Solution Manual For
Engineering Economy 3rd Edition by Hipolito Sta Maria
Chapter 2-11
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 of𝑃15,000?
Solution:
Let „n‟ be the number of interest periods. Thus, on the basis of 1 year (12 mos.), the
interest period will be,
4 1
𝑛= =
12 3
𝐼
Hence, the rate of interest given by the formula, i= , is computed as
𝑃𝑛
𝑃265
= = 0.053 or 5.3%
1
( )
𝑃15,000 (3)
Thus, the annual rate of interest is 5.3%
Problem 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 future amount is due at the end of
the loan period? algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and technology. The primary goal of
mathematics exams is to test students' understanding of mathematical concepts, their problem-solving abilities
Solution:
For the period of 13 months, the number of interest periods „n‟ on the basis of 1 year
(12 mos.) is calculated as
13
𝑛=
12
Using the formula for future worth, 𝐹 = 𝑃(1 + 𝑛𝑖) with interest and principal
given, the future amount is computed as
13
𝐹 = 𝑃2, 000[1 + ( ) (0.2)]
12
𝐹 = 𝑃2, 433. 33 is the amount due at the end of the loan period.
, Problem 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.
Solution:
algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and technology. The primary goal of mathematics exams is to test students' understanding of mathematical concepts, their
problem-solving abilities
The present worth of the borrowed money at the end of nine months is computed
using the formula,
𝐹 = 𝑃(1 + 𝑛𝑖)−1
The number of interest periods on the basis of 1 year (12mos.) is,
9 3
𝑛= =
12 4
Then, with the simple interest, number of periods, and the future amount given,
substituting these values to the present worth formula, the principal amount is
calculated as,
3
( ) −1
𝑃 = 𝑃20, 000[1 + ( ) 0.12 ]
4
Hence,
𝑃 = P18, 348. 62 is the principal amount/borrowed money
Problem 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%.
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
, In exact simple interest, 1 interest period is equal to 366 days for 1 leap year.
Thus,
318
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠 = days
366
Using the formula, 𝐼 = 𝑃𝑛𝑖, the exact simple interest is computed as
318
I = (𝑃5,000)( 𝑑𝑎𝑦𝑠)(0.22)
366
I = P955.74
algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and technology. The primary goal of mathematics exams is to test students' understanding of mathematical concepts, their
problem-solving abilities
Problem 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?
Solution:
𝑃2 = 𝐹(1 + 𝑖)−𝑛
= 200000 (1+0.03)-20
P2= P110,735.15
P1= P2 ( 1+i )-n
= 110,735.15 (1+0.10)-5
P1= P68,757.82
Problem 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?
Solution:
For compound interest, the rate of interest per interest period is given by the formula
𝑟
𝑖= ,
𝑚
If the nominal rate of interest is 8% compounded quarterly,
8% .08
then, 𝑖= = = 0.02
4 4
, Hence, the formula to be used is
𝐹 = 𝑃(1 + 𝑖)𝑚𝑛
Substituting the given values to the formula where 𝑚 = 4 (compounded quarterly),
algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and technology. The primary goal of mathematics exams is to test students' understanding of mathematical concepts, their
problem-solving abilities
𝑃45000 = 𝑃25000 (1 + 0.02)4𝑛
𝑃45000 ∕ 𝑃25000 = (1.02)4𝑛
1.8 = (1.02)4𝑛
Using algebra, multiply „ln‟ on both sides
𝑙𝑛 (1.8) = 4𝑛𝑙𝑛 (1.02)
29.682 = 4𝑛
Thus, the number of years the girl will receive the money is
𝑛 = 7. 42 𝑦𝑒𝑎𝑟𝑠
Problem 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?
Solution:
First, compute for the interest rate that is compounded semiannually (𝑚 = 2) using
the formula,
𝑖
𝐹 = 𝑃(1 + )𝑚𝑛
𝑚
With the given values of𝑃 = 𝑃5000, 𝐹 = 𝑃20000, and 𝑛 = 10 (after 10 years),
i 2(n1)
𝐹1 = 𝑃 (1 + 2)
𝑖 2(10)
𝑃20,000 = 𝑃5,000 (1 + )
2
thus,
i = 14.35%
at the end of 15 years (𝑛 = 15), the future worth can be computed as
Engineering Economy 3rd Edition by Hipolito Sta Maria
Chapter 2-11
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 of𝑃15,000?
Solution:
Let „n‟ be the number of interest periods. Thus, on the basis of 1 year (12 mos.), the
interest period will be,
4 1
𝑛= =
12 3
𝐼
Hence, the rate of interest given by the formula, i= , is computed as
𝑃𝑛
𝑃265
= = 0.053 or 5.3%
1
( )
𝑃15,000 (3)
Thus, the annual rate of interest is 5.3%
Problem 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 future amount is due at the end of
the loan period? algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and technology. The primary goal of
mathematics exams is to test students' understanding of mathematical concepts, their problem-solving abilities
Solution:
For the period of 13 months, the number of interest periods „n‟ on the basis of 1 year
(12 mos.) is calculated as
13
𝑛=
12
Using the formula for future worth, 𝐹 = 𝑃(1 + 𝑛𝑖) with interest and principal
given, the future amount is computed as
13
𝐹 = 𝑃2, 000[1 + ( ) (0.2)]
12
𝐹 = 𝑃2, 433. 33 is the amount due at the end of the loan period.
, Problem 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.
Solution:
algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and technology. The primary goal of mathematics exams is to test students' understanding of mathematical concepts, their
problem-solving abilities
The present worth of the borrowed money at the end of nine months is computed
using the formula,
𝐹 = 𝑃(1 + 𝑛𝑖)−1
The number of interest periods on the basis of 1 year (12mos.) is,
9 3
𝑛= =
12 4
Then, with the simple interest, number of periods, and the future amount given,
substituting these values to the present worth formula, the principal amount is
calculated as,
3
( ) −1
𝑃 = 𝑃20, 000[1 + ( ) 0.12 ]
4
Hence,
𝑃 = P18, 348. 62 is the principal amount/borrowed money
Problem 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%.
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
, In exact simple interest, 1 interest period is equal to 366 days for 1 leap year.
Thus,
318
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠 = days
366
Using the formula, 𝐼 = 𝑃𝑛𝑖, the exact simple interest is computed as
318
I = (𝑃5,000)( 𝑑𝑎𝑦𝑠)(0.22)
366
I = P955.74
algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and technology. The primary goal of mathematics exams is to test students' understanding of mathematical concepts, their
problem-solving abilities
Problem 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?
Solution:
𝑃2 = 𝐹(1 + 𝑖)−𝑛
= 200000 (1+0.03)-20
P2= P110,735.15
P1= P2 ( 1+i )-n
= 110,735.15 (1+0.10)-5
P1= P68,757.82
Problem 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?
Solution:
For compound interest, the rate of interest per interest period is given by the formula
𝑟
𝑖= ,
𝑚
If the nominal rate of interest is 8% compounded quarterly,
8% .08
then, 𝑖= = = 0.02
4 4
, Hence, the formula to be used is
𝐹 = 𝑃(1 + 𝑖)𝑚𝑛
Substituting the given values to the formula where 𝑚 = 4 (compounded quarterly),
algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and technology. The primary goal of mathematics exams is to test students' understanding of mathematical concepts, their
problem-solving abilities
𝑃45000 = 𝑃25000 (1 + 0.02)4𝑛
𝑃45000 ∕ 𝑃25000 = (1.02)4𝑛
1.8 = (1.02)4𝑛
Using algebra, multiply „ln‟ on both sides
𝑙𝑛 (1.8) = 4𝑛𝑙𝑛 (1.02)
29.682 = 4𝑛
Thus, the number of years the girl will receive the money is
𝑛 = 7. 42 𝑦𝑒𝑎𝑟𝑠
Problem 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?
Solution:
First, compute for the interest rate that is compounded semiannually (𝑚 = 2) using
the formula,
𝑖
𝐹 = 𝑃(1 + )𝑚𝑛
𝑚
With the given values of𝑃 = 𝑃5000, 𝐹 = 𝑃20000, and 𝑛 = 10 (after 10 years),
i 2(n1)
𝐹1 = 𝑃 (1 + 2)
𝑖 2(10)
𝑃20,000 = 𝑃5,000 (1 + )
2
thus,
i = 14.35%
at the end of 15 years (𝑛 = 15), the future worth can be computed as
