a critical metric in engineering economy, representing the profitability of an investment relative to its cost. It helps engineers and decision-makers assess the financial viability of projects by comparing expected returns against the initial investment. Understanding how to calculate and interpret the rate of return allows professionals to make informed decisions about resource allocation and project prioritization.