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Engineering Economics Board Exam Reviewer 2
Civil Engineering (De La Salle University – Dasmariñas)
Studocu is not sponsored or endorsed by any college or university
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ENGINEERING ECONOMICS
ASSESSMENT NO. 2
Annuity and Amortization
Arithmetic Gradient
Geometric Gradient
Bond Value
Capitalized Cost
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1. Find the accumulated amount of the ordinary annuity paying an amortization of 1000P
per month at a rate of 12% compounded monthly for 5 years.
Formula:
(1 + 𝑖)𝑛 − 1
}
𝐹 = 𝐴{
𝑖
Where:
j = 12% compounded monthly (n1 = 12) --- i = 1% = 0.01
n = 5 years = 60 months
A = 1,000P
F = ???
Solving for F:
𝐹 = 𝐴{
(1+𝑖)𝑛 − 1
𝐹 = 1000 {
𝑖
}
(1+0.01)60 − 1
0.01
𝐹 = 81,669.67𝑃
}
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2. What present sum is equivalent to a series of 1000P annual end-of-year payments, if a total
of 20 payments are made and interest is 12%?
Formula:
1 − (1 + 𝑖)−𝑛
]
𝑃 = 𝐴[
𝑖
Where:
𝑖=12%
A = 1,000P
n = 20 annual end of year
P = ???
Solving for P:
𝑃 = 𝐴[
1−(1+𝑖)−𝑛
𝑖
]
1 − (1 + 0.12)−20
]
0.12
𝑃 = 7,470𝑃
𝑃 = 1,000 [
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3. A man made ten annual-end-of year purchases of 1000P common stock. At the end of 10th
year, he sold all the stock for 12000P. What interest rate did he obtain on his investment?
Formula:
(1+𝑖)𝑛 −1
F=𝐴{
}
𝑖
Where:
F = 2000P
A= 1,000P
n = 10 years
i = ???
Solving for i:
𝐹 = 𝐴{
(1+𝑖)𝑛 −1
𝑖
}
12,000 = 1,000 {
𝑖 = 0.04 𝑜𝑟 4%
(1+𝑖)10 −1
𝑖
}
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4. A piece of property is purchased for 10000P and yields a 1000P yearly profit. If the
property is sold after 5 years, what is the maximum price to break-even if the interest is 6%
per annum?
Formula:
𝑃 = 𝐴[
1−(1+𝑖)−𝑛
𝑖
]
;
Set up EV at 0
∑↑= ∑↓
10000 P = P + x(1 + 𝑖)−𝑛
Where:
A = 1000 P
n = 5 years
i = 6% = 0.06
P = ???
x = ???
Solving for P:
𝑃 = 1000 [
1−(1+0.06)−5
P = 4 212.36 P
0.06
]
Solving for x:
∑↑= ∑↓
10000 P = 4 212.36 P + x(1 + 0.06)−5
10000 P – 4 212.36 P = x(1 + 0.06)−5
5787.64 𝑃
(1.06)−5
=
𝑥 (1.06)−5
(1.06)−5
x = 7 745.17 P
x = 7,745 P
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5. A condominium unit can be bought at a down payment of 150000P and a monthly payment
of 10000P for 10 years starting at the end of 5th year from the date of purchase. If money is
worth 12% compounded monthly, what is the cash price of the condominium unit?
Formula:
1 − (1 + 𝑖)−𝑛
𝑃 = 𝐴{
} (1 + 𝑖)−𝑚
𝑖
P = Price of condominium
n= 121
m= 49
1
2
3
4
11⁄12 1⁄12
150,000
6
7
8
A A
A
A
10
A
Where:
Down payment = 150,000P
A = 10,000P
𝑖=
0.12
12
= 0.01
𝑛 = (10𝑦𝑟𝑠)(12𝑚𝑜𝑠) + 1 𝑚𝑜𝑛𝑡ℎ 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 5𝑡ℎ 𝑦𝑒𝑎𝑟 = 121 𝑚𝑜𝑛𝑡ℎ𝑠
𝑚 = (4 𝑦𝑒𝑎𝑟𝑠)(12 𝑚𝑜𝑛𝑡ℎ𝑠) + 11 𝑚𝑜𝑛𝑡ℎ𝑠 𝑜𝑓 𝑡ℎ𝑒 5𝑡ℎ 𝑦𝑒𝑎𝑟 = 59 𝑚𝑜𝑛𝑡ℎ𝑠
Solving for P:
𝑃 = 𝐴{
1 − (1 + 𝑖)−𝑛
} (1 + 𝑖)−𝑚
𝑖
𝑃 = 150,000 + 𝐴 {
1 − (1 + 𝑖)−𝑛
} (1 + 𝑖)−𝑚
𝑖
𝑃 = 150,000 + 10,000 {
𝑃 = 539,171𝑃
1 − (1 + 0.01)−121
} (1 + 0.01)−59
0.01
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6. The owner of the quarry signs a contract to sell his stone on the following basis. The
purchaser is to remove the stone from the certain portion of the pit according to a fixed
schedule of volume, price and time. The contract is to run 18 years as follows. Eight years
excavating a total of 20,000 m per year at 10P per meter, the remaining ten years, excavating
a total of 50,000 m per year at 15P per meter. On the basis of equal yearend payments during
each period by the purchaser, what is the present worth of the pit to the owner on the basis
of 15% interest?
Formula:
𝑃 = 𝐴1 {
1 − (1 + 𝑖)−𝑛2
1 − (1 + 𝑖)−𝑛1
} + 𝐴2 {
} (1 + 𝑖)−𝑚
𝑖
𝑖
P=?
n1=8
1
2
3
n2=10
m=8
4
5
9
10
11
12
13
A2 A2 A2 A2
A1 A1 A1 A1 A1
14
18
A2
Where:
𝐴1 = (10) (20,000) = 200,000𝑃
𝐴2 = (15) (50,000) = 750,000𝑃
𝑖 = 0.15
𝑛1 = 8
𝑛2 = 10
𝑚=8
P = ???
Solving for P:
1−(1+𝑖)−𝑛1
𝑃 = 𝐴1 {
𝑖
1−(1+𝑖)−𝑛2
} + 𝐴2 {
𝑖
} (1 + 𝑖)−𝑚
1 − (1 + 0.15)−8
1 − (1 + 0.15)−10
𝑃 = 200,000 {
} + 750,000 {
} (1 + 0.15)−8
0.15
0.15
𝑃 = 2,127,948𝑃
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7. A wealthy man donated a certain amount of money to provide scholarship grants to
deserving students. The fund will grant 10,000P per year for the first 10 years and 20,000P
per year on the years thereafter. The scholarship grants started one year after the money
was donated. How much was donated by the man if the fund earns 12% interest.
Formula:
𝑃 = 𝐴{
𝑃=
𝐴
𝑖
1 − (1 + 𝑖)−𝑛
}
𝑖
Where:
A1 = 10,000P
A2 = 20,000P
n = 10 years
i = 12% = 0.12
P = ???
Solving for P:
1− (1+𝑖)−𝑛
𝑃 = 𝐴1 {
𝑖
}+
𝐴2
𝑖
(1 + 𝑖)−𝑛
1− (1+0.12)−10
𝑃 = 10,000 {
0.12
𝑃 = 110,164.44𝑃
}+
20,000
0.12
(1 + 0.12)−10
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8. What amount of money deposited 40 years ago at 12% interest would now provide a
perpetual payment of 10,000P per annum?
Solution:
𝑃=
𝐴
𝑖
Where:
A= 10,000P
i = 12% = 0.12
n = 40 years
P = ???
Solving for P:
𝑃=
𝐴
𝑖
𝑃=
10,000
𝑃=
83,333.3333
0.12
(1 + 0.12)40 𝑃 = 83,333.3333
(1+0.12)40
𝑃 = 896𝑃
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9. A company rent a building for 50,000P per month for a period of 10 years. Find the
accumulated amount of the rentals if the rental for each month is being paid at the start of
each month and money is worth 12% compounded monthly.
Formula:
𝐹 = 𝐴(
(1 + 𝑖)(𝑛+1) − 1
− 1)
𝑖
Where:
A = 50,000P
n = 30 years = 120 months
𝑖=
. 12
= 0.01
12
F = ???
Solving for F:
𝐹 = 𝐴(
(1+𝑖)(𝑛+1) −1
𝑖
𝐹 = 50000 (
− 1)
(1+0.01)(120+1) −1
𝐹 = 11,616,954𝑃
0.01
− 1)
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10. The amount of the perspective investor pay for a bond if he desires an 8% return on his
investment and the bond will return 1000P per year for 20 years and 20,000P after 20 years
is.
Formula:
P=𝐼 [
1−(1+𝑖)−𝑛
𝑖
]+𝐶(1 + 𝑖)−𝑛
Where:
C = 20 000 P
I = 1000 P
n = 20 years
i = 8% = 0.08
P = ???
Solving for P:
P=𝐼 [
1−(1+𝑖)−𝑛
𝑖
P = 1 000𝑃 [
]+𝐶(1 + 𝑖)−𝑛
1−(1+0.08)−20
P = 14 109.11 P
0.08
]+20 000 𝑃(1 + 0.08)−20
P = 14,109 P
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11. A machine costs 50,000P. Find the capitalized cost if the annual maintenance and
operational is 5,000P and money worth 15% per annum.
Formula:
𝐶𝐶 = 𝐹𝐶 +
𝑀𝐶
𝑖
𝐶𝑅
+
(1+𝑖)𝐿 −1
𝐶𝑅
+
(1+𝑖)𝐿 −1
+
(1+𝑖)𝑘 −1
+
(1+𝑖)𝑘 −1
𝑅𝐶
Where:
FC = 50,000P
MC = 5,000P
i = 15% = 0.15
CC = ???
Solving for CC:
𝐶𝐶 = 𝐹𝐶 +
𝑀𝐶
𝑖
𝐶𝐶 = 50,000 +
5,000
0.15
𝐶𝐶 = 83,333.33𝑃
𝑅𝐶
+ 0+0
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12. A machine cost 50,000P. Find the capitalized cost if the annual maintenance cost is 5000P
and cost of repair is 4000P every 4 years and money worth 12% per annum.
Formula:
𝐶𝐶 = 𝐹𝐶 +
𝑀𝐶
𝐶𝑅
+
𝑖
(1 + 𝑖)𝑘 − 1
Where:
𝐹𝐶 = 50000𝑃
𝑀𝐶 = 5000𝑃
𝐶𝑅 = 4000𝑃
𝑘 = 4 𝑦𝑒𝑎𝑟𝑠
𝑖 = 0.12
CC = ??
Solution:
𝐶𝐶 = 𝐹𝐶 +
𝑀𝐶
𝑖
+
𝐶𝐶 = 50 000 𝑃 +
𝐶𝐶 = 98,641𝑃
𝐶𝑅
(1+𝑖)𝑘 −1
4 000 𝑃
5 000 𝑃
+
(1 + 0.12)4 − 1
0.12
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13. A building cost 10 million and the salvage value is 150,000P after 25 years. The annual
maintenance cost is 60,000P costs of repair is 200,000P every 5 years. Find the capitalized
cost if money worth 15% per annum.
Solution:
𝐶𝐶 = 𝐹𝐶 +
𝑀𝐶
𝐶𝑅
𝑅𝐶
+
+
𝑖
(1 + 𝑖)𝑘 − 1 (1 + 𝑖)𝐿 − 1
𝑅𝐶 = 𝐹𝐶 − 𝐶𝑅 − 𝑆𝑉
Where:
FC = 10,000,000P
SV = 150,000P
MC = 60,000P
CR = 200,000P
L = 25 years
k = 5 years
RC = ???
CC = ???
Solving for RC:
𝑅𝐶 = 10,000,000𝑃 − 200,000 − 150,000
𝑅𝐶 = 9,650,000𝑃
Solving for CC:
𝐶𝐶 = 𝐹𝐶 +
𝑀𝐶
𝑖
+
𝐶𝑅
(1+𝑖)𝑘 −1
𝐶𝐶 = 10,000,000 +
60,000
0.15
𝐶𝐶 = 10,900,082.29𝑃
+
+
𝑅𝐶
(1+𝑖)𝐿 −1
200,000
(1+0.15)5 −1
+
9,650,000
(1+0.15)25 −1
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14. A salesman earns 1000P on the 1st month, 1500P on the 2nd month, 2000P on the 3rd
month and so on. Find the accumulated amount of his income at the 10th month if money
worth 12% compounded monthly.
Solution:
(1 + 𝑖)𝑛 − 1
]
𝐹𝐴 = 𝐴 [
𝑖
(1 + 𝑖)𝑛 − 1
𝐹𝐴 = 𝐴 [
]
𝑖
Where:
𝑖=12%
G = 500P
A = 1,000P
n = 10 years
F = ???
Solving for F:
𝐹𝐴 = 𝐴 [
(1+𝑖)𝑛 −1
𝑖
𝐹𝐴 = 1,000𝑃 [
]
(1+0.01)10 −1
0.01
𝐹𝐴 = 10,462.21254𝑃
𝐹𝐴 = 𝐴 [
]
(1+𝑖)𝑛 −1
𝑖
]
500𝑃 (1 + 0.01)10 − 1
𝐹𝐺 =
[
− 10]
0.01
0.01
𝐹𝐺 = 23,110.62706𝑃
𝐹 = 𝐹𝐴 + 𝐹𝐺
𝐹 = 10,462.21254 + 23,110.62706
𝐹 = 33,573𝑃
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15. A man wishes to accumulate a total of 500,000P at the age of 30. On his 20th birthday, he
deposited a certain amount of money at a rate of 12% per annum. If he increases his deposit
by 10% each year until the 30th birthday, how much should his initial deposit be?
Formula:
(1 − 𝑤 𝑛
𝐴
)
(
𝑃=
1+𝑖 1−𝑤
20
0
500kP
21
1
22
2
30
10
x=?
A=x(1.1)
x(1.1)2
P
x(1.1)10
Where:
A = 500,000P
n = 30 years
𝑖 = 12%
𝑟 = 10%
P = ???
Solving for P:
𝑃=
𝐴
1+𝑖
𝑤=(
(
1+𝑟
1+𝑖
(1−𝑤 𝑛
1−𝑤
)=
)
(1.1)
(1.12)
= 0.98214
𝑃 + 𝑃2 = 500000(1.12)−10
𝑃+
𝑃2 (1.1) 1 − (0.98214)10
{
} = 500,000(1.12)−10
1.12
1 − 0.98214
𝑃 = 15,988.92 𝑃
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16. If 2000P is deposited in a savings account at the beginning of each of 15 years and the
account draws interest at 7% per year, compounded annually. Find the value of the account
at the end of 15 years.
Formula:
𝐹 = 𝐴[
(1+𝑖)(𝑛+1) −1
𝑖
− 1]
Where:
A = 2000 P ; deposited in a savings account at the beginning of each of 15 years
n = 15 years
i = 7% = 0.07
F = ???
Solving for F:
𝐹 = 𝐴[
(1+𝑖)(𝑛+1) −1
𝑖
𝐹 = 2000𝑃 [
− 1]
(1+0.07)(15+1) −1
F = 53 776.11 P
0.07
− 1]
F = 53,776 P
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17. A man deposits 1,000P every year for 10 years in a bank. He makes no deposit during the
subsequent 5 years. If the bank pays 8% interest, find the amount of the account at the end
of 15 years.
Formula:
𝐹 = 𝐴(
𝑒 𝑗𝑛 −1
𝑒 𝑗 −1
)
Where:
A = 1,000P
n = 10 years
n2 = 5 years
i = 8% = 0.08
F = ???
Solving for F:
𝐹 = 𝐴{
(1+𝑖)𝑛 − 1
𝐹 = 1000 {
}
𝑖
(1+0.08)15 − 1
}
0.08
𝐹1 = 27,152.11𝑃
𝐹 = 𝐴{
(1+𝑖)𝑛 − 1
𝐹 = 1000 {
}
𝑖
(1+0.08)5 − 1
0.08
𝐹2 = 5,866.60𝑃
}
𝐹 = 𝐹1 − 𝐹2
𝐹 = 27,152.11𝑃 − 5,866.60𝑃
𝐹 = 21,285.51𝑃
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18. Twenty-five thousand pesos is deposited in a savings account that pays 5% interest,
compounded semi-annually. Equal annual withdrawals are to be made from the account,
beginning one year from now and continuing forever. Find the maximum amount of the equal
annual withdrawal.
Formula:
𝑗
(1 + )𝑛 − 1 = (1 + 𝑖)𝑛 − 1
𝑛
𝐴 = 𝑖𝑃
Where:
𝑃 = 25000𝑃
𝑗 = 0.05
𝑛=2
i = ???
A = ???
Solving for i:
𝑗
(1 + )𝑛 − 1 = (1 + 𝑖)𝑛 − 1
(1 +
𝑛
0.05 2
)
2
− 1 = (1 + 𝑖)1 − 1
𝑖 = 0.0506
Solving for A:
𝐴 = 𝑖𝑃
𝐴 = (0.0506)(25000)
𝐴 = 1,265P per year
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19. What amount of money deposited 50 years ago at 8% interest would now provide a
perpetual payment of 10,000P per year?
Solution:
𝐴
𝑖
𝑃=
∑↑=∑↓
𝑃𝑃 =
i = 8% per year
10,000
0.08
A
A
A = 10,000P per year
1
2
3
50
0
P (1.08)50
P
Where:
A = 10,000P
i = 8% = 0.08
n = 50 years
P = ???
Solving for P:
∑↑=∑↓
𝑃 = 𝑃𝑃
𝑃 (1 + 𝑖)𝑛 =
𝐴
𝑖
𝑃 (1 + 0.08)50 =
𝑃=
125,000
(1.08)50
10,000
0.08
𝑃 = 2,665.15𝑃
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20. A man buys a motor cycle. There will be no maintenance cost the first year as the motor
cycle is sold with one-year free maintenance. The 2nd year the maintenance is estimated at
2000P. In subsequent years the maintenance cost will increase by 2000P per year. How
much would need to be set aside now at 5% interest to pay the maintenance costs of the
motor cycle for the first 6 years of ownership?
𝐺 1−(1+𝑖)−𝑛
Solution: 𝑃 = 𝑃𝐺 = [
𝑖
𝑖
− 𝑛(1 + 𝑖)−𝑛 ]
Where: 𝑖=5%
𝐺 = 2,000P
𝑛 = 6 years
P = ???
Solving for P:
𝐺 1−(1+𝑖)−𝑛
𝑃 = 𝑃𝐺 = [
𝑖
𝑖
− 𝑛(1 + 𝑖)−𝑛 ]
2,000 1 − (1 + 0.05)−6
𝑃=
[
− 6(1 + 0.05)−6 ]
0.05
0.05
𝑃 = 23,936𝑃
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