ENGINEERING ECONOMY – PRACTICE PROBLEMS
with the first payment to be made one year
after his retirement. The supervisor, instead,
requested that he be paid a lump sum on the
date of his retirement less interest that the
company would have earned if the gratuity is to
be paid on yearly basis. If interest is 15%, what
is the equivalent lump sum that he could get?
a. P100,375.37
b. P137,375.37
c. P107,375.37
d. P237,375.37
Solution:
 (1 + i )n − 1
P = A
n 
 i (1 + i ) 
 (1 + 0.15)10 − 1 
P = 20
= P100,375.37 (a)
10 
 0.15(1 + 0.15) 
1. Maintenance costs for a machine are P500 each
year. What is the present worth of these
maintenance costs over a 12 year period if the
interest rate is 8%?
a. P1,768
b. P2,768
c. P3,768
d. P5,768
Solution:
 (1 + i )n − 1
P = A
n 
 i (1 + i ) 
 (1 + 0.08)12 − 1 
P = 500
= P3,768.04 (ans).
12 
 0.08(1 + 0.08) 
2. A P15,000 simple interest load is taken out at
15% per annum interest rate. The loan matures
in two years with no intermediate payments.
How much will be due at the end of the second
year?
a. P17,250.00
b. P19,500.00
c. P22,500.00
d. P30,000.00
Solution:
The interest each year is
I = (0.15)(15,000) = P2250
The total amount due in two years is
S = 15,000 + (2)(2250) = P19,500.00 (b)
5. A unit welding machine cost P45,000 with an
estimated life of 5 years. Its salvage value is
P2500. Find its depreciation using straight
depreciation.
a. 18.9%
b. 12.1%
c. 16.7%
d. 10.2%
Solution:
FC − SV 45,000 − 2500
Annual Dep =
=
= 8500
n
5
Annual dep 8500
Depreciation rate =
=
FC
45,000
Depreciati on rate = 0.1889 = 18.89%
3. For a loan acquired six years ago, a man paid
out the amount of P75,000. The interest was
computed at 18% compounded annually. How
much was the borrowed amount?
a. P24,782.36
b. P26,782.36
c. P25,782.36
d. P27,782.36
Solution:
F = P (1 + i )n
6. Ed who was then 26 years old planned to retire
at the age of 56 years and so he purchased an
annuity. He agreed to pay equal annual
payments, the payment was made on his 26th
birthday and the last was made on his 55
birthday. At present, Ed is receiving a yearly
income of P120,000 per which he started
receiving on his 56th birthday. Compute for the
annual amount that he paid for the annuity if
interest rate is 10%.
a. P2,295.10
b. P6,295.10
75,000 = P (1 + 0.18)6
P = P 27,782.36 (d)
4. For having been, loyal, trustworthy and
efficient, the company has offered a supervisor
a yearly gratuity pay of P20,000 for 10 years
1
the depreciation charged during the 5th year
using the sum of the years digit method.
a. P101,107.11
b. P107,110.11
c. P170,110.11
d. P100,711.11
Solution:
8
SYD = (1 + 8) = 36
2
FC = 1,000,000(1.03) = P1,030,000
SV = 0.12(1,030,000) = P123,600
c. P4,295.10
d. P7,295.10
Solution:
A = annual payment of annuity
F = future value of annuity
P = present value of perpetuity
F=P
 (1 + i )n − 1 A1
A
=
i

 i
A = P 7 ,295.09
 (1 + 0.10)30 − 1 120,000
A
=
0.10
0.10


(d)
n−4
D5 = (FC − SV )

 D 
 8− 4 
D5 = (1,030,000 − 123,600)

 36 
D5 = P100,711.11
7. The initial cost of a paint sand mill, including its
installation, is P800,000. The BIR approved life
of this machine is 10 years for depreciation. The
estimated salvage value of the mill is P50,000,
and the cost of dismantling is estimated to be
P15,000. Using straight line depreciation, what
is the annual depreciation charge?
a. P75,000
b. P85,000
c. P76,500
d. P90,000
Solution:
FC − SV
Annual Dep =
n
800,000 + 15,000 − 50,000
Annual Dep =
= 76,500 (c)
10
10. An engineer is planning for his 15-year
retirement. In order to supplement his pension
and offset the anticipated effects of inflation,
he intends to withdraw $5000 at the end of first
year, and to increase the withdrawal by $1000
at the end of each successive year. How much
money must the engineer have in his savings
account at the start of his retirement, if money
earns 6% per year, compounded annually?
a. $206,116.59
b. $126,116.59
c. $146,116.59
d. $106,116.59
Solution:
 (1 + i )n − 1
 (1 + i )n − 1
n 
+
−
P = A
G
 2

n 
n
i (1 + i )n 
 i (1 + i ) 
 i (1 + i )
A = 5000, A = 1000, i = 6%, n = 15
 (1 + 0.06)15 − 1 
 (1 + 0.06)15 − 1

15
P = 5000
+ 1000
−
15 
15
15 
2
0
.
06
(
1
+
0
.
06
)
0
.
06
(
1
+
0
.
06
)
0
.
06
(
1
+
0
.
06
)




P = $106,116.59 (d)
11. An asset is purchased for P9000. Its estimated
economic life is ten years, after which it will be
sold for P400. Find the depreciation in the first
three years using sum-of-the-years’ digit
depreciation methods.
a. P4221.81
b. P5221.81
c. P6221.81
d. P7221.81
Solution:
8. Find the nominal rate which if converted
quarterly could be used instead of 12%
compounded semi-annually.
a. 10.58%
b. 11.28%
c. 9.38%
d. 11.82%
Solution:
Effective rate quarterly = Effective rate semi-annually
4
2
 in 
 0.12 
1 +  − 1 = 1 +
 −1
2 
 4

in = 11.82% (d)
9. The corporation purchased a machine for 1
million. Freight and installation charges
amounted to 3% of the purchase price. If the
machine shall be depreciated over a period of 8
years with a salvage value of 12%, determine
2
income is P430,000 and a fixed cost of P190,000
and variable cost of P0.348 per unit. What is the
break-even point? (ME Board April 2002).
a. 259,940
b. 288,490
c. 295,490
d. 305,490
Solution:
Actual Production = 700,000(0.62) = 434,000 units
Selling price per unit = 430,000 / 434,000 = 0.991
Let x = no. of units.
To break even:
Income = Expenses
0.991x = 190,000 + 0.348x
X = 295,490 units (c)
10
(1+ 10) = 55
2
10
D1 = (9000 − 400 ) = P1,563.63
55
9
D2 = (9000 − 400) = P1,407.27
55
8
D2 = (9000 − 400) = P1,250.90
55
Total Depreciation = D1 + D2 + D3 = P4,221.81 (a)
12. What is the uninflated present worth of
P12,000 in two years if the average inflation
rate is 8% and interest is 12%?
a. P5,201.60
b. P6,201.60
c. P7,201.60
d. P8,201.60
Solution:
12,000
P=
= P8,201.58 (d)
(1.12)2 (1.08)2
T=
16. If the authorized capital stock of a corporation
is P2,000,000. How much must the paid-up
capital be?
a. P125,000
b. P60,500
c. P90,000
d. P150,000
Solution:
From The Corporation Code of the Philippines, Sec. 13
Subscribed Capital = Authorized Capital Stock / 4
Subscribed Capital = 2,000,000 / 4 = 500,000
Paid-up Capital = Subscribed Capital / 4
Paid-up Capital = 500,000 / 4 = P125,000 (a)
17. The balance sheet of Oriental Services Inc. is as
follows:
Asset
Liabilities
Cash
P10,000 Payables
P17,000
Receivables
12,000 Notes due
6,000
Inventory
7,000
Long
term 3,000
debt
Capital
20,000 Owner’s
23,000
equipment
equity
What is acid test ratio?
a. 0.846
b. 0.592
c. 1.11
d. 0.385
Solution:
Cash + Accounts Re ceivables
Acid Test Ratio =
Total Liabilities
10,000 + 12,000
Acid Test Ratio =
= 0.846 (a)
17,000 + 6,000 + 3,000
18. A large sewer system will cost P8,750.00
annually. There will be favorable consequences
13. A ski-resort installs two new ski lifts at a cost of
P2,000,000. The resort expects annual gross
revenue to increase P600,000 while it incurs an
annual expense of P75,000 for lift operation
and maintenance. What is the pay-back period?
a. 2.5 years
b. 3.0 years
c. 3.8 years
d. 4.5 years
Solution:
2,000,000
Pay − back Period =
= 3.8 years
600,000 − 75,000
14. A factory is running at 80% efficiency with a
fixed cost of P78,000.00 variable cost per unit of
P130.00, selling price per unit of P416.00, and
production capacity of 5000 units. What is the
current profit of the factory if all products
manufactured are sold? (ME Board October
2001).
a. P1,352,000.00
b. P1,066,000.00
c. P1,430,000.00
d. P1,144,000.00
Solution:
Income = 5000(0.80)(416) = P1,664,000
Expenses = 130(5000)(0.80) + 78,000 = P598,000
Profit = P1,664,000 – P598,000 = P1,066,000 (b)
15. In a small factory with a capacity of 700,000
units per year with a 62% efficiency, the annual
3
to the general public of P25,000,000 annually,
and adverse consequences to a small segment
of the public of P2,500,000 annually. What is
the benefit to cost ratio?
a. 2.57
b. 2.56
c. 2.55
d. 2.58
Solution:
B = annual benefits = P25,000,000
OM = annual operation and maintenance costs =
P2,500,000
C = annual cost = 8,750,000
Benefit B − OM
=
Cost
C
Benefit 25,000,000 − 2,500,000
=
= 2.57 (a)
Cost
8,750,000
r = 0.08
n = 10
Bond interest rate = Fr = (P 1000)(0.08) = P 80
P = P 1030
R = F = P 1000
 (1 + i )10 − 1
P 1030 = (P 80)
+ (P 1000)(1 + i )−10
10 
(
)
i
1
+
i


By trial and error
i = 0.0757 = 7.57%
Ans. (c) 7.57%
-
19. A P1 M issue of 3%, 15-year bond was sold at
95%. What is the rate of interest of this
investment?
a. 3.4%
b. 4.4%
c. 5.4%
d. 6.4%
Solution:
F = P 1,000,000
Fr = (P 1,000,000)(0.03) = P 30,000
P = 0.95(P 1,000,000) = P 950,000
 (1 + i )n − 1
P = Fr 
+ R(1 + i )−n
n 
(
)
i
1
+
i


 (1 + i )15 − 1 
950,000 = 30,000
+ 1,000,000(1 + i )−15
15 
i
(
1
+
i
)


i = 0.034 = 3.4%
Ans. (a) 3.4%
20. A P1,000-bond, which will mature in 10 years
and with a bond rate of 8% payable annually, is
to be redeemed at par at the end of this period.
If it is sold at P1,030, determine the yield at this
price.
a. 5.57%
b. 6.57%
c. 7.57%
d. 8.88%
Solution:
P = Fr (P A , i %, n) + R(P F , i%, n)
F = P 1000
4
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