CONTENTs
Ques
Question Bank .l
2
Question Bank 3
Question Bank 4
Question Bank 5
Question Bank 6
Question Bank 7
Question Bank g
Question Bank g
present
Question Bank
Question Bank
1-24
Simple lnterest and
Discount
Compound lnterest
25_40
41 _ 102
Annuity
103
Uniform Gradient
_
A bookstore
should this b
194
195 _ 212
A. P 2o0.oo
B. P 3oo.oo
Bonds
213 _ 220
Depreciation and
Depletion
10 Objective
Theories and Formulas
Glossary
Economy
and AnnualCost
sis
euestions
tion
221
_240
241
_260
261
_ 280
281
_ 302
c. P 4oo.oo
D.
P 5oo.o0
Letx=sellingprice
303
324
-
-
Note: The profit of
30o/
th"-r!'rri"g;.t""'"^ffi
Profit =
323
0.30x
_+
::t#:Tf
Eq.
X?i#^"";1;innor#"or,n"capitaror3oyoor
1
Also,
337
5ffi :1'[%']3?;"*,'*r-
=
F'rotrt = 0.90x _
;:1f
200
i%:'#lo;111
_+ Eq.
discount
2
and 2;
200 = g.5x
x=400
;
The selting price
of the book is p 400.00
A businessma
tncome from a
what minimum
to be justified?
A. B.o2 %
B. 12.07
c. 10.89 %%
D.
11
.08
0/o
2
Question Bank
-
Present Economy 3
Engineering Economics by Jaime R. 'l'iong
,%*Let x = capital
y
-
Dalisay Corporation's gross margin is 45o/o of sales. Operating expenses such as
sales and administration are 15oh of sales. Dalisay Corporation is in 40% tax
bracket. What percent of sales is their profit after taxes?
gross income
Projected earnings = 0.07x
A. 21%
B. 20
c. 19%
D. 18
Taxableincome=y-x
tax
= 0.42 (y
o/o
- x)
o/o
,(*rt,"..
Net earnings = Taxable income - tax
Net earnings = (y - x)- o.+z(v - *)
Net earnings = 0.58(y
Let x = sales of Dalisay Corporation
- x)
Gross margin = 0.45x
Operating expenses = 0.15x
Projected income = net earnings
0.07x=0.58(y-x)
0.07x:0.58y-0.58x
Net income before taxes = 0.45x
= 0.30x
0.65x = 0.58y
\r-_ 0.65x
o.s8
Net profit = 0.60(0.30x)
Net profit = 0.18x
=1.1207x
.'. Rate of return before payment of taxes = 0.1207 or
12.07o/o
of the capital
A manufacturing flrm maintains one product assembly line to produce signal
generators. Weekly demand for the generators is 35 units. The line operates for 7
hours per day, 5 days per week. What is the maximum production time per unit in
hours required of the line to meet the demand?
.'. Net profit is 18% of the sales
ln determining the cost involved in fabricating sub-assembly B within a company,
the following data have been gathered:
.... P 0.30 per unit
.... P 0.50 per unit
.... P 300.00 per set up
A. t hour
B. t hour and 10 minutes
C. t hour and 15 minutes
D. t hour and 20 minutes
It is decided to subcontract the manufacturing of assembly B to an outside
company. For an order of 100 units, what is the cost per unit that is acceptable to
the company?
,%/,*
A.
B.
c.
Let t = maximum production time per unit
')
. _ (, z nours )(s oays)( l week
'
day jt *""k ,1.35 ,",tr.l
[
t = t hour per unit
.'. The maximum production time per unit is
0.15x
Since tax is 40%, net profit after taxes will be:
'-
Y
-
t hour.
P 3.80
P 4.00
P 4.10
Present Economy 5
Question []ank
-
[!ngint,t'rirrg l,)r'onomir:s by,]airnu ll,. 'l'iong
Cost per unit = 0.30 * O.so * 1-09
100
Cost per unit = P 3.80
drr;-_.'8 h;rr/day work, *itn so men workins-
.'. The cost per unit is P 3.80.
By selling balut at P 5 per dozen, a vendor gains 20%. The cost of the eggs rises
by 12.Soh.lf he sells at the same price as before, find his new gain in %.
A.
B.
C.
D.
in 50
An equipment installation job in the completion stage can be complete.d
6.89 %
6.67 %
6.58 o/o
6.12 o/o
yll
t!9,P1tr.::t-"llil11s,i40
job,
Jri", in" mechanical engineer contractor decided to add 15 men on the
overtime not being Permitted.
are paid P150
lf the liquidated damages is P 5000 per day of delay, and the men
workers?
additional
the
re
with
sa
he
p"r OrV, how much m6ney would
A.
B.
c.
P 43,450
P 43,750
P 44.250
S/.k"o...
,c/7,.l,t,,..
With 50 men only:
Let x = original cost of a dozen of balut
No. of days delaYed = 50
cost+profit=5
Liquidated damages = 5000(10) = 59999
Salary of 50 men = 50(50)150) = 375000
x+0.20x=5
1.2Ox =5
-
40 = 10 days
Total exPenses = 50000 + 375000
Total exPenses = 425000
x = 4.17
Cost per dozen of balut = P4.17
With 65 men (an additional of 15 men):
Let y = new cost of dozen of balut
Solving for the man-day to finish the job' N:
y = 1.'125x
y =1.125(.17)
N
:
'
N=
y = 4.69125
(50 men)(50 daYs)
2500 man - day
(a total of 65)
Let x = number of days to finish the job with 15 more men
cost+Profit=5
4.69125 + (% gain)(4.6912s) = 5
ohgain 0.0658
=
o/o
gain: 6.58%
.'. The new gain of the balot vendor is 6.58 %.
(50+15)x=2500
x = 38.46
Sayx=39daYs
paid by the
Since number of days is 39 < 40, thus no penalty will be
workers:
the
of
salary
the
on
solely
Will
contractor. So expenses
Salary of 65 men = 65(150X39) = 380250
.'. The amount saved = 425000
PROPERTY TAG
21432 n
DO NOT REMOVE
-
380250 = P 44,750'00
6
Quest,ion
lllrnk
It)rrgirrct,rirrg l,l,orrornir:s by Jainro Ii,. 'l'iong
Present Econorny 7
ln a eertain department store, the monthly salary of a saleslady is parfly constant
and partly varies as the value of her sales for the month. When the value of her
sales for the month is P 10,000.00, her salary for the month is P g00.00. When
her monthly sales go up to P 12,000.00, her monthly salary goes up to
P1,000.00. What must be the value of her sales for the month so that her salary
for the month will be P 2,000.00?
A.
B.
c.
D.
P 30,000
P 31,000
P 32,000
P 33,000
Jojo bought a second-hand Betamax VCR and then sold it to Rudy at a profit of
40%. Rudy then sold the VCR to Noel at a profit ol20%. lf Noel paid P 2,856
more than it cost Jojo, how much did Jojo pay for the unit?
A.
B.
c.
D.
'
P 4,100
P 3,900
P 4,000
P 4,200
/ /,/,t,,,
.
Let x = amount Jojo paid for the VCR unit
,'/./,r/rz"r..
Amount Noel paid for the VCR unit = x +
Let x=fixedsalary
900
-+ Eq. 1
1
-->
000
I
Amount Rudy paid for the VCR unit = x + 0.40x = 1.40x
Amount Noel paid for the VCR unit =
Eq. 2
(
.+Ox)(t
.20)= 1 .68x -+ Eq.
2
Equating equations 1 and2'.
Subtract Eq. 1 from Eq.2:
x+2856=1.68x
0.68x = 2856
x= 4200
2000y = 1 00
100
v=' 2000
Y
-+ Eq. 1
I
k = constant of proportionality of salesgirl's variable salary
z = sales for salesgirl to earn P2,000.00
x + k(10000) =
x + k(12000) =
2856
.'.
Jojo paid
P 4,200 for the VCR unit.
= 0'05
Substitute the value of y in Eq.
x + 0.05(1 0000) = 900
x=400
To earn a salary of P 2,000.00:
x+y(z)=2000
400+0.052=2000
z=32000
1
The selling price of a TV set is double that of its net cost. lf the W set is sold to a
customer at a profit of 25Yo of the net cost, how much discount was given to the
customer?
A.
B.
c.
D.
37.5
37.9
o/o
o/o
38.2%
38.5 %
Let x=netcost
Selling price = 2x
Profit = 0.25x
Profit = Selling price
-
net cost
-
discount
(,lrrt,sl
i.lr llrrrrk l,),grrrr,r,ri,g l,l.rrr.r.r,.:,
0.25x 2x x
O.25x -.
x
I)rcsent Economy 9
lr1,,lr1rrrrr, lf 'l'iorrg
disr:ounl
discount
discount = 0.75x
The quarrying cost of marble and granite blocks plus delivery cost to the
processing plant each is P 2,400.00 per cubic meter. Processing cost of marble
lnto tile is P 200.00 per square meter and that of granite into tiles also is P 600.00
per square meter.
Discount is75% of the net cost.
Solving for discount in terms of percent of selling price;
o'75x
discount 2x
discount = 0.375
discount = 375%
.'.
lf marble has a net yield of 40 square meters of tiles per cubic meter of block and
eells at P 400 per square meter, and granite gives a net yield of 50 square meters
of tiles per cubic meter of block and sells at P 1000 per square meter, Considering
all other costs to be the same, the granite is more profitable than the marble by
how much?
The discount given to the customer is 3T.S %.
A.
B.
C.
D.
A Mechanical Engineer who was awarded a p 450,000 contract
to install the
machineries of an oil mill failed to finish the work on time. As nrovirtecr far in rho
contract, hE ha
enalty
t to on
er
day for the
first
%
per
per day for eve
lf the t
days was the completion of the contract
A.
B.
G.
e
next
t
ty was
?
'
P 12,000 per cubic
P 13,000 per cubic
P 14,000 per cubic
P 15,000 per cubic
meter
meter
meter
meter
/ /./,t.,
For Marble (Per cubic meter):
Quarrying cost & delivery cost to processing plant = 2400
26 days
processing cost into tit"" =
27 days
200
[( sq.m ]1+o ,q. *1= aooo
J'
Total production cost =2400 + 8000 = 10400
28 days
Total income = 40(400) = 16000
Daily penatty for the first 1 0 days =
Total penalty for the first
1
O
days
I
(O.OrX+SOOO0) =
1
12s
= 1125(10) = 1 12SO
Profit = Total income
-
Total production cost
Profit=16000-10400
Profit = 5600
Daily penalty for the next 10 days = (O.OO5X45OO00) 2259
=
For Granite (Per cubic meter):
Total penalty for the next 10 days 22SO(1O) = 22SOO
=
Quarrying cost & delivery cost to processing plant = 2400
Daily penalty for the succeeding days = (0.01)(450,000)
= 45gO
solving for the number of days (x) beyond 20 days from end of contract
that
would amount the penalty to 60750.0b
4500x + 22500 +11250 :60750
x=6
.'. The completion of the contract was delayed for 26 days.
processins cost into tir" =
[!99)@o
sq. m)= soooo
Total production cost = 2400 + 30000 = 32400
Total income = 50(1000) = 50000
Profit = Total income
-
Profit=50000-32400
Profit = 17600
Total production cost
10
(luestion Br.ruJt_ l,)rrgirroor.rrrg lrlcononrir:s by
Jairrro Il 'l'iong
I)resent Economy
11
Solving for the difference in profit per cubic meter.
Difference = 1 7600 _ 5600
Difference = '12,000
.'. Granite is the more than marble by p,|2,000 per cubic meter.
A man would like to invest p 50,000 in government bonds and stocks
that will
give an overall annual return of about 5%. The money to be invested
in
government bonds will give an annual return of 4.5o/o and the
stocks of about 6%.
The investments are in units of p 100.00 each. lf he desires to klep r,isito"t
investment to minimum in order to reduce his rrsk, determine now riany
stocts
should be purchased.
A.
B.
c.
D.
165
166
167
168
A22OV 2 hp motor has an efficiency of 80%. lf power costsP3.00 per.kw-hr for
the first 50 kw-hr, P 2.90 per kw-hr for the second 50 kw-hr, P 2.80 for the third
kw-hr and so on until a minimum of P 2.50 per kw-hr is reached. How much does
It cost to run this rhotor continuously for 7 days?
A. P 800
B. P 820
c. P 840
D. P 860
Solving for the input power of motor:
outDut
' =Linput
Eftrclencv
o.8o
2
=input
input = 2.5 hP
Let x = number of government bonds
input=rttr[,ffit)
y = number ofstocks
input = 1.865 kw
'100x+100y=5gggg
x+y=500
0.0a5(1 00x) + 0.06(1 00y)
:
-+ Eq
'1
0.os(so0oo)
4.5x+6Y=2500
-+ Eq.2
Let P = total power consumption for 7 days
p=(1
865**Xro*.{{$t)
P = 313.32 kw - hr
Multiply Eq. 1 by 6:
6x+6Y=3000
+
Table of Power Cost:
Eq. 1a
Subtract Eq. 2 from Eq. 1a:
1.5x = 500
x = 333.33
Substitute value of x in Eq.
1,:
333.33{y=SOO
Y
= 166'67
since there was a desire to keep stock investment minimum, y must be
Total Power Cost
=
P 860.00
,'. lt will cost P 860 to run the motor continuously for 7 days'
166 instead ot 167.
TIP ABLEEUILIBBARY
12
Quest,ion tsank , l,)ngirrccrirrg lrl<:onomics by Jairno lt. Tiong
Present, Economy 13
Cost of Project = Total cost + contingency
Cost of Project = 1050000 + 157500
Cost of Project = P 1,207'500
'
An 8-meter wide concrete road pavement s00 meters long is desired to
be
constructed over a well-compacted gravel road, together wii-h the necessary
concrete curbs and gutters on both sides. ln orderio put the subgrade on
an
even level grade, a 500 cubic meters of sand fiiling is nece"."ry,-or"i *hich
the
10-inch concrete pavement will be placed?
-
Assume the following data:
Sand fill, including rolling and watering = p 100 per cubic meter
concrete pavement, 10 inches thick (iabor and materials) including currng
= P 220 per sq. meter
Curbs and gutters = p 12 per linear meter
4
B.
C.
The project will cost P 1,207,500.
en etectric utility purchases 2,300,000 kw-hr per month of electric energy from
National Power Cbrporation at P 2.OO per kw-hr and sells all this to consumers
per kw-hr should
a-fteiJeOucting distribution loss es of 2OYo. At what average rate
expenses in
monthly
other
are
be-sold to break even if the following
tnii
"n"rgy
Ite operation:
A.
B.
c.
2.5Yo of qross revenue
P 750,000
P 2.250.000
P 700,000
Taxes
Salaries
How much will the project cost allowing 15yo for contingency?
lleoreciation
lnterest
P 1,207,000
P 1,207,500
P 1,208,000
P 300,000
P 200.000
Maintenance
Miscellaneous
"V.ur.rLet Cr = cost of sand filling, rolling and watering
Cz = cost of curbs and gutter (both sides)
Co = cost of concrete pavement
A. P 4.90
B. P 5.20
c. P 5.90
D. P 6.30
,'/,/nt
n.
Let x = rate per kw-hr of energy sold to consumers
",=[#)('oom3)
Cr = P50,000
Solving for ExPenses:
.,=[T)(soomtz)
AmountpaidtoNPc=[,
P1
1kw - hr
)1r.ooooo**
/'
-nr;
= 4600000
Cz = P120,000
With 20% losses, only 80% is sold to consumers'
.,
=[TP)(soom[am)
Totalnumber of kw - hr sold = 0.80(2300000)
= 1840000 kw - hr
Ca = P880,000
Total cost = C1 + Q2 + Q.
Total cost = 50000 + 120000 +
Totalcost=P1,050,000
Contingency = 0. 1 5(1 050000)
Contingency = P 157,500
Taxes = 0.025(1 8a0000)(x)
BSOOOO
= 46000x
+
Total expenses = Amt. paid to NPC + Taxes-+ Salary + Depreciation
lnterest + Maintenance + Miscellaneous
14
Qrr.st;iolr
Ilnrrk
l,)rrgirrr.r,rirrg l,)r:.,unrir:s [ry.lairrrt,
ll, ,l,iong
Total expenses = 4600000 +-46000x
+ 750000
300000 + 2OO00O
*
Present Economy
,ruOOOO + 700000 +
Re nta, =
Total expenses = gg00000 + 46000x
tT#)(. -!#)r, weeks)
= 280000
Solving for lncome:
Sarary or 2 nasmen =
lncome = 1,g40,000x
To break-even; income expenses
=
'1840000x
Total expenses = 12252450 + 280000 +
57600
Total expenses = p 12,590,050
= 8800000 + 46000x
x = 4.90
For Site B:
.'. The average rate the energy
be sold is p 4.90 per kw_hr.
An engineer biddino
weers)(z)
= 57600
17940OOx = 88OO0OO
'
(#{ffi}sz
oozo ms
[ f+zk,.liff](2'75kmx7km)
)f
),,,
" 1;5
-
cost or haurins =
r
= 13477695
the aspharting of 7 km stretch of road
is confronted
_on
y:'Jr:,ffilJinT orcho-osino
sites on w.rt-rr'io-"""i,p *,"
oltd;;'il;
il#"
Cost of installing & dismanfling of
machine = 10000
Rentat _
r\errrdr
=f
{.
p6,5oo
'"'th
l( l montn ),^^ weeks)
14 *""k.J(32
= 52000
Total expenses = 13477695 + 1OOO0 +
52000
Total expenses = p 13,539,695
The
res
to
is
A.
B.
P 949,645
P 962,10.1
c. P 956,807
D. P 974,090
Solving for the difference in expenses;
Difference = 1 3539695 _ 12596050
Difference = 949645
.'. Site B is more expensive than site
A by p 949,645.
A fixed capital
manufacturing
For Site A:
cost of hauting
E
retu
depreciation
the rate of
(z.stm[zr<m)
= 12252450
Cost of installing & dismanfling
of machine = 20000
A. 28.33%
B. 29.34%
c. 30.12%
D.30.78%
al
ne
16
()trr,sliorr llirrrk
l,)ngrrr,r,r'rtr1i l,l.onorrrr,.., lrt,lrrrrrr,, l( 'l'iorrg
Present Economy 17
Profit
Total capitalinvested
Rate of return
'50-
Number of oost
2500000
+
1
= 101
iooooooo , 2oooooo
_-
5ooo
0 2833
.,=lffi)(rorno,,,)
The rate of return is 28.33 %.
Cr = 505000
A call to bid was advertised in the philippine Daily lnquirer for the construction
of a transmission line from a mini-hydroeiectric power piant to the substation
Number of days to dig and erect posts =
TH*
= 33.67days ,say 34 days
day
",=[H)(a+oays[s)
Cz = 30600
Number of days to strung wires =
ffi
= 33.57 days , say 34 days
day
..=(H)(s+oays)(+)
A.
B
P 745,890.23
P 817,692.00
c. P 789,120.80
Cs = 34000
D. P 829,592.50
..=[H)(s+aavs)
Let Cr = cost of posts
C4 = salary of foreman
C2 = salary of 5 laborers
C3 = salary of
Cr
= 13600
cs
=
Cs = Cost of wire
4 electricians
Solving for the number of posts:
(pa\.
[-J(soss
m)
Cs =20140
Hydro-electric
station
Power
. = 505000 + 30600 + 34000 + 1 3600 + 20140
.-,]
50m
50 nt
Total directcost=Cr +C2+ C3 +Ca +C5
= 603340
50m
ContingencY = 0. 1 0(603340)
5035 m
Distance between first and last post
:
= 60334
5035 _ 15 _ 20
:5000
m
Total cost including contingency = 603340 + 60334
= 663674
18
Quest,ion
Pr
llank
I)rcsont Economy 19
lr)ngirrct,r'irrg l,)<:onornics by Jairut, ll. 'l'iorrg
ofit = 0.25(663674)
Pr ofit
:
16591 8.50
Total cost of Project = 663674 + 165918.50
Total cost of Project = P 829,592.50
is_3200
The monthly demand for ice cans being manufactured by Mr. Alarde
nlcces With a manual ooerated quillotine, the unit cutting cost is P 25'00' An
ne was offered
the unit cutting
Alarde be able
.'. The estimated cost of the project is P 829,592.50
Upon her retirement after 44 years in government service, Mrs. Safud Araowa
lump sum of P 2,300,000. As a hedge against
inflation
rt of it invested in real estate at pagadian City
and the
as follows:
arao
A.
B.
C.
30% in T-bills earning 12% interest
35% in money market placement earning
35% in blue chip stock earning 13%
14o/o
lf her annual earnings from the t-bills, money market and stock is p 50,000 how
much did she invests in real estate?
A.
B.
c.
D.
A.
B.
C.
D.
10 months
11 months
12 months
13 months
/r,/,/,;.r"
For manuallY operated guillotine:
Monthty expenses =
[ffi),.r.O
units)
= P 80,000
P 2,091,639.12
For electrically operated hydraulic guillotine:
P 1,916,858.24
P 1,856,120.53
P 1,790,274.78
Cost Per unit = 0.70(25)
= 17.50
Monthry expenses =
Let x = amount invested in real estate
y = amount invested in T-bills, money market placement and stock
Since annual income of y is P 50000, thus
0.30y(0.1 2) + 0.35y(0. 1a) + 0.35y(0.1 3) : 50000
0.1305y = 50000
Y
= 383141.76
x=2300000-y
x = 2300000 - 383141 .76
x = 1916858.24
.'. The amount she invested in real estate is P 1,916,858.24.
tTfl)0200
units)
= P 56,000
Amount saved per month = 80000 - 56000
=P
24,OOO
cost of machine
Number of months to recover the machine = amount saved
Per month
_275000
24000
=11.46 months
The machine can be recovered in 12 months'
t
20
Quest,io' Ba,k
- l)ngi'e.rirrg
llco,onrics by J:rrr,r,
ll
,l'iong
Present EconomY 21
Net income = 28000 - 400
a new gold mining area in Southern Leyte, the ore contains on the average
-ln
of.ten ounces of gold per ton. Different methods of processing
t.ort"t"o
"r"
follows:
Processinq Method
Cost per ton
P 5,500
P 2.500
P 400
A
B
C
A.
B.
C.
D.
".
= P 27600
.j. Processing Method A yields the biggest return'
% Recoverv
90
80
70
r\^ ^aaaccanr crarinn of the enoineS. lRf
MotOrS lnC. will pay freight On
Processing method A
Processing method B
Processing method C
Either of the processing methods B or C
,%zn*
For Processing Method A:
lncome per ton
A.
B.
C. P 650 Per engine
D. P 610 Per engine
P 670 Per engine
P 630 Per engine
+ooo'][ to ounces)fo.got
- [p
\ ounce /(
ton )'
lf shipped in container:
= P 36,000
Net income = 36000 - 5500
cost or rreight
= P 30,500
-[
\
p +ooo
)[
t"".(!#)
t o ounces
/\
ounce
[#}..
= 1 20000
For Processing Method B:
rncome per ton
=
ton )')ro.aot
= P 32,000
= 20 engines
Net income = 32000 - 2500
= P 29,500
.
cost per englne =
For Processing Method C:
cost of freight
nr-m0", of ensine"
'n20000
1
lncome per ton
- [p
\
+ooo'1( t o ounces
ounce
= P 28000
/\
ton )(o.zot
)' ,
= P 6,000
lf shiPPed in crates:
TIP ARLEEUI LEBARY
22
Quesl,ion Bank -- l,)nginecrirrg F.lconomics by Jaime R. Tiong
Present Economy 23
so
szo kg
cost of freight per ensine - [eg
][t
)
Totat cost =
\ t(g /\ engrne ,
= 5320
[t,'oo
l.ks \ru
)'
*n)
= 5000
Cost of crate per englne = 50
Cost per engine = cost of freight + cost of crate
Pr oflt = Total income - total cost
= 5320 + 50
= P 5,370
B.
Using ceramics:
Solving for the difference of cost:
14-!l9rsYq}.Y=Sl(s
\ hour /1. day /\ month /'
rotar output in 2 morrths time = f
Difference=6000-5370
Difference = 630
.'.
montns)
= 72000liters
Shipping the engine by crates is more economical by P 630 per engine
than shipping by container.
rotar income
=
[*9]t
I liter /
rooo liters)
= 1 080000
A paint manufacturing company uses a sand mill for fine grinding of paint with an
output of 100 liters per hour using glass beads as grinding media. Media load in
the mill is 25 kg costing P 200.00 per kg and is fully replenished in 2 months time
at 8 hours per day operation, 25 days a month. A ceramic grinding media is
offered to this paint company, costing P 400 per kg and needs 30 kg load in the
sand mill, but guarantees an output of 120 liters per hour and full replenishment
of media in 3 months. lf profit on paint production is P 15 per liter how much is
the difference in profit between the two media?
A. P 436,900
B. P 462,000
c. P 473,000
D.
,9"2',2,*.
Using Glass Beads:
rotaroutput
=
[',090']eo *nl
Iks
)'
= 12000
Profit = Total income - total cost
=1080000-12000
= 1 068000
The ceramic is more economical as a grinding media with a difference
in profit of P1,068,000 - P 595,000 = P 473,000
P 498,200
A.
rotat cost
i
in 2 months time =
[,1qq]lgrsYq$trlf
\ hour day ./\ month /'
=+)(z
^
= 40000liters
rotarincome
=
([99'ltooo00
liter /'
titers)
= 600000
ii ,ii[t:
;i r,'
t'1
montns)
24
Question Bank
-
Engineoring Economics by Jaime R. Tiong
Question Bank
2
It P1000 accumulates to P15P0 when invested at a simple interest for three years,
what is the rate of interest?
A.
B.
c.
D.
t(r*t
14.12%
15.89%
16.67 %
16.97 %
',,
F = P(1+ in)
15oo=tooo[t+i(s)]
i= 0.1667
i
=16 67%
.'. The rate of interest is 16.67 %.
You loan from a loan firm an amount of P 100,000 with a rate of simple interest of
2g7o but the interest was deducted from the loan at the time the money was
5fiowed. lf at the end of one year, you have to pay the full amount oiP 100,000,
what is the actual rate of interest?
A.
B.
c.
D.
23.5Yo
24 7 nto
25.0%
25.8%
,'/n7)zrn.'
lnterest = 0.20(1 00,000)
lnterest = 20,000
Amount received = 100,000
Amount received = 80,000
-
20,000
26
Question Bank
l=
-
\
Simple Interest & Discount 27
Engineering Economics by Jaime R. Tiong
Pin
20,000 = 80,000(i)(1)
i = 0.025
J. Reyes borrowed money from the bank. He received from the bank P 1,842
promised to repay P 2,000 at the end of 10 months. Determine the rate of
i=25%
.'. The actual rate of interest is 25
12.19%
o/".
12.030h
11.54%
10.29%
A loan of P 5,000 is made for a period of 15 months, at a simple interest rate of
'1
5%, what future amount is due at the end of the loan period?
F = P(1+ in)
A. P 5,937.50
B. P 5,873.20
c. P 5,712.40
D.
2ooo =
P 5,690.12
n+zlt.'l'19)l
L
\12i1
i= 0.1029
i=10.29%
,9urr*"
The simple interest rate is 10.29 %.
F = P(1+ in)
= sooolr.0.',u[E)l
'r --""1'
"'"[tz)]
F = 5,937.50
.'. The future amount at the end of the loan period is P 5,937.50.
annum, what is the
borrowed P 10,000 from a bank with 18% interest per
---amount to be repaid at the end of one
r
year?
P 10,900
P 11;200
P 11,800
P 12,000
lf you borrowed money from your friend with simple interest ol 12o/o, find the
present worth of P 50,000, which is due at the end of 7 months.
A.
B.
c.
D.
F = P(1+ in)
P 46,728.97
P 47,098.12
P 47,890.12
F=1O,ooo(+0.1S)
F = 11,800
P 48,090.21
The total amount to be repaid at the end of one year. is P
11
,800.
c'*r"rn
F=P(1 +in)
so,ooo=P[1
..,r(#)]
P = 46,728.97
.'. The present worth is P 46,728.97.
tag of P 1,200 is payable in 60 days but if paid within 30 days it will have a
discount. Find the rate of interest.
28
()uestion
llnnk
l,)ngirrooring lllconorni<:s by Jaime R. Tiong
Simple Interest & Discount 29
h,,,
/,/,,/,i,,,..
0.03(1,200)
Discount =
Discount = 36
Solving for rate of discount, d
.80
o=-
i
2,000
d = 0.04
Amount payable in 30 days = 1,200 - 36
Amount payable in 30 days = 1,164
l=
d = 4o/o
Pin
Solving for the rate of interest,
.d
36=1,164(i)[#)
i=
i
.
0.3711
= 37.11o/o
i
1-d
0.04
1- 0.04
The rate of interest is 37.11 %.
I
= 0.0417
t= 4.17%
,', The rate of interest is 4.1T
o/o.
A man borrowed P 2,000 from a bank and promised to pay the amount for one
year. He received only the amount of P 1,920 after the bank collected an advance
interest of P 80.00. What was the rate of discount?
tIVhrl wlll be the future worth of money after 12 months, if the sum of p 2s,ooo is
lltvoatcd today at simple interest rate of 1o/o pet leat?
A. 3.67 %
B. 4.OO
c. 4.15
D. 4.25
o/o
A,
3,
c.
0/o
o/o
P 25,168
P 2s,175
P 25,18e
P 25,250
D,
Solving for rate of discount, d
.80
d=-
'4ta,,
F = P(1+ in)
2,000
d = 0.04
F = 2s,0oo[1* (o.or)!99.1
L '
d = 4o/o
',360]
F =25,250
.'. The rate of discount is 4 %.
;. The future worth is P 25,250.
A man borrowed P 2,000 from a bank and promised to pay the amount for one
year. He received only the amount of P 1,920 after the bank collected an advance
interest of P 80.00. What was the rate of interest that the bank collected in
advance?
4.OO
o/o
4.O7
o/o
4.17
4.25
o/o
lltlrt
witt be the future worth of money after 12 months, if the sum of p 25,000 is
lnYrcteo today at simple interest rate of 1%o per month?
A.
B,
c.
D.
o/o
- ,)L
P 28,000
P 28,165
P 28,289
P 28,250
30
Questiorr
llrrrrk
l,)rrgirrorrringltrconornir:s lrv.Jrrirno
lt.'l'iong
Sirnple lntcrest & Discount 31
Net interest = l- 0.201
Nel interest :0.801
.22,t ,,.
F=
P(1+ in)
F=
25,000[1+ o.o1(r z)]
890.36 :0.80t
I
Bul
F = 28,000
l=
.'. The future worth is P 28,000.
1,1
12 s5=
i=
It is the practice of almost all banks in the Philippines that when they grant a
loan, the interest for one year is automatically deducted from the principal
amount upon release of money to a borrower. Let us therefore assume that you
applied for a loan with a bank and the P 80,000 was approved at an interest rate
of 14 o/o of which P 11 ,2AO was deducted and you were given a check of P
68,800. Since you have to pay the amount of P 80,000 one year after, what then
will be the effective interest rate?
A.
B.
c.
D.
A.
B.
c,
D,
o/o
(
11.50
o/o
11.75%
11.95%
12.32
o/o
P 5,066.67
P 5,133.33
P 5,050.00
P 5,166.67
L
F = 5,166.67
,'. The amount due at the end of 75 days is P 5,166.67.
A bueinessman wishes to earn 7% on his capital after payment of taxes. lf the
lnoome from an available investment will be tafed at an average rate of 42Yo,
what minimum rate of return, before payment of taxes, must the investment offer
lo be justified?
A.
B.
c,
D.
,%dz,*.
Yo.
r = s,ooolr. (0.16)f=.)l
',(360/l
i= 0.1628
i=16.28%
A.
B.
c.
D.
is 11.75
F = P(1+ in)
11,2oo = 68,800(i)(1)
A deposit of P 110,000 was made for 31 days. The net interest after deducting
20% withholding tax is P 890.36. Find the rate of return annually.
o.1175
rt/,,,,.
Pin
.'. The effective rate of interest is 16.28 %.
ooo(i)(#)
lt 0,000 is borrowed for 75 days at 16% per annum simple interest How much
wlll be due at the end of 75 days?
o/o
l=
1 1 o,
,'. The rate of return annually
16.32%
16.47
Pin
i = 11.75o/o
16.02o/o
16.28
= 1,112.95
12.07 %
12.34%
12.67
12.87
0/o
0h
32
(lrrcst,irrrr
llrrrrk
Simple Interest & Discount 33
l,)rrgrru,r,t'ittg I!<:ottoutit's lry.lairne It. Tiong
3,200 = (2o,ooo
profft
Rate of return =
invested caPital
i:0.1905
D
i= 19.05%
I
Rate of return =
C
,,, The actual rate
- 3,200)(i)(1)
of interest is 19.05 %
where: p = profit
C = invested capital
? .1,000 is borrowed for 75 days at 16
bc due at the end of 75 days?
After taxes are paid,
net Profit
invested capital
Rate of return = 7 oh = O.07
Rate of l9turn
-
='
D,
D
07: --- 0.42P
0.07
-
C
P 4,666.67
'(b^
0'58P
F=P(1 +in)
C
D
L
per annum simple interest. How much
A, P 4,033.33
l, P 4,333.33
C, P 4,133.33
Substituting values:
o
o/o
Wlll
F=4ooo[.,.r,.(#)]
= 0.1207
o
-C =lz.ot%
.'. The rate of return before payment of taxes is 12.07
F = 4,133.33
A mgn borrowed P 20,000 from a local commercial bank which has a simple
interest of 16% but the interest is to be deducted from the loan at the time that
the money was borrowed and the loan is payable at the end of one year. How
much is the actual rate of interest.
A.
B.
c.
D.
,,,
o/o
Thoamount due in 75 days is
P 4,133.33.
Agntr Abanilla was granted a loan of P 20,000 by her employer CPM lndustrial
lfbrlcator and Construction Corporation with an interest bf 6 % for 180 days on
ho prlncipal collected in advance. The corporation would accept a promissory
iOh for P20,000 non-interest for 180 days. lf discounted at once, find the
pDoreds of the note.
A, P 18,800
l, P 18,900
C. P 19,ooo
19.05 %
19.34 0/o
19.67 %
19.87 %
D,
P 19,100
,'L.*
"V;/'r.o,,
lnterest = 0.06 (20,000) = 1,200
r= 0.16(20,000)
I
Proceeds on the note = 20,000
Proceeds on the note = 18,800
= 3,200
But l= Pin
-
12OO
r'. The proceeds on the notes is P 18,800.
Substituting the values:
{
34
Qucsl,iorr
Ilrrrrk
lf you borrow money from your friend with simple interest of 12%, find the present
worth of P 20,000, which is due at the end of nine months.
A.
B.
c.
D.
Simple Interest & Discount 35
I,)rtgirtrrrring Econottrics by Jaimtr Il..'lliong
P 18,992.08
P 18,782.18
P 18,348.62
P 18,'t20.45
Mr, Danilo conde borrowed money from a bank. He receives from the bank
P1,340.00 and promised to pay P 1,5oo.oo at the end of 9 months. Determine
Tala of simple interest
A.
B.
c.
D.
15.92o/o
15.75%
15.45%
15.08 %
(*h,t
F = P(1+ in)
F = P(1+ in)
20 ooo = p[r
2f3zq).l
-' L'* 0.1
" '-(aooJl
1,500 =
t.scol t.,r?29)l
L
P = 18,348.62
(360/l
'
o.11g4 -270
360
i.= 0.1592
The presentwo;th is P 18,348.52.
i =15.92o/o
A man borrowed from a bank under a promissory note that he signed in the
aniount of P 25,000.00 for a period of one year. He receives only the amount of
P21 ,915.00 after the bank collected the advance interest and an additional of P
85.00 for notarial and inspection fees. What was the rate of interest that the bank
collected in advance?
A.
B.
C.
D.
13.05%
13.22%
,', The rate of simple interest is 15.92 %.
Ml, J, Dela Cruz borrowed money
P1,340.00 and promised to pay P
lh! corresponding discount rate or
A,
B,
C,
D,
13.46 o/o
13.64 %
.7;7)a-,,-.
Amount received = 21 ,915 + 85
Amount received = 22,000
13.15 0/o
13.32 o/o
13.46 o/o
13.73Yo
(tf,.b,.
Solving for simple interest rate,
lnterest = 25,000 -22,000
lnterest = 3,000
l=
Pin
F = p(1+ in)
1,500=1,340[1+(#)]
3,000 = (22,000xi)(1)
i= 0.1364
i=13.64%
.'. The rate of interest the bank collected in advance is 13.64 %.
rec€ive
nd of g
as the
0.1194
-27o
i
360
i= 0.1592
i=15.92%
i
ine
nt,,.
36
9::'tu' lly!
li:l*l'1""1,,*
Ecr)norrrir:s tr.y Jairne R. Tiong
Simple Interest & Discount
Solving for the discount, d
.t
d=1- |
i+i
d=1-
lnterest = 1,500.00 _1,342.00
lnterest = 15g.00
'
Also,
1
1 + 0.1 592
d = 0.1373
d =13.73%
.'. The discount rate
is
l=
Pin
rsa
=
r,s+z1nf!)
\4)
i = 0.'1569
1S.gZ%.
i= 15.69%
ffii:ffnil:FJ:',13i'"t
ca
s
h p rice on
rh
,', The simple interest rate
is 15.69%.
n"m a merchant who ask p 1,2soat
the end o160
TTidll"]l
",.",, -?,
-: l.d
th
"
T "l91,r
ri
;rf;' to ;;i
ute
e
th
Ddormine the exact simpre interest
on p 5,000 invested ror ttre perioo
rronof interest is 180/o.
J'nuery 15' 1996 to october rz, rsisi,
P 1,'t24.67
P 1,233.5s
P 1,289.08
P 1.302.67
;;ffi:
B, P 664.89
c, P 665.21
0, P 666.3e
F = p(1+ in)
1,250 =
l=
pL.o oa[jq)l
(
1
soo/]
Pin
+Eq.1
Solving for the total number
of days money was invested,
d
P = 1,233.55
.'. The cash price of the television
set is p 1,233.55.
A man borrowed money from a
roan shark. He rgleryes from
the roan shark and
;'rHI[
A.
B.
C.
D.
t",,ti"hfi:'fl?"?:3,1.i",g""0 '" '"p"v
i''r,soo^oo
"i
ir,l'","0'o',", 0,"n",.,
15.47 %
15.69 %
15.80 o/o
15.96 o/o
June
July
August
September
Octoberl-12
= 30
= 31
= 31
= 30
=12
271 days
Substitute values in Eq.
t= s,ooo(o.1s,("1\
. 1366 /
l= 666.39
u^*
1
Sinrplc lnt,erotrl, & l)iscourlt' 39
38
(,lut'strorr
llrrrrk
l,)rtgtttocriltg [lconottticr lrv ,lrtirtto lt. Tiong
. The exact slmple lntereet is P 666.39.
,'fi(/,,,
F = P(1+
The exact simple interest of P 5,000 invested from June
25, 1995 is P '100. What is the rate of interest?
A.
B.
c.
D.
21
December
' 1995 to
o/o
3.95 %
3.98 %
.'/./ot;,,
l=
Pin
-+
Eq.
1
Solving for the total number of days, d
June21-30
=$
July
=31
August
September
October
November
Eq'
1
was invested' d
Solving for the total number of days the money
- 30 = $
= 31
May
= 30
June
= 31
July
= 31
August
September = 30
October = 31
November = 30
December 1-25 =E-
April 22
3.90 %
3 92
-+
in)
=
=
=
=
31
247 daYs
Substitute in Eq.
1o,ooo=P[1.(oro)(#)]
30
P = 8,807.92
31
30
Decemberl-25=25
187 dayt
1
her mother on her birthday is P
,', The amount Nicole received from
8,807.92.
Substitute values:
1oo
: s.ooo(i)f]!Zl
' "t365/
i= 0.0390
i= 3.90%
.'. The rate of interest is 3.90 %.
On her recent birthday, April 22,2001, Nicole was given by her mother a certain
t on
sum of money a
Present.
an
lf-the ac
20% exact simp
her
w much
amount of P 10,
birthday?
A.
B.
c.
D.
P 8,807.92
P 8,827.56
P 8,832.17
P 8,845.78
182 days at 5'2%?
t rhat it the ordinary interest on P 1,500'50 for
P 39.01
P 39.45
P 39.82
P 39.99
,'&,an'
l=
I
Pin
= (1,500,
l= 39.45
50)(o.o5r)[#)
40
Quest iorr
llrrrrk
lrltrgrrroorirrg lj<:onorrrrr.x lry ,Inirno R. Tiong
Question Bank 3
Nicole has P 20,400 in cash. She invested ll alTo/o from March 1, 2006 to
November 1, 2006 at 7% interest. How much is the interest using the Banker's
Rule?
A.
B.
c.
D.
P 972.12
P 970.78
P 973.',t2
P 971.83
Tho amount of P 20,000 was deposited in a bank earning an interest of 6.5% per
tnnum. Determine the total amount at the end of 7 years if the principal and
1rrl,,2*.
htlrrcet were not withdrawn during this period.
ln Banker's Rule, use exact number of days the money invested and 360
days for the total days in a year.
Solving for the exact number of days:
A.
t,
c,
0,
P 30,890.22
P 30,980.22
P 31,079.73
P 31,179.37
Marchl-31=30
April= 30
MaY = 31
June = 30
July = 31
August = 31
r=e(t+i)n
F = 20,OOO(1+
0.065r
F = 31,079.73
September:30
October = 31
November 1 = 1
,'. The total amount at the end of 7 years is P 31,079.73.
Total = 246
A Joan for P 50,000 is to be paid in 3 years.at the amount of P 65,000. What is the
dhcrtive rate of money?
l= Pln
t= (2o.4oo)(o
\ ' /\
I
= 971.83
07)r?€)
,[360/
A. 9.O1%
B. 9.14
c. 9.31Yo
D. 9.41%
o/o
r=e(+i)n
65,000=SO,OOO(1+i)3
t.a=(t+i)3
1.0914 = 1+ i
i = 0.0914
47
Quost,ion
llank
lr)rrginooring Econorrrir.r by.Jaime R. Tiong
(
The present worth is P 40,530.49.
1.0914 = 1+ i
i= 0.0914
Since nothing was mentioned about how the interest was compounded, it
presumed that the mode of compounding is annually.
.'. The effed.t rate is 9.14 %.
is
$flut
lg the effective rate corresponding to 18% compounded daily? Take 1 year
aqu.l to 360 daYs.
A
I
c,
D
The amount of P 50,000 was deposited in the bank eaming an interest ol7.So/o
annum. Determine the total arnount at the end of S years, if the principal and
interest were not withdrawn during the period.
A.
B.
c.
D.
19,61
0/o
19,440h
1s.31%
19.72%
(**ER=(1+if-t
P 71,78',t.47
P 71,187.47
en
P71,817.47
P 71,718.47
ER =19.72%
(tzz.
)ortrpottrttl I rrl,t,rtttl nnd Cont.iuuous C<lnrpounding 43
0.18)360 _1
=[r+ 360/
\
,', The effective rate corresponding to 18% compounded daily is 19.72o/o.
F = P(1+ i)n
F=
5o,o0o(1+o.o75f
F =71,781.47
.'. The tota! amount at the end of 5 years is P 71,781.47.
Find the present worth of a future payment of P 80,000 to be made in six (6)
with an interest of 12Yo compounded annually.
A.
B.
c.
D.
.7"2*.*
P
P
P
P
40,540.49
40,450.49
40,350.49
40,530.49
yields the same amount as
\Mltt nominal rate,.compounded semi-annually,
a
Slnpounded
quarterly?
A, 16.09 %
l, 16.32
c, 16.45%
D, '16.78%
0/o
12,h.
ER,
= EP,
24
[,.T)'-,=(,.0i)'-,
o'ro)o
l..,*!B)'( 4)
( 2/ [.,*
[.,
\
*
rB)'
2)
1*
NR
2
='r.16e8556
= 1.0816
16%
36
()ucstiorr
llrruk
llrrgirrrrorirrg [!<:orrorrrir.* lrv.lrrirrro It. Tiong
Solving for the discount, d
lnlerest
.
Simple Interest & Discount
37
1,500 00
- .t,342.00
lnterest = 158.00
d=1_1
1+i
Also,
d=1-
l=
1 + 0.1592
d = 0.1373
Pin
158 =
d=13.73%
1,342({;)
i= 0.1569
i= 15.69%
,'. The discount rate is 15.92 %.
The simple interest rate is
15.69%.
Annie buys a terevision set from a merchant who
ask p l,zsoat the end of 60
ty, n J ir,
orers to comiute tne
1'\,ll:1vlil-^1"
ry^,,lmed.iare
cash price
on the assumption
"'rn ","nani
thar mor"y ir *ortn
aL
,inr, ,. ,n"
cash price?
::f
liil;ft;#.
for the perioo rromorinterest is 18%.
A. P 1,124.67
B. P 1,233.55
c. P 1,289.08
D.
$ermine the exact simpre interest on p 5,000
fnuary 15, 1e96 to october rz,1sigo, ;tiJI]t;invested
P 664.89
",B.
c. P 665.21
D. P 666.3e
P 1,302.67
F=P(1+in)
l=
1,250 =p[r.
1
o.oa[i9)l
(s60i.l
Pin
Solving for the total number
of days money was invested,
d
P = 1,233.55
January
.'. The cash price of the television set is p 1,233.55.
A.
B.
C.
D.
.V;z*--
15.47 o/o
15.69 %
15.80 o/o
15.96 %
to repay
1,500.00
"i
1S-31
February
March
April
man.bollory"-d
Iamount
rgney from a roan shark. He receives from the roan shark and
of p 1,u2.00 and^promised
p
What is the simple interest rate?
-+ Eq. 1
tn" Lna
otl
quarters.
May
June
July
August
September
October 1 - 12
= 16
=
29 (since 19g6 is a leap
=31
=30
= 31
= 30
= 3.1
= 31
=30
=12
271 days
Substitute values in Eq.
r
=
s,ooo(o.18,{!1\
' '\366'
t= 666.39
1
year)
38
(1)ucstrorr
llrrrrlr
Simplc lnt,ereel, & Discoupt 39
lr)rrgrrrrrr.rirrg l,lt:ortorrricx by ,lnime R. Tiong
The exact clmplo lnterest ls P 666.39.
'Aaa
F = P(1+
The exact simple interest of P 5,000 invested from June 21, 1995 to.December
25, 1995 is P 100. What is the rate of interest?
A.
B.
c.
D.
-)
April22-3O
3.95 %
3.98 %
-+
Eq.
1
=$
= 31
= 30
= 31
= 31
= 30
= 31
= 30
May
June
July
August
September
October
November
3.92%
Pin
Eq.
Solving for the total number of days the money was invested, d
3.90 %
l=
in)
Decemberl-25=25
1
247 days
Solving for the total number of days, d
June21-30
JulY
August
September
October
November
December
1-25
Substitute in Eq.
1
=g
= 31
= 31
= 30
= 31
= 30
=.25
1o,ooo=P[1.rr.,(#)]
P = 8,807.92
The amount Nicole received from her mother on her birthday is P
187 day$
8,807.92.
Substitute values:
1oo =
i=
s,ooo(i)fEz)
"\
365
/
What is the ordindry interest on P 1 ,500.50 for 182 days at 5.2 %?
0.0390
A. P 39.01
B. P 39.45
c. P 39.82
i= 3.90%
.'. The rate of interest is 3.90 %.
On her recent birthday, April 22,2001, Nicole was given by her mother a certain
sum of money as birthday present. She decided to invest the said amount on
20o/o exact simple, interest. lf-the account will mature on Christmas day at an
amount of P 10,000.00, how much did Nicole receive from her mother on her
birthday?
A.
B.
c.
D.
D.
P 39.99
,ttdlL/,an
P 8,807.92
l=
Pin
r
(1,
=
5oo, 50)
t= 39.45
P 8,827.56
P 8,832.17
P 8,845.78
--,.It,.
(o.o5r)(ffi)
40
Quesliorr
llrrrk
l')rrgrrroorirrg l!)<xrrtorttu x lry JaiIno R. 'liong
Question Bank 3
Nicole has P 20,400 in cash. She invested rl al7% from March 1, 2006 to
November 1, 2006 at 7% interest. How much is the interest using the Banker's
Rule?
A.
B.
c.
D.
P
P
P
P
972.12
970.78
973.12
971.83
Thc amount of P 20,000 was deposited in a bank earning an interest of 6.5% per
annum. Determine the total amount at the end of 7 years if the principal and
lntcrest were not withdrawn during this period.
,'/;Zr,ln Banker's Rule, use exact number of days the money invested and 360
days for the total days in a year.
Solving for the exact number of days:
A.
B.
c,
D.
P 30,890.22
P 30,980.22
P 31,079.73
P 31,179.37
Marchl-31=30
April= 30
MaY = 31
r=e(t+if
June = 30
F = 2o,ooo(1+
JulY = 31
o.o65f
F = 31,079.73
August = 31
September:30
October = 31
November 1= 1
Iotal=246
l= Pln
t= (2o.4oo)(0.07)f?€)
.",/[360J
I
= 971.83
,'. The total amount at the end of 7 years is P 31,079.73.
A loan for P 50,000 is to be paid in 3 years.at the amount of P 65,000. What is the
dlbaive rate of money?
A.
B.
9.01
9.14
D.
9.31%
9.41%
c.
o/o
o/o
r=e(t+i)n
65,000=so,ooo(t+i)3
1.3=(1+i)3
1.0914 = 1+ i
i= 0.0914
50,000
......_
61rr0lw
11]lrr_oil1,1,11r,k. _U,,rinooring Economics by Jaime
t.s = 1r+i;
R. Tiong
1.0914 = 1+ i
i= 0.091a
tfre pres"nt worth is p 40,530.49.
since nothing was mentioned about how the
interest was compounded, it is
presumed that the mode of compounding
is
annually.
I
.'. The effect rate is 9.i4 %.
The amount of p 50,000 was depositeo
in tne bank eaming an interest of 7.5%
annum. Derermine rhe rotar amount at *,e
eno oi s
i; il;'pri"ipri
interest were not withdrawn Ouring ifre p"ii"il -
G;;
A.
B.
c.
D.
/
I
""0
A
I
c
D
19.610/o
19.44%
19.31%
15.72%
(*aEn=(r+if_r
P 71,78't.47
P 71,187.47
P 71,817.47
P 71,718.47
en
=[r+S)'uo
\ 360/ -,
ER =19.72o/o
"'
The effective rate corresponding
to 1g% compounded dairy
is
1g.72%.
F = P(1+ i)n
F = so,ooo(1+o.o7s)s
Wtll
F:71,781.47
.'. The total amount at the end of
D.
P 40,530.49
o-
E
|
(1+i)n
P
_
80,000
(1+ 0.12)6
P = 40,530.49
r"t",
semi-annually,
"o.no1ilffi,a,,..
A, 16.09 %
!, 16.32%
c, 16.45 %
D, 16.78 %
71,781.47.
Find the present worth of a rrtrr" pryrr,"r,t
of p g0,000 to be made ln six
1o;
with an interest of 12o/o compounded'"nnrrlfy.
A. P 40,540.49
B. P 40,450.49
c. P 40,350.49
nominai
ompounded qr"rt"1y?pounded
yea)
ER, =5P,
(,.T)'-,=[,.T)'-,
[,.T)'=(,.T)'
[,.T)'=116e8s56
1.T
= 1.0816
._:_,
yields the same amount
as 16%
44
llrrrk
()trcstiorr
(
l,lrrgirrr,r.r'irrg l,lr:rrrrorrrir:s by .Jaime R. Tiong
l!!
o 0816
2
NR = 0.1632
(r , i)o
o
1,,
i
But,
l=
NR
-m
8.07
o/o
0/o
Y
NR = 0.'t183
NR = 11.83%
What rate of interest compounded anrtually is the same as the rate of interest of
8% compounded qua(erly?
8.12
8.16
1
.0.029563
o o2s563 =
A.
B.
C.
D.
t')'
(1ri)a 11236
NR = 16.s2 %
The nominal rate is 16.32o/o.
't
lorrtltottttrl l rtl.t'r'rl rrtrtl ( lrttl itttttttts ()orrtpotrnding 45
. Ths nominal rate's 11.83%'
o/o
8.24%
interest?
ffil-O,, of tn".e gives the lowest effective rate of
.V;Z*r-".
ER' =EP'
(1+i)1-
t=(t*IB)o -',
4,l
[
(
.
(1+i)=11+
\
o.o8
.)4
4,
1+i=1.0824
i= 0.0824
i = 8.24
Yo
.'. The rate of interest compounded annually is 8.24%.
A, 12.35% comPounded annuallY
a 11.90 % compounded semi-annually
C, 12.20% comPounded quarterlY
monthll
D,
11.60 % comPounded
'(,J*-,
Solving for the effective rates:
A.
ER = 12.35 % (if compounded annually, NR = ER
B. ER=(1+i)"-t
en=[r.9-119)'-r
ER = 0.1225
Flnd the nominal rate, which if converted quarterly could be used instead oI 12oh
compounded semi-annually.
A
u
(:
I)
'|.1.93
ER=12.25%
0/o
rn=[r,.9J??)'-r
11.09 %
11,85 %
11.26 0/
ER=0.1277
ER = 12.77o/o
tRr
1'
-ER1
D
en=(r*9119)"-r
(lrrcslion
llrrnk
l,)rrgirrr,r,r.irrg I,l.orrorrrir.s lry ,lirinrc R. Tiong
(
ER. 0.1224
lrrrrrltrrttntl
l ttl
r,tr"rt ttrttl ()orrt ilrutttts Ooltlpoun dtng 47
But,
.NR
l=-
ER =12.24%
m
.'. Thschoice D provides the lowest rate of
interest.
002=!E
6
NR = 0.12
Find the compound amount if p 2,500 is invested
at g% compounded quarterry for
5 years and 6 months.
A.
B.
c.
D.
NR = 127o
P 3,864.95
P 3,846.59
P 3,889.95
P 3,844.95
,', The nominal rate is 12"/o.
5 years & 6 months
=
5.5 years
,
r=n(t+if
.15
F
=
2'5oo(1*
\
!'oe
A, P 6,080.40
!, P 6,020 40
c. P 6,040 20
utol
4)
F = 3,864.95
pi OOo shall accumulate
for 10 years at 87o compounded quarterly, then what
lha compound interest at the end of 10 years?
ii
D,
2,s00
i....................
""'> F
P 6,060.20
l*oth',,
r=eft+if
The compound amount of p 2,000 after 5 years
and 6 months is
P3,864.95.
F=s,ooo[1*ofu)'o'''
F =11,04020
An amount of P 1,000 becomes p 1,60g.44 after 4 years
compounded bimonthry
Find the nominal interest.
A.
B.
C.
D.
,
lnterest=F-P
lnterest = 1 1,040.20
lnterest = 6,040.20
11.89 o/o
12.00 0/o
12.08 o/o
12.32 o/o
-
5'000
6'040'20'
,'. The compound interest at the end of 10 years is P
l-J2z
-/../t//rn
18% compounded semi-quarterly?
W-nrt i" the corresponding effective rate of
r=e(+i)n
A.
B.
c.
D.
1,608.44= 1,s66(1 * ;)+to)
1.AOAI+*
=t+i
1.02 = 1+i
i=o.o2
19.24%
19.48
o/o
19.84%
19.92%
f k)*,,.
gp = (r+i)m._1
__l-.
i
f
48_ ,,),::Il iorr
llrrrk
()ornllotttrtl I ttlrrrtrrl, nnrl Continuoue Compounding 4!)
-, lr)nginocring li)conornir:x by Jaime R. Tiong
t---
rn=lr*o,rtlt
( 8,
F=P(1
+i)2n
,
2ooo + 3ooo
ER = 0.1948
ER = 19.48 %
.'. The corresponding effective rate is
= 2ooo(1*
ol'l'"
[
2)
2.5 = (1.06)2n
.
19.4go/o.
Take log on both sides:
log2.5 = tog(t.oo)2^
log2.5 = 2nlog1.06
n = 7.86 years
present worth of a future payment of p 100,000 to be
made in 1 0 years
interest of 12o/o compounded quarterly.
)
B. P 30,656.86
c. P 30,556.86
D.
,,, Thc money
will increase by P 3,000 at in 7'86 years'
P 30,655.68
ifr-r,mount of P 150,000 was deposited i2the bank earning q1
interqs!9f-]
L
o_
F
(1
+ i)n
P 215,U4.40
P 213,il4.40
P 2Uj53.40
P 255.43.10
100,000
P=
^
f, *
o.rz'14(10)
(
l
4/
P = 30,655.68
tb
.'. The present worth is p 30,655.6g.
r=e(t+if
F = 1 50,
OOO (1
+ 0.075)s
F =215,344.4O
ln how many years is required for p 2,000 to increase by p 3,000 if interest
at 12%
compounded semi-annually?
,'. The total amount at the end of 5 years is P 215'344'40'
A. 7.86 years
B. 7.65 years
C.7.23years
D.
8.12 years
HOn, tong
will it take money to double itself if invested at 5% compounded
f,lnuallY?
A.
B.
C.
D.
,%*^
__..L
13.7 Years
14.2 years
14.7 years
15.3 years
-S-oZ
r -!tF'
50
Quesl,iurr
llntrk
I,)rrginooring l,lt:onuurit:x by Jairne Il,. Ti<lng
r=n(t+if
()ontJxrurrtl Ittit'rr,xl rrntl (lont,inuorrs Oompounding 51
,;-
To double the money, F = 2P
ER= (1+i)"-1
2P = P(1+ i)n
z=(r+o.osf
en =
Take log on both sides
o
[r*
]o)' -,
l. 2)
ER = 0.1449
ER = 14.49 %
log2=nlog1.05
loo2
i,
log1.05
Thc effective rate is 14.49%.
n = 14.2 years
.'. The money if invested at 5% compounded annually will double in 14.2
yeats.
what
Aan lnterest rate of 10% compounded annually, how much will a deposit of
ll,tOO be in 15 Years?
P 6,265.87
P 6.256.78
P 6,526.87
P 6,652.78
is the corresponding effective interest rate of 1g% compounded semi-
monthly?
,/
,'fu'
A. 19.35%
B. 19.84%
c. 19.48%
D. 19.64%
r=P(1+i)"
F = 1,500 (1 + 0.10)15
F = 6,265.87
ER=(1+i)'-r
,.. The amount of P 1,500
will become
P 6,265.87 after 15 years'
en=[r+0.18)24_1
\
ER
24)
:0.1964
ER = 19.64%
.'. The corresponding effective rate is 19.64 %.
What is the effective rate of 14% compounded semi-annually?
A.
B.
c.
D.
14.49%
14.59%
14.690A
14.79 o/o
P 25,000 in 8 years. How much is that money worth
considering interest at 8% compounded quarterly?
fiman expects to receive
i[r
lA. P 13,256.83
'iB. P 13,655.28
,l c. P 13,625.83
fl o. P 13,205.83
52
()uort,iorr
llrrrk
l{ltrgirxrorilrg lii:onornir:r bv Jairne lt. Tiong
0r23
'b*
ERquarterty = ERsemi -annually
o.oo
( * 2 ))2(8)
1.,
()orrrporurrl Inlorr,nt rrnd (lontinuous Compounding 53
''
P-n"""""'
[,nl)o -t=(t+o06)2-1
4)
2
\
P ='13,265.83
\
)
i= 0.0596.
i= 5.96%
.'. The present worth is P 13,265.83.
,,, Thc equivalent rate is 5.96 %.
About how many years will P 100,000 earn a compound interest of p 50,000 if
the interest rate is g% compounded quarterly?
A. 4 years
B. 5 years
C.
D.
t{hfl lr the amount
A,
t,
0,
0,
6 years
7 years
of P 12,800 in 4 years at 5 % compounded quarterly?
P 15,461.59
P 15,146.e5
P 15,641.59
P'15,614.59
9"2*,
r=e(r+if
F=P(1 +i)n
loo,ooo + 50,000 = 1oo,ooof1+ o.oe)4N
\.
4)
F
1.5 = (1.0225)4N
F = 15,614.59
Take log on both sides
log1.5 = 4N1o91.0225
N=
,', The amount after 4 years will be P 15,614.59.
-']991'5
4(1o91.0225)
N = 4.56 years
.'. The money will double in 4.56 yearsi or about 5 years.
ly he condition of a will, the sum of P 20,000 is left to a girl to be held in trust
fund Oy her guardian until it amounts to P 50,000. When willthe girl receive the
lltollcy if the fund is invested at 8 % compounded quarterly?
A.
!.
compute the equivalent rate of 6% compounded semi-annually to a rate
compounded quarterly.
A.
B.
c.
D.
5.12%
5.96 %
5.78%
6.12%
o'05
= 12.Boo(( 't* 4) )4(4)
C.
D.
11.23
11.46
11.57
11.87
years
years
years
years
F
54
Qucst,ir-rn
lltnk
l,)rrgirxroring Iimnomics by Jaime R. Tiong
(
50,ooo = zo,ooo Ir + 9]?8
2.5 =
(1
Oornglounrl lrrlr,r'oni nnd Continuous Compounding 55
.-
b.
\4n
4-.J
01
.o2)4"
Take log on both sides:
20,000
...'...-
50,000
1,000
log 2.5 = log(1.02)an
log 2.5 = 4n log 1.02
o= loo" 2.5
4log1.O2
n = 11.57 years
Fr
. P(1+ l)n
Fr
-
1-
.'. The girl will receive the money in 11.57 years.
:
1,ooo(1+ o.o5)8
1,477.46
Money left after the principal is withdrawn = 1,477.46
lf P 5,000.00 shall accumulate for 10 years at 4 % compounded quarterly, find
the compounded interest at the end of 10 years.
A.
B.
c.
D.
P
P
P
P
0123891016
-
1000 = 477.46
2,333.32
2,444.32
2,555.32
2,666.32
Fe = P(1+ i)n
.V.z*r-
F2
F=p(1 +i)n
- 477.46(+
0.05)8
Fz -7o5.42
o'04)4(10)
r = s,ooo
-'--- [t( + 4 )
,',
?hr wlthdrawal amount at the end of the 16th year is P 705.42.
F = 7,444.32
lnterest=F-P
lnterest = 7,444.32- 5,000
lnterest = 2,444.32
,
1,t00.00 was deposited in a bank account, 20 years ago. Today it is worth
10,000,00. lnterest is paid semi-annually. Determine the interest rate paid on this
aount,
.'. The compound interest at the end of 10 years ts P 2,444.32
A sum of P 1,000 is invested now and left for eight years, at which time the
principal is withdrawn. The interest has accrued is left for another eight years. lf
the effective annual interest rate is 5 %, what will be the withdrawal amount at
the end of the 16th year?
A. P 693.12
B. P 700.12
c. P 702.15
D.
F=P(1 +i)n
3,ooo = 1,500
2=
P 705.42
t4-
11 +
(1
ilao
+ i)z(zo1
56
(|uost,iorr
Ilirrrh
1.0175 = 'l+
lr)ngiruror.rrrg l,lr:orrorrrir.s by Jainre
Il. Tiong
Oorrrporrrrrl lnlr,r'r'sl rrrtrl ( ionl,irruorrs Oompounding 57
l1**
i
i= 0.0175
ER (t+i)'-1
Solving for the nominal rate:
.NR
En
m
0.0175 =
.=l
I
+
o'12)4
4)
-
1
ER = 0.1255
ER = 12.55 %
!E
2
NR = 0.035
NR = 3.5%
Thc effective rate is 12.55 %.
.'. The interest rate paid on the account is 3.5 %.
l{fndrrln Bank advertises 9.5 % account that yields 9.84 % annually. Find how
fifn
s
investm
the
A merchant puts in his P 2,000.00 to a
With a given interest rate on the
annually, how much will he collect at
A.
B.
c.
D.
r a period of six years.
year, compounded
year?
A
l,
0,
D,
P 4,626.12
P 4.262.12
P 4,383.12
P 4,444.12
thc interest is compounded.
Monthly
Bimonthly
Quarterly
Annually
(h*
en = (1+i)m _t
0.0984 = (',
(
F = P(1+ i)"
1.0984 =
F = 2,000 (1 + 0.1 5)6
F = 4,626.12
*
o'ogs
"l'
m) -,
[.,* o'ogs')'
(
m)
By trial and error, m = 4.
2,(N0,"""""""""""""""'-. 4
,', Since there are 4 interest periods per year, the interest is compounded
quarterly.
.'. The amount the merchant will collect at the end of the 6h year is
P4,626.12.
I[tcn
A man borrowed P 100,000 at the interest rate of 12Yo per annum, compounded
quarterly. What is the effective rate?
A.
B.
C.
D.
',t2.750/o
12.55o/o
12.45
olo
12.35%
witt an amount be tripled with an interest of 11.56%?
A. 9 years
B. 1o years
C. 11 years
D. 12years
58
Quost,ion llank
-
(
Cngineering ltrconornir:s by Jaime R. Tiong
F = P(1+ i)n
3P:P(+0.1156)n
3=1.1156n
F2
3,ooo(1 +
F2
= 4,392.30
o
lotttllottttrl
I
ttl
lrt,xl rurrl (lorttLinuotrs Oompounding
1o)a
Fs = 5,000(1+ 0.1)1
F3 = 5,500.00
Take log on both sides
log3=n1o91.1156
Substitute values in Eq.
loo3
1o91 .1 1 56
n = 10.04 years
1
P + 5,500 =2,657.34 + 4,392.30
P = 1,549.64
,', The amount of money left in the bank 1 year after withdrawal is
.'. The amount will triple in 10 years.
P1,549.64.
A student plan to deposit P 1,500 in the bank now and another p 3,OOO for the
next 2 years. lf he plans to withdraw P 5,000 three years from after his last
deposit for the purpose of buying shoes, what will be the amount of money left in
the bank after one year of his withdrawal? Effective annual interest rate is 10%.
A.
B.
c.
D.
P 1,549.64
P 1,459.64
P 1,345.98
P 1,945.64
must be invested on January 1 , 1998 in order to accumulate P 2,000
Jrnuary 1,2003? MoneY is worth 6%?
l{CrV much
dl
A,
!,
c,
0,
P 1,509.34
P 1,249.64
P 1,378.98
P 1,494.52
)
98
P
+F3 = F, 1P,
-+ Eq.
99
00
01
02
03
1
5,000---.- Ft
P,<.'.."..'-.'.-.-.-.-.--...-....- 2,000
,'. The amount invested on January 1, 1998 is P 1,494.52.
F2
F1
F = P(1+ i)n
i
I
I
I
t_
F1
= 1,500(1 + 0.1 0)6
1
=2,657.34
A nominal interest of 3% compounded continuously is given on the arcount.
What is the accumulated amount of P 10,000 after 10 years?
A. P 13,498.59
B. P 13,489.59
c. P 13,789.98
D. P 13,494.52
59
60 (lrrcsl iorr llirrrk
l,,)
rr
gr rrr,r,
(lorttgrorrnrl Itrlr,roxl rrrrrl Oonlirtrrous Oompoundilrg 61
rin g l,)rrrrronr rr.s by .la rrnc tt. Tiong
F = Pe(NR)N
o
F = 10,000e0
'
03)10
F:'10,000e(o
3
10,000
(1+ 0OElo
P = 4,631.93
F = 13,498.59
.'.
_
',
The initial amount that must be deposited is P 4,631.93.
The accumulated amount is P 13,498.59.
ll P 000,000 is deposited at a rate of 11.25 % compounded monthly, determine
A mechani
account at
what shoul
A.
B.
c.
D.
P
P
P
P
umulate a total of P 1O,OOO in a savings
bank pays only 4Yo compounded quarterly,
hl
Ompounded interest after 7 years and 9 months
A, P 690,848.73
!, P 670,65'r.23
c, P 680,649.56
6,176.35
T
D, P 685,781.25
6,761.35
6,716.53
)
%..'
6.167.35
,
I
I
F=P(1 +i)n
---'---(,
F = 5oO,OO0[1+
F
P=
(1
+ i)n
^
7 yr
+
9 mo.
=
93 months
0.1125)s3
12
)
lnterest = 1,190,848.73
Jaterest = 690,848.73
-
I
F = 1,190,848.73
10.000
(
.
lnterest=F-P
o.04 \10(4)
[.'. o
J
P:6,716.53
500,000
,', The compound interest alf'ar 7 years and 9 months is p 690,848.73.
The initial deposit should be P 6,716.53.
Funds are deposited in a saving account at an interest of 8% per annum. what is
the initial amount that must be deposited to yield a total of p 1o,ooo in 10 years?
A.
B.
c.
D.
P 4,196.30
P 4,721.39
P 4,796.03
P 4,631.93
&t lnterest rate is quoted as being 7.5% compounded quarterly. What is the
lfrcilve annual interest rate?
A, 7.91%
B. 7.51
c. 7.7',1
o/o
o/o
D. 7.31%
'l/.*
,'y'.at*.
fn=
P=
(1
F
rn
+ i)n
L--b-
=
(1+i)m_r
o
[r. Tu)' -,
\
62
Question Bank
-
Engineering Econornit:n lly Jaiuro
It
'l'rortg
(
ER = 0.0771
ER=7.71
.'.
r
o/o
3,500.00
of taking
pay bY b
1,
I
2
3 -_...
l0
oo0 (1 + 0.08)10
F =2.158.92
o/o per
from a friend at an interest rate of 1'5
much
per
year'
How
18o/o
rate
of
a
at
a bank
moneY from the bank?
The money in the acdount after 10 years will be P 2,158.92.
A. 7.91%
B. 7.51
C. 7.71
D. 7.31%
flllocn years ago P 1,000.00 was deposited in a bank account, and today it is
Worth P 2,370.OO. The bank pays interest semi-annually. What was the interest
o/o
o/o
tala peid in this account?
A,
B,
C,
D,
9"2*r;.
a.
0
= P(1+ i)"
F=
The effective interest is7.71 %.
lrrrrpouurl Irrlr,rr,nl rrnrl Oontinuous Cornpounding 63
Borrow money from a friend
5.72%
5.78%
5.84 o/o
5.90 %
F=P(1 +i)n
r=s,boo(i
+0.015)12
F = 4,184.66
b.
F = P(1+ i)"
Borrow money from a bank
F=P(1 +i)n
2,970 =1000
r=9,'5oO(i +0.018)1
F = 4,130.00
.'. You will pay
.02918 = 'l + i
i = 0.02918
- 4, 1 30.00
But: i=
P 54'66 less by borrowing the money from the bank'
7,000'"""""""""""'-"">
NR
m
Substitute the value of i to get the nominal rate:
o.o2e18 =
IE
2
.NR = 0.05836
A deposit of P 1,000 is made in a bank account that pays 8 7o interest
comiounaed annually. Approximately how much money will be in the account
after 10 years?
NR = 5.836%
The interest rate paid on the account is 5.84 %.
A. P 2,187.39
B. P 2,145.78
c. P 2,176.45
D.
+ i)2(15)
2.37 =(t+1)3o
1
Solving for the difference:
Difference = 4,1 84.66
Difference = 54.66
(1
P 2,158.92
V.e*-.
Eiii
2,370
64
Questi<lD Bauk
-
.liuginocring
li)<:()nor||r(.3 lrv
,lttttttt,
ll
'l'rorrg
P 200,000 was deposited on January 1 , 1988 at an interest rate of 24 %o
1' 1993?
compounded semi-annually. How much would the sum be on January
A.
B.
c.
D.
oolttlxluntl
'-'----=
P 631,627.78
Conrlder a deposit of P 600.00 to be paid back in one year by p 700.00. what is
lho rrte of interest, io/o per year compounded annually such that the net present
Uorlh of the investment is positive? Assume i > 0.
A, 16.50 %
!, ft.75%
c, 16.33 %
P 612,890.76
P 621,169.64
P 611,672.18
D,
7"/-s,*
90
'18.67
F = P(1+ i)"
2oo,ooo[,.0:o)"
7oo = 600(1+
20a,000
The amount on January 1, 1993 would be P 621,169.64.
what
,', The rate of interest is 16.67
year and
is the present worth of two P 100 payments at the end of the third
fourth year? The annual interest rate is 8%.
A.
B.
c.
P 150.56
P 152.88
P 153.89
P 151.09
D.
i)1
i= 0.1667
i=16.67%
F
F = 621,169.64
.'.
0/o
:(,..*
91
r=e(t+if
F=
Irrl,rrroxl, nrrrl Conl,irruous Compounding 65
%.
A flrm borrows P 2,000 for 6 years alg o/o. At the end of 6 years, it renews the
lorn for the amount due plus p 2,000 more for 2 years ata %. what isthe lump
lUm due?
A. P 3,260.34
B, P 3,280.34
c. P 3,270.34
D,
9"2;:
10.0
t00
P 3,250.34
r=e(t+if
Fr = 2,000(1+ o.Oa)B
Pt
Fr = 3,701.86
Pz
F2
=2,ooo(1+o.o8r
F2
=2,332.80
Present worth = Pr + Pz
present worth = 79.38 + 73.50
Present worth = 152.88
The present worth of two P 100 payments is P 1 52'8&'
Total future worth = Ft + Fz
Total future worth = 6,034.66
F2
Ft
66
Question llank
Solving for its present worth:
P=
ComJrounrl lrrloroxl rrntl Oontirruous Compounding 67
l,)ngirrccrirtg lt)trrtttttltirs lrv .laime Il. Tiong
F
With inflation,
Capitalto accumulate = 16,000(1 r 0.03)5
01
Capital to accumulate = 18,548.38
(1 + i)n
'.
6,034.66
_
-=(*oo8)3
P --3,260.34
The capital that must be accumulated is P 18,548.38.
6,034.66
P <'.."""""
i
Whd
r
, i 4-,n
or.,'/lZz,i*.
A,
E,
c,
D,
2,000
2 000
le
the effective rate corresponding to 16% compounding daily? Take 1 year
360 days.
17.35 %
',17.4s%
17.55%
17.65%
'lth.
en=(t+i)r-t
rn=[r+0.16)360_1
\ 360/
Present lumP sum due = 2,000 + P
Present lump sum due = 2,000 + 1,260.34
Present lumP sum due = 3,260.34
.'.
ER = 0.1735
ER = 17.35%
The present lump sum due is P 3,260.34
,'. The effective rate is
17.35o/o.
By the condition of a will, the sum of P 25,OOO is left to a girl to Oe nJO in a trust
fund by her guardian until it amounts to P 45,000. When will the girl ieceive the
money if the fund is invested at 8% compounded quarterly?
A.
B.
accumulated?
C.
D.
A. P 18,854.38
B. P 18,548.38
c. P 18,458.38
D. P 18,845.38
,V"zz>-.
7.42years
7.67 years
7.85 years
7.98 years
r=e(t+if
he machin
ll need the
of 2,000, t
cate 5 Years from now' the
8,000' Since there will be a
d the amount of P 16'000 only'
45.000
= 25.OOO( 1+
-urvvv-4vrvvv[\r1.8 =1.024^
Without inflation,
Capital to accumulate = 16,000
o'08
4
)4n
)
Itl!
68
(luasl.ion
Ilrrnk llngintlcring
lrlconotttit'a by .lairtrc
ll
'l'rrrrrg
Take log on both sides
()orrrgrorurrl IrrLr,roxl, rrlrrl
Orlutinuous Compounding 69
It*,-
log1.8 = 4nlog1.02
O=
n
."
F
loo1.8
"
41o91.02
=,
P(1+ i)n
F=10,000(1+0.15)s
=7.4?years
F = 20,113.57
The girl will receive the money after 7.42 years.
Let P = present worth of the account due to inflation:
P 2O0,OOO was deposited at an interest rate of 24o/o compounded semi-annually
o _ 20,113.57
C. 6 years
D. 7 years
("".'...................... 20,1
"'
o= 2oo,ooo[1 -
ry)'^
Take log on both sides
0585 -- Znlog1.12
loo 3.10585
The amount at maturity in terms of year zero pesos
is p 15,030.03.
A Empany invests
T
p 1 o,ooo today to be rgpaid
in 5 years in one rump sum
no* -rcn protrii'n-pl."nt day pesos is rearized?at
Gmpounded annuary.
A, P 7,563.29
t, P 7,4e8.20
C,
D,
P 7,340.12
P 7,623.42
21o91.12
n = 5 years
r=e(r+i)^
.'. The amount wil! become P 621,170 after 5 years'
F=10,000(1+0.12)5
F =17,623.42
ln year zero, you invest P 10,000.00 in a 15Yo security for 5 years. During that
time, the average annual inflation is 6 9/o. How much, in terms of year zero pesos
will be in the account at maturitY?
A. P 15,030.03
B. P 20,113.57
c. P 18,289.05
D.
13.57
P = 15,030.03
3.10585 = 1 .122n
1
P
(r + o.oo)5
r=e(t+i)n
log 3.
+ i)n
where: i = rate of inflation
A. 4 years
B. 5 years
621,1 7
|
(1
After how many years will the sum be P 621 ,170?
0l234
E
o-
P 16,892.34
Profit=F-P
Profit = 17,623.42 -1
Profit = 7,623.42
0,
000
,'. The profit in present day pesos is p 1,623.42.
12
ll'
70
I
Question lJank
-
Engineering Economicc by Jaime lt. 'l'iong
How long (in nearest years) will it take money to quadruple if it earns 7
compounded semi-annually?
Oo-urllounrl
,-'__
lrrlrrorl tn<l (jont.inuous Compounding
A mrn expects to receive P-2-0,000 in 10 years. How much is that money worth
tlovv conqidering interest at 6% compounded quarterly?
0zo
A, P',t1,042.89
!, P 11,035.12
c. P 11,025.25
A. 20 years
B. 18 years
C. 21 years
D. 19 years
D,
P 11,012.52
J;**-.
F=p(1 +i)2n
E
|
o-
+ = p(t* o'07)2n
\ 2)
(1
+ i)n
1.,
*
P=
4 = 11.035)2n
(
Take log on both sides:
o.oo
"1o(10)
4,
P =11025.25
tog4 = tog(1.m5)2n
,', The worth of the money now is p 11,025.25.
log4=2n1og1.035
n = 20.15
.'. The money invested at 7% compounded semi-annually will quadruple
about 20 years.
How much should you put into a 10% savings account in order to have
P10,000.00 in five years?
A.
B.
c.
D.
P
P
P
P
in
I
tO0,ooo was depositeg
?o 15
years ago at an interest rate of 7% com.pounded
Itltll-annually. How much is the sum now?
A,
!,
c,
D,
P
P
P
P
2,000,033.33
2!000,166.28
2,001 ,450.23
2,002,820.12
6,216.21
6,212.12
20,1 5 years
6,218.21
6,209.21
F=P(1 +i)n
r = soo,ooo(,*o:,)"'
.V"zrz-
F = 2,000,166.28
E
P
=---l(1 +
i)n
-
10,000
(1 + 0.1
0)5
P =6,209.21
.'. You should put into a 10o/o savings account an amount of P 6,209.21.
."
500,000
The sum of themoney now is p 2,000,166.2g.
,
71
72
Qur:sl,ion
Ilrrrrk
Ir)rrgirror,rirrg Iijrrrrronr ir:r by .laime R Tiong
Solving for interest rate
annually:
p.ays one percent interest
ftheb,*
effective anhual interest
rate?
A.
B.
C.
D,
on savings accounts four times
a year.what is
ERannuatty = ERsemiannua,y
(r+i)-r=f,-g)'_,
4.06%
\
4.12%
4.16
2)
i= o.dsoo
o/o
4.28%
D-
F
(1
rn=
(1+i)m_r
Pl
sp = (r+o.or4 _1
P,
ER = 0.0406
ER = 4.06 %
_
-
+ i)n
25,000
(1+ 0.0506)
F
:23,795.93
fr
.'. The effective annuat interest rate
is 4.06%.
fl,.
Alexander Michael owes p 25,000.00 due
He agrees to pay p SO,OOO.0O today and t
he pay at the end of two years if money is
D-
Ps =
A. P 39,015.23
B. P 39,026.25
c. P 39,056.21
D.
0'906P
i#
Substitute values in
Eq. 1:
P 39,089.78
50,000 + O.9O6p 23,795.93
=
+ 61,561.85
P = 39,026.25
Il;,;ffi;t
Atexander Michaet must pay
at the end of two years
is
FlndthedifferenceoP H il; ;. i;"i.,;Hf
"i
depos it
A. P 1,510
B. P 1,530
c. P 1,550
25,000
D.
75,000
P 1,570
Solving simple interest:
L
t;
P
'3-A+OoSO6p
L,
JT1"",.i
il:
74
Question Bank
-
Engineoring Economics by Jaime R. Tiong
Compound Intereet and Continuous Compounding 75
l= Pin
r= 50,000(0.10x3)
I
='15,000
SolVing for compounded interest:
F = P(1+ i)n
Ps=
F = 5O,OO0(1 + 0.1 0)3
F = 66,550
150,000 + 280,000
1,,
\.
lnterest=F-P
lnlerest = 66,550
lnterest = 16,550
^'s =
-
*
o.os'14(n)
4/
430,000
1.or2s4n
50,000
Substitute values in Eq.
Difference = 16,550- 15,000
Difference = 1,550
142,728.64
+
.'. The difference between simple interest and compound interest is
1
25g,51 1 sO = -1SO'OO9
1.0125*"
396,240.2-
P1,550.
430',000^
1.01254n
1.01254n = 1.0852
Take log on both sides
lf money is worth 5% compounded quarterly, find the equated time for paying a
loan of P 150,000 due in 1 year and P 280,000 due in 2 years.
A.
B.
C.
D.
4nlog1 .0125 -- logl.0852
1.52 years
1.64 years
1.69 years
n
_
n
:1.64 years
1o91.0852
41o91.0125
1.72years
The equated time for paying is 1.64 yeans.
.V;z*-..
P1+Pr=P,
1
For a loan acquired six years ago, a man paid out the amount of P 75'000.00.
The interest was computed al18% compounded annually. How much was the
F
D_
(1
_
P-=
'r
-+Eq.
+ i)n
(.
A. P 27,367.28
B. P 27,278.36
c. P 27,782.36
150.000
',
o.o5\4(1)
['* ,
D.
J
P 27,872.63
Pt =142,728.64
280,000
Solving for the borrowed amount, P
76 ('lrrcst iorr llrrrrk l')ttgtrtclt'ittg lt)txrltotltl"t lry'lrrrrrro ll 'liong
('r rrn lrrrrr rrrl
tLet P
PF
(1
_
9-
I
-
Inl
lrl'sl
ir
nrl ( brr( irrtrorrs ()ornpoun
tling 77
prcscnl verlue of inherrtance when boy is 6 years old
012...6789
i)n
...
2021
75.000
(1+ 0.18)o
P =27,782.36
.'. The borrowed money was
P 27,782-36.
P=
I
be born to them theY will Place a
birthday, the child will receive the
an interest of 18% comPounded'
e to make on the birth of a child to
I
li
x
i
I
rl
_
i)n
10.000
(r +
ie
o.o+)5
P = 5,552.64
,F
them?
A.
B.
c.
D.
r,Il
The value of the inheritance as of the boy,s 6th birthday is p 5,552.64.
P 15,367.18
P 15,249.13
lLr
I
P 15,722.16
,.
P 15,482.64
A man who won P 300,000 in a lottery decided to place so% of his winning in a
trust fund for the college education of his son. lf the money will earn M% per
year compounded quarterly, how much will the man have at the end of 10 years
when his son will be starting his college education?
.72r,*Solving for the deposit on the child's birth, P
D_
(1
A.
B.
c.
D.
F
+ i)n
300.000
-
9:_L
(1
P 593.12012
P 593,452.12
.'. The couple will deposit the amount ol P 15,249'13'
F
:
(
A. P 5,552.64
B. P 5,549.10
c. P 5,522.12
D. P 5,582.63
inheritance will be paid in a
the present value of the
st is compounded annually?
F:593,888
iN
::.t
,,\
il
P(1+ i)n
F = 1so.ooof1-
I
Lr<
I l,z,zn
P =15,249.13
I
t.
P 592,739.96
P s93,888.96
+ 0.1 8)18
On his 6,h birthday a boy is
lump sum of p t O,OOO on n
inheritance as of the boy's
Assume i = 4oh.
I
lriill
;l
il
F
('l+
o'14l4(10)
4)
96
The man will have P 593,888.96 at the end of 10 years.
78 (lrrt'sri,. Il:r.k
l,)rrgrrrr,r'r'r.g ri<:...lrrir:s rry Jairne R.
Tio.g
(
I
l,
1l r l)rl
lf the sum of P 15,000 is deposited in an account earning 4o/o plt
annum
compounded quarterly, what will be the deposited amount at the end of 5 years./
A.
B.
c.
D.
,,(x),000
(l
P 18,302.85
P 18,450.89
P 18,512.83
P 18,638.29
I'
Itrrrnrrl wrlrth = 300,000 + g0,375.51
rtornrrl
worth = 3g0,375.51
trrlvtng lor the difference of present
worth of proposals A and B:
r=e(r+i)"
I
o.o4
lltornrrco 400,000 _ 3g0,375.51
lltltororr<;c 19,624.49
---1.' 4 )
15,ooofl+
r {) 20)5
,r0,. t /5 51
f
.Vz*r,
F=
lorrrporrrrrl lrrlr,rrrst and
Continuous Compounding 79
)4(5)
llrr;rropoeal B is more economical than proposal
A by p 19,624.49.
F = 18,302.85
.'. The deposited amount at the end of 5 years will
be p ig,302.g5.
^
The Philippine Society of Mechanical Engineers is planning to put up its
own
building. Two proposals being considered are:
A.
B.
The construction of the building now to cost p 400,000.
The construction of a smaller building now to cost p 300,000 and at the
end of 5 years, an extension to be added to cost p 200,000.
,,llrr
T
li
a, 21 0/,
a2:t0%
4241%
It
1? 57
A
'*"
A.
B.
c.
D.
,,*
0/"
(r+i),
-r
t R [r*9!9)" -,
\ 12)
much is proposal B rnore economical than proposal A if interest rate is
and depreciation to be neglected?
!o*
2Oo/o
Py
lrnrr charges interest rate of 36% compounded monthry.
Find its effective
P 19,122.15
P 19,423.69
P 19,518.03
l:R -0.4257
ER = 42.57%
P 19,624.49
The effective rate
is
42.57%.
.9,1---.
Proposal A has a present worth
of P 400,000.
For proposal B;
Present worth = 300,000 + p
a maller carc,i compounlll_o{ntY and
ch_arges- an interest of 1.So/oper
month.
Whrl le the effective interest rate per year?
"
200,001
300,000
A
!,
c
P.
19.23 %
19.45 0/o
19.56 0/o
1e.65 %
80
()utrst,ion
ll:rrrk
llrrgirrcorirrg l,Ixrnorrrrcs by Jaime R. Tiong
Compound Inlr,rost,
.'/n/r.z,r*.
45
ER=(1+i)m_1
ER = (1 + 0.01 5)12 ER = 0.1956
=
25( 1*
lnd
Cont,iuuous Compounding 81
o'9t
t
4/)o'
1.8 = (1.02)4"
1
Take ln on both siqes
ER = 19.56%
.'. The effective interest rate per year is 19.56%.
25
M
-...-.-.........-.............,, 15
M
|n1.8=4nln1.02
n=- |n1.8
A man expects to receive P 20,000 in 10 years. lf interest is computed at 6%
compounded quarterly, how much is it worth today?
A.
B.
c.
D.
;.
"
4ln1.O2
n
=7.4zyeas
F
The child wil! receive the money in7.42 years.
P 11,025.25
P 11,035.25
fi
P 11,045.25
P 11,055.25
at the
Flnd the present value of installment payments of P 1,000 nbw, P 2,000
end of
at
the
year,
4,000
P
second
ofthe
end
at
the
year,
P
3,000
first
inO oftnl
ltrittriro year and P s,ogo at the end of the fourth year, if money is worth 10%
@mpounded annuallY.
I
i
:l
A. P 11,411.10
B. P 11,530.98
c. P 11,621.67
D.
lill
,ll
ik
P 11,717.85
ilI
p
". The worth of the money today is 11,025.25.
Microsoft CEO, billionaire Bill Gates willed that a sum of g 25 million be given to
a child but will be held in trust by the child's mother until it amounts to g 45
million. lf the amount is invested and earns 8% compounded quarterly, when will
the child receive the money?
A.
B.
C.
D.
8.1 1 years
7.90 years
7.42 years
7.24 years
Present value of investment = 1,000 + Pr
3;zzr.,
F = P(1+ i)n
\
L
+
Pz + P3
+
Pa
82
Question Bank
-
Ii)ngintruriug llconorrrics by Jaime R. Ti<lng
Oompound lnl,orost. and Continuous Compounding 83
p=1ooo* F, * Fr.* F, * Fo
(1+i) (1+i)2 (1+i)3 (1+i)a
P=1OOO+2ooo+
3ooo
(1.1) (.\2
+
4ooo
0123
F = P(1+ i)n
F = 1(1+ O.O8)2oo
+ 5ooo
(1.1)s (.\4
F = 4,838,949.58
P =11,717.85
.'. The present value of the installment payments is P 11,717.85.
How long will it take money to triple itself if invested at 8% compounded
annually?
A.
B.
a.
D.
14.27
14.56
14.78
14.98
years
years
years
years
.'. The value of the account today is
P 4,838,949.58.
,Whet is the present worth of a future payment of P 200,000 to be rnade in 10
y.rrs with an interest of 1Oo/o compounded annually?
A.
B.
c,
D,
P 76.901.21
;:
fr
t:
P 77,108.66
P 78,109.32
P79,667.32
f,t
rt!
D-
F = P(1+ i)^
(1
3P = P(1+ 0.08)n
,,1
--=tr*orlo
P
Take ln (or log) on both sides:
ln3=nIn1.08
,1f
+ i)n
;.
:
1l:l
77,108.66
The present worth is P 77,108.66.
ln3
ll
-
1n1.08
n = 14.27 years
-
.'. The money will triple itself in 14.27 years.
A deposit of P 1,000.00 is made in a bank account that pays 8% interest
@mpounded annually. How much money will be in the account after '10 years?
A.
B.
c.
Two hundred years ago, your great, great, great grandfather deposited P 1 in a
saving account. Today, the bank notified you that you are the sole heir to this
account. How much is the account today if it earns 8% per annum?
A.
B.
c.
D.
7"r.**
P 4,002,450.78
P 4j02305.90
P 4,838,949.58
P 4,909.289.45
D.
P 2,374.21
P 2,158.92
P 2,734.12
P 2,400.12
P = P(1+ i)n
F = 1,000(1+ 0.08)10
F =2,158.92
.'. The account after
I
,,, J
200,000
3 = 1.08n
I
:l
E
t
10 years
wil! be P 2,1 58.92.
84
Oompound lnl,oroet nnd Continuous Compounding 85
Question Bank - Enginooring Econorrricc by Jaime R. Tiong
Suppose that P100,000 is invesied at a certain rate of interest compounded
What nominal rate compounded annually would quadruple the principal in 4
years?
A.
B.
41.42%
D.
40.45
lhnually for 2 years. lf the accumulated interest at the end of 2 years is p 21,000.
Flnd the rate of interest.
A.
B.
c.
40.81%
c.41.790h
D.
o/o
10.12%
10.00 %
10.92%
10.32
o/o
,t/Zrz,,.
F = P(1+ i)n
F = P(1+ i)"
100,000 + 21,000 = 100,000(1+ i)2
4P = P(1+ i)a
(1+i)2 =1'21
4 = (1+l)a
1.i =(.1T.21f
1+i = 1.21
1+i=Ta
0.4142
i = 41.42%
i=
F
i= 0.1
i=10%
.'. The nomina! rate is 41.42%.
Itrl
,tl
.'. The rate of interest is 10%.
,lt
Ir
lt
Five years ago, you paid P 34,000 for a residential lot. Today you sell it at
P50,000. What is your annual rate of appreciation?
A.
B.
C.
D.
8.12%
to break even?
8.00 %
7
.92
An investment of P 20,000 will be required at the end of the year. The project
\fould terminate at the end of the 5'n year and the assets are estimated to have a
atlvage value of P 25,000 at that time. What is the rate of interest for this projecl
o/o
A.
B.
c.
8.320/,
D.
5.74%
5.43%
5.91%
5.67 %
F = P(1+ i)n
5O,OO0 =
25,000
34,000(1+ i)5
(1+ i)s = 5o'ooo
34,000
,*,=u@
1134,ooo
i=
34,000
0.08
i = 8o/o
.'. The nominal rate of appreqiation is 8%.
20,000
:
Pi -"':
86
Queetion Bank
-
Engineoring Ilconornicc by Jaime R. Tiong
Compound lnlorort and Continuous Compounding 87
Equating:
20,000 _ 25,000
1+i (1+i)s
2.5
-^,:(1+i)a
(1+i)a = 1.25
1+i = {1,25
1+i =1.0574
boirowed a certain amount on June 1990 from Romeo. Two years later'
borrowed again from Romeo an amount of P 5,000. Sonny paid P 1'000
June 1993 and discharged his balance by paying P 7'500 on June 1995'
|at was the amount borrowed by Sonny on June 1990 if the interest rate is 8%
A, P 1,511.61
B. P 1,611.51
c, P 1,711.41
D.
P 1,811.31
i= 0.0574
i=5.74%
The rate of interest for this project to brcakeven is 5.74o/o.
P+P,=P2+P3 -+Eq.
E
o-
|
(1
Frank Medina possesses a promissory note, due in 2 years hence, whose
maturity value is P 3,200, What is the discount value of this note, based on an
interest rale ol 7o/o?
A.
B.
c.
D.
2,795.00
2,975.00
2,579.00
2,759.00
+ i)n
5'000
'' _ 1t*0.08;2
D
P1
P
P
P
P
1
:4,286.69
P
1,000
_
' = (1+ 0.08)'
%
5,000
P1
Pz = 793.83
,72;*-..
The discgunt value is the same as the present worth.
Solving for the present worth, P
F
(1
_
=
Pa = 5,104'37
o_
D
l--------
7'5oo
" =(1+ 0.08)3
P"
+ i)n
3'200
(1+ 0.07)z
P = 2,795.00
The discount value of this note is P 2,795.00.
Substituting values in Eq.
1
P + 4;286.69 = 793.83 + 5,104.37
P = 1,611.51
The amount borrowed by Sonny on June 1990 is P 1,611.51.
88
Question Ilank
-
Unginr,r.ring l,)urnonricH by.lairne R. Tiong
()olttpourtrl IrrIr,n,rl rrrrrl (lrnl,irrrrous Oornpounding 89
F = P(1+ i)n
Dr. Leopold Lucero invests P 50,000 in a time deposit that yields 10% for his
retirement 30 years from now. lf the inflation rate is 5%, what will be the value of
the account at maturity in terms of today's peso?
A.
B.
c.
D.
F = 1OO,OO0(1+ 0.12)s
l
F =176,234.17
Solving for the equivalent future amount in today's peso due to inflation
(rate = 4ok) tP
P 20'1,689.91
P201,571.91
P 201,U5.91
F
D_
P 201,869.91
(1
^
,V.2,*r*.
+ i)n
176,234.17
l--
(1+ 0.04)5
Solving for the future amount of the account, F
P = 144,851.64
F = P(1+ i)n
Profit = 144,851.64
Profit = 44,851.64
F=50,OOO(1+0.10)30
F =872,470.11
tF
-
f
100,000
F
l;;rr
lrr
.'. The profit realized, in terms of today's peso, is
P
4,851.64.
872,47l.',t1
(1+ 0.05)30
quarterly:
(1
^
I
A manufacturing firm contemplates retiring an existing machine at the end of
2002. The new machine to replace the existing one will have an estimated cost of
P400,000. This expense will be partially defrayed by sale of the old machine as
acrap for P30,000. To accumulate the balance of the required capital, the firm will
depos[t the following sum in an account earning interest at 5% compounded
P=
+ i)n
P = 201,869.91
.'. The value of the account at maturity.in terms of today's peso ia
P201,869.91.
A.
B.
c.
D.
P 44,512.89
P 44,672.10
P 44,851.64
P 44,901.23
P 155,890.12
P 153,085.56
P 154,200.12
P 156,930.38
,VzzrP+
F1
+ F2 +F3 + 30,000 = 400,000
Note that SV = 30,000
-V"Zr*Solving for the future amount of the account, F
I
,i ,:l
,t(
;i:l
new machine?
A.
B.
c.
D.
I
t,,.
P 60,000 atthe end of 1999
P 60,000 at the end of 2000
P 80,000 at the end of 2001
What cash disbursement will be necessary at the end of 2002 to purchase the
First Benchmark Publisher, lnc. invests P 100,000 today to be repaid in five
years in one lump sum at 12o/o compounded annually. lf the rate of inflation is 4%
compounded annually, how much profit, in today's pesos, is realized over the
five-year period?
l,i:rl
I
'a
Solving for the equivalent future amount in today's peso due to inflation
(rate = 5%) ,P
F
i
-+ Eq. 1
,
90
Question llank
-
Compound lnterest and Continuous Compounding 91
Enginoering Economir:s by Jaime R. Tiong
400,000
F = P(1+ i)n
F1
F1
=oo,ooo(r.T)'"'
1
=69,645.27
2001
2000
999
l
F = P(1r i)n
I
Fr = 4,400(1+ i)l
I
F1
2002
F2 = 4,400(1+ t)2
F, = 60.00011*
\.
F2
o'05
.14(2)
\
F3 =
999
2000
30,004
4,
60,000
Fs = 4,200(1+ i)s
60,000
80,000
=66,269.17
F. = 8o.ooof1 +
1
o.o5
4,)4(1)
',
.Ft
1................r.r..),
:il
::l
:tt
i
:
Substituting values in Eq. 1:
81000.00
144, OOO(1 + i)a = 4, 999
F2
*
1
52, OOO
F3
1
P + 69,645.27 + 66,269.1 7 + 81,000 + 30,000 = 400,000
P = 153,085.56
152,000
'"'t F'
i................-
+ 4,400(1+ i)
Substituting values in Eq.
0
::t
;il
F+ = 144,000(1+ i)a
P
+ 4,400(1+ i)2
F2
+ 4,200(1+ i)3
.. Ft
By trial and error:
i= 0.0426
,'. The cash disbursement necessary at the end of 2002 is p 153,095.56.
i = 4.260/o (semi - annual)
NR = 4.26(2)
NR = 8.52%
on June 1 , 1998, Mrs. Emelie Roe purchased stock of San Miguel corporation
a total cost of P144,000. she then received the following semiannual dividends:
P
P
P
P
seminannually.
4,200 on December 1, 1998
4,400 on June 1, 1999
4,4OO on December 1, 1999
4,000 on June 1, 2000
After receiving the last dividend, Mrs. Roe sold her stock, receiving p152,000 after
deduction of brokerage fees: what semiannual rate did thie dividend realize on
investment?
A. '4.26%
B. 4.54
C. 4.87
D. 4.91 o/o
o/o
o/o
Engr. Altarejos has P 13,760 in cash and he would like to invest it in business.
His estimates of the year-by-year receipts and disbursements for all purposes
are shown in the tabulation below:
Year
00
4
5
6
7
Receipts
P 5,000
P 6,200
P 7,500
P 8,800
Disbursements
- P 13,760
+
1,000
+
1,200
+
1,500
+
1,800
P
P
P
P
He estimates that his equipment will have a salvage value of P 2,000 at the end of
useful life. Find the rate of return of the prospective investment.
.l/"a*rFr = 4,000 + 152,000 +F.1+F2 + F3 -+ Eq.
The interest is 4.26% per semiannual period or 8.52Yo compounded
1
A.
B.
10.11o/o
c.11.10%
11.80%
D.10.51
o/o
E*ttF
9t {unlf|lllrlndneering
Econornir:n by Jaime R. Tiong
Compound Interesf. and Continuous Compounding 93
4*eq
.13.760_ 4'000
ItltJt?low
6'009
9'009
+9'=0q+
+
(1ri)* -(1+i)5 (1+i)o (1+i)'
'
Equivalent (Reduced) Cash Flow
By trial and error:
'l-
i=
i
'rl,2A0
0.1011
.
= 10j1%
The rate of return of the investment is 10.11
ot
''What interest ra(e, compounded monthly, is equivalent to a 10% effective rate?
A.
9.45%
8.9.26o/o
8,800
'
ifffilng
9,000
I
t
P3
2,000
Pa
c.
D.
9.65 %
9.56 %
fA
dq
,tl
tolqulvalent cash flow:
ER=(1+i)'-1
o.1o = (1+ i)12 -
,ll
1
1.10=(1+i)12
tlltsrPr
+Pz +P3
+Pa
..-L
(l
+ l)n
Eq.
1.00797 = 1+
i
i= 0.00797
Solving for nominal rate, NR
.NR
t=
Pl'
m
,t.ffi
,n.fffi
9,ooo
' - (1+l)'
pr
-+
tubrtltutc ln Eq. 1
10,7t0 -Pt+Pr+P. +Po
0.00797 =
NR
12
NR = 0.0956
NR = 9.56%
The interest rate is 9.56%.
94
Quort,ion llank
-
lNngineering Economicr by Jaime R. 'l'iong
Excel Flrct Rcvlew & Training Center, lnc. plans to purchase a piece of land
and to bulld a school building on this land. However, since the sehgol building is
not an lmmedlate requirement, the inetitute is considering whether it should
purchaae th6 land and build the building now or defer this acti'on for 3 years. The
current costs are as follows:
Land : P 800,000
Building : P 12,000,000
The purchase price of the land and the cost of the school building are expected to
appreciate at the rate of 15% and 4o/o per annum, respectively. What will be the
total cost ofthe land and structure 3 years hence?
A.
B.
c.
D.
P
P
P
P
14,520.120
Compound lnterest and Continuous Compoundins 95
Mr. Ramos owes Mr. Alarde the following amounts:
P 40,000 due 2 years hence
P 60,000 due 3 years hence
P 72,000 due 4 years hence
Having won the loftery, Mr. Ramos decided to liquidate the debts at the present
tlme. lf the two parties agree on a 5% interest rate, what sum must Mr. Ramos
Pay?
A.
B.
c.
D.
14,715,068
14,902,189
15,021,781
P 147,520.20
P 147,U6.O2
P 147,902.89
P 147,021.81
,_y';Z;.,r.
Let A = Amount Mr. Ramos pay
at present time
I'u'
Solving for total cost 3 year hence:
Total cost =
F,r
+
Fz
-+ Eq.
I'
A=Pt+Pz+Ps
1
1
,lh
F
D_
Note: Rate of appreciation of land
and building and not the same.
-+ Eq.
(1
I
ril
40,000
^11' = ---------------(1+ 0.05)'
For land:
I:l
Il
R = 36,281.18
60,ooo
F = P(1+ i)n
Fl =8oo,0oo(1+0.15)3
_
P^ :
'
1=1,216,700
72,'000
60.000
----------:--
(1+ o.o5)3
Pz = 51,830.26
For building:
Fz = 12,00Q,000(1+ 0.04)3
^'
ra -_-
Fz = 13,498,368
Substitute in Eq.
z
72,000
1t +
o.os;a
Po = 59,234.58
1
Substitute values in Eq.
Total cost = 1,216,700 + 1 3,498,368
Total cost = 14,715,068
.'. The tofal cost of land and structure
3
yearr hcnco
lr
P 14,71$068.
A = 36,281 .18 +
A = 147,346.02
frl
;t
+ i)n
51
1
,830.26 + 59,234.58
The sum that Mr. Ramos must pay to Mr. Alarde is P 147,346.02.
'rA lfrrr,rlIrtt llrrrrk
Mr lllnr
l,)n1irrrr,r,r
Oornpottrttl l nlcrt'sl rrrtrl Oottl,irrtrotts ()tlrnpounding 97
ing l,)r'orrorrrrlr lry ,l:rrrno ll. 'l'ronpi
1-
F2 = 8o,ooof
\.2)
I ubay working in the United States planned of returning to the
I l,rlrlrlrtrrrr al lho end of 2001 . He established a fund starting in
1,995
qpl"
Fz = 91,910's+
with the
I'rll'rwltU rerordod deposits and withdrawals:
lirlity I ltm5
lrrrllty I l(X)7
lrtly I lUtl/
lrrly I lUUlt
ltrlrrV I llX)g
Fs=12,ooof1.oPul'
[
Deposit of P 40,000
Deposit of P 80,000
Withdrawal of P 12,000
Deposit of P 64,000
Withdrawal of P 48,000
2)
Fs = 13,549.46
Note that starting 1998, the interest rate was augmented to 4%
compounded semiannually.
rrr lrrrrl rrrnorl lnlerest at the rate of 3.54/o compounded semiannually until the
,'l lltlt, nt lltat date, the interest was augmentedlo 4o/o compounded
- iilrrilililtrlly What will be the principal in the fund at the end of 2001?
I
,.,',1
l. ltn, tr3,08
c. P '146,846
il^ l, l{n utt2 09
F" =64.00011+
I
0'04)5
ir.tt
2)
';,r
;';t
F3 = 70,661.17
92
D. P 146,02282
I
F, = 48.ooof1+
\
o'04)4
I
I
2)
I
I
Ft
F+ = 51'956.74
F2
Substitute values in Eq.
F.1
F = 49,257.57 + 91 ,910.54 + 70,661 .17
F = 146,323.08
I
I
64.000
10,000
""""""""'-
0
,,,1
000
t;t
rl t -
l']rlncipalat end of 2001
| .lr I Fr+Fs-F+-Fs -)
t_1,1trl)n
rr - ro,ooo(r *
rt -
19,257.57
e'912
5)"
I
-
51,956.74
-
13,549.46
The principal in the fund at the end of 2001 is P 146,323.08.
JRT Publishers is contemplating of installing a labor-saving printing equi
T,'I
I
1
Eq1
It has a choice between two different models. Model A will cost 1,460,000 while
nlodel B will cost 1,452;000. The anticipated repair costs for each model are as
follows:
Model
A:
Model
B:
P 60,000 at the end of Sth year
P 80,OOO at the end of 1Oth Year
P 152,OOO at the end of
9th
year
The two models are alike in all other respects. lf this publisher is earning a 7%
return of its capltal, which model should be purchased? How much savings will be
accrued if the publisher will purchase the more economical model?
A.
B.
c.
D.
P 8,769.18
P 8,918.23
P 9,012.53
P 9,341.11
I
I
ra
'l
-
98
Question Bank
-
Compound Interest and Continuous Compounding 99
Engineering Economics by Jaime R' Tiong
,7"f-*r.".
What is the future amount of P 50,000 if the single payment compound amount
factor of this invegtment is 1.23?
Analyzing model A:
A. P 61,700
B. P 61,900
c. P 61,200
Let A = Total present cost
A = 1,460,000 + Pr + P2
+
Eq.
1
D.
'
P 61,500
ffiffi
ffi
r=p(i+i)n
But
(1
+ i)n = single payment compound amount factor
F = 50,000(1.23)
F = 61,500
Substitute values in Eq.
The future amount is P 51,500.
1
A = 1,460,000 + 42,779.17 + 40,667.94
A= 1,il3,447.11
An investment indicates a single compound amount factor of 1.32 if invested for
up"
years. lf the interest rale is 4.73o/o per annum, find the value of .ni.
Analyzing model B:
O
Let B = Total present cost
B = 1,452,000 +
-P^=:
'r
P3
-+ Eq. 2
152.000
(1+ o.o7)e
Ps =82,677 '93
(1
(1+ 0.0473)n = 1.32
Substitute in Eq. 2
1
B = 1,452,000 + 82,677.93
B = 1,534,677.93
Difference=A-B
Difference = 1,543,447.11
Difference = 8,769.18
.0473n = 1.32
Take log on both sides
-
1,534,677,9C
The amount saved by purchasing tho
P8,769.18.
+ i)n = single amount factor
n log
1
.0473 = log 1 .32
log1.32
motl
1o91.0473
n=6
The value of n i5 6.
tfi)
-
Qucltiorr llrrtk
Mru, Tambangan invests P 50,000 today. Several years later it becomes
lf
P6O,OOO, Wnit is the single payment present worth factor of this investment?
lhl amount was invested for 5 years, what is the rate of interest?
A.
B,
3.1
D.
3.7
c.
Oompound lnlcrusl irnd ()onlinuous Compounding
l!rrginooling lt)conornics by Jairne R' Tiorrg
Money is deposited in a certain account for which the interest is compounded
continuously. lf the balance doubles in 6 years, what is the annual percentage
rlale?
A.
B.
c.
D.
o/o
3.3%
3.5%
o/o
11
.55
0/o
11.66%'
11.77 %
11.88 o/o
/, /r/rr'.n
Solving for the single payment present worth factor
2P = Pso(NR)
tr
t
o(1
2 = s6(Nn1
+ i)n
But
1
5q000
-
(1 +
1
(1
F = Pe(NR)N
-
i)"
= single payment present worth factor
Take ln on both sides:
In2 = Ine6(NR)
60'00-0
(1
+ i)"
ln2 = 6(NR)lne
ln2 = 6(NR)
= 0.833
+ i)n
NR
lf the amount was invested for 5 years
1
(1
NR = 0.1155
NR = 11.5s%
=:0.833
+ i)5
1.20 = (1+ i)s
1.037 = 1+
=13
6
.'. The annual percentage rate is 11.55%.
i
i= 0.037
i=3.7Yo
The rate of interest is 3.7 %.
On January 1 , 1999, Mrs. Jocelyn De Gala opened an account at Bank of
Philippine lslands with an initial deposit of P 1,000,000.00. On March 1, 2000,
she opened an additional P 1 ,000,000.00. lf the bank pays 12% interest
compounded monthly, how much will be in the account on April 1,2OOO?
A.
B.
c.
D.
,9"2r--
?)1,7.
P
P
P
P
2,180,968.95
2,190,968.95
2,160,968.95
2,170,968.95
'101
102
Question Bank
r = e(t+
Fr
-
.Engineering Economics by Jaime R. Tiong
if
Question Bank
4
15 months
=1,ooo,ooo(,.+f)"
.lan.
I'99
Mar.
l'00
Apr.1'00
Fr = 1,160,968.95
F2
:1,ooo,ooo(,.r#)'
I,000,000
What is the present worth of a P500 annuity starting at the end of the third year
and continuing to the end of the fourth year, if the annual interest rate is 10%?
F2 = 1,010,000.00
Total amount
=1
A.
B.
c.
D.
+Fz
Total amount ='1, 1 60,968.95 + 1,01 0,000
Total amount = 2, 1 70,968.95
P 727.17
P 717.1V
P 714.71
P 731.17
nlit
,,F.
Tr
i:il
ft,
The amount in the account on April 1, 2000 is P 2,170,968.95.
!
I
a
t
I
5oo[(1+ o.r)'z - r]
I
^r=-;@
\
, t
'],
= 867.77
t,ra
;:lra
E
P= '
(1
.
+ i)n
o
t1
o(1
+ i)"
867.77
' _ (1+ 0.1)'z
o
P =717.17
J&,,-a.7;Zt*..
Present worth = Pr + Pz
..
I
lr
tr
ri,
I,
f{
-+ Eq.
0
1
F
D_
' - (1 +'r)n
500
D_
''-{r*o.r;'
Pt.'"""""""""" 500
a
.Pr',i,
= 375.66
t"
""
""
500
"""""""""""";
li
(
/
104
Quection Bank
^
'
P2
-
Annuit,y 105
l,]nginooring Economics by Jaime R. Tiong
500
(1+ 0.1)a
of
ty is required over 12 years to equate with a future amount
annually.
6%
i=
Assume
= 341.51
Substituting values of Pt anWz in Equation
Present worth = 375.66 + 341.S.1
Present wo(h = 717.17
1:
The present worth is P 117.17.
f
F
Today, a businessman borrowed money to be paid in 10 equal pavments for 10
quarters. lf the interest rate is 1OYo compounddrl quJrterly ano thaquarterlya
7
payment is P 2,000, how much did he borrow?
A.
B.
c.
D.
irt
P 17,304.78
P 17,404.12
P 17,5U.13
P 17,604.34
a
.'. The annuity required is P 1,185'54'
FindtheannualpaymenttoextinguishadebtofPl0,000payablefor6yearsat
Amount borrowed = P
-+ Eq.
12% interest annuallY.
1
A. P 2,324.62
B. P 2,234.26
c. P 2,432.26
'i(t+i)'
zoool(u9l!)"
L( 4'l
i
tiil
I
A = 1,185.54
A = 2,@0
P-
I!i:I
D.
P 2,342.26
_,1
.J
(T)(,.T)'
AAAAAAAAAA
P =17,504.13
Substituting in equation 'l:
Amount borrowed = P 17,504.13
.'. The amount the businessman borrowed is p 17,604.i3.
10,000=
n[tr*o.rz)'-r]
-;1r(d.o12(-
A=2,432.26
':"1
The annual PaYment is P 2,432.26'
_
- \-
I
106
Quosl,iorr
lla,k
I').girrooring Ecorr.rrrics by Jair,e It. 'l'i,rrg
Annuity
D_ F
'z -,n;y
A manufacturer desires to set aside a certain sum of money to provide funds
cover the yearly operating expenses and the cost of replacing every year the
dyes of a stamping machine used in making radio chassis aimooet changes for
a period of 10 years.
Operating cost per ye-ai
Cost of dye
Salvage value of dye
=
p
=p
=p
_
p--
'
P2 --
5OO.OO 6
1,
100
(t + o.oof
o
55.84
Rmount to set aside = A + Pr - Pz
Amount to set aside = 1,200 + 7,481.86
Amount to set aside = 8,626.02
200.00 ^
600.00 L
The money will be deposited in a saving account which earns 6% interest.
Determine the sum of money that must be provided, including the cost of the initial
107
-
55.84
,'. The amount to set aside is P 8,626.02.
dye.
A.
B.
c.
f,
P 8,626.02
A factory operator bought a diesel generator set for P 10,000.00 and agreed to
pay the dealer uniform sum at the end of each year for 5 years at 8% interest
compounded annually, that the final payment will cancel the debt for principal and
interest. What is the'annual payment?
P 8,662.02
P 8,226.02
ln the cash flow shown:
A = 1,200
l=
Q=
500
600
A. P 2,500.57
B. P 2,544.45
c. P 2,540.56
CCCCCCCCCC
11t11ltltt
0246810
D.
F
!i
h
ll'll
ri;
lr
Fri
,rl
'l
I
ll
l,l
,,.1
[''
lrl
P 2,504.57
rt[
LetD=A+B-C
D=1,200+500-600
tt
prl
AAAAAAAAAA
ilililtilt
D = 1,100
*tl
i(1 +
BBBBBBBBBB
i)"
But P = 10,000
Using equivalent (reduced) cash flow:
n[{r*o.oa)'-r]
t00
I
10,000 =
0.08(1+ 0.08)5
A =2,504.57
10
.'. The annua! payment is P 2,504.57.
A
Equivalenr cash flow
Pt
AAAAA
108
(lrr.sr i.rr
llirrl<
l,)rrgi.t,r,r'i.g 0trrrr.rrrir:s by .lairnt, R. ,l'i,rg
Arrrrrrity 109
P
A man paid 10% downpayment of p 200,000 for a house and lot and agreed to
pay the 90% balance on monthly installment for 60 months at an intereit
rate of
15% compounded monthry. compute the amount of the monthry payment.
A.
B.
c.
D.
P 42,821.86
P 42,128.67
P 42,218.57
P 42,812.68
1,800,000
42.0346A : 1,800,000
A: 42,821.86
.'. The monthly payment is P 42'821.86'
what
is the present worth of a 3 year annuity paying P 3,000.00 at the end of
year, with interest at 8% compounded annually?
A. P 7,654.04
B. P 7,731.29
c. P 7,420.89
D. P 7,590.12
Bdruce
p-
il!
l;il
D-
i(1
AL44444444L4AL4444L44AL4 AAA4AL4444L4
^
iii,l
nltr*if-r]
r
L
a
+ i)n
r
3,0001 (1 +
I
AAA
tt -t'l
0.08,
.l
0.08(1+ 0.08)3
P.
Downpayment = 0.1O(Total Cost)
200,000 = 0.1 0(Totat Cost)
Total cost = 2,000,000
Balance = Total Cost
Balance = 2,000,000
Balance = 1,800,000
Rlrt+ir"-r]
DL"l
i(1
+ i)n
-
-
Downpayment
200,000
I
I
,P -.'"""""""''"""':
= 7,731 .29
t,a
:,
llrt
.'. The present worth is P 7,731.29.
,i
u
'.'ra
J
of
is the accumulated amount of five-year annuity paying P 6,000 at the end
annually?
compounded
15%
al
interest
year,
with
each
what
A. P 40,519.21
B. P 40,681.29
c. P 40,454.29
D. P 40,329.10
P = 42.0346A
But:
.l
AAAA
:
A
F
110
(lrrrrsl,i,rr
llrrrrk
l,).grrt,r,rirrg r4r..rr.rrrit:s
b.y
J,i,r.
[t. ,l'i,rrs
Arrrrrrit.y 111
.'. The accumulated amount is p 40,454.29.
A debt of P 10,000 with 10 % interest compounded semi-annuaily
is to be
amortized by semi-annuar payment ore," tirn"rt 5 years. The
first due in 6
months. Determine the semi-annual payment.
\
...
A, P 1,234.09
B. P 1,255.90
c. P 1,275.68
D.
1*AAAA444LL4444
P 1,295.05
Ft=Fz
i
,
i)
,_r]
g(r + i)"
f.,
,l
\
n=2(5)=10
'
:
p
t
a
t
I
ERrontnty = ER"nnratty
I
<-r.r.r.ii.ri.rr-...........................i
A = 1,295.05
(r+i)" -1:(1+o.oa)l-t
1+
.'. The semi-annual payment is p 1,295.05
a lending institution which willbe paid after 10
lZV3
compounded
annually lf mon-ey is worth g% per
.o!
ld he deposit to a bank monthly in ordei to discharge his
OO-0.t1-o.m
A. P 5,174.23
B. P 5,162.89
c. P 5,190.12
D.
;I
=e31,754.46
I
J
(r + o.os)10 (o.os)
:;l
' rl
For annuity, the interest is 8% per annum. Since the payment is monthly,
solve the interest per month.
AAAAAAAAAA
R[(r
L' + o.os)10-rl
at
p!t
=e(t+i)"
Fl = 3oo,ooo(1+ 0.12)10
i
2
10,000=
,ttlllt
jts
where:i= o'10=0.05
I
-+ Eq. 1
10,000
I
_ n[(r *
AA444A444LM
P 5.194.23
Ir
i= 1.00643
i= 0.00643
,.(
'rt
.tl
Fz:
But
F2
= Fr = 931 ,754.46
n[{r * o.ooo+r)""0'
931754.46
'
A
-'']
0 00643
==
5,174.23
.'. The monthly deposit in the bank is P 5,174.23.
.7;7*,.
Please refer to the cash flow on the next page
112
(lrrcnl
iorr\lLrrrk
l,)rrginr,r.r.irrg l,l:orrorrir:s by Jaimc Il.. 'l'itrng
Annuity
il
.'/n/-tr*
A man loans P 187,400 from a bank with interest at S% compounded an ually. He
agrees to pay his obligations by paying 8 equal annual payments, the fir
due at the end of 10 years. Find the annual payments.
A.
B.
c.
D.
,
P 43,600.10
P 43,489.47
_
n[(r * i) "-r]
where:
(t+i)"i
'i=
o'12=0.03
4
n=6
P 43,263.91
P 43,763.20
p=
,9/2,n,,
Pz= 187,400
-+ Eq.
1
P = 10,834.38
187,400
.'. The amount initially borrowed was P 10,834.38.
Pr = 6.643 A
q
D'z -,1*y
_ 6.6434
' (1+ 0.05)'g
Pz
ce of lot for P 1 00,000 downpayment and 10 diferred semiP 8,000 each, starting three years from now. WhaI is the
present value of the investment if the rate of interest is 12 % compounded semi-
A AAAAAA
Pt n""""""""""""""":
i
P2 *""""'!
= 4'282A
annually?
A. P 142,999.08
B. P'143,104.89
c. P 142,189.67
D.
P 143,999.08
Substitute values in Eq. 1:
I0 payments
4.2824 =187,400
A = 43,763.20
The annual payment is P 43,763.20.
Pr=
Pr = 58,880.69
Money borrowed today is to be paid in 6 equal paymenta et the end of 6 quarters.
lf the interest is 12 o/o compounded quarterly. How much wer lnltlally borrowed if
quarterly payment is P 2,000.00?
A.
B.
c.
D.
P 10,834.38
P 10,278.12
P 10,450.00
P 10,672.90
P1
=P2(1 +i)n
58,880.69 = Pz (1+.0.06)5
Pz =
'Pz n'
43,999.078-
Total amount = 100,000 + Pz
Total amount = 100,000 + 43,999.08
Total amount = 143,999.08
)-
114
Quest,i,n
13tr,k Ilrrgrrr.r,ri^g
l,lt:,rrrruir:s by ,lairne
Annuity
ll. Ti'rrg
.'. The present value of the investment is p 143,999.0g.
.'. The annual dePosit is P 716.81
How much must you invest today In order to withdraw p 2,000 annually
years if the interest rate is 9%?
A.
B.
c.
D.
for\0
A.
B.
4.61
4.71
c.
0/o
4.51
0/o
9;/,t.,.
Amount invested
{qrrl
Cash price = 2,000 +
(t+i)"i
P_
0/o
4.41%
D.
.1./;/;,.r,r,,
p=
.
A piece of machinery can be bought for P 10,000 Lsn or for P 2,000 down and
payments of P 750 per year for 15 years. what is the annual interest rate for the
time payments?
\.
P 12,835.32
P 12,992.22
P 12,562.Q9
P 12,004.s9
P = Amount invested today
115
P
-+ Eq.
1
Solving for P:
2ooo[(1+ 9.oe) 1o-1]
(1+ o.oe)10 (o.oe)
P
P = 12,835.32
P=
AAAAAAAAAA
:
..........................i
The arnount invested is P 12,835.32.
How much must be depo-sited at 60lo each year beginning on January 1, year 1, in
order to accumulate P 5,000 on the date of the lasi Oepolit, January .l y-ear 6?
,
A.
B.
c.
D.
7so[(1+ i)'5 - r]
1
0,000 = 2,000 +
--ilr;--
(1+i\15
P 728.99
P 742.09
P 716.81
P 702.00
10.6667
r----l---1
i(1
+
i)''
By trial and error:
,9o/,2,rn
i= 0.0461
i= 4.61%
n
5,000 =
[(r
+
o.oo)
Since the mode of compounding is annually, the nominal rate i5 equal
to i = 4.610/o.
u-r]
0.06
A = 716.81
,F
116
Annuity
Question llank -. I4nginecling hkxlnonrics by Jainrc [t. 'l'iorrg
.
A company issued 50 bonds of P 1,000.00 face value each, redeemab[at par at
the end of 15 years to accumulate the funds required for redemption. Th{irm
establiqhed a sinking fund consisting of annual deposits, the interest rate df the
fund being 4 %. What was the principal in the fund at the end of the 12th
A.
B.
c.
D.
117
/ /uz,;,,
Cash Price = 500,000 +
Pz
-+ Eq.
1
_'roo,ooo[{r*o.r+)'o-r]
-,-@
_
P 38,120.00
P 37,520.34
P 37,250.34
P 37,002.00
Pt = 521,61't.56
AAAAA,AA
500,000
t-/o1rtrn.
Pr=
-
F = 50(1,000)
F = 50,0000
D
:
'1
Pt
(1+i)"
^
_521.611.56
Pz
= 308,835.92
Pz '"""'-"'
"- (1*oi4f
Solving for A:
AAAAAAAA
F=
AAA
A
Substitute values in Eq. 1:
I
50,000 =
F
n[tr+o.o+I'-r]
A =2,497.06
ilI t[fIIt
12
With A = 2,497.06
Solving for Fe:
2,4e7.06[(1+
Ftz =
O.O+)1'?
0.04
- r]
AA/1,'1.'1,4A.4
AAAA
:..-.-...... ., -..-----....- Fp
7z = 37,520.34
.'. The principal in the fund at the end of the 120' yorr
lr
P 31 ,520.34.
A house and lot can be acquired by a downpaymenl ol P 500,000 and a yearly
payment of P 100,000 at the end of each year for a porlod of 10 years, starting at
the end of 5 years from the date of purchase lf money ln worlh 14% compounded
annually, what is the cash price of the prope(y?
A.
B.
c.
D.
Cash price = 500,000 + Pz
Cash price = 500,000 + 308,835.92
Cash price = 808,835.92
.'. The cash price of the property is P 808,835.92.
A parent on the day the child is born wishes to determine what lump sum would
have to be paid into an account bearing interest at 5 %,compoundedennually, in
order to withdraw P 20,000 each on the child's 18'n, 19'n, 20'n and 21"' birthdays.
How much is the lumP sum amount?
A. P 30,119.73
B. P 30,941.73.
c. P 30,149 37
D. P 30,419.73
tl/,,/,2,,.-
Let: Pz = lumP sum amount
Solving for Pr:
P 806,899 33
P 807,100 12
P 807,778 12
P 808,835.92
,r1-
n[1r*i;"
a
i(1+i)"
-rlI
118
Qrrostion
lltnk
I,lnginocling
,'t _- 20, ooo [(r + o.os)a - r]
03u,1*gp5y
q
= 70,919.01
Solving for P2:
ru
l,l<:onornir:s by
Jairnc It.
1 ... l415 16 1718 19 2021
0---------+-
:
Et
'2 --
^'
Pz
D
r1
(1+Dn
70,919.01
:
Pz
: AAAA
i
ii::
Arrnui[y
'l'iorr11
P, <"""""""'i
'''""""""""""""j
A manufacturing firm wishes to give each 80 employees a holiday bonus. How
much is needed to invest monthly for a year al 12 % nominal interest rate
compounded monthly, so that each employee will receive a P 2,000 bonus?
A.
B.
c.
D.
P
P
P
P
12,615.80
12,516.80
12,611.80
12,510.80
,'.y'n/1,1,-.
(1+ 0.05)''
F=80(2,000)=160,000
=30,941.74
E-
n[(r
L\
+ i)
.'. The lump sum amount is P 30,941 .73.
160,000 =
A. P 242,860.22
B. P 242,680.22
c. P 242,608.22
D. P 242,806.22
'
f,
ul
"-r'])
[!r
A[r1+
An instructor plans to retire in exaclly one year and want an account that will pay
him P 25,000 a year for the next 15 years. Assuming a 6 % annual effective
interest rate, what is the amount he would need to deposit now? (The fund will be
depleted after 15 years).
119
it
0'12)1,-1.l
il
L( 12) l
,I
o.12
A = 12,615.80
The amount needed to invest monthly"is P 12,615.80.
A man purchased on monthly installment a P 100,000 worth of-land. The interest
rate is 12 % nominal and payable in 20 years. What is the monthly amortization?
P 1,101.08
P 1,202.08
P 1,303.08
P 1,404.08
':*ir**..
P = 242,806.22
.'. The amount to deposit now is P 242,806.22.
',
il
I
12
A.
B.
c.
D.
i"l
AAAAAAAAAAAAAL44A44+L4M AL44A4L4LLL4
....'.'...'.,....'.....' j
1ZO Question Ilank - ltrnginooring
Ecolrotni<:s by Jaime R.
Tiong
Annuity
/
:
121
0,000
F = 6,922.93
6,07 4.7O + 0.567F
1
The amount the engineer has to pay at the end of the
where: i=
5th
year is
P6,922.93.
o'12=0.01
12
n=12QO)=24O
n[1t
100,000 =
+
An investment of P 350,000 is made today and is equivalent to payments of
P200,000 each year for 3 years. What is the annual rate of return on investment
for the project?
o.or) '?40-r]
---E-
(1+ o.o1)240 (0.01)
A = 1,101.0d
A.
B.
c.
D.
,'. The monthly amortization is P 1,101.08.
A young engineer borrowed P 10,000 al 12 % interest and paid P 2,000 per
annum for the last 4 years. What does he have to pay at the end of the fifth year
in order to pay off his loan?
A.
B.
c.
D.
32.7 %
33.8 %
33.2%
33.6 %
,?"e*r",
. _ n[1r*i;"-r]
(r+i)"i
P
P
P
P
6,999.39
6,292.93
6,222.39
6,922.93
2oo,ooo[(r
.--D_
(t +
+ i)
3-r]
i)'(i)
But P = 350,000
ty';rs.*-..
zoo,ooo[(r+i)3
B +Pz =10,000 -+ Eq. 1
350,000=
10,000
t'-'
i(1 +
i)'
-r]J
AAA
By trial and error:
:
i=0.327
i
2,ooo[(1+ o.rz)a
^
L
=_
Pt
- r]
o,, -
(,t+0.12\5
Pz =
0'567F
F
Substitute values in Eq. 1:
I
.'. The annual rate of return of the investment is 32.7%.
AAAA
O.12(1+ 0.12)a
=6,074'70
= 32.70/.
Pt
P2
:
Maintenance cost of an equipment is P 200,000 for 2 years, P 40,000 at the end
of 4 years and P 80,000 at the end of g r,sars. €ompute the semi-annual amount
that will be set aside for this equipment. Money worth 10% compounded
annually.
A.
B.
c.
D.
P 7,425.72
P 7,329.67
P7,245.89
P 7,176.89
Annuity
122- Quostiorr llrrrrk l')rtgirtt'trt'irlg l')<',rttotttit:s lly Jairntr ll'l'iorlg
123
/,/,2,,,,
Solving for interest semiannual:
ERsemiannuatly
= ERmnratty
(1+i)2 -1=(1+0.10)1-1
1+ i = 1.0488
i= 0.0488
P2
i
Pl
= 4.88o/o
Also, P = 81,170.06
Solving for present worth of maintenance cost, P
P=Pr+Pz+Pe -+
I
Substituting values:
1
F
o-
I
I
Eq.
(1
81.170.06
+ i)n
0.0488)16
20'000-
' =(1 + 0.1 0)'
P.
n[1r+o.o+ea)'u -rl
L
r
= 0.0488(1+
A =7,425.72
.'. The semi-annual amount that will be set aside is p 7,425.72.
Pt =16,528'92
40,000
^'
(1+ 0.10)"
Pz
=27'320-54
^
'r
Y^=-
Pg =
Mr. cruz plans to deposit for the education of his 5 years old son, p 500 at the
end of each month for 1 0 years al 12o/o annual interest compounded monthly.
The amount that will be available in two years is
80,000
P 13,100.60
(1+0.10)U
P 13,589.50
37,320'59
Substitute in Eq.
P 13,982.80
P 13,486.7p
1
P = 16,528.93 + 27,320.54 + 37,320 59
P = 81,170.06
.Vbr,rn.
Solving for the semiannual amount to set aside for thc equipment' A
0123451r78
AAAAAAAA
I I II-I
I II11,,lAA^,
AA44A44444L4
A.4
P
)
174
Qucst.ion Uank
-
Annuity
lr)nginccl'ing lir:ottottrics[.ry Jaime lt.'l'iorrg
Equivalent (reduced) cash flow:
o.12
where: i= -"-=0.01
12
n=2(12)=24
=
_
5oo[(1+ o.or)'z4-r]
5500 5500
0.01
s50a
ss00 5500
F = 13,486.70
.'. The amount that will be available in two years is P 13,486.70.
A small machine has an initial cost of P 20,000, a salvage value of P 2,000 and a
life of 10 years. lf your cost of operation per year is P 3,500 and your revenues
per year is P 9,000, what is the approximate rate of return (ROR) on the
investment?
A
B.
c.
D
Pr + Pz =
242o/o
24.8 o/o
20,000 -+
Eq.
1
Solving for Pr:.
25.1%
25.4 %
2000
I
9000 9000
0123
3500 3500
1l
9
10
TI
3500.1500
o_'z
F
*;y
_
2.000
P-:---:'2
(1+i)10
Substitute values in Eq. 1:
5,500[(1+
i)10
- r]
i)10 - #F
.
20,000
i(1+
= 2o'ooo
By trial and error:
i=0.248
i
= 24.8o/o
.'. The rate of return on the investment is 24.8 %.
125
176
Question
llank
[,)ngirtccrirrg I']trtlttotttitrs
Lrv
Annuity
.lairnc It. 'l'iorrg
'|27
I
AAAAAAAAAAA
A
:
:
D.
n[1r*i;"
P 1,263.71
F
-rl
J
L
F-
.'.....'....>,|,-.
I
A --3,942.44
iAAAAA
;
.'. The proceeds the employee obtains is P 1,263.71.
The amount needed to invest monthly is P 3,942.44.
Mr. Robles plans a deposit of P 500 at the end of each month for 10 years at
12o/o ennual interest, compounded monthly. What will be the amount that will be
available in two years?
The preside+t of a growing engineering firm wishes to give each of 50 employees
a holiday bonus. How much is needed to invest monthly for a year al 12o/o
nominal rate compounded monthly, so that each employee will receive a P
1,000 00 bonus?
A.
B,
c.
D.
P 3,942.44
P 3,271.22
P 3,600.12
P 3,080.32
A.
B.
c.
D.
P
P
P
P
13,941.44
13,272.22
13,486.73
13,089.32
9;1,,r,r",
A
Let F = Total holiday bonus for 50 employees at P1,000 each, the firm will
need at the end ofthe year (12 months)
:....,..,..,.....,,..
F = 50(1,000)
F = 50,000
Solving for monthly saving, A:
AAAAAAAAAAA A AAAAAAAAAAA
F=
n[tr*i)"
L\') -t'l
I
AAAAAAAAAAAA
128
Question tlank - It)ngineering Itrconornics by Jaime R. f iong
Annuity
179
After paying 2 annual payments:
12
F = 13,486.73
.'. The amount that will be available in two years is P 13,486.73.
AAAA
Mr. Ramirez borrowed P 15,000 two years.ago. The terms of the loan are lOyo
interest for 10 years with uniform payments. He just made his second annual
payment. How much principal does he still owe?
A.
B.
c.
D.
F = P(1+ i)"
Fl = 15,000(1+ 0.10)2
Fr = 18,150'00
P 13,841.34
P 13,472.22
P 13,286.63
P 13,023.52
Fz=
.V;/,tr...
n[o+u"
-r]
i
Fz:
Fz
2,441.181(1+ o.r)'?
- r]
01
='5'126'48
Balance = Ft -Fz
Balance = 18,150.00 -5,126.48
Balance = 13,023.52
AAAAA/1//1/1/
.'. The amount Mr. Ramirez still owe is P 13,023.52.
Solving for the annual amortization, A
A man inherited a regular endowment of P 100,000 every end of 3 months for 10
years. However, he may choose to get a single lump sum payment at the end of
4 years. How much is this lump sum if the cost of money is 14% compounded
quarterly?
A. P 3,702,939.73
B. P 3,607,562.16
c. P 3,799.801.23
D.
P 3,676,590.12
130
Question Bank
-
Annuity
Engineering lJconornica by Jaime R. 'l'iorrg
I
6 payments
131
24 payments
01 2 3
1515171819...
3940
A man paid 10% down payment of P 200,000 for a house and lot and agreed to
pay the balance on monthly installments for "x" years at an interest rale ol 15o/o
compounded monthly. lf the monthly installment was P 42,821.87, find the value
of x?
A A,a A A A A A
D
A
:
A.3
B.4
c.5
D.6
,.-F
Lumpsum amount= F +
P
-+ Eq.1
Solving for F:
_
L_
o[(''+i)" -r'l
I
0.035
A:
F = 2,097,102.97
42,821 .87
Solving for P:
Equivalent (reduced) cash flow:
,_n[(r*i)"-r]
i(1
1
P=
+ i)"
oo,ooo[(1 + o.oss)'?4
- r]
0.035(1+ 0.035)24
P = 1,605,836.76
Substitute values in Eq. 1:
AAA
A
Lump sum amount = 2,097,102.97 + 1,605,836.76
Lump sum amount = 3,702,939.73
.'. The equivalent lump sum amount is P 3,702,939.73.
Down payment = 10% of Cost of house and lot
200,000 = 01O(Cost)
Cost = 2,000,000
Balance = CosJ - Downpayment
Balance = 2,000,000 - 200,000
Balance = 1,800,000
A-- 42,821 .87
13i-
Question Bank
- Engintrcring
l')corrorrrir:s by Jaime
Annuity
[l' 'lirtrtg
133
Solving for P:
5,000
where:
i#=0.0125
n
=12x
=P,l(1+i)n
F.,
= 5oo(1+ o.o7)21
Also, P = 1,800,000
Substituting values:
AAA
q
;......,...,,,.,.,...'.
A
"..t
F2
i.... Fj
F1=2O,7O2'81
42,821 4z[1r + o'o1 25)l2'
1,800,000=ffi
(1 .01 25)12^
- 1 = 0.52543(1.025)1?'
(1.0125)12" --2.1072
o.o7
Fg:Fz(1+i)"
2x log 1 .0125 = log2.1O72
x = 5 Years
Fe
=
F3 =
The value of x'is 5 Years'
17,759.72 (t + O.ozf
19'002.95
Money left = Fr
I Y ar ru
1-6.
P4,0OO at the end of your 1 7. ,
"i year't
in the account at the end ofthe 21"'
4-1]
L-
-
Fz= 17,759.77
Take logarithm on both sides:
1
4ooo[(1+ o.o7)
-t z
E
zu
A. P 1,666.98
B. P 1,699.86
c. P 1,623.8e
D. P 1,645.67
'/./rrl,i.n
colleoe. Your father invested P
u *"16 born lf You withdraw
h:+r-,,.ta,
much will be left
how mtteh
birthday' hmrr
-
Fs
Money lefl=2O,7O2.81
Money left = 1,699.86
-
19.002.95
The money left at the end of the 21"t year is P 1,699'86'
p1 Ayata borrows P 1oo,0oo al lOo/o effective annual interest. He must pay back
on the first day of
the loan over 30 y""r" *ith uniform monthly payments due
month?
pay
each
eatn montn. Wnit Ooes Mr. Ayala
A. P 839.19
B. P 842.38
c. P 807.16
D. P 814.75
page
Refer to the cash flow on the next
A+P=
100,000
-+ Eq.
1
134
Question Bank
-
Annuity
Engineering Economics by Jaime R. Tiong
135
Solving for monthly interest rate:
A mdchine is under consideration for investment. The cost of the machine is
P25,000.00. Each year it operates, the machine will generate a savings of
P15,000.00. Given the effective annual interest rate of 18%, what is the
discounted payback period, in years, on the investment of the machine?
A. 3.15
B. 1.75
c. 2.15
D.
2.75
,'/n1r.rlrr,.,
f
Cost of machine = 25,000
r
Let P = Present worth of the
total amount generated bY the
ftl
U
ER=(1+i)'-t
'f
For machine to PaY off for
itself, P = 25,000
1.19=(1+i)u
i
iI
machine
o.to.=(r+i)12-1
= o.007974
ra
I
:l
Solving for number of years, n
for machine to pay off for itself:
Solving for P:
il
:i!
rI
ll,,
i(1
lru
+ i)"
l-
.a<o
1
Al(1+0.007e74)"- -11
L
)
D_
O.OO7 97 4(1
+ 0.007974
25.0C0
)35e
P = 118.163A
Substitute values in Eq.
=
15,ooo[(1+ 0.18)"
L
0.1 8(1 + 0.
1
- r]
r lll
rtl
8)"
0.3(.18)n = (1.18)" -1
0.7(1.18)n =
1:
1
('18)" =1'428
A+118.163..A=100,000
1 19.163A = 100,000
A = 839.19
Take log on both sides:
nlog1.18 =1o91.428
.'. Mr. Ayala pays each month an amount of P
ttC,ll,
"
1oo1.428
logl .18
n = 2.15 years
The payback period of the machine is 2.15 years'
I
,-
136
Question Bank
-
Annuity
Engineering .Economics by Jaime R. 'Iiorrg
40,87178 =
A machine costs P 20,000.00 today and has an estimated scrap value of
P2,000.00 after 8 years. lnflation is 2o/o per year. The effective annual interest
rate earned on money invested is 8%. How much money needs to be set aside
each year to replace the machine with an identical model 8 years from now?
A. P 3,345.77
B. P 3,389.32
c. P 3,489.11
D.
n[tr*o.ro)'-r]
-L-10
A = 3,573.99
.'. The money to set aside each year is P 3,573.99.
pay
Engr. Sison borrows P 1OO,OO0.OO at10% effective annual interest. He must
of
day
first
payment
on
the
due
monthly
years
uniform
with
30
over
loan
tne
oaJr
each month. What does Engr. Sison pay each month?
P 3,573.99
A.
B.
c.
D.
Cosl = 20,000
P 838.86
P 849 12
P 850.12
P 840.21
Let F = future amount needed for the replacement of the old machine
F=Fr-2,000
r = p(1+ i)"
Fr
= 20,OOO(1+ 0.'t0)B
1=
A L4444AA4 ALA
42,871.78
A
LL4LL4LMA A
P
Note that interest used was 10% instead of 8%. This is due to inflation rate
which was added to the regular rate. The inflation rate must be added
because money's purchasing power with inflation becomes smaller, thus
bigger amount is needed for the replacement of the old machine.
F
137
A+P=
100,000 -+
Eq.
1
= 42,871.78 -2,000
F = 40,871.78
Solving for the annual amount to be set aside for future replacement of the
machine, A
012
Aiso, solving for the equivalent monthly interest,
ER.on,L1, = ER
('1
Fz=
AAA
But Fz = 35.018.60
+ i)12
- 1:
(1+i112
1+
(1+ o.'10)
=1.16
i= 1 .00797
i= 0.00797
-1
AAL4AL4L4L4
138
. l)nginccring
(luosl.ion []tnk
Annuity
.Economics by Jaime R. 'I.iong
Substitute values of P and i in Eq.
139
solving for the annual income which is the 100,000 rental plus the interest
1
per year:
n [1r + o.oozoz)"' -
A* '
0.00797(1
+
r]
=i
0.00797)""'
Annual income = 1 00,000(1 .18)
Annual income = 118,000
=100.000
A+118.21A = 100,000
Difference = 128,205.13
119.21A = 100,000
Difference = 10,205.13
A = 83836
-
118,000
.'. The difference between the firm's annual revenue and expenses is
.'. Mr. Sison pays each month an amount of p 83g.g6.
P10,205.13.
lnstead of
,000.00 in annual rent for office space at the beginning of
each year
0 years, an engineering firm has decided to take out a
10-year P
loan for a new building at 6% interest. The firm will invest
P 100,000.00 of the rent saved and earn 18o/o annual interest on that amount.
what will be the difference between the firm's annual revenue and expenses?
A.
B.
c.
D.
A service car whose cash price was P 540,000 was bought with a downpayment
of P 162,000 and monthly installment of P 10,874.29 for 5 years. what was the
rate of interest if compounded monthly?
P 10,200.12
P 10,205.13
P 10,210.67
P 10,215.56
P+ 162,000 = 540,000
P = 378,000
The expenses will be that of
the amortization of the loan.
Solving for the equivalent
annual expenses, A
A + P = 100,000 -+ Eq.
n[1r*i1^
ButP= L'
i(1
+
4
?
39
tl
/
AA
1
-r'lJ
i)'
Substitute in Eq.
I
1
n[(r*o.oo)n-rl
..
___u__t__J
0.06(1+ 0.06)'
= 1 00o,ooo
A+6.84=1,000,000
A =128,2A5.12
i(1
+ i)n
140
378.000
'
10,874.291fi+
-
L'
i(1
+
i)5(12)
i)5(12)
A. P 3,919.52
B. P 3,871.23
c. P 3,781.32
.NR
t-_
D.
m
IE
P 3,199.52
.71;zn,
12
NR = 0.24
NR = 24%
.'. The rate of interest is
1t.,fiu"y"",",P18'000willbeneededtopayforabuildingrenovation.lnorderto
gl;Li"i"lni" .rr, sinkins fund consisting 9L1'::T:,?1"11T:1"J3.
*,.oo
maOe aft-er three
"
furthLr payment
established now. For tax pu-rposes' no
ryt]lpe
is worth 15o/o pet annum?
V""ii. Wnrt payments ,r" n""".t"ry if money
i= 0.02
o.02 =
Fz
24Yo
compounded monthly.
=
18,000 -+ Eq.
1
Solving for Fr:
F=
annual amount should be deposited each year in order to
100,000.00 at the end of the 5'n annual deposit if money earns'
A.
B.
c.
D.
AA
A
F1>F2
P'16,002.18
P 15,890.12
1=3.47254
P 16,379.75
Solving for Fz:
P 15,980.12
F = P(1+ i)n
.V"zr*-
F, = Fl('l+ i)^
Fz =
Fz
1
oo, ooo
141
- 1'l
By trial and error:
But
Annuity
llank - Dngirrooring Economice by Jaime R. Tiong
Quest,ion
nltt*o.ro)u -r'l
A = 16,379.75
.'. The uniform annual amount to be deposited each year is p 16,379.75.
3.4725A('l+ 0.15)2
= 4'59244
Substituting value of Fz in Eq.
1
4.5924A = 18,000
A = 3,919.52
The annual Payment necessary is P 3,919.52'
(lrrtrsliolr
lllnk
Find the present value in.pesos, of a perpftuity of pl 5,000 payable
semi_
annually if money is worth 8% compounOpO quarterfy
A.
B.
c.
D.
P
P
Arrntrity 143
Iinginct,ring Uconomics by Jairnc ll.'l'iorrg
371,719.52
371,287.13
per
How much money must you invest today in order to withdraw P 1,000.00
year for 10 years if the interest rale is 12o/o?
A. P 5,467.12
B. P 5,560.22
c. P 5,650.22
D. P 5,780.12
t
r
P 371,670.34
P 371,802,63
V1,.to,,
ERsemi
(1 + i)2
annually
= ERquarteay
LetP=amountinvested
I
- t =( t *o'08')4 -1
4)
\
i_-
f,
o-
P=4
I
D
_ 15,000
,|
P=
0.0404
looo[(1+o.tz)10
0.12(1 + 0.1 2)10
AAA
0.0404
P =371,287.13
rl
I
-t]
AAA
AA
I,f
P = 5,650.22
.'. The amount you must invest today is P 5,650'22'
.'. The present value of the perpetuity is p 371,2g1 .13.
tl
irr
lf P 500.00 is invested at the end of each year for 6 years, at an annual
interest
rate of 7Yo, what is the total peso amount available upon tne deposit
oi the sinth
payment?
A.
B.
c.
D.
Amanpaidal0%downpaymentofP200,000forahouseandlotandagreedto
rate of 15ok
pay the balance on .onit''tv installments f9r.5.vegr;
"t3i,il!"]::l
pesos?
in
installment
monthlv
the
was
Whit
monthly.
;;;iil;;;
A. P 42,821.87
B. P 42,980.00
c. P 43,102.23
P 3,671.71
P 3,712.87
P 3,450.12
P 3,576.64
D.
P 43,189.03
.fu,ro.,
F=
,_soo[(r+o.oz)u-r]
o.07
F = 3,576.64
.'. The total peso amount avairabre upon the 6ir'dcpollt rr p 3,576.64.
AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAA4:
atl
ll
il
144
Question Bank
-
Annuity
Engineering Economics by Jaime R. Tiong
Cost of house and
1ot
=
145
2oo'ooo
0.10
Cost of house and lot = 2,000,000
Balance = Cost - Downpayment
Balance = 2,000,000 - 200,000
Balance = 1,800,000
-+ Eq.
But P = 1,800,000
1,
800,000 =
1
t, -_ 3,702,939.80
" (.+ 0.1414(4)
l.
^[(,,.T)'"''-,]
0.15)5(1',)
[9]!)[.'*
(12r( 12.)
3,702,939.80
4/
,,
Pt n"
Pt =2,135,507.273
.'....'......:
t,
I
I
Ii.
ti
ll
il
Equating P and Pr:
A = 42,821.87
.'. The monthly installment is P 42,821.87
.
= 2,135,507 .273
I
li
A man inherited a regular endowment of P 100,000.00 every end of 3 months for
x years. However, he may choose to get a single lump sum of P 3,702,93g.80 at
the end of 4 years. lf the rate of interest was 14o/o compounded quarterly, what is
the value of x?
A.
B.
c.
D.
,l%r,-"
(1.035)4"
-1 =0.7474
(1.035)4"
O'2526(1'035)4n =
10
11
12
13
1
1'0354n = 3'9588
Take log on both sides:
4n1og1.035 = 1o93.9588
1oo3.9588
" 41o91.035
n = 10 Years
.'. The value of x = n = 10 Years'
146
Queetion Bank
-
lf you obtain a loan of P 1M at the rate ol112Yo compounded annually in order to
build a house, how much must you pay rlnonthly to amortize the loan within a
period of ten years?
P
P
P
P
Annuity
Engineering Econodics by Jaime R. Tiong
13,994.17
14,801.12
13,720.15
14,078.78
Rainer wandrew borrowed P 50,000.00 from Social security System, in the form
of calamity loan, with interest at 8% compounded quarterly payable in equal
quarterly installments for 10 years. Find the quarterly payments'
A. P 1,590.83
B. P 1,609.23
c. P 1,778.17
D. P 1,827.79
9"zar-
AL44AL44L44A
AAAAAAAA
AA
AA
Since money is compounded annually while payment is monthly, it is
necessary to determine first the interest rate per month.
ERmontnty = ERannuattv
1t +
i1t'
- 1 = (1 + o.tz)l - t
(t+i)" =0:tz
i= 0.009489
But P = 50,000
. 0.08\
1+_
4)
t(
411 0)
|
n=12(10)=120
h-
R
[(r
L\
+ i)
'
"-rl)
(t+i)"i
n[(r * o.oos+as)l'zo-r]
1,000,000 =
(1 +
A=
o.oo9a89;"0 1o.oos+ao1
13,994.17
.'. The amount you must pay monthly is P 13,994.17.
147
A =\827.79
The quarterly payment of Mr. Rainer Wandrew'is P 1,827 '79'
148
Question Bank
-
Annuity
Engineering Economit:s by Jaime lt. 'l'iorrg
149
1.000.000
For having been loyal, trustworthy and efficient, the company has offered a
supervisor a yearly gratuity pay of P 20,000.00 for 10 years 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. lf
interest is 15%, what is the equivalent lump sum that he could get?
A.
B.
c.
But F = 4,000,000
n[{r*o.rs)'o
4,000,000=
P'100,357.37
P 100,537.73
P 100,375.37
-r]
j___L15
A = 197,008.25
A
p
.'. The company's deposit each year is P 197,008.25.
'-/"2:2r,,
A man purchased a car with a cash price of P 350,000. He was able to negotiate
with the seller to allow hinr to pay only a downpaymenl of 20o/o and the balance
payable in equal 48 end of the month installment al'l .5% interest per month. One
the day he paid the 20th installment, he decided to pay monthly payment. How
much is his monthly payment?
Equivalent lump sum = present worth of annuity, P
P=
Equivalent
Lump Sum
A.
B.
c.
D.
AAA
AA
P 8,929.29
P 8,225.00
P 8,552.00
P 8,008.20
.'/,fr,>n
Total price = 350,000
Downpayment = 0.20(350,000)
Downpayment = 70,000
Balance to amortize = 350,000 - 70,000
Balance to amortize = 280,000
Solving for equivalent monthly payment, A
ln anticipation of a much bigger volume of business after 10 years, a fabrication
company purchased an adjacent lot for its expansion program where it hopes to
put up a building projected to cost P 4,000,000.00 when it will be constructed 10
years after. To provide the required capital expense, it plans to put up a sinking
fund for the purpose. How much must the company deposit each year if interest
to be earned is computed at 15%?
A.
B.
c.
D.
P
P
P
P
194,089.17
195,780.12
196,801.56
197,008.25
280,400
AAAAAA
150
Qut:sl.ion
Ilrrnk
Arrnuitv
I,)ttgirtcct'ittg [!txrnotnit:s by Jairnc It. 'l'irrrrg
,. -t'l'
(1 + 0.015)45(0.015)
n[1t+o.ots;o8
280.000=
'
'
n[1t+o.otsyo8
280.000=--+
(1
+ 0.01
-tl
5)"'(0.0'l 5)
A =8,225.00
A = 8,225.00
.'. The man's monthly payment is P 8,225.00.
After the 20th payment:
280,000
......-,'"F)
A man purchased a car with a cash price of P 350,000. He was able to negotiate
with the seller to allow him to pay only a downpayment of 20o/o and the balance
payable in equal 48 end of the month installment at 1.SYo interest per month. One
the day he paid the 20th installment, he decided to pay monthly payment. What is
the remaining balance that he paid?
A.
B.
c.
D.
P
P
P
P
151
t
a
I
I
I
I
186,927.24
188,225.00
187,701.26
185,900.20
AAA
AAA
J
I
a
t
:
i,.,.,.,..,...,.........,.
1.
Ft
't
;l
,7.L;,t2.,,
,l
Total price = 350,000
Downpayment = 0.20(350,000)
Downpayment = 70,000
Balance to emortize = 350,000 - 70,000
Balance to amortize = 280,000
Solving for equivalent monthly payment, A
280,AAA
rd
Fl=
rl
ri
F,l=
B,2zsl(1+ o.ors)'z. - r]
rl
rl
0.015
Fr = 190,192.16
tr
Solving for F:
Fz
= P(1+
F2
= 280, 000('t + 0.01 5)20
Fz =
i)n
377,119.40
Balance to be paid = Fz- F.r
Balance to be paid = 377 ,119.40
Balance to be paid = 186,927.24
-
'1
90,192.16
The remaining balance to be paid is P 186,927.24.
152
Quest,ion
lllnk
Annuity
l,)ugineoring Flxrnomics by Jainre lt. 'l'rorrg
153
.?2a.,.
A company purchased for a cash price of P 500,000.00 a machine which is
estimated to have a salvage value of P 50,000.00 at the endott!110 years
economic life. How much yearly deposit must the company deposif in a sinking
fund that will pay 18%
to accumulate the needed
fund to purchase the n
10th year economic life of the
machine it purchased if a new machine will cost 75o/o morc by that time?
A.
B.
c.
D.
P 34,859.78
P 35,890.12
P 35,074.58
P 34,074.85
AAAAAAAA'AAA
AA
A A A A A A A'A A
A
A
A
:
'/.Zz.ru.
Balance to amortize = cash price - downpayment
Balance to amortize = 280,000 - 0.15(280,000)
Balance to amoritze = 238,000
Let F = Amount needed to purchase the machine 10 years from now
F = 500,000(1.75)
F = 875,000
F,+50,000=875,000
E = 825'000
But P = 238,000
Also,
Substituting values:
AAA
E=
825,000
AA
:
n[{r*o.ra)'' -r]
ol8
=
A = 35,074.58
238,000 =
A = 15,185.78
I
.'. The required monthly payment is P 15,185.78.
The yearly deposit is P 35,074.58.
A car dealer advertises the sale of a car model for a cash price of P 280,000. lf
purchased on installment, the regular downpayment is 15% and balance payable
in 18 equal monthly installments at an interest rale oI 1.5o/o per month. How
much will be required monthly payments?
A.
B.
c.
D.
P 15,185.78
P 15,289.12
P 15,783.90
P 15,632.11
A machinery supplier is offering a certain machinery on a 10% downpayment and
the balance payable in equql end of the year payments without interest for 2
years. Under this arrangery'ent, the price is pegged to be P 250,000. However for
cash purchase, the machitle would only cost P 195,000. What is the equivalent
interest rate that is being charged on the 2-year payment plan if interest is
compounded quarterly?
A.
B.
c.
D.
18.47
o/o
19.21%
19.47 %
19.74%
154
Question Bank
Annuity
Ilngineering lJconomics by Jaime R. 'l'iong
-
155
,72r.".,Cost of machine = 250,000
Downpayment = 0.1 0(250,000)
Downpayment = 25,000
Since no interest is considered for the balance, the annual payment, A is
A=_
Balance
Cash
2
250,000
^
Price
25,000
2
A company has approved a car plan for its six senior officers in which the
company will shoulder 25o/o of lhe cost and the difference payable by each officer
to a financing company in 48 equal end of the month installments at an interest
rate of 1.5% per month. lf the cost of each car is P 350,000, determine the
amount each officer has to pay the financing company per month?
A.
B.
c.
D.
P
P
P
P
7,523.90
7,619.22
7,190.00
7,710.94
A = 112,500
,7"2*r-"
Solving for interest:
P + 25,000 = cash price
Let P = total amount each officer will shoulder
P+25,000=195,000
P = 170,000
L
i(1
J
Let A = monthly amortization of each officer
+ i)"
112,500[(1*i)'-rl
170,000
i('1
+
P = 0.75(350,000)
P = 262,500
-rl
n[1r*i;"
D_
AA
262,504
i)'
By trial and error:
i= 0.209
i=20.9%
NR = 20.9% compounded annually
Solving for NR compounded quafterly:
ERqr"rt".ty
[',-Y)'
[ 4)
-
1
:
ERannuatty
= (1+ o 2oe)l
'
i('l + i)^
-'r
262,500
[,.,*rE)'=1.20s
\
4)
t+E
4
P
0.015)48
A =7,710.94
= 1.0486
.'. The amount each officer has to pay the financing company per month is
NR = 0.1947
NR = 19.47% compounded quarterly
.'. The equivalent interest rate is 19.47
n[1r*oorsyou-rl
L
r
- 0.015(1+
o/o.
P 7,71094.
156
Question Bank
-
Engineering Economics by Jaime R. 'I'iorrg
lf P 1
get a
P 34,467.21
P 34,567.81
P 34,675.18
P 34,867.37
0.14(1+ 0.14)B
B = 34,675.18
.'. The annuity the percon can get from the bank each year is P 34,675.18.
i" """"""""""""
" ".'..'..t,1' |rfl-r
An employee is earning P 12,000.00 a month and he can afford to purchase a
car which will require a downpayment of P 10,000.00 and a monthly amortization
of not more than 30% of his monthly salary. What would be the maximum cash
value of a car he can purchase if the seller will agree to a downpayment of P
10,000.00 and the balance payable in four years at 18o/o per year payable on
monthly basis? The first payment will be due at the end of the first month?
r'
:
AAAAAAAAA
A. P 135,267.21
B. P 135,507.42
c. P'135,605.48
.BBBBBBBB
'P ."""""
D.
P 135,807.30
I
i
.V"zr*-
Referring to the cash flow, F = P
Solving for F:
0.14
10,000
F = 160,853.46
P
Solving for P:
-'..
Cash value = 10,000 + P
,_n[(r*i)"-r]
i(1
' l14G or4f
L
Equating:
\
-+ Eq. 1
Downpayment = 10,000
+ i)n
_- s[(''+ o r+)B - r'j
D-
157
F:P
a[{r*o.r+)'-r]
160.853.46 - L
ch year for g years, how much annuity can a person
every year for 8 years starting 1 year after the 9th
oney is 14%.
depo
A.
B.
c.
D.
Annuity
J
Let A = Maximum monthly the employee can afford
A = 0.30(12,000)
A = 3,600
158
\
Question llank
-
It)nginecring Economics by Jaime R. Iighg
Annuity
F
ELI
8,ooo[(1+ o.os)5 - t'l
0.09
AAAA
F = 47,877.46
Solving for equivalent monthly interest
:
ERrontnty = ERannually
(r + i)1'z -1= (1+ 0.18)l
i
No. of stocks
-1
-
47877'46
A
F
100
No. of stocks = 478.776 x 478
= 0.0't389
.'. The employee will be able to purchase 478 shares of stocks.
3,600[(1+ o.orsse)a.
P=
- t]
0.01389(1 + 0.01389)48
Mr. Pablo Catimbang borrows P 100,000 al 10o/o compounded annually,
agreeing to repay the loan in twenty equal annual payments. How much of the
original principal is still unpaid after he has made the tenth payment?
P = 125,507 .42
Substitute in Eq.
1
A.
B.
c.
D.
cash value = 10,000 + 125,507.42
Cash value = 135,507.42
The maximum cash value of the car is P 135,507.42.
"'-/"/r,t
A new company developed a program in which the employees will be allowed to
purphase shares of stocks of the company at the end of its fifth year of operation,
when the company's thought to have gained stability already. The stock has a
par value of P 100.00 per shqre.
Believing in the good potential of the company, an employee decided to save in
bank the amount of P 8,000.00 at the end of every year which will earn for hirn
9% interest, compounded annuallY.
P 69,890.42
P 72,000.80
P 72,173.90
P 72,311.44
.
rn
100,000
a
How much shares of stocks will he be able to purchase at the end of the fifth year
of his yearly deposits?
A.
B.
c.
476
478
480
Solving for the annual payment, A
Solving for the future worth of annuity:
100,000=#
(1+
9,2**
n[(t+o't)20-l"l
0.1)'"(0.1)
159
160
Question Bank
Annuiry
Engineering Economics by Jaime R. 'l'iong
-
161
A = 11,745.96
A debt of P 12,000 with interest of 2Oo/o compounded quarterly is to be amortized
by equal semi-annual payments over the next three years, the first due in 6
months. How much is the semi-annual payments?
After the tenth payment is made;
100,000
A.
B.
c.
D.
P 2,775.50
P 2,662.89
P 2,590.04
P 2,409.78
.'-/./r,,h..,.
AAA
A
i
Solving for the equivalent semi-annual interest rate:
'n
ERsemi
,..,.,1,rt;:.,,ii
"nnratty
Solving for the future worth of the annuity, Fr
E_
(1+i\2 -1
\ ,
nltr+lr
-r']J
L
't't,7 4s.96l
(+
L'
o. t)10
'
- tl-l
11+
\
o'20
)4
4) -
1
= 0.1 025
Solving for semi-annual payment, A
0.1
E = 187,200.35
hL)
n ltr + il"
Fz = P(1+ i)n
1
F2:1OO,OOO(1+0.1)10
2,000 =
A = 2,775.50
= 259,374'25
Balance = Fz-Ft
Balance = 259,37 1.25
Balance =72,173.90
- r'.l
(1+ i)ni
Solving for future worth of 100,000, Fz
Fz
-
1+i = 1.1025
i
=-
= ERquarterty
.'. The semi-annual payment is P 2,775.50.
-
187,200.35
The original principal that is still unpaid after the
P72,173.90.
1Oth
payment is
A fund donated by a benefactor to PICE to provide annual scholarships to
deserving CE students. The fund will grant P 5,000 for each of the first five years,
P 8,000 for the next five years and P 10,000 each year thereafter. The
scholarship will start one year after the fund is established. lf the fund earns 8%
interest, what is the amount of the donation?
A.
B.
c.
D.
,'/;/,)r,ru
P 99,490.00
P 99,507.35
P 99,601.71
P 99,723.54
162
Question Ilank
-
Annuity
Engineering Economics by Jaime R. 'I'xrrrg
D-
P4
rE-.....-........:
(1 +
"
i)''
D _ 125'000
'5 - (1r+0.o8f
Ps = 57,899'186
The amount of donation, P is
P=?*P.+P.
P = 19,963.55 + 21,738.97 + 57,899.19
P = 99,601.71
Amount of donation = Pl + P3 +
5,ooo[(1+ o.oa)s
- r]
P5
.'. The amount of the donation is P 99,601.71.
lf a low cost house and lot worth P 87,000 were offered al 10o/o down payment
and P 500 per month for 25 years, what is the effective monthly interest rate on
the diminishing balance?
P1
(1+ 0.08)5(0.08)
P,l
A.
:19,963.55
B.
C.
D.
8,ooo[(1+ o.oa)s
_
P2
= 31,941.68
o/o
O.O492
49.2
o/o
o/o
,9*i**
(r+l)'l
P2
0.492%
4.92
- r]
(1+ 0.08)5(0.08)
IALL44444LLMAL444LL4LLM AL444LL4{4M
,
P4
-A
I
10,000
P4 _
0.08
P4 = 125,000
Balance = 0.90(87,000)
Balance = 78,300
Balance=P=78,300
('1+ i)"i
163
164
Question Bank
-
Itrngineering Ucononrics by Jairnc ll. 'l'irrrrg
5oo[(1+i)12('z5)
,*----!(1 + i)''t"';
78,300 =
Annuity
-1]
1o91.4164 = nlog1.08
1oo1.4164
n = ----:z--
log1.08
By trial and error:
n=
i= 0.00492
4.523 years
.'. The dam will pay for itself in 4.5 years.
i = O.492oh
.'. The effective monthly interest rate is 0.492 %.
A businessman borrowed P 10,000.00 from a bank al 12Yo interest, and paid
P2,000.00 per annum for the flrst 4 years. What does he pay at the end of the
fifth year.in order to pay-off the loan?
The average annual cost of damages caused by floods at Dona Rosario Village
located along Butuanon river is estimated at P 700,000. To build a gravity dam to
protect the area from the floods, would cost P 2,500,000 and would involve an
annual maintenance cost of P 20,000. With interest at 8% compounded annually,
how many years will it take for the dam to pay for itselP
A. 4.0 years
B. 4.5 years
C. 5.0 years
D. 5.5 years
A.
B.
' c.
D.
P
P
P
P
6,812.54
6,782.31
6,917.72
6,660.90
P,,+Pr=
BBB
tl1
x=A-B
x=700,000-20,000
0123
x = 680,000
Using equivalent (reduced) cash flow:
B
1
10,000
Pt
=6'074'70
D_ F
,r-{r*,1n
E
o-'
'
[(t * i)" (1+i)n
680, OOo
2.500.000
1
[tr*o.rz)o -r]
^'t = -2,ooo
(l*olz)\olz)
Cost of dam = 2,500,000 = P
nLl
--+ Eq.
t"l
i
ftr * o.oal" - r]
''-
(r+0.12)5
Pz
= 0'567427 F
Substitute values in Eq. 1:
(1+ 0.08)"(0.08)
(1.08)" -1 = 0.294(1.08)"
1 = (1.08)n(0.706)
1.4164 = (.1:08),
P|+P2 = 10,000
6,07
P
Equivalent cash flow
4.70 + 0.567427 F = 1 0,000
F -- 6,917.72
The amount the businessman will pay at the end of the 5th year is
P6,917.72.
165
166
Annuity
Question Bank - Dngineering Econo,mics by Jaime R. 'l'iorrg
Engr Richard Flores agreed to pay the loan he is borrowing from a developmbnt
bank in six annual end-of-the-year payments of P 71 ,477 .7O. lnterest is 18% per
annum compounded annually and is included in the yearly amount he will be
paying the bank. How much money Engr. Flores is borrowing from the bank?
A.
B.
c.
D.
P
P
P
P
_/
18000
nlrt*i)"
p= L' '- -t'lI
i(1 +
--+Eq.
1
i)"
Solving for equivalent
monthly interest:
250,000
ERmonmty =
260,000
270,000
280,000
AAAAA
ERannuatly
;
P
1+ i = 1.00949
i= 0.00949
Substitute the value of P and i in Eq' 1:
P = loan
1
P=
7
1,477 .7ol(+ o. r a)6
P=
- r]
5'
P = 250,000
n[(r*o.oos+uF'-1]
olfur4r(1. ooo%nr
A:701.82
ooo =
AAAAAA
0.1 8(1 + 0.1 8)6
Solving for the monthly financing charge, B:
^'1215,000(1.12)
H=
.'. The amount of money Engr. Flores is borrowing from the bank is
P250,000.00
B = 150.00
Total monthly installment = 7O1.82 + 150
Total monthly installment = 851 .82
Barbette wishes to purchase a 29-inch flat-screened colored W at Gillamac
Appliance Center an amount of P 20,000.00. She made a lownpayment of
P5,000.00 and the balance payable in 24 equal monthly installments. lf financing
charge is 12% for each year computed on the total balance to be paid by
installment and interest rale 12Yo, how much would Barbette pay every month for
the colored TV? What will be the actual cost of the money?
A.
B.
c.
Solving for actual cost of money:
851.82[(1+i)'?4
36.71%
36.21%
15,000 =
i(1
35.89 %
D.35.23o/o
Zz"a)
By trial and error:
i=0.0264
Solving for the monthly installment without taking into account the
financing:
But
.NR
l=m
P = 15,000
167
+ i)24
-1]
:
'l
68
Annuity
Question lJank - Engineering liconornics by Jairnc ll,. 'l'iorrs
169
Let SV = salvage value of the machine 5 years from now:
0.0264 = IE
12
F=P(1+i)"
NR = 0.3168
NR = 31.68%
1,683,062.077
SV :80,oo0(1+ 0.07)5
sv=112,204.138
lf converted to annual effective rate;
ER"nnu"tty = ERmontny
(1
:
(t * o.ozu)l2 - t
:0.3671
i
+ i)i -.1
AAAA
i....................
i:36.710/o
F
+112,204.138 = 1,683,062.077
F = 1,570,857.939
Engr. Omayan of Main Engineering decided to purofrase a machine which is to
be used foi their refrigeration and airconditionlng wonfts at an amount of P
'1
,200,000. The useful life of the machine is estimated b be 5 years with a
salvage value of P 80,000 as based on current prices. The average annual rate
of inflation during the next 5 years will be 7%. The machine will be replaced with
a duplicate and the firm will accumulate the necessary capital by making equal
end-of-year deposits in a reserve fund that earns 6% per annum. Determine the
annual deposit.
P 278,664.U
I
!12244p8
But
F
nftr*o.oo)'-r]
1,570,857.939
=
o06
A = 278,664.54
P 277,189.56
P 279,180.00
F
Referring to the constructed cash flow:
.'. The actual cost (effective rate) of the money is 36.71 %.
A.
B.
c.
D.
A
.'. The annualdeposit is P 278,664.54.
P 280.673.12
Cost of the machine at present = 1 ,200,000
Let C = Cost of machine 5,years from now:
F = P(1+ i)"
C = 1,200,000(1+ o.o7)s
c
= 1,683,062.077
Note that the interest used was due to inflation only since nothing was
mentioned about its interest. The interest of 6% stated in the problbm was
implied to be that of the annuity (fund) only.
Salvage value at present = 80,000
Engr. Omayan of Main Engineering decided to purchase a machine which is to
be used for their refrigeration and airconditioning works at an amount of P
1,200,000. The useful life of the machine is estimated to be 5 years with a
salvage value of P 80,000 as based on current prices. The average annual rate
of inflation during the next 5 years will be 7%. The machine will be replaced with
a duplicate and the firm will accumulate the necessary capital by making equal
end-of-year deposits in a reserve fund. lf money is worth 6% per annum,
determine the annual deposit.
A.
B.
c.
D.
P 367,890.12
P 366,062.33
P 365,089.34
P 364,890.43
170
Quest,ion
llank
Annuity
lt)ngincering !)conornics by Jairrru It. 'l'iorrg
This problem is similar to the previous problem only that all amounts this
time is to earn an interest of 6%.
Cost of the machine at present = 1 ,200,000
Let C = Cost of machine 5 years from now:
C = 1,200,000
(1
, a car costing P 150,000 is to be purchased
ad of paying cash. lf the buyer is required to pay
4,000 each month for 4 years, what is the
effective interest rate on the diminishing balance?
A.
B.
c.
F = P(1+ i)n
+ o. 1 3)s
D.
C =2,210,922.215
35.28%
35.82o/o
34.89%
34.29o/o
Note that the interest 13% used was actually the sum of the interest rate
and the inflation rate.
Balance = 1 50,000
Balance = 1 10,000
Salvage value at present = 80,000
-
40,000
Let SV = salvage value of the machine 5 years from now:
F = P(1+ i)"
SV = 80,000(1+ 0.13)5
SV =
147,394.81
. ira A.r.rE
Referring to the constructed cash flow:
i
0
F + 1 47,394.81
:
2,210,922.21 5
F = 2,063,527.40
A AA,4AAAAAAAA A
:..,........................,.,....-
F
n[(r*i)"
P=
I
2,063,527.40 =
-r]
0.06
147,394.81
4,ooo[(1+
110,000=
iffi
By trial and error:
A = 366,062.33
i= 0.0255
.'. The annua! deposit is P 356,062.33.
-r]
+ i)"
i(1
I
n[{r*o.oo)u
AA,UAAAAAAA
AAAAA
But,
171
.NR
m
0.0255 =
!B
12
NR = 0.306
NR = 30.6%
i)4(12)
- 1]
AAA,+4AAAAAAA
172
Question llank -. llngineering Economics by Jaime R. ,l,iong
Arrnrrity 173
Converting into effective rate:
ER=(1*!)'-1
Present worth = Pr + Pz
Present worth = 1,277,O34.56 + '148,643.63
Present worth = 1,425,678.19
En=1,r+0.306)12_1
B.
12)
\
Leasing the building:
ER = 0.3528
ER = 35.28%
AAA
Effective rate is 35,28% or 30.6% compounded monthly
.'. The effective interest rate on the diminishing balance is 35.2g %.
Solving for the difference in present worth, D
Engr. Clyde Vasquez plans to purchase a new office building costing
P1,000,000. He can raise the building by issuing 10%,2}-year bond that would
D = 1,425,678.1
D = 233,779.27
I-
1,1
91,898.92
... The difference in amount between buying the building and leasing the
building is P 233,779.27.
A.
B.
c.
D.
P
P
P
P
233,779.27
233,070.12
JJ Construction Firm had put up for sale of some of their heavy equipments for
construction works. There were two interested buyers submitting their respective
'bids for the heavy equipments. The bids are as follows:
234,070.U
234,560.12
.V.l,*rn
P 10,000,000 PaY
nually for 8 years.
A.
the balance
Buying the building:
PaYa
the balance PaYable
00 payable P2,000,000
allY for 7 Years.
How much is the difference between the two bids if money is worth 10%
effective?
A. P 346,520.05
B. P 346,980.12
c. P 347,019.45
D. P 347,733.29
P2
-
C
------------:-
(t + i)"
C = 1.000.000
To compare the two bid, compare/their present worths:
P2
1,000,000
-Gor,
For Buyer A's Bid:
P2
= 148,643.63
Downpayrnent = 0.20(1 0,000,000)
Downpayment = 2,000,000
174
(lrrcstiou
llunk
l,)nginr-.ering lJconornir:s by.lairnc
ll
Anrruity
,l,rorrg
175
Solving for the semi-annual interest rate:
A's Bld
ER""r;.nnr"1 = 0 10
(1+i)2
-1=0.10
1+i =1.04881
i= 0.04881
hL)
2,000,000
n[1r*i;"-rl
i(1
+ i)n
0.04881(1+ 0.04881)14
Present worth of A's Bid = 2,000,000 + p
-+ Eq.
P = 4,987,192.91
1
Substitute the value of P in Eq. 2
P
D-
present worth of B,s bid = 2,000,000 + 4,997,192.91
Present worth of A's bid = 6,987,1 92.91
r]
' - l
1,ooo,ooo[(1+ o.ro)B
L'
Difference between two bids = 7 ,334,926.20
Difference between two bids = 347,733.29
0.10(1+ 0.'10)B
P = 5,334,926.20
Substitute the value of P in Eq.
-
6,987,192.91
.'. The difference in amount between the two bids is P 347,733.29.
1
Present worth of A's bid = 2,000,000 + 5,334,926.20
Present worth of A's bid = 7 ,334,926.20
Excel First Review & Training Center, lnc. is expanding its school facilities
starting 2001. The program requires the following estimated expenditures:
For Buyer B's Bid:
P 1,000,000 at the end of 2001
P 1,200,000 at the end of 2002
P 1,500,000 at the end of 2003
B's Bid
To accumulate the required funds, it establish a sinking fund constituting of 15
uniform annual deposits, the first deposit has been made at the end of 1992. The
interest rate of the fund is 2o/o per annum. Calculate the annual deposit.
A.
B.
c.
D.
2,000,000
P
<.............
.
Present worth of B's Bid = 2,000,000 +
p
>
l
t1 2
/).t
P 346,520.05
P 346,980.12
P 347,019.45
P 347,733.29
176
Question Bank
-
Engineering Economics by Jaime R. Tions
Annuity
Substitute values in Eq.
177
1
\21062A + A = 788,493.j2 + 927639.03 + 1,136,8j2.54
1
3.
1
062A = 2,852,944.69
A= 217,679.01
Solving for the balance on January 1 , 2002 (i.e. end of 2001)
The annual deposit is P 217,679.01
xcel First Review & Training Center, lnc. is expanding its school facilities
starting 2001. The program requires the following estimated expenditures:
P 1,000,000 at the end of 2001
P '1,200,000 at the end of 2002
P 1,500,000 at the end of 2003
Solving for annual deposit, A:
+A =P, +P,
Po
o_
F
(1
o_
+P,
+ i)"
1000,000
'1- (+o.W
P1
= 788,493.18
o_
1,200,000
Pz =
927,639.03
-+
Eq.
1
To accumulate the required funds, it establish a sinking fund constituting of 15
uniform annual deposits, the fiist deposit has been made at the end of 1992. The
interest rate of the fund is 2o/o pet annum. Calculate the balance in the fund on
January 1,2OO2.
A.
B.
c.
D.
P
P
P
P
2,185,902.11
2,195,600.03
2,165,399.54
2,175,380.00
.%2r.,.
'2-(1+orop'f
D _ 1'500'000
'3 - (1+ooa14
Ps = 1,136,812.54
o_
t4-
n[1r.i)'-r]
i(1
+ i)n
jPr,"i<"":.'""
p.
al
.
t.2M
<.............
"""""', l,M
.A
il,l
:l
,t
,89
'91
'93
'9s
',97
'99
'0t
03
178 fluestion llank -
l!ngineering li]conornics bv Jairnc ll.'l'iorrg
Annuity
Solving for annual deposit, A
-+
Po+A=P,,+Pr+P.
o_
^'
End of 2001, or
1
Jan. 1, 2002
F
(1
l-1i=-
Eq.
+ i)n
1,000,000
(1
+ O.02)''
q
= 788,493.18
^'
1,200,000
AA
A A A A A A A A A
A
i
(1+ 0.02)'"
Pz = 927,639.03
Balance=F-1,000,000
1,500,000
P3
(1+ 0.02)1a
P3
= 1,136,812.54
0.o2
e[1r*i1'-rl
nL)
r, =-
'
F = 3,195,600.03
i11+i1"
n[1r*o.oz;'o
P-LJ
'4
-
P4
=12.10624
Balance = 3,195,600.03 - 1,000,000
Ealancs.- 2,1 95, 600. 03
-rl
o.o2(1+0.02)la
Substitute values in Eq.
.'. The balance in the fund on January 1 ,
2OO2
is 2,195,500.03.
1
12.1062A + A = 788,493.12 + 927639.03 +
1 3. 1 062A = 2,852,944.69
A= 217,679.01
1, 1 36,81
2.54
J & N Pawnshop sells jewelries on either of the following arrangements:
Cash
price:
Discount of
1Oo/o
of the marked price
lnstallment: Downpaymenl of 20o/o of the marked price and the'balance
payable in equal annual installments for the next 4 years.
Solving for the balance on January 1 ,2002 (i.e. end of 200'1):
lf you are buying a necklace with a marked price of P 5,000, How much is the
difference between buying in cash and buying in installment? Assume that
money is worth 5%.
I
A.
B.
c.
D.
P 40.76
P 41.90
P4354
P 45.95
,%l*on
Compare their present worths:
179
180
Queslion
llank
Arrnui(,y
llngineoring Economics by Jaime R. 'l'iorrg
01
Cash basis:
The present worth of his earnings must be equal to the present price of the
property.
P = 0.90(5,000)
Prfue
P = 4,500
Present worth = 4,500.00
lnstallment basis:
I,A00A
A
A
A
P -"""""""""""""""""'..
Downpayment = 0.20(5,000)
Downpayment = 1,000
r=-
n[1r*i;"
hLJ
AAAAA
-rl
AA
PI
i(1 + i)n
r,ooo[(r+o.o.ef
P=
-r]
P2
0.05(l + 0.05)a
Pz
P = 3,545.95
Cash price = Pt +
Present worth = 3,545.95 + 1,000
Present worth = 4,545.95
Solving for annual net income, A:
Difference = 4,545.95-- 4,500
Difference = 45.95
Eq.
A=350,000-135,000
A = 215,000
.'. Buying in cash is economical by P 45.95.
Bon Fing Y Hermanos, lnc. has offered for sale its two-storey building in the
commercial district of Cebu City. The building contains two stores on the ground
floor and a number of offices on the second floor.
A prospective buyer estimates thai if he buys this property, he will hold it for
about 10 years. He estimates that the average receipts from the rental during this
period to be P 350,000.00 and the average expenses for all purpose in
connection with its ownership and operation (maintenance and repairs, janitorial
services, insuFance, etc.) to be P 135,000.00 He believes that the property can
be soid for a net of P 2,000,000 at the end of the 10'n year. lf the rate of return on
this type of investment is 7oh, determine the cash price of this property for the
buyer to recover his investment with a 7o/o return before income taxes.
A.
B.
c.
D.
P
P
P
P
2,526,768.61
2,490,156.34
2,390,189.00
2,230J20.56
181
21
5,ooo[(1 + O.oz)10
- r]
P1
0.07(1+ 0.07)10
P1
= 1,510,070.03
D_
F
'z-,r*y
D _ 2'000'000
'2 - (+o.W
P2 = 1,016,698.58
Substitute values in Eq.
1
Cash price = 1,5'10,070.03 + 1,016,698.58
Cash price = 2,526,768.61
.'. The cash price of the property is P 2,526,768.61.
h
182
Question Bank
-
Engineering Econornics by Jairne R. 'l'iorrg
Annuity
P=
E
,
(1
^r--
183
+ i)n
36.000
(1+ 0.10)
P =32,727.27
net present value of the low price strategy.
A.
B.
c.
D.
.'. The net present value of the high price strategy is P 32,721.27.
P 34,389.12
P 34,490.10
P 34,518.89
P 34,710.74
i(1
+ i)n
slq
20;
P=
D_
Jan Michael plans to purchase a new house costing P 1,000,000. He can raise
the building by issuing a 1Oo/o,20 year old bond that would pay P '150,000
interest per year and repay the face amount at maturity. lnstead of buying the
new house, Jan Michael has an option of leasing it for P 140,000 per year, the
first payment due one year from now. The building has an expected life of 20
years. lf interest charge for leasing is 12o/o, which of the following is true?
+ 0. 1oF -11
---"-=
20,000
0.10(1 + 0. 10)'
A.
B.
C.
D.
Lease is morei economical than borrow and buy.
Lease has same result with borrow and buy.
Borrow and buy is half the value than lease.
Borrow and buy is economical by almost a hundred thousand than lease.
34,710.74
,9;2,r.,.
.'. The net present value of the low price strategy is P 34'710'74'
Compare present worth of borrow and buy and lease..The Smaller the
present worth the better for Jan Michael.
For borrow and buy: (A = 150,000, i = '10%)
net present value of the high price strategy.
A.
B.
c.
D.
P 32,727.27
P 33,737.34
P 33,747.20
P 33,757.89
'/-/,2--
z=
150,ooo[(1 + o.r o)'zo
0.10(1+ 0.10)20
P1=1,277,034.56
- r]
184
Qucst,ion 13rrrrk I!)rrginotrrirrg llt:onornics by Jaime It. 'l'iorrg
Annuity
For lease: (A = 140,000, i = 12oh)
185
Gross income per year = 150,000
Expenses per year = 40,000 +
Expenses per year = 140,000
Net income per year = 150,000
Net income per year = 10,000
A
70,OOO
-
+
1O,OOO
+ 0.1 O(1 SO,OOO) + 5,000
14O,OOO
olz
Pz -'
AAAAAAA
;
P2 = 1,374,540.64
.'. The borrow and buy option is economical by 1,374,540.64
= P 97,506.08 than lease.
-
1,277,034.56
1
=' =
Pt
Engr. Harold Landicho is considering establishing his own business. An
investment of P100,000 will be required which must be recovered in 15 years. lt
is estimated that sales will be P 150,000 per year and that annual operating
expenses will be as follows:
Materials
P 40,000
Labor
P 70.000
P 10,000 + 10% of sales
P 5,000
Overhead
Sellino exDenses
Engr. Landicho will give up his regular job paying P 15,000 per year and devote
full time to the operation of the business. This will result in decreasing labor cost
by P 10,000 per year, material cost by P 7,000 per year and overhead cost by P
8,000 per year. lf he expects to earn at least 2Oo/o of his capital, is investing in
this business a sound idea?
A. Yes, it is a sound idea.
B. No, it is not a sound idea.
C. Neither yes nor no because it simply breakeven.
D. lt depends on the current demand of the market.
.VZ.*r,"
CASE 1: He does not resign from his job.
o,ooo[(1 + o.zo)15
/
L'
- r'l'l
0.20(1+ 0.20)15
= 46,754.73
CASE 2: He resigned his job.
Gross income per year = 150,000
The new annual operating expenses are now as follows:
Material
=40,000-7,000
33,000
Labor = 70,000 - 10,000== 60,000
Overhead = 10,000-8,000 + 15,OOO = 17,000 + 10% of sates
Selling = 5,000
Expenses per year = 33,000 + 60,000 + 17,000 + O.1O(150,000) +
Expenses peryear = 130,000
o
1 2
AAAAA
3
4
s
1t
AA
1s
S,OOO
186
Question Bank
-
Annuity
lt)ngineering Economics by Jaime R. '['iorrg
187
A=150,000-130,000
A = 20,000
A Civil Engineering student borrowed P 2,000.00 to meet college expenses
during his senior year. He promised to repay the loan with interest al 4.5 % in 10
equal semi-annual installments, the first payment to be made 3 years after the
date of the loan. How much will this payment be?
20,ooo[(1+ o.zo)15
Pz=
P2 =
A. .P 252.12
B. P 261.89
c. P 273.90
D. P 280.94
- r]
0.20(1+ 0.20)15
93,509'45
.'. lt is not a sound idea since the present worth of the income for both
case 1 and case 2 are less than the investment which is P 100'000.00.
A housewife bo\ght a brand new washing machine costing P 12'000 if paid in
cash. However, she can purchase it on installment basis to be paid within 5
years. lf money is worth 8% compounded annually, what is her yearly
imortization if all payments are to be made at the beginning of each year?
A.
B.
c.
D.
P
P
P
P
2,617.65
2,782.93
2,890.13
2,589.90
P=F
--) Eq.
n[(1.
P=
')"
(1
P=
1
- 1]
+ i)"
A AA
8.866A
I
I
Vi/,,2,-"
-+
Eq.
F = P(1+ i)n
1
F = 2.ooo( 1+
(
F
I
0'045
2) )5
=2,235.355
Substitute values in Eq.
'
P:
n[1r*o.oalo -r.l
J
L
O.OS(1+ 0.08)a
P=F
AAAA
P = 3.312A
Substitute values in Eq. 1:
'1
:
P ...'..".."""""""""""""i
12,000=A+3.3124
12,000 = 4.312A.
A --2,782.93
.'. The yearly amortization of the housewife is P 2'782.93.
8.866A =2,235.355
A = 252.12
The paymentis P 252.12.
188
Question Bank
-
Engineering Economics by Jaime R- '['iong
Annuity
A father wishes to provide P 4O,OO0 for his son on his son'sZ1"t birthday. How
much should he deposit every 6 months in a savings account which pays 3%
compounded semi-annually, if the first deposit was made when the son was 3 %
years old?
A.
B.
c.
D.
The investment that gives a bigger rate of return is the better investment
Compare the rate of return of the two situations:
A.
Domestic operation:
P 829.68
P 815.80
P 830.12
P 846.10
AAAAAAAAAA
0 1 2
40,000 =
l.
4 '..
2g
21
^[(,.T)"-,i
--j3-5
2
Pt + Pz =
8,000,000 +
Eq.
1
A = 846.10
.'. The deposit every six months in a saving account is P 846.10.
'
A group of Filipino-[4echanical Engineers formed a corporation and the
opportunity to invest P 8,000,000 in either of the two situations. The first is to
expand a domestic operation. lt is estimated that this
net year end cash flow of P 2,000,000 each year for 1
at
that time, the physieal assets, which would no longer
a
operating
and
involve
building
would
opportunity
The
alternative
P 5,000,000.
P-=
E
t
' ('t+i)"
^ 5,000,000
-'=
*,)*
Substitute values in Eq.
2,ooo,ooo[(1+ i)'o
is true?
A.
B.
c.
D.
Foreign operation yields bigger rate of return.
Domestic and foreign operations yield the same rate of return.
Domestic operation has double the rate of return of the foreign operation.
Foreign operation yields approximately 3% less rate of return than
- r]
-----* Dt.-.
By trial and error:
i= 0.2381
i
= 23.81o/o
1
Tffi
= 8,ooo,ooo
189
190
Question Bank
B.
-
Engineering Economics by Jaime It. 'l'iorrg
Annuity
Foreign operation:
A
th
at
at
8,000;000
I pay P 90,OOO cash, P
60,000 at
nnual payments starting with one
es as to the principal and interest
?yment which must be made for 6
years.
;
.
Pa .""1
A.
B.
c.
D.
A AAAAAA
P' 1."
P 66,204.14
P 65,701.67
P 67,901.34
P 68,900.12
:
-+ Eq.2
P+ = 8,000,000
Solving for P1:
4,ooo,ooo[(1+ i)? -
r]
P3
i(1
+ i)7
P,+P.+90,000=400,000
P1
4,ooo,ooo[(1+i)?
i(1
-r]
+ i)7 (1 + i)3
+ P. =
310,000
= 8,000,000
By trial and error:
i= 0.2080
i=20.80%
The foreign operation has a rate of return whlch ls 3% less than the
domestic operation.
I
191
P2
_
o[t,+o
oz)6
-r]
0.07(1+ 0.07)6
P2
=
4.7665A
--r Eq.
1
192
Quostion Bank
o- '3
_
-
[trnginoering Economics by Jaime
ll. 'l'iorrg
Annuity
F
(1
+Dn
15,ooo[(1+ o.oa)s
4,7665A
0.08(1+ 0.08)5
--
Ps =
- r]
P1
't - {t*0.97;'
I
193
P1
3'89094
Substitute values in Eq.
o-
F
'
'z-*y
1
52,4o6.32 + 3. 8ebeA
:
= 59,890.65
o
_ 59,890.65
'z-,r*ooty
:::::.,,.
Pz = 51,346.58
.'. The annual payment is P 66,204.14.
Fair prepayment = 51,346.58
Two years ago, the rental for the use of equipment and facilities as paid 5 years
in advance, with option to renew the rent for another years by payment of P
15,000 annually at the start of each year for the renewal period. Now, the lessor
asks the lessee if it could be possible to prepay the rental that wilt be paid
annually in the renewed 5 years period. lf the lessee will consider the request,
whatwould be the fair prepayment to be made to the lessor if interest is now
figured alSo/o?
A.
B.
c.
D.
P 51,84330
P 51,346.58
P 52,OO2.45
P 52,740.20
.'. The fair prepayment to be made to the lessor is P 51,346.58.
NCT Bui
an estim
.equal ye
n additional building at the end of 10 years for
00. To accumulate the amount, it will have
eaming 13%. However, at the end of the Sth
year, it was decided to have a larger building that originally intended to an
estimated cost of P8,000,000.00. what should be the annual deposit for the last
5 years?
A.
B.
c.
D.
,7"22*..
P 734,391.48
P 742,890jt0
P 738,900.45
P 740,010.86
Let P2 = fair prepayment to the lessor
n[{r*i)"
P=
i(1
-r]
F = 5,000,000
+ i)"
Solving for annual deposit to
accumulate 5,000,000 in 10
years, A:
AAAAA
n[{r*o.rs)'o
5,000,000 =
0.13
A =271,447.78
-r]
194
Question Bank
- Engineering Economics
by Jaime R' 'I'iong
Let B = last 5 annual deposits
-+ Eq.
Fz+Fs=9000,000
n[{r*i)"
F,r:
-r]
Question Bank
5
8,000,000
1
89
0
1
10
I
zt t,++t .tel(r
F,r=
+ o. r s)5
- r]
0.13
An investment, consisting.of deposits of P 1,000, P 1,500 and p2,000 are made
the end of the 2nd year, 3-'d year and 4th year, respectively. lf money is worth 10%
what is the equivalent present wo(h of the investment?
BBBBB
Fr = 1;759'055'07
F. =F,r(1+i)'
Fa
= 1,759,055.07(1+ 0.13)5
\
= 3,240,944.94
Ft t"'- """"n
i"""""""""'-
Fz
l7.r
A.
B.
c.
D.
P 3,129.89
P 3,319.45
P 3,372.12
P 3,490.09
e[1r*i;"-rl
J
L
E-
i
a[{r*o.ra)'-r]
Fz=
Fz
0.13
=6480278
Substitute in Eq.
1
6.480178 + 3,240,944.94 = 8,000,000
B = 734,391.48
.'. The annual deposit for the last 5 years is P 734'391.48.
Let P = Equivalent present worth
P =Pt
+Pz
where:
P1
P2
., _ c [1-
-+
Eq.
1
= present worth of uniform arithmetic gradient
= present worth of annuity
(1+
i)-" n
'l
"-tL--(1-if]
D _ soo [r-(r*0.10)-4 4
'1- oroL olo -1r;oro)il
1
a
196
Question Bank
Pr =
-
Ilnilbrnr Arithrnetic and Geometric Gradients 197
Engineering Economics by Jaime R. Tiong
2'189'06
^
"
0
Ps
500
500
1,500
(1+ 0.10)'
=1,126'97
2,000
D_
'6 -
500
o+0.'to)a
;
P6 = 1,366.03
Substitute values in Eq. 2
P = 826.45 + 1,126.97 +
1,
366.03
P = 3,319.45 it checks!
5oo
P3
[(1
+ o. r o)3
- r]
.'. The equivalent present worth is P 3,319.45.
0.10(1+ 0.10)3
P3
=1,243.42
o 'z -
Miss Calledo deposited P 1,000, P 1
3'd year and 4th year, respectively in
annum. How much is in the account
D
t3
11+i)
1,243.42
r^=' (1+ 0.10)
^
A.
B.
c.
D.
P2 = 1,130.39
Substitute values in Eq.
P 4,880.00
P 4,820.00
P 4,860.00
P 4,840.00
1
,177o/,.n.
P=2,189.06+1,130.39
P = 3,319.45
Let F = equivalent future worth
F=Fl+F2
-ct*,no* S/Zor*.
-+Eq.3
Using compound interest
formula:
where: Fr = future worth of uniform arithmetic gradient
P=Pa+P.+Pu -+ Eq.2
clrr+i)n -1 I
-i =rl-i-"1
Fz = future
F
.'' _- soo [(t+o.to)4 -t _ol
P4
1,000
P4
=-(1 + 0.10)'
P4
=826.45
worth of annuity
o.1o
Pa
I,500
Pt
P6
I o.to
Fl = 3'205
I
of the 2nd year, .
rned 10% per
198
Question Bank
-
Engineering Economics by Jaime R. Tiong
Llnifonrr Arithmetic and Geometric Gradients 199
Miss Calledo deposited p
500
500
sa0
>F2
Solving for F2
.t2---_ o[(l+i)" -r]
i
1,000,
respectivel
u
A.
B.
c.
D.
only?
P 670.81
P 690.58
P 660.53
P 680.12
-
5oo[(1+ o.ro)q - r]
_
Fz=
.
'
2nd year,
O% per
3'o year and 4h year,
annum. What is the equivalent
Let A = equivalent periodic payment for the
uniform gradient
o.10
n
n=cilI
Fz = 1,655
(1+ i)n _
L
Substitute valUes in Eq. 3
n=sool1
F=3,205+1,655
-
I
1.1
4
1
10.10 (1+0.r0)4 _1_l
F = 4;860
A = 690.s8
The equivalent uniform deposit is p 690.5g.
-M,& 9"2*r^
F=2,000+Fa+Fr -+ Eq.4
Fe
= P(1+ i)n
F3
=
1,
An amortization of a debt is in a
years, p 4,500 on masecono
fourth year. What is the
ye
equivale
500(l + 0.10)1
A. P 15,093.28
B. P 15,178.34
c. P 15,890.12
Fs = 1'6so
Fg = 1,000(1 + 0.1 0)1
Fz
D.
=1'210
Substitute values in Eq. 4
F
:2,000
+ 1,650.+ 1,210
F = 4,860 it checks
!
.'. The equivalent future worth is P 4,860.00.
P 15,389.82
ffil;
the
f
J3""11T"
I
Sy".
200
Question Bank
-
Uniform Arithmetic and G€ometric Gradients 201
Engineering Economics by Jaime R. Tiong
The cash flow in the previous page is equivalent to the figure below:
n.=c[1"
'
(1+i)"
Li
'i
rt
-<ir.r!...!.
,t.
ll,,
I
'.
I.500
I
-1.1
1-l
10.05
(1+ 0.05)"
- 1.1
Ar = 719'53
Let A = equivalent uniform payment of the. amortization:
A:5,000-A1
A=5,000-719.53
A = 4,280.47
s,000 5,000 s,000
s,000
:
P
-+
=Pt-Pz
Eq.
1
P,r=
,
4280.474280.474,280.47 4,28.0.47
:
7
P;:,<"""""'
=17,729.75
n-.|
i -fl.'r]
o_c[t-1t*i;"
''-TL
D_soo[t-1t*o.os)4
''-
oosL oJ5
Pz =
P = 17,729.7 5
:
1
(1+oGf]
2'551'41
^'
4,280.47l
P = 1 5,1
Substitute values in Eq.
P
4
-
1
2,55',t.4',1
15,178.34
.92-24,. ,%:;*,.".
Let 41 = equivalent uniform payment due to uniform arithmetic series only
T.
.a
1
(t+o.os) -t.1
O.O5(1+ 0.05)a
78.3a it checks
I
The equivalent present worth of the debt is P tl5'178.34.
"""""""j
Unilbrm Arit,hnrot,ic and Geometric Gradients 203
2OZ Question Bank - Engineering Econnmics by Jaime R. 'l'iong
An amortization of a debt is in a form of a gradient series of P 5,000 on the first
year, P 4,500 on the second year, P 4,000 on the third year, P 3,500 on the
fourth year. Determine future amount of the amortization if interest is 5%.
A.
B.
c.
D.
.t'l"Zz,i,u.
P
P
P
P
18,030.56
18,290.12
18,621.89
18,449.37
n[1r*i;'-rl
J
L
oI1-
i(1+i)n
_
t1-
s, ooo
+ o.os)a
[(1
- r]
o.o5(1+0.05)a
P1=17'729.75
F, =P,(1+i)"
Ft = 17,729.75(1+ 0.05)a
1=21,550.62
_c[t-(t+i)*o
''-iL
-l
"
* o.o5)-4 4 I
-' - 5oo [t - (t o.os
(1+ o.o5)']
' o.o5 [
i
Pz
(1+i)".1
=2'551'41
Fz = Pz(1+ i)n
Fz=2,551.41(1+0.05)a
Fz = 3'1O1'25
Solving for future worth, F:
F
=1-Fz
-+ Eq. 1
Substituting values to Eq.
1:
F =21,550.62-3,101.25
F = 18,449.37
.'. The future amount of the amortization is P 18'449'37'
ffiofadebtisinaformofagradientseriesofP5,oooonthefirst
g:. f :1 ?^what
s"Lo no vqa r,. P
G;; ilEoil; ",h "tn"
1;999 :i -11i, pavment
l;lll"is ?l"tn
the "
equiialent u1it9r1 periodic
i;dh'G;;. wn"ri.
is 5%'
equirai"nt uniform periodic payment if interest
A. P 4,280.47
B. P 4,378.17
c. P 4,259.68
D.
P 4,325.12
204
Question Bank
-
tlnif<rrm Arithmetic and Geometric Gradients 205
Engineering Economics by Jaime R. Tiong
Comparf the equivalent annual salary of the two ball clubs:
7"2r.,.'-
For LoqiAngeles Lakers, annual salary = $ 5,000,000
Solvir/q for the equivalent annual salary of Chicago Bulls:
n
o.=c[1^t-"Li (r*D"-rl]
Ar=6oo,ooot#
\
f;jful]
=2,235,276.31
Let 41 = equivalent uniform payment due to uniform arithmetic series only
Total annual salary of the NBA superstar with the Chicago Bulls, A:
n
n.=c[]'
(1+i)"
A=3,000,000+Ar
Li
l
-1.1
' -
A = 3,000,000 + 2,235,276.31
+
A= 5,235,276.31
'l
o.=uoo[
'
10.05 (1+ 0.05)4 _ 1_l
... The offer of Chicago Buits is $235,276.31 more annually'than that
of LA
Lakers'
Ar = 719'53
Let A = equivalent uniform payment of the amortization:
A=5,000-41
A=5,000-719.53
John Grisham, author of the best selling novel "The Chamber" sold its
copyright to warner Bros. for the rights to make it into a motion picture. Mr.
Cristram's nas options between the following Warner Bros' proposals:
An immediate lump sum payment of $ 5'000'000.
B. An initial payment of $ 2,500,000 plus 4% of the movie's gross receipts for
the next 5 years which is forecasted as follows:
A.
A = 4,280.47
.'. The equivalent uniform payment is p 4,280.47.
End of year
1
2
3
4
5
A. Chicago Bulls offer is smaller than that of l-A Lakers
B. Chicago Bulls offer is exactly the same as l-A Lakers,
chicago Bulls offer is just few dollars more p6r year than that of l-A Lakers,
ID. Chicago
Bulls offer is
per
over $150,000
year than that of LA Lakers,
Gross Receipt
$ 10,000,000.00
4% of Gross ReceiPt
$ 400,000.00
$ 8,000,000.00
$ 6,000,000.00
$ 4,000,000.00
$ 2,000,000.00
$ 320,000.00
$ 240,000.00
$ 160,000.00
$
80,000.00
lf money is worth 10% and Mr. Grisham will not receive any royalty after the fifth
year of Lxhibition of the movie, by how much is proposal A bigger than proposal
B?
A.
B.
9"2r2.,,
P 1,532,630
P 1,390.090
c. P 1,478,100
D. P 1,289.450
206
Question Bank
-
Llnifurrn Alit,lrrnotic antl Geometric Gradients 207
Engineering Economics by Jaime lt. 'l'iorrg
Proposal A will provide the author a present worth of g 5M.
Solving for the present worth of proposal B:
Let 41 = equivalent uniform payment of the uniform arithmetic gradient
n I
n.=c[1(1+i)"
-1l
Li
,1-
A.=8o,ooo[
10.10
What is thq pregent worth of
the given cash flow diagram
if interest rate is i10%?
i
5=
I
(1+0.10)5 _1j
A. P 2,200.34
B. P 2,089.37
c. P 2,191.49
D.
P 2,119.49
Al :144,810.08
Let A = equivalent uniform payment of proposal B:
A = 400,000 -'144,81 0.08
A = 255,189.92
680.24448
This is an example of uniform geometric gradient since the next term in the
series is a fixed percentage of the previous term.
Solving for the fixed percentage, 1+ r:
.
1+l=-
540
500
'l+r=1.08
Solving for the present worth of the royalty
Thus, r = 0.08
Since r is not equal to i, then x + 1 . Use the following formula:
D_
P=
zss,r as.sz[(r + o.ro)5
G f1-x")
o_
'
1+i[ 1-x
- r]
0.01(1 + 0.1 0)5
1+r
P =967,370.57
1+i
1+ 0.08
Total present worth of proposal B = 2,500,000 + gOT,37O.S7
1+0.10
= 3,467,370.57
Solving for the difference between the two proposals:
Difference =
Difference --
5, 000,
1,
J
000
- 3,467,37
0.57
532, 629.4
Proposal A is P 1,532,630 bigger than propoeal B.
x=
0.9818
-r =
5oo
Ir - o.osreu I
u*s!10)L 1-or818 ]
P =2,191.49
.'. The presentworth of the given cash flow is P 2,191.49'
D
208
Question Bank
-
Engineering Economics by Jaime R. Tiong
1
[lrrrlbrrrr Arit,ltrno[it: ttnd (]oomettic Gradients 209
Solving fQr its future worth, F:
\
What isithe equfualent future
worth of'the-gi'ven cash flow
diagram if interest rate is
F = P(1+ i)\
10o/o?
F = 3,529.54"
A. P 3,529.54
B. P 3,259.34
c. P 3,925.64
D.
F
o
=2,19L+qt*O.tO;u
.'. The equivalent future worth of the given cash flow is P 3'529'54'
P 3,295.54
7;zt-,,
the other sister?
This is an example of uniform geometric gradient since the next term in the
series is a fixed percentage of the previous term.
A.
B.
C.
Jocelyn, P 671.'18
Jocelyn, P 763.27
Joan, P 671 .18
Solving for the fixed percentage, 1+ r:
1+r=540
,9oz*r.,.
500
'l+r=1.08
To compare, solve the future amount of the two investments:
Thus, r = 0.08
For Joan:
Since r is not equal to i, then x + 1. Use the following formula:
G (t-r"')
D_
' - t+rIr-r.1
x=-1+r
1+i
x=
1+ 0.08
1+ 0.10
x = 0.9818
'_
D
5oo
[r - o.sar au I
0*olo)L 1-oss1s l
P =2,191.49
2,357.9s
Solving for the fixed percentage, 1+ r:
1+r -
1.100
--:-1,000
1+r=1.10
210
Question Bank
-
L.lnilbrnr Arithmotic and Goometric
Engineering Economics by Jaime R. Tiong
Gradients
211
Thus, r = 0.10
Since r = 0.10 is not equal to I = 0.14, then x+ 1. Use the following formula:
- c (t-r")
H=-t
1+i[ 1-x
J
-l
1+r
x=
1+O.14
_
1t{-e!!!!!
A.
B.
c.
1+i
1+0'10
D.
x = 0.9&t9
o
Engr. Herqro, be\eving that life begins at 40, decided to retire a1{
life-at ase +0. neVisnis to have upon his retirement the sum of P 5,000,000. On
his 21"t-birthday, he deposited a certain amount and increased his deposit by
15% each yeai until he will be 40. lf the money is deposited in a super savings
account which earns 15% interest compounded annually, how much was his
initial deposit?
1,000 [t - o.so+s,o I
(1.oJOL 1-0s64, ]
P 17,253.18
P 17,566.33
P 17,672.77
P 17,490.21
P =7,508.45
since the first deposit was made when Engr. Hermo was 21, then the value
of n is 20.
F = P(1+ i)n
r --15%
F = 7,508.45(1+ 0.14)10
F=
i=15%
27,835.48
Since, r = i, then
For Jocelyn:
o'
Gn
,=
= 1. Use the following formula:
H
+Eq.1
1+r
Solving for the present worth of P 5,000,000'
E
o-
'
('1
+ i)"
_
5.000.000
P==
(1 + 0.1
5)"
P = 305,501.39
F-
Substitute values in Eq.
305,501
1
3e=dtrb
0.14
F =27,072.21
Joan has P 763.27 more than Jocelyn.
G:17,566.33
.'. Engr. Hermo's initial depositwas P 17'566.33.
212
Question Bank
-
Engineering Economics by Jaime R. Tiong
Question Bank 5
A newly-agq.uired equipment requires an annuar maintenance
costs of p 1or0o0.
lf the annubt rndintenance cost is increased by 2oo/o each
v"r,
v"ar for 10
years, what is the estimated present worth of ihe maintenince
costs it money is
"*w
worlh 15Yo?
A.
B.
c.
D.
P',t0,s,712.33
P 106,101.37
P 107,490.12
P 108,890.11
r=
2OVo
i=
15o/o
A.
B.
c.
D.
x=1+r
,
A company issued 50 bonds of P 1,000.00 face value each, redeemable at par at
the end of '15 years. To accumulate the funds required for redemption the firm
established a sinking fund consisting of annual deposits, the interest rate of the
fund being 4%. What was the principal in the fund at the end ol ,l2h year?
P 37.002.54
P 37,520.34
P 38,010.23
P 38,782.34
9;z*r...
1+i
_1+ O.20
Amount needed by the company after fifteen years = 50(1,000) = 50,000
1+ 0.15
x =1.04348
LetA=annualdeposit
c
D_
' - 1-l[t-r")
i-r.,J
1o,ooo
(t
-t.o+s+a,o')
' _- 1*ols[-1-l.rrn4l
)
D
P = 106,101.37
AAAAAAAAAAAAA
.'. The estimated presentworth of the maintenance cost is p l06,1ol
.37.
AA
:
n[{r*i)'-r]
,F=
But F = 50,000
Substituting values:
n[1r*o.o+';'u
50,000
-r'l
A=2,497.06
J
2'14 Question Bank - Engineering Economics
by Jaime R. Tiong
Bonds
Solving for the principal at the end of 12th year.
0
2
4
6
8
1011
12
A P 1,000,000 issue of 3%, 1S-year bond was sold at 95%. What is the rate of
A.
B.
C.
D.
3.O
o/o
3.4%
3.7
o/o
4.O
o/o
n[tr*i)"-r]
F'rz
=
I
24e7.06[(1+ o.o+)1'?
Ftz =
,, _rrl(+i)"
-r]-(1+ir
c
+Eq.1
"n-- ;6*y
- r]
0.04
Fe = 37,520'34
.'. The principal in the fund at the end of the 12th year is P
37
,520-34.
A local firm is establishing a sinking fund for the purpose of accumulating a
sufficient capital to retire its outstanding bonds at maturity. The bonds are
redeemable in 10 years and their maturity value is P 150,000.00. How much
should be deposited each year if the fund pays interest at the rate of 3%?
A.
B.
c.
D.
P
P
P
P
13,084.58
13,048.85
13,408.58
13,480.58
Zr Zr Zr Zr Zr Zr Zr Zr Zr Zr Zr Zr
Vn =
Vn
0.95(1,000,000)
= 95,000
C = 1,000,000
.V"z*...
Zr = 1,000,000(0.03)
Zr = 30,000
Substituting values to Eq.
150,000
0.03
A = 13,084.58
.'. The annual deposit is P 13,084.58.
zr[1r * i)"]
,"" =-lnY
-
95,000
30, ooo
[(1
* iI
1
c
(1+i)n
u
_
------+
i(1+ i)15
1]
,,,
ooo, ooo
-11;
Dl5-
21 5
216
Question Bank
-
Bonds 217
Engineering Economics by Jaime R. Tiong
By trial and error:
i:0.03
i
%=
=3Yo
.'. The rate of interest of the investment is 3%.
\
\
vn = 90,614.93'.
.'. The man must pay the ceilificate the amount of P 90,614.93.
A man was offered a Land Bank c,ertificate with a face value of P 100,000 which is
bearing interest of 6% per year payable semiannually and due in 6 years. lf he
wants to earn 8o/o semiannually, how much must he pay the certificate?
A
P 1,000 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. lt is sold at p1,030.
Determine the yield at this price.
A. P 90,123.09
B. P 90,614.93
c. P 90,590.12
D. P 90,333.25
A. 7.56%,
B. 7.65
c. 7.75%
o/o
9"2a.".
D.
,,v"=-
zrf(r*i)" -1]
,
c
;1'1*;y -(1+ir
-+ Eq.
7.86%
1
Zr ZrZr Zr Zr Zr Zr Zr Zr Zr
Zr ZrZr Zr Zr Zr Zr Zr Zr Zr Zr Zr
,, _rrl(*i)'-1]
, c
-ll;;n--,r*;y,
-+Eq
"n
Zr = 1,000(0.08)
Zr=80
Zr = 3,000
C = 1,000
Substitute values in Eq.
i;^l
1
zrf1r *
c
:vn-______!__________ '(1+i)"
i(1+i)"
\,
Substitute the values in Eq.
., _zr[{r*i)"]
, c
--@!n--(t*n
"n
1
1
I
2'l8
Question Bank
-
Bonds 219
Engineering Economics by Jaime R. Tiong
i
r,
oooto.oe)
'
r..dao =
kr+
i)10
i(1+i)"
- r]
+
- 0.0385
i= 3.85%
1009_
(1+i)''
of interest is 3.85 %.
By trial and error:
i= 0.0756
i--7.56%
A P1,000, 6% Uond pays dividend semiannually and will be redeemed al 11Oo/o on
June 21, 2004. ll is bought on June 21,2OO1 to yield 4% interest. Find the price of
the bond.
The yield is 7.56 %.
You purchase a bond at P 5,100. The bond pays P 200 per year. lt is redeemable
for P 5,050 at the end of 10 years. What is the net rate of interest on your
A.
B.
c.
D.
P
P
P
P
1,122.70
1,144.81
1,133.78
1,155.06
investment?
A.
B.
c.
D.
3.56%
3.85
o/o
Fr=1,ooo[ry)
3.75%
Bought on
Will be redeemed
lune 21, 2001
June 21, 2004
3.68 %
Fr=30
,V/or,r-
Vn: 5,100
30 3A
30
I
I
I
200 200
C = 1,100
200
I
I
V" = 1,144'81
C = 5,050
.'. The price of the bond is P 1 ,144.81.
220
Question Bank
-
Engineering Economics by Jaime R. 'l'iorrg
Question Bank
,-1;;
7
A voM has a sellingprice of P 400. lf its selling price is expected to decline at a
rate of 10o/o per 4nrfum due to obsolescence, what will be its selling price after 5
years?
A. P 222.67
B. P 212.90
c. P 236.20
D.
P 231.s6
Using Matheson Formula:
k=',-,8
\lco
o.1o=1-FE
v
400
cu
:(o.gou
4o0/ *T
Cs =236'20
.'. The selling price at the end of the
Sth
year is p 236.20.
A machine costs P 8,000 and an estimated life of 10 years with a salvage value of
P 500. What is its book value after 8 years using straight line method?
A.
B.
c.
D.
P
P
P
P
2,000.00
2,100.00
2,200.00
2.300.00
,9zo*Solving for annual depreciation charge, d
222
Question Bank
-
l)trprcr:rnl iorr arrd Depletion 223
tr)ngineering Economics by Jaime R. 'l'iorrg
. Co-Cn
d=n
.
ABC Corporation makes it a policy that for any new equipment purchased, the
annual depreciation cost should not exceed 20% of the first cost at any time with
no salvage value. Determine the length of service life necessary if the depreciati
used is the SYD method.
8.000-500
10
d=750
Solving for book value after 8 years, C6
C. =Co-Dm
c8 =8,000-750(8)
Ce = 2,000
/
A. 7 years
,/
B. 8 years
C. '9 years ./
D. 10 years ./
The book value after 8 years of using is P 2,000.00.
ln SYD method, the largest depreciation is on the first year.
Analyze first year depreciation, d1
A telephone company purchased a microwave radio equipment for P 6 million,
freight and installation charges amounted to 4% of the purchased price. lf the
equipment will be depreciated over a period of 10 years with a salvage value of
8%, determine the depreciation cost during the 5"'year using SYD.
A.
B.
c.
D.
d, =(co
-.")r,:"*
But:rvear=n(n*1)
2
P
P
P
P
626,269.09
d, =(co
623,209.09
625,129.09
624,069.89
d" =
o 2oc^ = c"
"
Solving for d6preciation during the 5th year, ds
n=
I
years
The length of service life necessary is 9 years.
But
Years =
r-r)
"[n+1]
n+1=10
-..)rEk
-
c^r-?-)
"In+1]
But: d1= 0.20C0
6,000,000 + 0.04(6,000,000) = 6,240,000
Cn= 0.08(6,240,000) = 499,200
Co =
=(co
^
2
../;/;r*,,
du
-o)--!
n(n+1I
-:=-
company purchases an asset for P 10,000.00 and plans to keep it for 20 years.
the salvige value is zero althe end of 20th year, what is the depreciation in the
third year? Use SYD method.
2
Substituting,
d5 = (6, 24o,ooo
- 499'2OO)!#
d5 = 626,269.09
The depreciation cost during the
Sth
year is P 626,269.09'
A.
B.
c.
D.
P 857.14
P 862.19
P 871.11
P 880.00
224
l)oplor:urliorr rrrtt.l l)cplution 225
Engineering Economics by Jaime R. 'l'iong
-
Question Bank
The total depreciation in the firCt three years is P 48,000.00.
'-/,./u1,;',,
Solving for depreciation at the end of 3'd year, d3
-+ Eq.
1
A machine has an initial cost of P 50,000 and a salvage value of P10,000 after 10
years. What is the book value after 5 years using straight line depreciation?
A.
B.
c.
r
vears -2o(2o
2
+1)
- r,o
Substitute values in Eq.
D.
P 30,000.00
P 31,000.00
P 30,500.00
P 31,500.00
1
Solving for the annual depreciation charge, d
q
=(1o,ooo
-o)4-3
6=
ds =857'14
Co-Cn
n
.50.000-10.000
.'. The length of service life necessary is 9 years.
'10
d = 4,000
Solving for the book value at the end of 5th year, Cs
An asset is purchased for P 500,000.00. The salvage value in 25 years is
P100,000.00. What is the total depreciation in the first three years using straightline method?
A. P 45,000.00
B. P 46,000.00
c. P 47,000.00
D.
Cs=Co-Ds
cs =50,000-4,000(5)
c5 = 3o,ooo
The book value after 5 years is P 30,000.00
P 48,000.00
?).,r^n.
Solving for the annual depreciation charge, d
c^
-c_
n
.
ct =
500.000
-=-
- 100.000
'
25
d = 16,000
Solving for the total depreciation in the first three Years, Dg
Ds = 16,000(3)
Ds = 48'000
An asset is purchased for P 9,000.00. lts estimated life is 10 years after which it
will be sold for P 1,000.00. Find the book value during the first year if sum-ofyears' digit (SYD) depreciation is used.
A.
B.
c.
D.
P 7,545,45
P 7,320.11
P 7,490.00
P 7,690.12
V/,,t;"n..
Solving for depreciation on the first year,
d,:(co-c")=+.years
z
d1
226
Question Bank
-
Engineering Economics by Jaime R. Tiong
Deprocintion and Depletion 227
But
)
vears
r
vears
n(n +
-
Cr=Co-D,
cr =530,000-96,000(4)
1)
2
1o(10 + 1)
2
-
Ca = 146'000
au
.'. The book value at the end of the 4th year is P 146,000.00
Substituting,
d1
=(9,000-1,000)#
dt =1,454'54
Solving for the book value after first year,
C1
C' = Co -d1
cr = 9,00o- 1,454.54
q
--7,545.45
c.
D.
P 1,344.21
P 1,245.45
.V;z;r,,,,
The book value during the first year is P 7,545.45.
F
Let dr = first year depreciation
d, = (co
The cost of equipment is P 500,000 and the cost of installation is P 30,000. lf the
salvage value is 10o/o ol the cost of equipment at the end of 5 years, determine the
book value at the end of the fourth year. Use straight-line method.
A.
B.
c.
D.
P
P
P
P
146,320.50
146,000.00
146,230.50
146,023.50
-.")r#"*
r
vear
)
vear =
)year
-
n(n +
1)
2
10(1'1)
------:-----!'
2
= 55
Substituting;
Vt,,,,,,
d1
Co = 500,000 + 30,000 = 530,000
Cn = 0.10(500,000) = 50,000
Solving for the annual depreciation charge, d:
.
= (2o,ooo
- 12'ooo)r+l
't55/
dt = 1'454'54
.'. The depreciation for the first year is P 1,454,54.
c^-c_
n
_ _ 530,000-50,000
A machine has an initial cost of P 50,000 and a salvage value of P 10,000 after
10 years. Find the book value after 5 years using straight-line depreciation.
5
d = 96,000
Solving for the book value at the end of the fourth !ear, Ca
A. P 31,000.00
B. P 31,500.00
c. P 30,000.00
D.
P 30,500.00
228
Question Bank
l)ngineering Econornics by Jaime R. 'l'iorrg
-
l)o;rrocint,iort and
Depletion 229
.'ll,/,r.,
Yo=
Let Cs = book value after 5 years
o/o
Solving for annual depreciation charge, d:
xtoo
= 8ok
.'. The straight line method depreciation rate is 8%.
aa
r-"0
4'ooo
50,000
vn
n
_
- 10,000
50,000
A machine costing P 45,000 is estimated to have a book value of P 4,350 when
10
d = 4,000
Solving for Cs
:
Cs=Co-Ds
cs =50,000-4,000(5)
B. 32.25o/o
32.00%
D. 32.75o/o
c.
C5 = 30'000
.'. The book value after 5 years of using is P
3O,OOO.OO
k=1-"8
!co
A machine has an initial cost of P 50,000.00 and a salvage value of P 10,000.00
after 10 years. What.is the straight line method depreciation rate as a percentage
of the initial cost?
A.
B.
c.
1O
\145,000
k = 0.3225
k = 32.25o/o
o/o
8%
7%
D. I
k=1-u@
.'. The annual rate of depreciation is 32.25
o/o
o/o.
V;1itr..
Let d = annual depreciation
d-co-cn
n
,{ _ 50,000
-
10,000
10
d = 4,000
Solving for the percentage of depreciation with respect to the initial cost:
%=
4x100
co
An asset is purchased for P 120,000.00. lts estimated life is 1 0 years, after which
it will be sold for P 12,000.00. Find the depreciation for the second year using the
sum-of-years' digit method.
A.
B.
c.
D.
P 17,578.13
P 17,412.43
P 17,344.67
P 17,672.73
V;z*,,,
d, =(co
-c,).4
jyears
230
Question Bank
-
n(n +
)
vears
)
vears
)
years = 55
Depreciatiort and Depletion 231
Engineering Economics by Jaime R. Tiong
1)
2
10(1
An asset is purchased for P 9,000.00. lts estimated economic life is 10 years
after which it will be sold for P 1,000.00. Find the depreciation in the first three
years using straight line method.
1)
2
/o\
+
'\55/
d2 = (120,000-12,000)l
I
dz =17,672'73
A. P 2,400.00
B. P 2,412.34
c. P 2,250.OO
D.
P 2,450.00
.%7i*--.
.'. The depreciation for the second year is P 17,612.73.
A machine costing P 720,000 is estimated to have a book value of P 40,545.73
when retired at the end of 10 years. Depreciation cost is computed using a
constant percentage of the declining book value. What is the annual rate of
depreciation in %?
A.
B.
C.
D.
25
26
27
d=800
o/o
Solving for total depreciation in the first three years, D:
o/o
o/o
D = d(3)
28%
D = 800(3)
D
.V;/;r,,Using Matheson Formula
=2,400
.'. The total depreciation in the firct three years is P 2'400.00.
k=1-"8
lco
Y
=
1-.o@'il\n
i/
720,000
k = 0.25
k=25%
.'. The annual rate of depreciation is 25%.
An engineer bought an equipment for P 500,000. He spent an additional amount
of P 3O,0OO for installation and other expenses. The estimated useful life of the
equipment is 10 years. The salvage value is x % of the first cost. Using the
stiaight line method of depreciation, the book value at the end of 5 years will be
P 291,500. What is the value of x?
A.
B.
O.2
0.4
c. 0.3
D.
0.1
Let Cs = book value at the end of 5 years
Cs=Co-Ds -r
Eq'1
232
Question Bank
-
Depreciation and DePletion 233
Engineering Economics by Jaime R. Tiong
Co :500,000 + 30,000
co = 530,000
n
d
Solving for annual depreciation, d:
-
800.000
---ro
- 35,000
d = 76,500
,..-Co-Cn
The annual depreciation charge is P 76'500'00'
n
r_
530,000-x(530,000)
10
d:53,000-53000x
Total depreciation for 5 years, Ds:
""l"rri"
iS
Ds = d(5)
D5
The
The initial cost of a paint sand mill, including its installation, is P 800,000.
Tle
years
for
10
is
depreciatio.l.'
Bli approved tife oi this machine
"tl]'?lg9;
ill is P 50,000 and the cost of dismantling is estimated to
of the
value
book
is
the
what
p
aight-line depreciation'
O"
of
six
end
machine at the
Years?
=(53,000-53,000x)5
Book value at the end of 5 years, Cs
D.
Cs=Co-Ds
291,500 = 530,00
A. P 341,000.00
B. P 343,Q00.00
c. P 340,000.00
-
P 342,000.00
5(s3,000 _ 53,000x)
- 53,000x) = 238,500
53,000 - 53,000x = 47,700
5(53,000
53,000x = 5,300
x=0.1
Co = 800,000
Cn = 50,000- 15,000 = 35,000
Solving for annual depreciation charge, d:
.'. The value of x is 0.1.
n
The initial cost of a paint sand mill, including its installation, is P 800,000. The
BIR approved life of this machine is 10 years for depreciation. The estimated
salvage value of the mill is P 50,000 and the cqst of dismanfling is estimated to
be P 15,000. Using straight-line depreciation, what is the annual depreciation
charge?
A.
B.
c.
D.
P 75,500.00
P 76,000.00
P 76,500.00
P 77,000.00
.V.z*-.
Co = 800,000
Cn = 50,000- 15,000 = 35,000
Solving for annual depreciation charge, d:
. 800,000 - 35,000
o=
io
d = 76,500
Solving for book value at the end of 6 years' C6
Co = Co
-Do
co =800,000-76,500(6)
Co = 341'000
The book value at the end of the 6th year is P 341,000'00'
234
Question Bank
-
Deprocial,ion and DePletion 235
Engineering Economics by Jaime R. Tiong
6=Co-Cn
n
A unit of welding machine cost P 45,000 with an estimated life of 5 years. lts
salvage value is P 2,500. Find its depreciation rate by straight-line method.
A.
B.
c.
D.
. 410,000 - 60,000
-20
18.89 %
d = 17,500
19.21%
19.58%
years, cro
Solving for the book value (appraisal value) at the end of 10
19.89 %
Cls=Co-D16
cro = 41o,0oo - 10(17,500)
Cro'= 235' o0o
6=Co-Cn
...Theappraisalvalueofthemillattheendofl0yearsisP235,000.00.
n
,_45,000-2,500
.J
E
d = 8,500
10
An equipment costs P 1O,OOO with a salvage value of P 500 at the end of
years. catcutate the annual depreciation cost by sinking fund method at 4 %
Depreciation rate = -9
interest.
co
Depreciation
,"t" = 9:590 x
45,000
loOo/o
Depreciation rate = 18.89 %
A. P 234,000.00
B. P 235,000.00
c. P 234,500.00
D.
P 235,500.00
.'. The depreciation rate is 18.89 %.
.V;z,t*,,
d = 791.26
A. P 234,000.00
B. P 235,000.00
c. P 234,500.00
D.
P 235,500.00
Co = 360,000 + 5,000 + 25,000 + 20,000
C" = 410,000
Solving for annual depreciation charge, d
.'. The annual depreciation cost is P 791'26'
year by 20% of its
A machine initially worth P 50,000 depreciates in value each
it is 9 years old.
when
its
book-value
year.
Find
of
that
value at the beginning
A. P 6,710.89
B. P 6,400.89
c. P 6,666.89
D. P 6,512.78
236
Question Bank
-
L)oprecial,ion and Depletion 237
Engineering Economics by Jaime R. Tiong
.V,z"arsince the percentage of depreciation is constant every year to be 20%
of
the book value, use the Matheson Formula
*=1_.8
!Co
o.2o=1-'E
50,000
life is 10 years
An asset is purchased for P 9,000.00. lts estimated economic
in the first three
depreciation
the
Find
1,000.00.
P
aftei wnicn it will be sotd for
method'
digit
sum-of-years'
years using
A. P 3,279.27
B. P 3,927.27
c. P 3,729.27
v
D.
Vffi=ouo
722*-
,tn
50,000
P 3.792.72
Solving for depreciation in the first three years, Da
= 0.80s
Cg = 6,710.89
.'. The book value when it is 9 years old is p 6,7i0.g9.
A consortium of internationaltelecommunication companies contrActed
for the
purchase and installation of fiber optic cable linking [ianila
citv anaCeou city at
a total cost of P 960 million. This amount includesireigtrt ana insiaiiaili:nehaiges
estimated at 10% of the above total contract'price. lf fie cable shal
be
depreciated over a period of 15 years withzero sarvage varue anJ money
is
worth 6% per annum, what isthe annual depreciation-charge?
A. P 41,04p'903.40
B. P 41,211,158.40
c. P 41,254,000.40
+
'"t':''l=55
- 10(10
1)
=
(e,000-1,ooo)*
= 1,454'54
-"")thk
=
(e,ooo-1,ooo)*
= 13oe.oe
-.,,)k
= @,ooo
dl = (co
-.^)t**"
d, = (co
d. = (co
Substitute in Eq.
P 41,24.253.40
1
)years=ry
lyears
D.
-+ Eq.
Dg=dr+dr+d.
- 1,000)*
= 1163.64
1
Ds =1,454.54+ 1'309.09 + 1,163'64
Ds =3,927 '27
since a rate of interest is given in the probrem, use qinking fund method.
.'. The depreciation in the first three years is P 3,927 '27 '
Solving for annual d€preciation charge, d
,n_(Co-Cn)i
(1+i)" -1
d
_ (s60,000,009 - 0x0.06)
(1+ 0.06)15
-
1
d= 41,244,253.40
.'. The annual depreciation charge is p 41,241,253.40.
resale value at the end of
A radio service panel truck initially cost P 560,000. lts
ui"irriir" is esiimated.at P 150,000 ?J T:iT,"l.declining
il?'-G;;fli"
oatancemethod,determinethedepreciationchargeforthesecondyear.
A. P 99,658.41
B. P 99,128.45
c. P 99,290.00
D. P 99,378.45
238
Question Bank
-
l)r:prot:iat,iort anrl Depletion 239
Engineering Economics by Jaime R. Tiong
9Z**
efinery developed an oil well which is estimated to
s of oil at an initial cost of $ 50,000,000. What is the
the year where it produces half million barrels of oil? Use
computing dePletion.
Using the Matheson formula:
k=
1-"8
!Co
A. $ 500,000.00
B. $ 510,000.00
c. $ 525,000.00
k = 1- uE!g,ogq
\l 560,000
k = 0.2316
D.
$ 550,000.00
Tabulating the depreciation charges:
Year
0
1
2
Depreciation
0
Book Value
560.000
6(560,000) =1 29,696
0.2316(430,304) = 99.658.41
330,645.59
0.231
430,304
Unit depletion 121"
=
$50'ooo'ooo
5,000,000
Unit depletion 131s = $10.00 per barrel
Totaldepletion during the year = (500'000)(10)
.'. The depreciation on the second year is P 99,658.41.
Total depletion during the year = $ 5,000,000
.'. The total depletion charge during the year is $5,000,000'00'
Shell Philippines,
particular y6ar of
except for depleti
for that particular
A.
B.
c.
D.
P 9,358.41
P 9,228.45
P 9,250.00
P 9,308.45
9,Zz*,"
iJsing Depletion Allowance Method:
Depletion charge = O.22(5O,000,000) = '1 1,000,000
Depletion charge = 0.50(18,500,000) = 9,250,000
Use the smaller value, P 9,250,000
,'. The allowable depletion allowance is P 9,250.00.
ome for a
all deductions
etion allowance
as 22o/o.
240
Question Bank
-
Engineering Economics by Jaime R. Tiong
Question Bank I
J,;*
The first cost of a certain equipment is P 324,000 and a salvage value of P 50'000
at the end of its life of 4 years. lf money is worth 6% compounded annually, find
the capitalized cost.
A. P 540,090.34
B. P 541,033.66
c. P 540,589.12
D. P 541,320.99
.7;za-.
Capitalized cost= Co+
P+
Eq.
1
The maintenance cost is due to the depreciation of the equipment only
Total depreciation = Co
Co
Co
-
Cn
-Cn =32a,000-50,000
- Cn =274,OOO
Since, there is a given interest rate,
use sinking fund method in solving
for annual depreciation charge.
i
'-[tr*ir-r]
^_(co-c")
\ '
.{- 274.000(0.06)
[t',.
o.oo)' - r]
d--62,634.07
Solving for the present worth of all depreciation charges, P
242
Question Bank
P_
-
(lrrpit,rrliz,rrrl
Engineering Economics by Jaime R. Tiong
62,634.07[(1+0.06)4
(bst & Anntral Cost 243
Capitalized cost = 325,000
-1]
.'. The capitalized cost is
0.06(1+ 0.06)a
,/ /
P 325,000.00.
P = 217,033.66
Another way to solve for the value of P
,
_
o[(r*i)'-r]
i(1
+ i)"
A piece of equipment was purchased for P 4,000,000. The useful life is
estimated at 5 years with no salvage value. The equipment is estimated to be
used for 1,000 hours per year. Operating costs are as follows:
Fuel
Oil
Grease
,_ffi[t'.'1"-']
i(1
t
: l0liters per hour, P 10 per liter
: 1 liter Per hour, P 20 Per liter
: 2 liters per hour, P 5 per liter
ti
insurance,
is Problem
+ i)n
nt uses 6
P 60,000 each. Cost of money is 15% per
age, etc. is 15% per year. The interest tabie
the figure.
Determine the cost of using the equipment and tires, in pesos per hour.
^r--
A. P 2,202.90
B. P 2,101.80
c. P 2,000.90
274.OOO
(1+ 0.06)"
P = 217,033.66
Substitute in Eq.
D.
P 2,303.70
,7"2r*-
1
Capitalized cost = 324,000 + 217,033.66
Capitalized cost = 541,033.66
A.
P = 4,000,000
.'. The capitalized cost is P 541,033.66.
An item is purchased for P 100,000. Annual cost is p 1g,ooo. Using interest rate
8%, what is the capitalized cost of perpetual service?
4.000.000 =
r
i
0.15(1+ 0.15)5
A = 1,193,262.21
P 325,000
ln terms of pesos Per hour,
Hourly cost of equiPment =
Capitalized cost = Co +
AAAAA
n[{r*o.rs)u-r]
A. P 310,000
B. P 315,000
c. P 320,000
D.
Annual cost of equiPment onlY
1,193,262.21
year
Yeat
-----.-.:-= 1.1 93.26
1,000hours
x
I
I
Capitalized cost = 100,000 +
1
8,000
0.08
B.
Annual cost of insurance, taxes' storage, etc':
-
Annual cost of insurance, etc = 1,193,262.21(0.15) = 178,989'33
244
(lrrgrit,ulizorl Oont & Anrrual Cosl 245
Question Bank
-
Engineering Economics by Jaime R. Tiong
Total annual cost = 1 1193.26 + 178.99 + 130 + 593.26 + 107.39
Total hourlY cost = 2,202.90
ln terms of pesos per hour,
Hourly cost of insurance, etc =
c
178.989.33
-'-----x
The total cost of using the equipment and tires is P 2,202.90 per hour.
vear
'--' =178.99
1,000hours
yeil
Hourly fuel, oil and grease costs:
A corporation uses a type of motor which costs P 5,000 with life 2 years and final
value of P 800. How much could the corporation 3fford-to pay.for another
""tr"b"
typ" Jt motor of the same purpose whose life is 3 years with a final salvage value
of P 1,000. MoneY is worth 4%.
Fuel cost = 10('!0) = 100 per hour
Oil cost
= 1(20) = 20 per hour
Grease cost = 2(5)
= 10 per hour
Hourly cost of fuel, oil and grease = 100 + 20 + tO =
D
A. P 7,892.13
B. P7,157.40
c. P7,489.21
D. P 7,300j2
130
hour
Annual cost of depreciation:
Using sinking fund method since the interest is given
SlnZr,r."
, (co-c")i
The two types of motors must have the same annual cost'
ftt*i)"-t]
For Motor 1:
Using sinking fund method in computing depreciation:
d= 593,262.21
Annuar
ln term of pesos per hour,
Hourly cost of depreciation =
E.
592.262.21
year
""o
=
ffi
vear
1,000hours
Annuat cost
+ co(i) +
-
(5'-000
- 800)(0
o.c.
94) + (5,oooXo.oa)
Itt*o.o+)'-t.1
Annual cost of tires (in pesos per hour):
Annualcost = 2,258.82
P = 60,000(6)
P = 360,000
For Motor 2:
Using sinking fund method in computing depreciation:
360,000-
n[(r*o.rs)'-_r]
AAA
'ffi.
.
0.15(1+ 0.15)5
ln terms of pesos per hour,
-
107'393'60x---yegl--
year
..
i
Annuat
A = 107,393.60
Hourty cost of tires
Annuar
1,000hours
= 107.39
"or,
=
*"1 =
j"' - 9") I + co(i) + o.c.
[tt*i)"
-'11
(?o -
000X0
1'
0j)
[(t+o'o+)" -t]
Annual cost = 0.32035C0
Annual cost = 0.36035C0
+ (co )(0.04)
- 320.35 + 0.04C0
- 320.35
246
Question Bank
-
()rrlritalizod Cost & Annual Cost 247
Engineering Economics by Jaime R. Tiong
Equating annual costs:
2,258.82 =.0.36035C0
- 320.35
Co =7'157 '4O
,'. The corporation could afford to pay P 7,157.40 for anot[er motor.
At 6%, find the capitalized cost of a bridge whose cost is P 250M and life is 20
years, if the bridge drust be partially rebuilt at a cost of P 100M at the end of each
20 years.
A.
B.
c.
D.
A motorcycle costs P 50,000 and has an expected life of 10 years. The salvage
value is estimated to be P 2,000 and annual operating cost is estimated at P
1,000. What is the appropriate rate of return on the investment if the annual
revenue is P 10,000?
A.
B.
C.
D.
P 297,308,323.10
P 298,308,323.10
P 296,308,323.10
P 295,308,323.10
,%h*.
12.12%
12.54
o/o
12.72o/o
12.ggo/o
,:%bTo get the rate of return, i, equate annual cost and annual revenue:
Using sinking fund method in computing depreciation:
Annuat
*r,
- c") I + c^(i) + o.c.
= Ico
l{t*i)"-tl
r50. 000 2. 000) (i)
10,000=. - - ,=' -," +(50,000Xi)+1,000
l(1+i)'"-t]
e,
ooo =
jg-qqq!= + 50, oooi
i[ - t]
[(t.
By trial and error:
i=
o.1272
i=12.72%
The appropriate rate of return of the investment is 12.72 %.
Capitalized cost = Co +
i
Solving for the equivalent interest rate in 20 years,
i
i = (1+ NR)20
i = (1+
-1
0.06)20 - 1
i--2.2071
i=220.71%
Capitalized cost = 250,000,000
+
109:qqq900
2.2071
Capitalized cost = 295,308,323.10
.'. The capitalized cost of the bridge is P 295,308,323.10.
248
Question Bank
-
Cnpitalized Coet & Annual Cost 249
Engineering Economics by Jaime R. 'liorrg
Total annual cost = 319,833.49(2)
Total annual cost = 639,666.98
A newly hired Mechanical Engineer of CPPI was made to evaluate on the
following condition.
FOR SECOND EQUIPMENT:
A packing equipment that cost P 500,000 will last for 20 years and have a scrap
value at that time of P 50,000. Repair will average P 30,000 per year in which the
e P 20,000 per month. The
as many units per year. lt costs
25 years at a cost of P
1,OOO,OOO. Repairs for this machine will average P 25,000 per year and operating
expenses will be P 30,000 a month.
Annuat
B.'
c.
D.
= Ico 9") i
o.c. -r Eq. 2
[tt*i)'-t.]
Substitute values in Eq.2
P120,978.20
P120,962.20
P120,988.20
P120,999.20
Annual cost
=
518,678.78
Difference = 639,666.98
Compare the annual cost of the two packaging equipments:
+ co(i) +
O.C. = 25,000 + 30,000(12)
O.C. = 385,000
lf money is worlh 8% effective, How much will be saved each year if the more
economical equipment is used?
A.
"o",
-
518,678.68 = 120,988.20
.'. The amount that will be saved each year if the more economical
equipment is used is P 120,988.20.
FOR FIRST EQUIPMENT:
Annual cost = depreciation + interest of first cost + operating cost
Using sinking fund method in computing depreciation:
Annuar
"o",
= Ico 9") i
[(t*i)"
+ co(i)+ o.c.
-+
Eq.
For its proposed expansion, an ice plant company is selectlng from two
offers of ice cans. The data on the offers are as follows:
1
Total cost
Annual maintenance
No. of vears. Life
-t.1
Solving for total operating cost, O.C.
= 30,000 + 20,000(12)
O.C. = 270,000
Annuat
"o.t
=
(509
P 640.000.00
P
12
1
00, - 50'9-00)(9' 08)
(r+o.oa)'?o
-r
+ (500, 000)(0. 08) + 270, 000
I
Annual cost 1'tg,gsz.qg
//
,/
Offer B
1.6 mm thick
90.000.00
I
For their replacements, the company is putting up a sinking fund to eam 16%
interest compounded annually. lf the money to purchase the ice cans is to be
borrowed at 20% annual interest and the tax on the first cost is 2%, what is the
difference in the annual cost of the two offers?
o.c.
Substituting values in Eq.
Offer A
1.8 mm thick
P 720,000.00
P 60.000.00
Since the,(econd equipment can produce twice as many units than the first
equipmCnt, we will consider acquiring 2 units for the first equipment to
balance the production, thus
A. P 34,489.01
B. P 34,OM.92
c. P 34,320.12
D.
P 34,617.22
,%)r,*.
For Offer A:
Using sinking fund method in computing depreciation
250
Qtrestion llank
-
Annuatcost =
jt'-:')
/
|
=
(lrst
251
r,,r,, ,.
For alternative A:
(220'000 -0)-(0' 1 9)
(t *
lrrpilrrlizorl ()ost & Attttuitl
l+co(i)+o.c.
f(t+r) -r]
Annual cost =
Annual cost
(
Engineeling Dcortrlrnics by Jaime Il. 'l'iorrg
+ (720, 000)(0 .2o) + o.o2(7 20, 000) + 60, 000
Using sinking fund method in computing depreciation
o.to)'' - t.]
Let x = number of days the equipment is used in one year period
241,738.61
Annuat cost
For Offer B:
-9") +co(i)+o.c.
= it'(1+i)
I
I
L") -1
|
Using sinking fund method in computing depreciation
(c^
Annual 6ss1 =
-c_)
l(t*i)
Annual cost
Annual cost =
i
+ Cn(i) + O.C.
@
-1
Annual cost = 151,508.78
-t.1
(6-40'000 -0X0' 1-6)
--
(700'000
+ (6a0, 000)(0.20) + 0.02(6a0,000) + 90, 000
00'000X0' 1 8)
+
+ (700,000)(0.1 8) + 500x
500x
For alternative B
ftt*o.to)'-t.]
Annual rental = 1,500x
Annual cost = 275,743.53
Equating:
Solving for the difference between the two offers:
Difference = 27 5,7 43.53
Difference =34,004.92
-
x = 151.5 days
The difference between offer A and offer B is P 34,004.92.
A company is considering two alternatives with regards to an equipment
which it needs. The alternatives are as follows:
Alternative A: PURCHASE
Cost of equipment
Economic life ............ ....
Salvage value ...
Daily operating cost ........
P 700,000.00
10 years
P 100,000
P 500.00
Alternative B: RENTAL at P 1,500 per day
At 18% interest, how many days per year must the equipment be in use if
alternative A is chosen?
A.
B.
C.
152 days
154 days
156 days
151,508.78 + 500x = 1,500x
1,000x = 151,508.78
241,7 38.61
.'. The equipment must be used lor 152 days per year.
A new broiler was installed by a textile plant at a total cost of P 300,000 and
projected to have a useful life of 15 years. At the end of its useful life, it is
estimated to have a salvage value of P 30,000. Determine its capitalized cost if
interest is 18% compounded annually.
A. P 323,500.33
B. P 322,549.33
c. P 332,509.33
D.
P 341,240.33
.Vn/oln.'
.
Capitalized cost = Co + present worth of all depreciation charges
Let P = present worth of all depreciation charges using sinking fund method
252
Question Bank
D
'-
-
Engineering Economics by Jaime R. Tiong
Cepitalized Coet & Annual Coet 253
_ 300'000-30'000
Annuatcost
(1+Ol8f
=
9-:")
[(t+i)"
P =22,549.33
=
Annual cost
Capitalized cost = 300,000 + 22,549.33
Capitalized cost = 322,549.33
l+co(r)+o.c.
-t]
+
ffiffi
({200)(0'12) + 1'200(0'01)
Annualcost = 224.38
.'. The capitalized cost is P 322,549.33.
FOR TREATED POLE:
Using sinking fund method in computing depreciation
A man planned of building a house. The cost of construction is P 500,000 while
annual maintenance cost is estimated at P 10,000. lf the interest rate is 6%, what
is the capitalized cost of the house?
A,
B.
c.
D.
P 666,000.00
P 666,666.67
I?'
Annuatcost
=
Annuat cost
=-
,!")l+co(i)+o.c.
f(t.r[ -t]
(co -0)(9--12)
- +(co)(0.12)+co(0.01)
[(t+o.tz)'" -t.]
P 633,333.33
P 650,000.00
Annual cost
=
0.14388 Co
Equating annual costs:
Capitalized cost = Co +
1
0.14388C0 =224.38
Co = 1'559'50
I
Capitalized cost = 5OO,0oO +
'
]lPqq
0.06
Capitalized cost = 666,666.67
The maximum justifiable amount that can be paid'for the treated pole is
P 1,559.50.
.'. The capitalized cost is P 666,666.67.
An untreated electrical wooden pole that will last 10 years under a ce(ain soil
condition, costs P 1,200.00. lf a treated pole will last for 20 years, what is the
maximum justifiable amount that can be paid for the treated pole, if the maximum
return on investment is 12o/o. Consider annual taxes and insurance amount to be
1 % of the first cost.
A.
B.
c.
D.
P 1,612.01
P 1,559.50
P 1,789.23
A granite quarry purchased for P 1,600,000 is expected to be exhausted at the
end of 4 years. lf the resale value of the land is P 100,000, what annual income
is required to yield an investment rate of 12%o? Use a sinking fund rate of 3%.
A. P 550,540.57
B. P 550,540.57
c. P 550,540.57
D.
P 550,540.57
92b.".
P 1,409.38
Using sinking fund method in computing depreciation
,V./,ir,rn
FOR UNTREATED POLE:
Using sinking fund method in computing depreciation
Annuat
"o.1
=
$5f")
ftt.i[
I
-t1
+ co(i)+
o.c.
254
Questiou Bank
,Annual
cost
-
Engineering Economics by Jaime R. Tiorrg
(1,600,ooo
=
-1
00,000)(0.03)
()npil,alizod Oost & Annual Cost 255
where:
+ (1,600,000X0.12)
Q = discharge in cublc meter / second
or
[{r*o.os)'-r]
Annual cost
=
550,540.57
= density of water = 9810
+
E = head in meters
For an assufance of rate of return of 12o/o,lhe one used in computing
annual cost, the annual income must be equal to annual cost.
For full load operation, efficiency =
7Oolo
(e'810x43)
P_ 4s.5
Annual income = Annual cost
Annual income = 550,540.57
P=
3,600 0.70
7,616.375 watts
P=
.'. The annual income required is P 550,540.57.
7.6'16375 kw
cost of power = (7.616375 rw1[2a hours][tso
An ice plant owner decided to install a water pump from a nearby clean fresh
water source to supply its water requirements. The capacity of the pump is 45.5
cubic meters per hour against a head of 43 m with an efficiency of 70% at full
load an 40% al half load. The total cost of pumping unit including the piping
system to the plant is P 700,000 with zero salvage value at the end of its
estimated life of 20 years. The prime mover of the pump is an electric motor and
cost of power is P 2.10 per kw-hr. Taxes, insurance and maintenance cost is
1.5% of the first cost per year. The pump will operate 200 days continuous
operation per year at half load and 150 days at full load. lf the cost of money is
12o/o, how much will it cost the owner to operate the pump per cubic meter?
Cost.of.power = P 57,579.80 per year
For half load operation, efficiency = 40yo
45.5 (e,810X43)-
' _- 3,600 o4o
D
1
^z
P = 6,664.328 watts
P = 6.664328
kw
cost of power = (6,664.328 my[24 hours )[ zoo oavs )f - P2' 10
'(
A. P 0.84
B. P 0.20
c. P 1.15
D.
oavs)rE!)
'( day /1. yeat./l.tw-nr/
day /( year /(kw-hr/)
Cost of pow€r = P 67,'1 70.60 per year
P 1.35
Total cost of power = 57,579.80 + 67,170.60
Total cost of power = 124,75O.4O
9"2a...
Using sinking fund method in computing depreciation
Annuar
"or1
-9") I + co(i)+ o.c
f(t+i)" -t.l
= Ico
+
Eq.1
Analyzing operating cost, O.C.
Solving for operating cost;
O.C. = 1 24,7 50.40 + 0.0 1 5(7OO,O0O)
o.c. = 135,250.40
Substituting values in Eq.
1
Operating cost = cost of power + cost of taxes, insurance and maintenance
Solving for cost of power:
P=QorE,
Annuar cost
=
Annual cost
=
tffiffi
+ (700,000)(0 .12) + 13s,2s0.40
228,965.546
Solving for the cost per cubic meter:
256
Question Bank
-
Engineering Economics by Jaime R. Tiong
Oepitalizod Cost & Annual Coet 257
Let V = total volume pumped per year
Annualcost = 2,716.67
V = volume during full load + volume during half load
(
For Machine B:
+s.sm')(z+nours)(rSodavs\ / t \(+s.sm3'\(z+nours\t. 200davs'l
'=[--.](.
., _ 273,000
d,y
]t v..-1.[zJ[ n' ]t *, ll.-y;)
Annual cost = depreciation + interest of first cost + operating cost
m3
year
Therefore,
=
Annual cost
=
a^ - vn
vo
+ c^i + o.c.
n"
(20.000 - 6.000)
---G-+(20,000)(0.08)+350
Annualcost = 4,283.33
228.965.546 P0.839
273,000
Annual cost
Difference = 4,283.33 -2,716.67
Difference = 1,566.66
m3
.'. The cost to operate the pump is 84 centavos per cubic meter.
.'. The machine to choose is machine A and the difference of annual costs
is P 1,566.66.
Make an economic analysis to determine which of the following two machines
capable of performing the same task in a given amount of time, should be
purchased. The minimum attractive return is 8%.
First cost
Estimated life
Salvaqe value
Annual maintenance
Machine A
P 10.000
6 years
0
P 250
Machine B
P 20.000
14 vears
P 6,000
P 300
Basing on annual cost, using straight line method of computing depreciation,
determine which machine you would choose and how much is the difference in
its annual cost?
A.
B.
C.
D.
Machine B, P
Machine B, P
Machine A, P
Machine A, P
1,205.06
1,566.67
1,205.06
1566.67
,'%h-.
A decision by the supervising mechanical engineer must be made whether to
replace a certain engine with a new one or to rebore the cylinders of the old
engine and thoroughly recondition it. The original cost of the engine 10 years ago
was P70,000, to rebore and recondition it now will cost P 28,000 but would
extend its useful life for 5 years. A new engine will have a first cost of P 62,000
and will have an estimated life of 10 years. lt is expected that the annual costs of
fuel and lubricants with the reconditioned engine will be about P 20,000 and this
cost will be 15% less with the new engine. lt is also believed that repairs will be P
2,500 a year less with the new engine than with the reconditioned one. Assume
that neither engine has any salvage value when retired. lf money is worth 10%,
how much money is saved each year if the old engine will be replaced?
A.
B.
c.
D.
P 2,796.12
P 2,123.45
P 2,340.12
P 2,555.30
Compare annual cost of either engine.
For Machine A:
For reconditioned engine:
cost: depreciation + interest of first cost + operating cost
to tn
Annual cost =
* coi + o.c.
-n
Annual
Annuat cost
(10.000 - 0)
= #+(10,000)(0.08)+2S0
6
Using sinking fund method in computing depreciation
Annual cost
=
(c^
-c-)
[(t.
i)"
i
- t]
-+C^i
+O.C.
258
Question Bank
-
Engineering Economics by Jaime R. Tiong
Annuat cost
=
Annual cost
=
(?8'000
0)-(0'1-0)
ftr.o rof -rl
Capitalized Cost & Annual
Cost 259
+ (28,000)(0.10)+ (20,000 + 2,500)
29,886.33
Using sinking fund method
in computing depreciation
Annual
cost
^rrrru'rr uosl
=
Using sinking fund method in computing depreciation
Annuat
rr ruar
^r
"o.,
jco - c") I
=
=
"""'
"o.,
[{t.i)" -t
+ c^i + o.c.
+o'c'
0)fi-T
Annuat cost
=
Annuar cosr
=
$2,ooo
_z
,2oo)(0.2o)
,
+ (52'000)(0 20) + (8'300
+ 27
.rl;',o"T-
'600)
For diesel driven unit:
1
(9''ooo - o)(o'10)
I;;r
-C ) i
FE+coi
For new engine:
Annuar
'(C"
+ (62' 000)(0' 1 0) + (20' 000)(0'85)
Annual cost = 27,090.21
Difference = 29,886.33 -22,090.2i
Difference = 2,796.12
.'. The amount that witl be saved each year is the old engine is replaced
with a new one is P 2,796.12.
Using sinking fund method
in computing depreciation
Annual wo'(
cost
=
lco
-C,)
i
fGf-f+coi+oc'
Annuat cost
=
(98,_250 _ 1i,9oo)(0.20)
ffr. ozol'o_l-
+ (e8'250)(0'20) + (5'800
+ 18,800)
Annualcost = 47,576.44
Solving for the difference:
'
Excel Review Center is considering the purchase of a generating set to
the electrical needs during brown outs.
The first offer he received is for a gasoline driven unit with an estimated life span
of 5 years and would cost the review center p 52,000. The annual maintenance
cost is estimated to be P 8,300, annual operating cost would be p 27,600 and
salvage value would be P 7,200. The second offer the review center received is
for a diesel engine driven unit with an estimated life span of l0 years. Annual
maintenance cost is P 5,800, annual operating cost is p 1g,goo and with salvage
value of P 1 1,900. The price of the unit is 98,250.
Difference = 52,320.21 _
47,S7 6.44
Difference = 4,243.22
]ff ,1i:::llffi,U:5;"" ffiJ#;,,"f ", p 4,7 43 7 t c hea pe r tha n the
sewa ge havin g .or"
crete, cast iron and steel.
Because of
-r 'J^^^
^!,,
"orro"r]h
How much cheaper is the annual cost of the diesel driven unit than the gasoline
driven unit?
A.
B.
c.
D.
P 4,612.04
P 4,743.77
P 4,490.12
P 4,500j2
For gasoline driven unit:
____---.ljv
q,,v vqDr rrun ptpe nave
the same annual costs.
I
/
260
Question Bank
-Ungineering Econotttit:s bv Jairne
-
ll. 'liong
Question Bank 9
,7Lr,"Compare annual cost of the three bids
For reinforced concrete PiPe:
Using sinking fund method in computing depreciation
Annuat cost
= Jt' -:") I +coi+o.c.
ftr*i)'-t1
Annuatcost
=
Annual cost
=
+(770,000)(0.05)+900
A manufacturer produces certain items at a labor cost per unit of p 315, material
cost per unit is P100, variable cost of P 3.00 each. lf the item has.a selling price of
P 995, how many units must be manufactured each month for the rnanufacturer to
breakeven if the monthly overhead is P 461 ,600?
A. 782
B. 800
c. 806
40,417.64
For cast iron piPe:
,7"2a,
Using sinking fund method in computing depreciation
Annuat cost
Annuar cosr
Annual e,ost
: Itt -:") I + coi+ o.c
Let x = number of units to be manufactured per month to break-even
ftt*i)" -t1
=
=
(190
000
Material cost = 100x
0)-(0'0-5)
Labor
+ (790,000)(0.05) +
600
[(t*o.os)" -t.]
cost
= 315x
Variable cost = 3x
Total expensss = (100x + 315x + 3x) + 461,500
Total expenses = 418x + 461,600
40,402.68
For steel piPe:
Using sinking fund method in computing depreciation
Annuar cost
=
f+"!
+ coi+ o.c.
[(t*i)'-t1
Annuatcost
Annual cost
=
=
577x = 461,600
+(580,000)(0,05)+3,600
35,370.51
The steel pipe has the least annual cost.
x=800
.'. The number of units to be manufactured each month to breakeven is
800 units.
262
Question Bank
-
Engineering Economics by Jaime R. Tiong
Break-evenAnalysis 263
Expenses = 1 1Sx + 76x + 2.32x +42g,OOO
Expenses = .1g3.32x + 42g,O0O
The annual maintenance cost of a machine shop is p 69,994:rf
the cost of
making a forging is p 56 per unit and its selling price is n f SS pei
unit, flnd
the number of units to be forged to break_evei.'
fpffi
To breakeven:
lncome = Expenses
A. 892
B. 870
c. 886
D.
600x=j93.32x+42g,000
406.68x = 428,000
x = 1,052.42 r 1,053 units
862
'.
,tEJXXlfr
of units to be manufactured each
month to breakeven !s
Let: x = number of units to be forged to break even
lncome = 135x
Expenses=69,994+S6x
To breakeven;
lncome = Expenses
135x=69,994+56x
79x = 69,994
x = 886 units
Steel
yearly fixed operating cost of
$ 200,000.
drum
- Each
produce and sells $ 200. What
is the
manu
manU.--.-,-, e vreo^-Eve, sales volume
i" dr.;;;::;;
A. 5,031
B. 5,0,17
c. 5,043
D.
5,000
.'. The number of units to be forged to breakeven is gg6 units.
Let: x = number of drums to be sold
out per year to breakeven
lncome = Expenses
A manufacturer produces certain items at a labor cost of p 1 15
each, material
cost of P 76 each and variabre cost of p2.32 each. rf the item
has a unit price of
P600, how many units must be manufactured each month tor
tn" ,anuticiiiel to
break even if the monthly overhead is p 42g,000.
A. 1,033
B. 1,037
c. 1,043
D.
200x=200,000+160x
40x = 200,000
x = 5,000 drums
"'
The number of drums to be sord
each year to breakeven is 5,000
drums.
1,053
kcases that sell for p 65.00 each.
lt costs
to operate its plant. This sum inclules
rent,
Let: x = number of units to be manufactured per month to breakeven
11_oe"**,;;fii;o:#,x,;"JlUi"i;,;y.ffi
Ttoid taking a loss?
,#il"ii,"^11[?,1ffi ;
lncome = 600x
Expenses:
Labor
Materials
= 115x
=
76x
Variable = 2.32x
Overhead = 428,000
9;2r.,..
II
264
Question Bank
-
Engineering Economics by Jaime R. f iong
Broak'cven Analysis 265
To avoid taking loss means that the minimum production must be enough to
breakeven.
Let: x = number of units to be forged to breakeven
lncome = Expenses
Let; x = number of cases to be sold each year to breakeven
lncome = Expenses
125x=56x+70,000
69x = 70,000
x = 1,014.49 units
65x=50x+35,000
15x = 35,000
x = 2,333.33 cases
!
2,334 cases
e
1,015 units
The number of units to be forged to breakeven is 1,015.
.'. The number of cases to be sold each year to breakeven is 2,334.
A company which manufactures electric motors has a production capacity of 200
motors a month. The variable costs are p 150.00 per motor. The average selling
price of the motors is P 275.00. Fixed costs of the company amount to p zo,ooo
per month which includes taxes. Find the number of motors that must be sold
each month to breakeven.
A.
B.
c.
D.
A manufacturing firm maintains one product assembly line to produce signal
generators. Weekly demand for the generators is 35 units. The line oper-ates for
7 hours per day, 5 days per week. What is the maximum production time per unit
in hours required of the line to meet the demand?
A.
B.
C.
D.
160
157
153
1.0
1.2
1.4
1.6
hour per unit
hours per unit
hours per unit
hours per unit
,(y'.Z,,/*,,
163
Let x = number of hours to produce 1 unit
v-l-ll-
Let: x = number of motors to be sold each month to breakeven
+ 20,000
)/ Sdavs\(Thours)
ll-l
\35units/\ week
x = t hour per unit
lncome = Expenses
275x= 150x
/ lweek
/\
day )
.'. The maximum production time is
t
hour per unit.
125x = 20,000
x = 160 motors
.'. The number of motors to be sold each month to breakeven is 160.
The annual maintenance cost of a machine is p 70,000. lf the cost of making a
forging is P 56 and its selling price is p 125 per forged unit. Find the numbeiof
units to be forged to break even.
A.
B.
c.
D.
.72,*,,
1,000
1,O12
1,015
1,018
Compute for the number of blocks that an ice plant must be able to sell per
month to break even based on the following data;
Cost of electricity per block
Tax to be paid per block
Real Estate Tax
Salaries and Wages
Others
Selling price of ice
A.
B.
c.
D.
1.220
1,224
1,228
1,302
- P 20.00
- P 2.00
- P 3,500.00 per month
- P 25,000.00 per month
- P 12,000.00 per month
- P 55.00 per block
266
Question Bank
-
l}roak'even Analysis 267
Engineering Economics by Jaime R. Tiong
.7'/u,,,,,
... The number of units to be producod and sold per month
to breakeven is
120.
Let: x = number of blocks to be sold per month to breakeven
lncome = 55x
Expenses:
ElectricitY
Tax
= 20x
=2x
Fixed charges = 3,500 + 25,000
i
tZ,OOO =
40,500
To breakeven:
lncome = Expenses
55x=20x+2x+40,500
33x = 40,500
x = 1,227 .3 blocks
s
1
,228 blocks
,'. The number of blocks to sel! per month to breafieven is 1,22g.
A.
100
B.
104
C.
110
D.
112
,92d,.,
Let x = number of motors to produce per month to break-even
Expenses=4000+78,000
JRT lndustries manufactures automatic voltage regulators at a labor cost of
P85.00 per unit and
fixed charges on the
business are P 15,000
e p 20.00 [er unit. lf
g0.00 each, how many
the automatic voltage
units must be produced an
Expenses = 82,000
materi
per
regu
A.
B.
c.
D.
lncome = 750x
To break-even,
lncome = Expenses
120
124
128
750x = 82,000
x=109.33*1'10 units
130
... The number of motors
.VZ,./,r,,.
Let: x = number of units to be produced per month to breakeven
lncome = 580x
Expenses:
Labor
= 85x
Materials = 350x
Variable = 20x
Fixed charges = 15,000
To breakeven:
lncome = Expenses
580x = 85x + 350x + 20x+ 15,000
A.
B.
C.
D.
125x= 15,000
x = 120 units
.V"Z*,*.
1,000
1,040
1,100
1,120
feet
feet
feet
feet
to produce each month to breakeven is 110.
268
Question Bank
-
Engineering gconomics by Jaime R. Tiong
Break-evenAnalysis 269
For enameled wire:
To breakeven,
lncome = Expenses
Let x = length of wire
1800x = 1,000x + 38,333.33
800x = 38,333.33
x = 47.92: 4Sunits
Cost of wire = 0.55x
Cost of soldering = '1.65(a00) = ggg
Total cost = 0.5Sx + 660
For tinned wire:
lf 200 units are produced per month, there wiil be profit since
it needs onry
48 units to break-even.
Let x = length of wire
Solving for profit:
Cost of wire = 0.75x
Cost of soldering = 1.15(400) = 460
Total cost = 0.75x + 460
lncome = 200(1,800) = 360,000
Expenses = 1,000(200) + 3g,333.33
Expenses = 238,333.33
ln order to get the rength of wire that wiil give the same cost,
equate totar
cost of enameled wire and total cost of tirined wire.
0.55x+660=9.75r*460
0.2x=200
Profit = lncome
- Expenses
Profit = 360,000 - 238,333.33
Profit = 121,666.61
.'. There is a profit of p 121,666.67..
x = 1,000
The length of cable to make cost of installation the same
is i,000 feet.
A local factory assembring carcurators produces 400 units per month
and sefls
them at P 1,800 each. Dividends are 8% on the 8,000 snaiei wrn paivarue p
ot
250 each.
!1eO operating cost per month is p 2O,OOO. Other 6sts are p
'1
per
unit.
lf 200 units weie produced per month, oetermine tne piont
,000
or toss.
fp
A.
B.
C.
D.
Profit of P 121,666.67
Profit of P 21,666.67
Loss of P 121,666.67
Loss of P 21,666.67
Nike shoes manufacturer produces a pair of Air Jordan shoes at
a labor cost of p
900.00 a pair and a material cost of p B0O.O0 a pair. The fixeri charoes.an
the
nth and the variaLle cos
pair.
1,000 per pair of shoes
sell at
must be produced each
A.
B.
c.
D.
2,590
2,632
2,712
2,890
VZaLet x = number ofcalculators produced per month to break_even
Expenses = 1,000x + 2S,OOO + g,000(250)
Expenses = 1,000x + 38,333.33
lncome = 1,800x
(H)
Let x = number of pairs of shoes produced each month to break_even
Expenses = 900x + 800x + 400x + 1,000x + 5,OO0,O0O
Expenses = 3,1 00x + 5,0OO,OOO
lncome = 5,000x
To break-even,
lncome = Expenses
,iI
270
Question Bank
-
Engineering Economics by Jaime R. Tiong
Break'evenAnalyeis
5,000x = 3,100x + 5,000,000
1,900x = 5,000,000
x = 2,631 .6 x 2,632 pairs of Air Jordan Shoes
271
0.75n = 0.0417n + 0.50n + 2000
0.75n=0.5417n+2000
0.2083n = 2000
n = 9601.54 holes
.'. The number of pairs of Air Jordan shoes to produce each month to
breakeven is 2,632.
Sayn=9601 holes
The number of rivet holes for the multiple-punch machine to pay for
itself is 9,601.
minute. This method requires P0.50 per holes to set multiple punch machine and
an installation cost of P 2,000.00. lf all other costs are the same, for what number
of rivet holes will the multiple punch machine pay for itself?
A.
B.
c.
D.
'
A new Civil Engineer produces a certain construclion material at a labor cost of P
per
7.4O
P
cost
of
per
piece
and
variable
per
piece,
P
38.50
cost
of
material
16.20
piece. The fixed charge on the business is P 100,000.00 per month. lf he sells the
hnished product at P 95.00 each, how many pieces must be manufactured in eacl"
month to breakeven?
A. 3,045
B. 3,035
c. 3,030
9,601
9,592
9,581
D.
9,566
3,040
9"2*""
,V"z:ro,-.
Let x = number of pieces to be manufactured per month to breakeven
Using first method, the Drill Press:
Cost of machinist per hole
-
lncome = 95.00x
20:25
27
Cost of machinist per hole = 0.75
Expenses:
Labor
Materials
Using the second method, the Multiple punch Machine:
Variable
Fixed charges
Cost of machinist per hole
= 8(60)
1-0-=
= 16.20x
= 38.50x
=
7.40x
= 100,000
To breakeven:
Cost of machinist per hole = 0.0417
Additional cost for setting up th€ machine per hole = 0.50
lnstallation cost = 2,000.00
lncome = Expenses
95.00x = 16.20x + 38.50x + 7.40x+ 100,000
32.90x = 100,000
x = 3039.5 say 3040 Pieces
Let n = number of rivets holes that will make the multiple punch machine
pay for itself
\
Total cost using drill pr€SS = total cost using multiple puhph machine
\
The number of pieces to be manufactured each month to breakeven is
3,040.
272
Question Bank
-
Engineering Economics by Jaime R. Tiong
Break'evenAnalyeie 273
To breakeven,
Co
uni
n Transmission
ume is 500,000
Fixed expenses tot
a year?
A.
B.
c.
D.
nt
lncome = Expenses
50)
O.Sx = 80,000
x = 160,000
The breakeven point is 160,000 units.
P 168,000
P 170,000
P 172,000
P 174,000
Pro&ction
pacity due to
portation of their
Solving for the total profit, P:
'l9O,0OO.O0 and
loss?
500,000 units P0.50
tncome
x -year
A. P 87,450
B. P 88,960
c. P 88,450
lncome = P250,000
D.
Expenses = P80,000
Profit = lncome
-
costs are P
variable costs are P 0.348 per unit. What is the current profit or
Expenses
P 87,960
,7o/,sru.,..
Profit=250,000-80,000
Profit = 170,000
Solving for the current profit or loss:
.'. The present total profit is P 170,000.
Pr oduction
:
0.62(700,000)
Production = €a,000
nsm
s 50
exp
nt in
A.
B.
c.
D.
160,000
162,000
165,000
170,000
.V./,:,r,r,,
sent
O.SO)
rotat expens"r
=
(ffi)(+s+,ooo
Total expenses = 341,032
Total income = 430,000
Pr qfit = total income - total expense
Profit= 430,000 -341032
Profit = 88,960
.'. The current profit is P 88,960.
Let x = number of units produced to breakeven
lncome = 0.5x
Expenses = 80,000
units)+ 1eo,ooo
274
Question Bank
-
Engineering Economics by Jaime R. Tiong
ation of an automobile part with a production
ear is only operating at62% of capacity due to
foreign currency to finance the importation of their
me is P 430,000.00. Annual fixed costs are
P190,000.00 and variable costs are P 0.348 per unit. what is the breakeven
point?
A. 294,560
B. 291,000
c. 290,780
D.
295,490
I
It'r'it k-rrvt'tt nnrilrll\ -275
.,af
At
A certain firm has the capacity to produce 650,000 units of product perl''
present, it is operating al62% capacity. The firm's annual income is. . ,sts"r"
P4,160,000.00. Annual fixed costs are P 1 920,000.00 and the variableni
equal to P 3.56 per unit of product. What is the firm's annual profit or lo5)
A.
B.
c.
D.
P 814,320
P 815,230
P 816,567
P 817.239
,%,n -.
.7;/)zr-.
Pr oduction = 0.62(650,000)
Let x = number of units to breakeven
Production = 403,000
Solving for the expenses:
Total expenses = 1,920,000 + 3.56(403,000)
Total expenses = 3,354,680
Variable cost = 0.348x
Fixed cost = 190,000
Total Income = 4, 1 69,000
Solving for the income:
Pr ofit = total income - total expenses
Pr oduction = O.OZIzOO,OOO;
Profit = 4,169,000 - 3,354,680
Profit = 814,320
Production = 434,000
Totat expens""
s+a)1oao,ooo
= [0unrt
\
,\
Tqtal expenses
=
units)+190,000
/
341,032
Totalincome = 430,000
settins prics
.'. The firm's annual profit is P 814,320.
= i:9'999
434,}Uo
0.00.
3.56
= o.es1 per unit
p r,gzo,oo'o.oo
in"
""0does ""ti"5;ec0sls
the firdl
at volume of sales
breakeven?
Total income = 0.99x
A.
B.
c.
D.
To breakeven:
lncome: Expenses
,re?r.
At
A certain firm has the capacity to produce 650,000 units of product peil
. The flrm's annual income is c0!
is op
P 3,354,680
P 3,534,880
P 3,155,690
P 3,254,680
__-
0.991x = 0.348x + 1 90,000
0.643x = 190,000
Solving for the flrm's annual profit or loss:
x = 295,490
Pr oduction = 0.62(650, 000)
The current breakeven point is 295,490.
Pr oduction = 403,000
are
276
Question Bank
-
Brenk'ovon Analyeis 277
Engineering Economics by Jaime R. Tiong
Total expenses = 1,920,0Q
Total expenses : 3,354,680
+ 3.56(403,000)
\ \
'\
Total income = 4,169,000
ln as much as the total expenses is P3,354,680, then the volume of sales
must be equal to this value in order breakeven.
Thus, volume of sales = P3,354,680
.'. The firm's volume of sales to breakevpn is P 3,354,680.
.'. The number of thresherg to be produced and sold annually to
breakeven is 25.
The direct labor cost and material cost of a certain product are P 300 and P 400
per unit, respectively. Fixed charges are P 100,000 per month and other variable
qosts are P 100 per unit. lf the product is sold at P 1,200 per unit, howmany
units must be produced and sold to breakeven?
A. 280
B. 250
c. 260
D.
A small shop in Bulacan fabricates threshers for palay producers in the locality.
The shop can produce each thresher at a labor cost of P1,800.00. The cost of
materials for each unit is P 2,500.00. The variable costs amounts to P 650.00 per
unit while fixed charges incurred per annum totals P 69,000.00. lf the portable
threshers are sold at P 7,800.00 per unit, how many units must be produced and
sold per annum to breakeven?
A.
B.
c.
D.
28
25
26
27
270
Let x = number of units produced per month to breakeven
Expenses:
Direct labor
= 300x
Direct material cost = 400x
Variable
= 100x
Fixed
= 100,000
cost
cost
charges
Total expenses = 300x + 400x + 100x + 100,000
Total expenses = 800x + 100,000
,7"22*^
Let x = number of threshers produced and sold in 1 year
Expenses:
Labdr
Total income =1,200x
To breakeven:
cost
= 1,800x
Materials cost = 2,500x
Variable cost'= 650x
Fixed charge = 69,000
Total expenses = 1,800x + 2,500x + 650x + 69,000
Total expenses = 4,950x + 69,000
Total income = 7,800x
To breakeven:
Total income =Total expenses
7,800x=4950x+69,000
2,850x = 69,000
x =24.2 say 25 threshers
Total income = total expenses
1,200x=800x+100,000
400x = 100,000
x=25O
.'. The number of units to be produced and sold monthly to breakeven is
250.
278
Question Bank
-
Engincering Ilcouourir:s by Jairnc
ll
llrr'nk-cvon Analvsis 279
'l'iorrg
'-/nl,t;,,
The following data for year 2000 are drqllable for Cagayan Automotive Company
which manufactures and sells a single arlto-motive product line:
Unit selling price
Unit variable cost
Unit contribution margin
Total fixed costs
.....
.....
.....
.....
P
P
P
P
Let x = number of units produced per month to breakeven
Expenses:
Labor
40.00
20.00
20.00
200,000.00
What is the breakeven point in units for the currentyear?
A.
B.
c.
D.
cost
= 230x
Material cost = 370x
Variable cost = 10x
Fixed charges = 1,000,000
Total expenses = 230x + 370x + 10 x + 1,000,000
Total expenses = 61 0x + 1,000,000
10,000
10,100
10,050
10,200
Total income = 750x
To breakeven:
Total income = total expenses
Let x = number of units produced per month to breakeven
Expenses:
Variable cost
Fixed
charges
750x=610x+1,000,000
140x = 1,000,000
= 2Ox
= 200,000
x=7,142.8 say 7,143 ynits
.'. The number of sets to produce per month to breakeven is 7,143.
Total expenses = 20x + 200,000
Total income = 40x
To breakeven:
Total income = totalexpenses
4Ox=2Ox+ 200,000
20x = 200,000
x = 10,000
.'. The breakeven point for the current year is 10,000.
The cost of producing a small transistor radio set consists of P 230.00 for labor
and P 370.00 for material. The fixed charges in operating the plant is P
1 ,000,000.00 per month The variable cost is P 10.00 per set. The radio set can
be sold for P750.00 each. Determine how many sets must be produced per
month to break even.
An item which can be sold for P 63.00 per unit wholesale is being produced with
the following coSt data:
Labor cost
Material cost
Fixed charges
Variable cost
= P 10.00 per unit
= P 15.00 per unit
= P 10,000.00
= P 8.00 per unit
What is the breakeven point sales volume if one out of every 10 units produced is
defective and is rejected with only full recovery on materials?
A.
B.
c.
D.
P 25,011
P 25,111
P 25,121
P 25,132
Let x = number of units produced per month to breakeven
280
Question Bank
-
Engineering Economics by Jaime R. Tiong
Question Bank
Expenses:
Laborcost
Material
L0
= 10x
cost = 15x
= 8x
Variable cost
,/
Fixed charges = 10,990
Total expenses = 10x + 15x + 8 x + 10,000
Total expenses = 33x + 10,000
Total income = 63x
A. T-bills
B. Bank note
C. Check
D. Coupon
To breakeven:
Total income = total expenses
63x=33x+10,000
30x = 10,000
x = 333.33
say 334 units
[Pffi)(ss+ rnit")
\ unit /'
Breakeven sales volume --P4,042
Breakeven sates votume =
lf 1 out of 10 (10%) is defective and rejected:
Number of units sold per month = 0.90x
Total expenses = 10x + 15x +8 x + 10,000 - 15(0.10x)
Total expenses = 31.5x + 10,000
A.
B.
C
D.
Devaluation
Deflation
lnflation
Depreciation
Posl ME Boord Exom
It is a series of equal payments
occurring at equal interval of time.
= 56.7x
To breakeven:
Posl illE Boord Exom
The place where buyers and sellers
come together.
Total income = total expenses
56.7x = 31 .5x + 10,000
A.
B.
C
D.
25.2x= 10,000
x = 396.8 say 397
Breakeven sates votume =
(ffi)t
nr
r"*1
Breakeven sales volume =P25,011
The breakeven sales volume is P 25,011.00.
Posr
Market
Business
Recreation center
Buy and sell section
ECE
Boqrd Exom
A market whereby there is only one
buyer of an item for which there are no
goods substitute.
Monopsony
Oligopoly
Monopoly
Oligopsony
6. Post E(E Boord Exom
It is a series of equal payments
occurring at equal interval of time
where the first payment is made after
several periods, afterthe beginning of
the payment.
Post ECE Boord Exsm
Reduction in the level of national
income and output usually
accompanied by the fall in the general
price level.
A. Annuity
B. Debt
C. Amortization
D. Deposit
Total income = 63(0.90)x
A.
B
C
D
Post E(E Boord Exom
The paper currency issued by the
Central Bank which forms part of the
country's money supply.
A. Perpetuity
B. Ordinary annuity
C. Annuity due
D. Deferred annuity
t.
Posl lhE Boord Exom
The total income'equals the total
operating cost.
A.
B.
C.
D.
Balanced Sheet
ln-place value
Check and balance
Break even - no gain no loss
8. Post IllE Boord Exom
Kind of obligation which has no
condition attached.
A. Analytic
B Pure
C. Gratuitous
D. Private
9. Posl ftlE Boord Exonr
Direct labor costs incurred in the
factory and direct material costs are
the costs of all materials that go into
production. The sum of these two
direct costs is known as
A.
B.
GS and A expenses
Operating and maintenance
costs
282
Question Bank
C.
D.
-
Engineering Economics by Jaime R. Tiong
Prime cost
OandMcosts
Posl tE Boord Exomm
An index of short tem paying ability is
called
10.
I
I.
lf.
receivable turn-over
profit margin ratio
current ratio
acid-test ratio
Post
ilE
Boord lxom
infinite period of time.
A
B
D
I7.
Demand
B.
Supply
c.
D.
Stocks
Goods
Posl
tE
A.
B.
c.
D.
President
Board of Directors
Chairman of the Board
Stockholders
Posl
tE
A.
B.
c.
19.
A.
B
Posl f,lE Boord Exom
Additional information of prospective
bidders on contract documents issued
prior to bidding date.
c
D.
Equitable
Public
Private
Pure
20.
Posl illE Boord Exom
Decrease in the value of a physical
property due to the passage of time.
Technological assessment
Bid bulletin
A.
B.
c.
lnflation
Depletion
Recession
Posl
ECE
26. Porr E(E Boord
Post E(E Boord Exom
It is defined to be the capacity of a
commodity to satisfy human want.
A.
B.
G.
D.
Discount
Luxury
Necessity
Utility
27.
A.
B.
C.
D.
25.
Fair value
Market value
Book value
Salvage value
A. Perfect competition
B. Oligopoly
C. Monopoly
D. Elastic demand
Post
ECE
Boord Exom
somewhat the same quantity even
though the price varies considerably.
28.
A.
B
C.
D.
Utilities
Necessities
Luxuries
Product goods and services
Post
ECE
Boord Erom
A condition where only few individuals
produce a certain product and that any
action of one will lead to almost the
same action of the others.
A.
B.
C.
D.
Oligopoly
Semi-monopoly
Monopoly
Perfect competition
29.
Post ME Boord Exon
Grand total of the assets and
operational capability of a corporation.
A. Authorized capital
B. lnvestment
C. Subscribed capital
D. Money market
Posl E(E Boord Exom
This occurs in a situation where a
commodity or service is supplied by a
number of vendors and there is nothing
to prevent additional vendors entering
the market.
Exom
These are products or services that are
required to support human life and
activities, that will be purchased in
24.
Post ECE Boord Exom
It is the amount which a willing buyer
will pay to a willing seller for a property
whbre each has equal advantage and
is under no compulsion to buy or sell.
283
A. Utilities
B . Necessities
C Luxuries
D. Product goods and services
Company
Partnership
Corporation
Boord Exom
Test
These are products or services that are
desired by human and will be
purchased if money is available after
the required necessities have been
obtained.
23.
Boord Exom
Type of ownership in business where
individuals exercise and enjoy the right
in their own interest.
14.
Sole proprietorship
A. nominal rate
B. rate of return
C. exact interest rate
D. effective rate
Boord Exom
an asset
a liability
an expenses
an owner's equity
A.
B.
C.
D.
We may classify an interest rate, which
specifies the actual rate of interest on
the principal for one year as
Post IllE Boord Exom
What is the highest position in the
corporation?
Post tlE Boord April Exom
Consists of the actual counting or
determination of the actual quantity of
the materials on hand as of a given
date.
Delict
Escalatory
22.
18.
13.
A.
B.
C.
D.
A.
D.
A. Physical inventory
B. Material update
C. Technological assessment
D. Material count
Posl lhE Bocrd Exom
An association of two or more
individuals for the purpose of operating
a business as co-owners for profit.
Work-in process is classified as
Posl tlE Boord Exom
Estimated value at the end of the
useful life.
Market value
Fair value
Salvage value
Book value
Depreciation
Annuity
Perpetuity
lnflation
Posl tE Boord Exom
The quantity of a certain commodity
that is offered for sale at a certain price
at a given place and time.
12.
A.
B.
C.
D.
Depreciation
21.
I6.
An artificial expenses that spreads the
purchase price of an asset or another
property over a number of years.
A. Depreciation
B. Sinking fund
C. Amnesty
D. Bond
D
Fost IilE Boord Exom
A series of uniform accounts over an
c
A
B.
C.
D.
Mull;ipla Ohoico
30.
Posl IIIE Boord Exom
The worth of the property equals to the
original cost less depreciation.
A
B
C.
D.
Scrap value
Face value
Market value
Book value
284
31.
Question Bank
Posl
tE
-
Engineering Economice by Jaime R. Tiong
Eoord Exom
-
- A---- profit margin
B. gross margin
C. net income
D. rate of return
Money paid for the use of borrowed
capital.
A.
B.
C.
D.
Discount
Credit
lnterest
Profit
Post tE Boord Exom
Liquid assets such as cash and other
assets that can be converted quickly
into cash, such as accounts receivable
and merchandise are called
37.
Posr
A. lnitial investment
B. Current accounts
C. Woking capital
D. Subscribed capital
A.
B.
C.
D.
39.
Queslion
A market situation where there is only
one seller with many buyer.
A Monopoly
B. Monopsony
C. Oligopoly
D. Oligopsony
tE
Boord Exom
The provision in the contract that
indicates the possible adjustment of
material cost and labor cost.
Secondary clause
Escalatory clause
Cohtingency clause
Main clause
tl0.
A.
B.
C.
D.
36.
book value
capital recovery
depreciation recovery
sinking fund
Post tlE Boord Exsm
Gross profrt, sales less cost of goods
sold, as a percentage of sales is called
Queslion
A market situation where there are few
sellers and few buyers.
A. Oligopoly
B. Oligopsony
C. Bilateral oligopoly
D. Bilateraloligopsony
35.
Posr tlE Boord Exom
The present worth of all depreciation
over the economic life of the item is
called
Fair value
Market value
Salvage value
Book value
Posi tlE Boord [xom
Those funds that are required to make
the enterprise or project a going
con@rn.
Posl tE Boord Exorn
The length of time which the property
may be operated at a profit.
3f.
A.
B.
C.
D.
38.
33.
A. Physical life
B. Economic life
C. Operating life
D. All of the above
tE
42. Qucrllon
A market situation where there are only
two buyers with many sellers.
A.
B.
C.
D.
Boord Exom
Worth of the property as shown in the
accounting records of an enterprise.
32.
A. total assets
B. fixed'assets
C. current assets
D. none ofthe above
Posl
Mult,iplo Choice
{1.
Queslion
A market situation where there is one
seller and one buyer.
A.
B
C.
D.
Monopoly
Monopsony
Bilateral monopoly
Bilateral monopsony
t[3.
Queslion
Qucrtlon
The payment for the use of borrowed
money is called
A.
B.
C.
D.
48.
ft[.
Queslion
Defined as the future value minus the
present value.
A. lnterest
B. Rate of return
C- Discount
D. Capital
15.
loan
maturity value
interest
principal
Oueslion
The interest rate at which the present
work of the cash flow on a project is
zero ofthe interest earned by an
investment.
earn.
A. Present worth factor
B. lnterest rate
C. Time value of money
D. Yield
285
f7.
Duopoly
Oligopoly
Duopsony
Oligopsony
The cumulative effect of elapsed time
on the money value of an event, based
on the earning power of equivalent
invested funds capital should or will
Test
A. Effective rate
B. Nominal rate
C. Rate of return
D. Yield
19.
Question
The ratio of the interest payment to the
principal for a given unit of time and
usually expressed as a percentage of
the principal.
A. lnterest
B. lnterest rate
C. lnvestment
D. All of the above
Question
The flow back of profit plus
depreciation from a given project is
called
A.
B.
C.
D.
50.
Oueslion
The true value of interest rate
computed by equations for compound
interest for a 1 year period is known as
capital recovery
cash flow
economic return
earning value
f5.
0ueslion
The profit derived from a project or
business enterprisg without
consideration of obligations to financial
contributors or claims of other based
on profit.
A. Economic return
B. Yield
C. Earning value
D. Expected yield
A.
B.
C.
D.
51.
expected return
interest
norninal interest
effective interest
Question
The intangible item of value from the
exclusive right of a company to provide
a specific product or service in a stated
region of the country.
A.
B.
C.
D.
Market value
Book value
Goodwill value
Franchise value
286
52.
Question Bank
-
A.
Oueslion
The recorded current value of an asset
is known as
A.
B.
C.
D.
53.
B.
c.
D.
scrap value
salvage value
book value
present worth
Queslion
Scrap value of an asset is sometimes
Ordinary annuity
Annuity due
Deferred annuity
Perpetuity
A.
value?
55.
B.
c.
D.
Deferred annuity
Delayed annuity
Progressive annuity
Simple annuity
A.
Book value
Going value
c.
Queslion
B.
D.
60.
63.
D.
57.
Market valu
Goodwill value
Fair value
Franchise vblue
Question
are made at the end ofeach payment
period starting from the first period.
Declining balance method
SYD method
67. 0ueslion
Ordinary annuity
Aniruity due
Deferred annuity
Perpetuity
B.
6f.
c.
D.
A.
B.
C.
D.
A
annuity
B.
c.
capital recovery
annuity factor
D.
amortization
68.
The reduction ofthe value of an asset
due to crnstant use and passage of
time
A.
B.
C.
D.
65.
Scrap value
Depletion
Depreciation
Book value
Question
an amount equal to the total
depreciation of an asset at the end of
the asset's estimated life.
A.
B.
C.
D
69.
Straight line method
Sinking fund rnethod
Declining balance method
SYD method
Question
The function of.interest rate and time
that determines the cumulative amount
of a sinking fund resulting from specific
periodic deposits.
A.
B.
C.
D.
Queslion
A method of computing depreciation in
which the annual charge is a fixed
percentage ofthe depreciated book
value at the beginning of the year to
which the depreciation applies.
Declining balance method
Sinking fund method
Straight line method
SYD method
A method of depreciation where a fixed
sum of money is regularly deposited at
compound interest in a real or
imaginary fund in order to accumulate
Queslion
The amounts of all payments are
equal.
The payments are made at equal
interval of time
The first payment is made at the
beginning of each period.
Compound interest is paid on all
amounts in the annuity.
methods cannot have a salvage value
o'f zero?
Queslion
Oueslion
A.
Book value
Fair value
Goodwill value
Going value
Question
The value which a disinterested third
party, different from the buyer and
seller, will determine in order to
establish a price acceptable to both
parties.
c.
demand factor
sinking fund factor
present worth factor
As applied to a capitalized asset, the
distribution of the initial cost by a
periodic changes to operation as in
depreciation or the reduction of a debt
by either periodic or irregular
prearranged program is called
Which is NOT an essential element of
an ordinary annuity?
56.
B.
B.
D.
operation.
A.
load factor
c.
Question
A type of annuity where the payments
are made at the start of each period,
beginning from the first period.
Scrap value
Salvage value
An intangible value which is actually
operating concern has due to its
A.
B.
C.
D.
A.
59.
Ouestion
What is sometimes called second hand
C
D
Questlon
A mathematical expression also known
as the present value of an annuity of
one is called
Which of the following depreciation
Ouestion
It is a series of equal payments
occurring at equal intervals of time
where the first payment is made after
several periods, after the beginning of
the payment.
book value
salvage value
replacement value
future value
51.
A.
B.
C.
D.
62.
58.
known as
A.
B.
C.
D.
Multiple Choice Test 287
Engineering Economics by Jaime R. Tiong
70.
Sinking fund factor
Present worth factor
Capacity factor
Demand factor
Queslion
The first cost of any property includes
61.
Queslion
A.
Sfaight line method
B.
Sinking fund method
SYD method
Declining balance method
A is a periodic payment and I is the
interest rate, then present worth of a
c.
perpetuity =
D.
A. Ai
B. AiN
c. An/
D. Ni
65.
Queslion
A method of depreciation whereby the
amount to recover is spread uniformly
over the estimated life of the asset in
terms of the periods or units of output.
A
B
Straight line method
Sinking fund method
A.
B.
C.
D.
the original purchase price and
freight and transportation
charges
installationexpenses
initial taxes and permits fee
all of the above
Tl.Oueslion
ln SYD method, the sum of years digit
is calculated using which formula with
n = number of useful years ofthe
equipment.
288
Question Bank
n.
"
n(n
o
C.
D.
n(n+1)
o.t
72.
-
76.
Queslion
A distinct legal entity which can
practically transact any business
transaction which a real person could
-1)
2
2
n(n +1)
n(n
do.
A.
B.
C.
D.
-1)
Queslion
Capitalized cost of any property is
equal to the
A.
B.
C.
D.
73.
Mult,iplo Choico
Engineering Economics by Jaime R. Tiong
Which is TRUE about PartnershiP?
Sole proprietorship
C.
Enterprise
Partnership
Corporation
D.
77.
annual cost
first cost + interest of the first
cost
first cost + cost ofperpetual
maintenance
first cost + salvage value
Question
Double taxation is a disadvantage of
which business organization?
A.
B.
C.
D.
A.
B.
C.
78.
Sole proprietorship
Partnership
Corporation
Enterprise
A.
B
79.
Sole proprietorship
Corporation
Enterprise
Partnership
Queslion
What is the minimum number of
incorporators in order that a
75.
Question
An association of two or more persons
for a purpose ofengaging in a
profitable business.
A.
B.
C.
D.
Sole proprietorship
4.3
B.5
c. 10
i
D.7
80.
Ouestion
I ln case of bankruptcy of a partnership,
A
B
Enterprise
Partnership
Corporation
The minimum number of
incorPorators to start a
Ouestion
Represent ownershiP, and enjoYs
certain Preferences than ordinary
stock.
A.
B.
C.
D.
Authorized capital stock
Preferred stock
Common stock
lncorporator's stock
partnership
the partners may sell stock to
generate additional capital.
A. Bond
B. T-bill
C. Preferred stock
D. Common stock
87.
Queslion
A form of fixed-interest security issued
by central or local governments,
companies, banks or other institutions.
They are usually a form of long-term
security, buY maY be irredeemable,
secured or unsecured
A.
B.
C.
D
Queslion
Represent the ownershiP of
stockholders who have a residual claim
on the assets ofthe corporation after
all other claims have been settled.
A. Authorized caPital stock
B. Preferred stock
C. lncorPorators stock
D. Common stock
85.
Question
The amount of comPanY's Profits that
the board of directors of the
corporation decides to distribute to
ordinary shareholders.
Bonds
T-bills
Certificate of dePosit
All of these
88.
Queslion
A type of bond where the corporation
pledges securities which it owns (i.e.
stocks, bonds of its subsidiaries).
A
B.
"C
D.
84.
liabilities.
D
mortgage on certain assets of the
corporation.
83.
the partners are not liable for the
liabilities of the partnership.
the partnership assets (excluding
the partners personal assets)
only will be used to pay the
the partners personal assets are
attached to the debt of the
A certificate of indebtness of a
corporation usually for a period not less
than '10 years and guaranteed bY a
corPoration are onlY liable to the
extent of their investments.
corporation be organized?
B. Partnership
C. Enterprise
D. Corporation
lt is the not best form of business
organization.
C. lts life is dependent on the lives
of the incorPorators.
D. The stockholders of the
organization?
A.
B.
C.
D.
lts capitalization must be equal
for each Partner.
corPoration is three.
Question
Which is NOT a type of business
Depreciation
Depletion
lnflation
another.
86. 0uestion
82.-Queslion
Which is TRUE about corporation?
Ouestion
The lessening ofthe value ofan asset
due to the decrease in the quantity
available (referring to the natural
resources, coal, oil, etc).
lt has a PerPetual life.
lt will be desolved if one of the
partners ceases to be connected
with the PartnershiP.
lt can be handed down from one
generation of Partners to
289
A. Dividend
B. Return
C. Share stock
D. Par value
81. Oucrtion
A.
B.
Test
Mortgage bond
Registered bond
CouPon bond
Collateral trust bond
89.
Question
A type of bond which does not have
security except a promise to pay by the
issuing corPoration.
A. Mortgage bond
B. Register bond
C. Collateral trust bond
D. Debenture bond
I
290
Question Bank
-
90. 0uestion
A type of bond issued jointly by two or
more corporations.
A.
B.
C.
D.
91.
5 years
6 years
7 years
B. lnvestment in subsidiarY
companles
C. Furnitures
D. Patents
l0I.
96.'Question
Debenture bond
Registered bond
Collateral trust bond
fhe 72 rule of thumb
Queslion
Lands, buildings, plant and machinery
are examples of
is used to
determine
A.
B.
C.
D.
Question
Equipment obligations bond
Debenture bond
Registered bond
lnfrastructure bond
how many
hew many
dquble
how many
million
how many
money
years money wlll triple
years money will
years to amass
1
years to quadruple the
97.
Question
To triple the principal, one musl,rse
Question
lf the security of the bond is a
mortgage on certain specified asset of
a corporation, this bond is classified as
A. integration
B. derivatives
C. logarithms
D. implicit functions
A. registered bond
B. mortgage bond
C. coupon bond
D. joint bond
98.
Queslion
A currency traded in a foreign
exchange market for which the
demand is consistently high in relation
to its supply
93.
Queslion
A type of bond where the corporation's
owners name are recorded and the
interest is paid periodically to the
owners with their asking for it.
A. Money mprket
B. Hard currency
C. Treasury bill
D. Certificate of deposit
99. Queslion
A Registered bond
B. Preferred bond
C. lncorporators bond
D. All of these
l
9tl.
Question
Bond to which are attached coupons
indicating the interest due and the date
when such interest is to be paid.
A.
B.
C.
D.
.95.
4 years
B
C
D
92.
.
A
Joint bond
A type of bond whose guaranty is in
lien on railroad equipments.
A.
B.
C.
O.
Multiple ()hoice'l'est
Engineering Econornics by Jaime R. Tiong
Registered bond
Coupon bond
Mortgage bond
Gollateral trust bond
Everything a company owns and which
has a money value is classified as an
asset. Which of the following is"
classified as an asset?
A. lntangible assets
B. Fixed assets
C. Trade investments
D. All of these
100. Queslion
Which an example of an intangible
Question
An amount of money invested at
interest per annum will double in
approximately
ASSCT?
12o/o
A.
Cash
A. current assets
B. trade investments
C. fixed assets
D. intangible assets
102. Oueslion
An increase in the value of a capital
asset is called.
A. Profit
B. caPital gain
C. caPital exPenditure
D. capital stock
103. 0ueslion
The reduction in the money value of a
capital asset is called
A. capital exPenditure
B. caPital loss
C. loss
D. deficit
104. 0uestion
It is a negotiable claim issued bY a
bank in lieu of a term dePosit.
A. Time dePosit
B. Bond
C. CaPital gain
D. Certificate of dePosit
106. Oucrtion
It denotes the fall in the exchange rate
of one currency in terms of others The
term usuallY aPPlies to floating
exchange rates.
A. CurrencYaPPreciation
B. CurrencYdevaluation
C. CurrencY float
D. CurrencYdePreciation
107. 0ueslion
The deliberate lowering of the price of
a nation's currency in terms of the
accepted standard (Gold, American
dollar or the British Pound).
A. CurrencY aPPreciation
B. CurrencY float
C. CurrencY devaluation
D. CurrencYdePreciation
108. Queslion
The residual value of a comPany's
assets after all outside liabilities
(shareholders excluded) have been
allowed for.
A. Dividend
B. Equity
C. Return
D. Par value
109. Question
A saving which takes Place because
goods are not available for
consumption rather than the consumer
really want to save.
A.
B.
C.
D.
105. Queslion
Any particular raw material or primary
product (e.9. cloth, wool, flour,
coffee...) is called
A. utility
B. necessitY
C. commoditY
D. stock
291
ComPulsory saving
Consumer saving
Forced saving
All of these
ll0.
Question
A document that shows proof of legal
ownershiP of a financial securitY.
A.
B.
C.
D.
Bond
Bank note
CouPon
Check
292
Question Bank
-
l.
II
Question
Defined as the capacity of commodity
to satisfy human want.
A.
B.
C.
D.
I
13.
increase, increasing the other factors
of production will result in a less than
proportionate increase in output',.
A.
B
c.
D.
II
Debenture
Goodwill
Capital gain
lnternal rate of return
Demand
Suppty
Utility
Market
B.
D.
I
I 15. Question
" When free competition exists, the
price of a product will be that value
where supply is equal to the demand,,.
A.
B.
C.
D.
Lawofdemand
Law of supply and demand
Sunk cost
B.
Opportunity cost
Replacement cost
lnitial cost
supply
demand
supply and demand
D.
Cost of goods sold
Variance
Overhead
Payback
Ouostion
The simplest economic order quantity
(EOQ) model is based on which of the
following assumptions?
A.
B.
C.
Shortages are not allowed.
Demand is constant with respect
to time.
Reordering is instantaneous. The
time between order placement
and receipt is zero.
All of the choices
ll9.
Question
ln economics, a "short-term,,
transaction usually has a lifetime of
122.
Quesrion
ln replacement studies, the existing
process or Piece of equiPment is
known as
B.
challenger
defender
c.
liability
D.
asset
l23.0uestion
ln replacement studies, the new
process or piece of equipment being
considered for purchase is known as
A. challenger
B. defender
C. asset
D. liabilitY
124. 0uestion
mgans that the cost of the
isset is divided into equal or unequal
parts, and olrly one of these parts is
taken as an bxPense each Year.
A.
B.
A.
B.
C.
D.
3 months or less
1 year or less
5 years or less
10 years or less
l20.Ouestion
Law of diminishing return
Law ofsupply
A.
c.
18.
Question
Demand
Supply
Market
Utitity
Oucclion
An imaginary cost representing what
will not be received if a particular
strategy is rejected
A.
A.
The quantity of a certain commodity
that is bought at a certain price at j
given time and place.
A.
B.
C.
D.
I2l.
diminishing return
Qucstion
An accounting term that represents an
inventory account adj ustment.
Question
A.
B.
C.
D.
Law of
Law of
Law of
Law of
7.
C
The quantity of a certain commoditv
that is offered for sale at a certEin price
at a given time and place.
llf.
I 16. Question
"When one of the factors of production
is fixed in quantity or is difficult to
Discount
Necessity
Luxuries
Utility
I 12. Question
It is the proflt obtained by selling stocks
at a higher price than its original
purchase price.
A.
B.
C.
D.
Multiple Choice Teet 293
Engineering Economics by Jaime R. Tiong
ln the cash flow, expenses incurred
before time = 0 is called
A.
receipts
B.
disbursements
c.
sunk costs
first costs
D.
c.
D.
Capitalizing the asset
Expensing the asset
Depreciating the asset
Artificial expense
125. Question
lndicate the CORRECT statemenl
about dePreciation.
A.
B.
The depreciation is not the same
each year in straight line method.
The declining balance method
can be used even ifthe salvage
value is zero.
c
The sum-of-years' digit rnethod
(SYD), the digits 1 to (n + 1) is
summed.
D.
Double declining balance
dePreciation is indePendent of
the salvage value.
126. Quegtion
An artificial deductible operating
expense designed to compensate
mining organizations for decreasing
mineral reserves.
A.
B.
C.
D.
Deflation
Reflation
DePletion
lnflation
127. Queslion
The change in cost per unit variable
change is known as
A.
sunk cost
B.
incrementa! cost
fixed cost
semi-variable cost
c.
D.
128. Question
What type Qf cost increases step-wise?
A. SuPervision cost
B. Direct labor cost
C. Semi-vaiiable cost
D. Operating and maintenance
cost
129. Oueslion
Which of the following is NOT a
variable cost?
A. Cost of miscellaneous
B. lncome taxes
C. Payroll benefit costs
D. lnsurance costs
suPPlies
130. Oucstion
Which of the following is NOT a fixed
cost?
A. Rent
B. Janitorial service exPenses
C. Supervision costs
D. DePreciationexPenses
294
Question Bank
l3l.
Question
-
The annual costs that are incurred due
to the functioning of a piece of
equipment is known as
A
c.
General, selling and
administrative expenses
Prime cost
Operating and maintenance
D.
Total cost
B.
Multiplo Choicc'l'est 295
Engineering Economics by Jaime R. 't'iong
COSTS
I36.
Ouestion
Which of the following is NOT a direct
labor expense?
A.
B.
C.
D.
B.
c.
D.
lnspection
Testing
Supervision
The sum ofthe direct labor cost and
the direct material cost is known as
A.
B.
c.
prime cost
total cost
indirect manufacturing expenses
total cost
\
I33. Oueslion
Resparch and development costs and
adm\nistrative expenses are added to
the fEctory cost to give the
of the product
A Asset accounts
B. Bank accounts
C. Liability accounts
D. Owner's equitY accounts
All are administrative expenses
A. total cost
B. marketing cost
C. manufacturing cost
D. prime cost
l3t[.
0uesrion
The sum of the prime cost and the
indirect manufacturing cost is known as
A.
B.
c.
D.
I35.
factory cost
research and development cost
manufqcturing cost
total cobt
Question
The manufacturing cost plus selling
expenses or marketing expenses
equals
A.
total cost
B.
indirect production cost
c.
administrative cost
miscellaneous cost
D.
Ouestion
The journal and the ledger together are
of the
known simply as
company.
A. accounting system
B. the books
C bookkeePing system
D. balance sheet
I44.
A. Testing
B. Drafting
C. Prototype
D. Laboratory
Which is not a factory overhead
expense?
A. Pension, medical, vacation
benefits
B. Expediting
C. Quality control and inspection
D. Testing
l4L Queslion
Bookkeeping consists of two steps,
namely recording the transactions and
categorization of transaction6 Where
are the transactions (receipts and
disbursements) recorded?
149. Oueslion
The ratio of the net income to the
owner's equity is known as
A. price-earning ratio
B. profit margin ratio
C. return of investment
D. gross margin
Question
A. Assets = Liability + Owner's
equitY
B. Liability = Assets + Owner's
equitY
C. Owner's equity = Assets +
LiabilitY
D. Owner's equitY = LiabilitY -
Research and development expenses
includes all EXCEPT one. Which one?
140. Queslion
A gross profit to net sales
B. net income before taxes to net
sales
C. quick assets to current liabilities
D. net income to owner's equitY
The basic accounting equation is
139. Question
I50.
tl5.
0uestion
Payback period is the ratio of
A. initial investment to net annual
profit
B cost ofgoods sold to average
cost of inventory on hand
C. gross profit to net sales
D. net income before taxes to net
Assets
I
Oueslion
An acid test ratio is a ratio of
-
One of the following is NOT a selling or
marketing expense. Which one?
Advertising
Commission
lnsurance
Transportation
I48.
lf3.
138. Queslion
A.
B.
C
D.
A Current ratio
B. Acid test ratio
C Gross margin
D. Return of investment
EXCEPT:
137. Question
A. Marketing
B. Accounting
C. Data processing
D. Office supplies
short{erm paying abilitY?
Columnar
Statement of account
The following are ledger accounts
Assembly
Qucction
What is considered as an index of
142. Queslion
EXCEPT:
132. Queslion
I{7.
Joqrnal
Lpdger
A.
sales
Queslion
The ability to convert assets to cash
quickly is known as
A.
B.
C.
D
I46.
solvency
liquidity
leverage
insolvencY
Queslion
The ability to meet debts as theY
become due is known as
A
B.
C.
D
solvencY
leverage
insolvencY
liquidity
I
5l
.
Ouestion
A secondary book of accounts, the
information of which is obtained from
the journal.
A.
B.
C.
D.
Balance sheet
Ledger
Worksheet
Trial balance
152. Queslion
The present worth of cost associated
with an asset for an infinite period of
time is referred to as
296
Question Bank
A.
B.
C.
D.
-
annual cost
capitalized cost
increment cost
operating cost
153. Oueslion
A stock of a product which is held by a
trade body or government as a means
of regulating the price of that product.
A.
B.
C.
D.
158. Quesrion
Refers to the order quantity that
minimizes the inventory cost per unit
time.
A.
B.
C.
D.
Economic order quantity
Social order quantity
Public order quantity
Private order quantity
159. Ouestion
Stock pile
Hoard stock
Buffer stock
Withheld stock
What is referred to as an individual
who organizes factors of production to
undertake a venture with a view to
proflt?
154. Ouestion
A negotiable claim issued by a bank in
lieu of a term deposit is called
A.
Cheque
B.
T-bills
Currency
Certificate of deposit
c.
D.
155. Question
firm which is owned
'
of individuals for
B.
C.
D.
Corporation
Enterprise
Partnership
156. Queslion
A document which shows the legal
ownership of financial security and
entitled to payments thereon.
A.
B.
C.
D.
Coupon
Contract
Bond
Consol
A. Agent
B. Enterpreneur
C. Salesman
D. Commissioner
l60.Ousstion
The money that is inactive and doeS
not contribute to productive effort in an
economy is known as
A.
B.
C.
D.
idle money
hard money
soft currency
frozen asset
16l.
Qucstion
ln counting the number of days when
computing simple interest,
A. the first day is included
B. the last day is excluded
C. the first day is included and the
last day is excluded
D. the first day is excluded and the
last day is included
162. Queslion
ln the so-called "Banker's Rule",
157. Queslion
A government bond which have an
indefinite life rather than a specific
maturity.
A.
B.
C.
D
Coupon
T-bi[
Debenture
Consol
Mtrlt,iplo (lhoice'l'est 297
Engineering Econornics by Jaime R. Tiong
A. the number of days in 1 year is
360 days
B. the number of days in 1 year is
365 days
C. the number of days in each
month is 30 days
D. the number of days in 'l year is
366 days
l67.0uortlon
163. 0ucstlon
To discount an amount F for n
conversion periods means
A.
B.
to find the present
which is n periods
to find the present
which is n periods
C.
to find the present value on a day
which is (n-1) periods before F is
D.
value on a day
after F is due
value on a day
before F is
due
due
to find the present value on a day
which is (n+'l) periods before F is
due
164. Question
ln the formula for compound interest, F
= P(1 + i)n, the value (1 + i)n is called
A. discount factor
B. interest factor
-C.
accumulation factor
D. increase factor
165. Oueslion
To find the present worth of a future
amount in compound interest, we use
the formula P = F(1 + i)-n . What do you
calt the
factor
(1 +
i)'n ?
A. discount factor
B. accumulation factor
C interest factor
D. reductioh factor
166. Queslion
What refers to an equation stating that
the sum of the values, on a certain
comparison date, of one set of
obligations is equal to the sum of the
value of another set of this date?
A. Equality of value
B. Equation of value
C. Equality equation
D. Similarity equation
What is an annuity whose payments
extend over a period of time whose
length cannot be foretold accurately?
A. Annuity ce(ain
B. Annuity uncertain
C. lncremental annuity
D. Contingent annuity
I68.Ouestion
What do you call the time between
successive payment dates of an
annuity?
A.
B.
C.
D.
Period interval
Annuity period
Payment interval
Annuity term
169. Question
The time from the beginning of the first
payment interval to the end of the last
of the annuity.
one is called the
_
A. period
B. term
C. nature
D. type
170. Question
What iefers to the extinction of the debt
by any satisfactory set of payments?
A. Liquidation
B. Liability discharge
C. Discharging of debt
D. Amortization of debt
I 71. Question
What do you call a fund, usually by
periodic deposits, to insure the
accumulation of money to provide for a
possible large payment?
A.
B.
C.
D.
Escrow fund
Sinking fund
Mutual fund
Corporate fund
298
I
72.
Question Bank
-
Queslion
B.
What is the term for the borrowed
principal usually mentioned in a typical
C.
bond?
A.
B.
C.
D.
I
73.
Mrrlt,ipkr ()hoicc
Engineering Economics by Jaime R. Tiong
D.
Bond rate
Face value or par value
Coupon rate
Coupon value
I
Queslion
Any date on which a coupon of a bond
becomes due will be referred to as a
When the price of the bond is
less than the redemption value
When the price of the bond is
equal to the redemption value.
When the price of the bond is
either equal to or greater than
the redemption value.
77.
Question
Which of the following will happen if a
bond is bough at a discoqnt?
C and-interest price
D. coupon price
I8l. 0ueslion
^
A
U.
A.
B.
C.
D.
maturity date
term of the bond
0ueslion
lf P is the price of a bond and V is its
redemption value, what do you call the
value P - V?
I
75.
Par value
Face value
Premium
Bond discount
D
The unpaid interest on each
coupon date will not be
considered as a new investment
in the bond.
The difference between the
coupon payment and the interest
due is a pa(ial repayment of
principal.
I
78. 0uestion
ln the sale of a bond, the actual
purchase price on any day is called
Queslion
When can we say that the bond is
A.
purchased at a discount?
B.
A.
B.
C.
I
When the price of the bond is
greater than the redemption
value.
When the price of the bond is
less than the redemption value
When the price of the bond is
equal to the redemption value.
When the price of the bond is
either equal to or greater than
the redemption value.
76. 0ueslion
When can we say that the bond is
purchased at a premium?
A.
the person dies
B
C
D
When the price of the bond is
greater than the redemption
value.
C.
D.
I
Face value
Quoted price
Accrued price
Flat price
per period?
A
B
C.
D
79.
Queslion
What do you call the difference
between the flat price of the bond and
the quoted price ofthe bond?
Accrued interest
c.
Bond rate
D
And-interest price
180. Queslion
The quoted price of a bond is
sometimes called
B
A
B.
What is the term for the sum of
depreciation charges to date?
Question
The difference between the value of an
asset and its salvage or scrap value at
the end of n year is called
C
^
U.
Par value
Face value
A.
B
C.
D.
depreciation
accrued value
book value
wearing value
185. Ouestion
What is a life annuity?
quick assets
current liabilities
net credit sales
average net receivables
gross profit
current liabilities
gross profit
net sales
I88.
-
0ueslion
Which of the following represents the
gross margin?
A
Par value
A.
B
A
The acid test ratio is also known as
quick ratio. Which one represents the
quick ratio?
183. Queslion
I84.
Yield
Nominal rate
Fixed rate
Net rate
187. Question
A. Annuity bond
B. Serial bond
C. Treasury bond
D. Government bond
A. Accrueddepreciation
B. Applied depreciation
C Accumulateddepreciation
D. All of the above
indefinitely.
It is the same as perpetuity
What term is usually used by the banks
to represent the effective interest rate
182. Question
What is a bond whose face value is
redeemable in installments, with
interest payable periodically as due on
outstanding principal?
A sequence of payment intended
for the life insurance of a person.
A sequence of payment for a
certain person which continues
186. Sueslion
coupon date
due date
l7tl.
A.
B.
C.
D.
average investment
average annual interest
average annual interest
average investment
par value
flat value
flat value
quoted value
299
A sequence of payment for a
certain person which stops when
The yield of a bond is obtained by
which of the following formulas:
^
'l'est
B
C
D
net income
owners'equity
net credit sales
average net receivables
gross profit
current liabilities
gross profit
net sales
300
Question Bank
-
189. Question
A receivable turnover is calculated
using which of the following formulas?
A.
B.
C
D.
A.
^
U.
owners'equity
net credit sales
average net receivables
gross profit
current liabilities
gross profit
net sales
The percentage of each peso of sales
that is net income is called
l9l.
Equilibrium market price
Fair market price
Real market price
Exact market price
net income
190. Ouestion
A.
B.
C.
D.
price-earning ratio
profit margin
profit margin ratio
return of investment ratio
194. Queslion
A principle that states that consumers
will tend to spend an increasing
proportion of any additional income
upon luxury goods and a smaller
proportional on staple goods; so that a
rise in income will lower the overall
share of consumer expenditures spent
on stable goods (such as basic
foodstuffs) and increase the share of
consumer expenditures on luxury
goods (such as motor cars)
A.
B.
C.
D.
Placibo effect
Luxury effect
Engel's law
Staple law
Question
Which one represents that priceearnings ratio?
A
market price per share
earnings per share
B
earnings per share
market price per share
C
net credit sales
average net receivables
gross profit
current liabilities
192. Ouestion
The book value per share of common
stock is the ratio of the common
shareholdersl equity to
A average shares
B. number of outstanding shares
C. total subscribed shares
O. authorized capital stock
I93.
Mult,iplo Ohoico
Engineering Economics by Jaime R. 'tiong
Question
What refers to the price at which the
quantity demanded of a good is exacfly
equal to the quantity supplied?
D
198. Question
A market where new entrants face
costs similar to those of established
firms and where, on leaving, firms are
able to recoup their capital costs, less
depreciation.
A.
B.
c
D
Th{ory of values
B.
Ec5nometrics
Economatics
Econoscience
c.
D.
l96.Oueslion
What refers to the fall in the general
price level, frequently accompanied by
a reduction in the level of national
income?
A.
B.
C.
D.
lnflationary gap
Dissavings
Disinflation
lnflation
I97.Ouestion
A price for a product which just covers
its production and distribution costs
with no profit margin added.
A.
B
C
Cost price
Actual price
Real price
Free market
Competitive market
Limited market
Contestable market
199. Question
What refers to a temporary grouping of
independent firms, organization and
governments, brought together to pool
their resources and skills in order to
undertake a particular project?
A.
B.
C.
D.
200.
A.
Original price
Consortium
Cartel
Cooperative
Union
Question
What refers to a market for buying and
selling of raw materials such as tea,
coffee, iron ore, etc.?
A.
B.
C.
D.
Commodity market
Raw market
Natural market
National market
Tost
301
302
Question Bank
-
Engineering Economics by Jaime R. Tiong
Chapter
1
ANSWERS
1.
2.
3.
4.
5.
6.
7.
8.
9.
B
B
A
A
A
D
D
C
C
'10. D
11. A
12. C
13. A
14. D
't5. c
16. B
't7. A
18. C
19. C
20. D
21. C
22. D
23. B
24. B
25. A
26. C
27. B
28. A
29. A
30. D
31. C
32. C
33. B
34. B
35. C
36. B
37. D
38. C
39. A
40. c
41.C
42.C
43.C
44.C
45.B
46.4
47.C
48.C
49.B
50. D
51. D
52.
C
B
B
D
C
53.
54.
55.
56.
57. A
58. A
59. B
60. c
61. D
62. D
63. D
64. C
65. D
66. A
67. A
68. B
69. A
70. o
71. B
72. C
73. B
74. A
75. C
76. D
77. C
78. C
79. B
80. c
81. B
82. D
83. B
84. D
85. A
86. A
87. A
88. D
89. D
90. A
91. A
92. B
93. A
94. B
9s. c
96. B
97. C
98. B
99. D
121.
122.
123. A
124.
125.
126.
127.
128.
129.
130.
131.
163. B
164. C
165. A
166. B
C
168.
169.
170.
171.
172.
173.
174.
D
C
C
D
143.
B
B
145.
B
144 A
't46. A
147. A
148. C
149. C
150. A
B
152. B
153. C
154. D
155. A
156. A
157. D
158. A
159. B
160. A
D
162. A
167
141. A
15't.
161.
B
135. A
136. C
137. A
138. C
139. D
142.
C
C
134. A
102. B
103. B
104. D
105. C
106. D
107. C
108. B
109. C
110. C
111. B
112. C
113. B
114. A
115. D
116. A
117. A
118. D
120.
D
133. C
140.
C
A
132. A
't00. D
't01. c
't19.
B
B
't75.
Di
C
B
D
B
B
C
C
B
176. A
177. A
178. D
179. B
180. C
181. B
182. B
183. A
184. D
185. A
186. A
187. A
188. D
189. B
190. C
191. A
192. B
't93. A
194. C
195. B
196. C
197. A
't98.
D
199. A
200. A
What is Economics?
Adam Smith (1723 - 1790), a Scottish
professor of philosophy at Glasgow
University, whose writings "An lnquiry into
the Nature and Causes of the Wealth of
Nations in 1776, formed the basis of
classical economics defined economics as
the study ofthe nature and causes of
national wealth or simply as the study of
wealth.
Alfred Marshall (1842 - 1924), an English
professor of politieal economy at
Cambridge University who depicted
precise mathematical relationships
between economic variables in his
textbook "Principles of Economics" in
'1890, defined economics as the study of
man in the ordinary business of life.
Arthur Cecil Pigou (1877 - 1959), an
English student of Professor Albert
Marshall, who succeeded him as professor
of political economy at Cambridge
University and devefoped the theory of
welfare economlcs in his book 'The
Economics of Welfare" in 1919, defined
economics as the study of economic
welfare which can be brought, directly or
indirectly, into relationship with the
measuring-rod of money.
Lionel Robbins, an economist in the
tradition of the English classical school
defined economics as the science which
studies human behavior as a relation
between ends and scarce means have
other alternative uses.
Collins dictionary of Economics defined
economics as the study of the problems of
using available factors of production as
efficiently as possible so as to attain the
maximum fulfillment of society's unlimited
demands for goods and services.
While the standpoints of well-known
economists and professors of economics
to the definition of the subject "economics"
may vary, the main idea remained that in
every human activity, in one way or
another, ie connected with either money or
time or both.
'ln its simplest definition, economics is the
wise use of both money and time.
What is Monev?
There are a lot of definitions of money.
However the most aceeptable was that of
Geoffrey Crowther who describes the
major functions of money.
According lo him, money is anything that is
generally acceptable as a means of
exchange (i.e. as a means of settling
debts) and at the same time acts as a
measure and a store of value.
The money which is now used as a
medium of exchange on local, national and
international markets has evolved from
shells, elephant tusks, animal skins to
metallic money (i.e. coins) to paper money
after the invention of printing press in 18tn
century to finally bank money (i e.
currency) with the development of the
banking system.
With the currency system, a standard unit
called "monetary unit" forms the basis of a
country's domestic money supply ln the
Philippines, the monetary unit is peso
while dollar and pounds for the United
States and United Kingdom, respectively.
The physical form of the money supply
(bank ncites, coins, etc.), the denomination
of the values of monetary units (peso and
centavos) and the total size of the money
supply of the country are regulated
through the policies and guidelines of what
is known aslhe "monetary system".
Money can fulfill the following functions:
A. As a medium ofexchange.
Goods and services can be
exchanged for money, which can be
exchanged for other goods and
servtces.
304
Question Bank
-
Engineering Economics by Jaime R. Tiong
As a unit of account.
The units in which money is
measured (pesos, dollars, pounds,
etc.) are used as the units in which
prices, financial assets, debts,
accounts, etc. are measured.
c
'l'hr'orir.s rttttl l'\rt'ntulas 305
Cowumer goods and services are those
As a store of value.
satisfy
much as
form in order to yield future
consumption. Since money grows
with time, the money held over a
period of time as a store of value or
purchasing power.
What is Enqineerino Economv?
goods and services. Exarnpres
goods and services are factory builOings,
machine tools, airplanes, ships, busesl
etc.
"iJffJ"",
What is the difference between
Necessitv and LuxuEfGoods and services are divided into two
types, necessities and luxuries.
Necessities refer to the goods and services
that are required to support human life,
needs and activities.
Engineering economy may also be defined
Necessity product or staple product is
defined as any product that has an
income-elasticity of demand less than one
This means that as income rises,
proportionately
What ar6 9o.nsumer and producer
Goods and Services?
Good or commodity is defined as any
tangible economic product (soap, cir,
shirts, tools, machines, etc.) that
contributes direcfly or indirecfly to the
satisfaction of human wants.
Seryice is defined as any tangible
economic activity (hairdressing, insurance,
banking, catering, etc.) that contribute
directly or indirecfly to the satisfaction of
human wants.
Econo
either
"produ
nd seruices as
services,'or
s',.
such products.
products includ
bread'and rice, clothing, etc.
n
Luxuries are those goods and services that
are de-sired by human and will be acquired
only after all the necessities have been
satisfied.
I.uxury producr is defined as any product
that has an income-elasticity of demand
greater than one. This means that as
income rises, proportionately more income
is spent on such products.
Examples of luxury products includes
consumer durables like electric
appliances, expensive cars, holidays and
entertainment, etc.
Necessities and luxuries are relative terms
because there are some goods and
seryices which may be cJnsidered by one
person as necessity but luxury to another
person.
For example, a man living in the
Metropolis finds a car as an absolute
necessity for him to be able to go to his
workplace and back to his home. lf the
same person lived and worked in another
city, less populated with adequate means
of public transportation available, then a
car will become a luxury for him.
@
Situalhons?
The term "market" refers to the exchange
mechanism that brings together the sellers
and the buyers of a product, factor of
production or financial security. lt may"also
refer to the place or area in which buyers
and sellers exchange a well-defined
commodity.
l'rt'f tt'l t tnqrclition (also known as
ttlttttti:;lit' t'omltetitionl refers to the market
situation in which any given product is
supplied by a very large number of
vendors and there is no restriction against
additional vendors from entering the
market
Perfect competition is a type of market
situation characterized by the following:
A.
Many sellers and manY buYers:
Since there is a large number of
sellers and a large number of
buyers, each seller and buYer will
become sufficientlY small to be
unable to influence the Price of the
Product transacted.
B.
Homogeneous Products: The
products offered bY the comPeting
sellers are identical not only in
physical attributes but are also
regarded as identical by the buyers
who have no Preference between
the products of various producers.
C.
Free market-entry and exit: There
are no barriers to entry or
impediments to the exit of the
existing sellers.
D.
Perfect information: All buyers and
all sellers have complete information
on the prices being asked and
offered in all other Parts of the
market
Biyer or consumer is defined as the basic
consuming or demanding unit of a
commodity. lt maY be an individual
purchaser of a good or seryice, a
household (a group of individuals who
make joint purchasing decisions)' or a
government.
Selleris defined as an entity which makes
product, good or service available to buyer
or consumer in exchange of monetary
consideration.
The price of any commodity or product will
depend largely on the market situation.
The following is a tabulation of the differenl
market situations:
E. Absence
of all economic friction:
There is a total absence of economic
friction including transport cost from
one Part of the market to another.
This market situation Provides an
assurance of complete freedom on the
part of both the vendors and the buyers
though the latter benefits more from the
reduced prices brought about by the
competition while more and better services
are afforded by the vendors or players in
the industry.
306
Question Bank
-
Engineering Economics by Jaime R. ,l,iorrg
Monopoly is the opposite of perfect
competition. This market situation is
characterized by the following:
A.
B.
C.
One seller and many buyers: A
market comprised of a single
supplier selling to a multitude of
small, independenfl y-acting buyers.
Lack of substitute products: There
are no close substitutes for the
monopolist's product.
Blockaded entry: Barriers to entry
are so severe that it is impossible for
other sellers to enter the market.
There exists a perfect monopoly if lhe
single vendor can prevent the entry of all
other vendors into the market. The
monopolist is in the position to set the
market price.
A natural monopoly is a market situation
where economies of scale are so
significant that costs are only minimized
when the entire output of an industry is
supplied by a single producer so that
supply costs are lower under monopoly
than under perfect competition and
differences are created through
advertising and sales promotion.
C. Difficult market entry:
High barriers
of entry which make it difficult for
new sellers to enter the market.
What is demand?
Demandis the need, want or desire for a
oney to purchase
demand is always
d abitity to pay,,
want or need for
the product.
'l'hr.orir,n irntl l,'ot'utttltts 307
lf the selling price for a product is high,
more producers will be willing to work
harder and risk more capital in order to
reap more profit. However if the selling
price for a product declines, capitalists will
not produce as much because of the
smaller profit they can obtain for their labor
and risk.
Therefore, the relationship between price
and supply is that they are directly
proportional, i.e. the bigger the selling
price, the more the supply; and the smaller
the selling press, the less is the supply.
Figure 2 below illustrates the price-supply
supply and demand at a given Price
Price
relationship.
The demand for a product is inversely
proportional to its selling price, i.e. as the
selling price is increased, there will be less
demand for the product; and as the selling
price is decreased, the demand will
increases. Figure 1 below illustrates the
relationship between price and demand.
Figure 3
When there is additional supply without an
additional demand, a new and lower price
is established as shown in Figure 4.
Previous supply
oligopoly.
Oligopoly exists when there are so few
suppliers of a product or service that the
action of one will inevitably result in a
similar action by the other suppliers. This
type of market situation is characterized by
the following:
A Few seller and many buyers: The
bulk of market supply is in the hands
of a relatively few sellers who sell to
many small buyers.
B
Homogeneous or differentiated
products: The products offered by
the suppliers may be identical or
more commbnly, differentiated from
each other in one or more respects.
These differences may be of a
physical nature, involving functional
features, or may be purely
"imaginary" in the sense that artificial
What is the Law of Supplv and
Demand?
Figure 4
Figure
1
The law of supply and demand maY be
stated as follows:
Under conditions of perfect competition,
price at which any given product will be
supplied and purchased is the price that will
result in the supply and the demand being
"
Assuming a linear function for the
relationship between price and demand, it
shows that at point 1, the selling price, pr
is high, thus there is less demand for the
product as compared to point 2, where the
demand, D2 is great because of lower
selling price, P2.
What is Suoolv?
Supply is the amount of a product made
available for sale.
When there is additional demand without
an additional supply, a new and higher
price is established as shown in Figure 5
the
equal."
The relationship between price-supplydemand may be illustrated by combining
figures 1 and 2. For general application,
the curves for supply and demand are no
longer represented in linear function.
Figure 3 illustrates the price-supplydemand relationship, showing equal
od
308
Question
lJank
l!rrginot'rirrg l,it:orrornir:s by .laime
I.1..
'l'rorrg
'l'hcorics and
Chapter 2
"
d
was invested
Suppose a debtor loans money from a
creditor. The debtor must pay the creditor
the original amount loaned plus an
additional sum called interest. For the
debtor, interest is the payment for the use
of the borrowed capital while for the
creditor, it is the income from invested
capital.
Simple Interest (denoted as l) is defined as
the interest on a loan or principal that is
based only on the original amount of the
loan or principal. This means that the
interest charges grow in linear function
over a period of time. This is usually used
for short{erm loans where the period of
the loan is measured in days rather than
years. lt can be calculated using the
following formula:
l=
Pin
where: P = principal / loan
| = interest
i = interest rate
n = period
The future amount of the principal may be
calculated by adding the interest (l) to the
principal (P).
F=P+l
F=P+Pin
Thus,
Exact simple interest is based on the exact
number of days in a given year. A normal
year has 365 days while a leap year
(which occurs once every 4 years) has 366
days. Unlike the ordinary simple interest
where each month has 30 days, in the
type of simple interest, the number of days
in a month is based on the actual number
of days each month contains in our
Gregorian calendar
To determine the year whether leap year
or not, one has to divide the year by 4. lf it
is exactly divisible by 4, the year is said to
be leap year otherwise it will be
considered just a normal year with 365
days. However, if the year is a century
year (ending with two zeros, e g. '1700,
I 800... ), the year must be divided by 400
instead of 4 to determine the year whether
or not a leap year Hence, year 1600 and
year 2000 are leap years
Under this method of computation of
interest, it must be noted that under
normal year, the month of February has 28
days while during leap years it has 29
days. Again, the values of n to be used in
the preceding formulas are as follows:
n=- d
There are two types of simple interest
namely, ordinary simple interest and exact
simple interest.
ln a bank loan, a person borrows P 100 for
1 year with an interest of 10%. The interest
as computed is P 10. The bank deducts
the interest, which is P10 from P 100 and
gives the borrower only P 90. The
borrower then agreed to repay P100 at the
end of the year. The Pl0 that was
deducted represent the interest paid in
advance. ln this case, discount also
represent the difference between the
present worth (i.e. P90) and the future
worth (i.e. P 100).
Discqrnt
= Fduevorth - peoentworttt
fhe
rate of discount is the discount on one
unit of principal per unit time.
d=1-:for leap years
366
DISCOUNT
Ordinary simple interesl is based on one
banker's year. A banker year is composed
of '12 months of 30 days each which is
equivalent to a total of 360 days in a year.
The value of n that is used in the
preceding formulas may be calculated as
CASE 2:
for normal years
F = P(1+ in)
n=- d
A person has a bond or financial security
that is not due yet but payable in some
future date, desires to exchange it into an
immediate cash. ln the process, he will
accept an amount in cash smaller that the
face value of the bond. The difference
between the amount he receives in cash
(present worth) and the face value of the
bond or financial security (future worth) is
known as discount. The process of
converting a glaim on a future amount of
money in the present is called discounting.
Let d = rate of discount
365
Consider the following cases where
discount is involved:
309
Chapter
360
where: d = number of days the principal
I,brlnulas
3
Compound interest is defined as the
interest of loan or principal which is based
not only on the oiiginal amount of the loan
or principal but the amdunt of loan or
principal plus the previous accumulated
interest. This means that aside from the
principal, the interest now earns interest as
well. Thus, the interest charges grow
exponentially over a period of time.
Compound interest is frequently used in
commercial practice than simple interest,
more especially if it is a longer period
which spans for more than a year.
The future amount of the principal may be
derried by the following tabulatlon:
Period
Principal
lnteresl
Total Amount
P
P
P+Pi
Pll + i'l
P('!+i)(1+i)
=
2
P(l
+
i1
P(1 + i)i
3
P(1 +
i)z
P(1 +i)zi
= P (1 +i)z
P(1 +i)z('l +i)
= P(1 + i):
n
P(1 + i)"
The tabulation above shows that the future
amount (total amount) is just the value p(1
+ i) with an exponent which is numerically
equal to the period.
It is also observed that compound interest
is based on the principles of geometric
progression and using such method, the
total amount after each period are as
follows:
Firstterm, ar = P(1 + i)
Second term, a2 = P(1 + i)2
Third term, a3 = P('l + ;;3
and so on.
Solving for the common ratio, r:
310
Question Bank
-
Engineering Economics by Jaime R. Tiong
a^
f =-L
where:
a,l
I
(1
worth factor
i)
Using the formula for nth term of a G.P.
?n
:3tr"^t
'
The concept of continuous compounding is
based on the assumption that cash
payments occur once per year but
an =P(1 +i)n
There are two types of rates of interest,
namely the nominal rate of interest and the
effective rate of interest.
compound iryterest is
FUTURE AMOUNT, F:
F = P(1+ i)n
Nominal rate of interest is defined as the
basic annual rate of interestwhile effective
rate ofinteresf is defined as the actual or
the exact rate of interest earned on the
principal during a one-year period.
but for m periods per year
r =p(t*!E)'*
I
Cash Flow
F = P(1+
let
m,J
NR
z
,
I
x./
ln this statement, the nominal rate is 5%
while the effective is greater than 5%
because of the compounding which occurs
four times a year. The following formula is
used to determine the effective rate of
interest:
rx(NR)N
.=,[1,.i)'l'.'
r-i, [r * 1)'
x
-->o\
x)
=
"
where: m = number of interest period per
therefore,
year
PRESENTWORTH,P
i = interest per period
.NR
F =PENR)N
F
P=
E
'
(1+i)n
where the payments are made at the
end of each period beginning from the
first period.
Derivation of formula for the sum of
ordinary annuity:
Let A be the periodic or uniform
payment and assuming only four
payments:
AA
A
A
;
:
where: p = priniipal
e = 271828...
NR = nominal rate
N = number of
years
e(NR)N - continuous
,/
compounding /
compound amouht
factor
J8r
D)
dz
A(1+
iF
_+
a3
A(1+
iro -+
a4
A(1+
:
factor
b.
0123n
1. Ordinary annuity is a type of annuity
:
but
+ i)n = single PaYment
compound amount
Annuity ls defined as a series of equal
payments occurring at equal interval of
time. When an annuity has a fixed time
span, it is known as annuity certain.The
following are annuity certain:
For example: A principal is invested at 5%
compounded quarterly.
m
F=Pl1+fl
i)n
where: P = Principal
i = interest per period (in decimal)
n = number of interest periods
(1
x=
eNR(N)
Rate of interest is the cost of borrowing
money. lt also refers to the amount earned
by a unit principal per unit time.
The basic equation for future worth of
4
E
What are the Tvpes of lnterest Rates?
compounding is continuous throughout the
year.
an =P(1 +iX1 +i)n-l
a.
ts:'
'
What is the concept of continuous
comDoundinq?
r=1+i
ChapEer
fhe present worth of continuous
compounding is
present
^ =single payment
+ i)n
P(1 + i)2
P(l+
Theorios and Formulas 31 1
m
NR = nominal rate of interest
Annuity is based on the principles of
compound interest. Hence computation of
the sum of annuity may be done using the
formulas for geometric progression.
Solving for common ratio:
aa1
-
.,/
I =-
A(1+i)
A
r='l +i
Note: i = NR if the mode of compounding is annually
Solving for the sum:
312
Question Bank
-
'l'ltcorics irrttl l,'ormrrlas 313
Engineering Economics by Jaime R. Tiong
PRESENT WORTH OF PERPETUITY:
n[1r*i1"-rl
r
But F= L
l?34
I I I I--TI
12341
I
n-1
AAAA
AA
Substituting the value of F:
A
a. SUM OF ORDINARY ANNUITYT
01
where:
ltr*if-rl
L
i(1
,
AAAA2
i.........................- F
ELf
nltr+if
+
i)n
=uniformseries
present worth
I
factor
Annuity due is a type of annuity where
the payments are made at the
beginning of each period starting from
the first period.
r.
P=+
(n-2)G
where: i = interest per period
A = uniform payment
Chapter
-r"l
5
I
where: i = interest per period
n = number of periods
A = uniform payment
ltr*if-r'] uniform series compound
*
=
i
amount factor
b.
P -'...'.....'.-.
.'......-.......i
:.........,..........
.,.,.....,>, , F
The total present worth P is equal to sum
of Pr and P2, mathematically,
P = P1+P2
3
Deferred annuity is a type of annuity
where the first payment does not b'egin
until some later date in the cash flow
Unifurm arithmetic gradient is a series of
disbursements or receipts that increases
or decreases in each succeeding period by
constant amount.
Also, the total future worth is
F
:FJ +Fz
Analyzing the uniform arithmetic gradient
cash flow only:
PRESENT WORTH OF ORDINARY
ANNUITY:
A.
PRESENTWORTH:
A+(n-2)G
Let
A
AAAA
When an annuity does not have a fixed
time span but continues indefinitely,
then it is referred to as a perpetuity.
The sum of a perpetuity is an infinite
value.
Using compound interest formula:
F
D_
(1
+
i)n
:
The above cash flow can be converted into
its equivalent annuity and uniform
arithmetic gradient whose present worth
are Pr and P2, respectively
P = present worth of the following
uniform arithmetic gradient
Please refer to the cash flow in
the following page
314
Question Bank
-
'l'lroorics
Engineering Economics by Jaime R. Tiong
F
n-l
-l,.,*,,
-]
(n-2)G
2G
where: G = arithmetic gradient change in
the periodic amounts at the
end of each period
i)3
JU
+
31 5
(1+i)"]'
(J
(1+i)2
(1
n_ n
trlas
r=eitr*iI-r_nl
rL ,
U
+
rrrl l'irlrrt
n
0
(1
i)n
F cliI 1 (1iri)
P(1. )-P=G
4 ..'
P(1 r
lr
C.
EQUIVALENT PERIODIC AMOUNT
i)a
Let A = equivalent periodic amount
(n
-
(1
+ i)n-1
234
2)G<.
(n-'l)G('1
+
^G2G3G
(1 + i)' (1 + i)'
-=-f
b
Substitute values in Eq. 3
2G
i)n
..............-T-f
' ',Pi G[t-rt*il'n
I i
'..
(1
+ i)*
(n-2)G (n-l)G
"'-(1
-
+il'r
1
'
(i*D"
-
4)--
(nJ)G
I
(1+i)"
P
]
.\-n
1-(1+i)"
n 1
'-iL
i
(1+i)n]
o_Gl
(1+i)'
I
So v ng for the common rat
The above cash flow is equivalent to the
figure below:
I
t---
o
r
a. (1+il2
\
,
a1 I -(1+i)
1
(1+i)
-,r-rTl-lr
01234...n-ln
where: G = arithmetic gradient change in
the periodic amounts at the
end of each period
AAAAAA
P
B. FUTUREWORTH:
1
Multiply Eq. 1 by (1 + i):
"""""''"""i
-""""""'
f
(1+i)
1
Factor out G:
n I
.,*..-.-........
................-'i
Using formula for annuity:
Let F = future worth of the uniform
arithmetic gradient
0
1 2
3 4 ... n-l
n
1
(1+i)3 _
1
(1
+ i)2
Subtract Eq. 1 from Eq. 2
Solving for the sum, S:
./
t
=
"'('-'/)
ltr
'l
n
._clr-(1+i)n
-(1.,f
' -TI
'
I
316
Question Bank
-
Engineeling Economics by Jaime R. Tiong
A.
Substituting:
'l'lrr.orrcs
PRESENTWORTH, P
ot=X
B. FUTURE
?z= *
0123
33=x3
0
G1t*r1"2
:
-\'r''
G(1+r)z
iclt
-.................i
(1-i)3
a.
a1
i
i
i(l
+
i)n
(1+
i[.]L(1+
ir
_1
ni I
n=9[ri
(1+i)"-t]
L
:
:
I
G(l +
r)n-2
G(l + r)n
I,",
i
iA=c[]^ I
(1+if_jl
,/
.
Li
'1.':l
Uniform geometrio gradient is a series
consisting of end-of-period payments,
where each payment increases or
decreases by a fixed percentage.
(1
r)n-2
+ifr -
G(1t)rt
-rn)
,(t-x")
/
\
-x
Q_
when x
'1
G G(1 +r), G(1 +r)2
D_
' -(1+D- (i+D2 * (1+i)p +"'
"'r, G(t+
*r)3
x
1-r
(1 + i)n
.l:
x2
G(1+r)"2
a,(1
^
i
{..................
Substitute in Eq.
1:
+rf
p=9f tr)[t--l"
1+r\1rilf t-x
,
I
]
,=erL{l
1+i( 1-x
P=G
*1'.
=1*'
1+i
F = P(1+ i)n
o x(1-x"
,- 1+r
(1-x)
G(1 + r)n-1
(1
c(t.+
Sum of geometric progression, S:
:
1
II
l=X
!
(1 + i)n-1
t?
t
Solving for the common ratio, r:
(1 + i)2
lrrtl l"orrurrlits 317
=-9-[L{],,
(1+i)l 1-x
l'
,
,r
x"
r = c1r * iY-'[1- )
1-,
(
J
Whenx=1,r=l
J
F = P(1+ i)n
Multiply by
where:
;#
x=
1+r
1+i
*1.thusr +i
G = first payment at the end of
the first period
(1 + r) = rate of increase or
decrease
Whenr=i,thenx=1.
F= Gn (1 +i)n
(1
F
+
i)'
=Gn('!+0n-l
Chapter
5
Substituting values in Eq. 1:
Gn
o' 1+i
Geometric progression
Bondis a financial security note issued by
businesses or corporation and by the
government as a means of borrowing longterm fund. lt may also be defined as a
long-term note issued by the lender by the
318
Question Bank
-
'l'htrorion and
Engineering Economics by Jaime R. Tiong
Bonds do not represent ownership of a
business or corporation and therefore not
entitled to share of the profits. The
bondholder has no voice in the affair of the
business However the bondholder has
Typor of Depreclatlon:
Let Vn = value of bond n years before
redemption
borrower stipulating the terms of
repayment and other conditions.
a
more stable and secured investment than
does holder of common or preferred stock
of the business or corporation.
Vn=Pl+P2
-+Eq
A.
1
Using formula for present worth of annuity
{w
B.
a
q=
z'[{r*
)"
+
i)n
i(1
I'hysicol Deprecialion is due to the
reduction of the physical ability of
an equipment or asset to produce
results.
Functional Depreciation is due to
the reduction in the demand for the
function that the equipment or
asset was designed to render. This
type of depreciation is often called
Lender
Bonds are issued in certain amounts
known as the /ace value or par value of the
bond. When the face value has been
repaid, normally at maturity, the bond is
said to be redeemed or retired. Bond rate is
the interest rate quoted in the bond
llorrnulns 319
-r]
obsolescence,
Methods of Gomputing Depreciation:
Using the formula for present worth of
compound interest:
A.
ln this method of computing depreciation,
it is assumed that the loss in value is
directly proportional to the age of the
equipment or asset.
E
P^=
' 11+i1n
p.= C
' (1+i)n
'
The following illustrates the normal life
cycle of a bond.
Value of a bond is the present worth of all
the amounts the bondholder will receive
through his possession ofthe bond. The
two payments that the bondholder will
Derivation of the formula for the value of a
bond:
fw
Substituting in Eq.
STRAIGHT LINE METHOD
Annual depreciation charge, d
,-Co-Cn
u--
1
n
,
_
z'[(r*i)" -r]
(1+i)n
c
*(1+i)n
where: Co = first cost
Cn = cost'after "n" years
(salvage/scrap value)
n = life of the property
where: Z = par value ofthe bond
r = rate of interest on the bond
per period
C = redemption price of bond
i = interest rate per period
n = number of years before
redemption
Chapter
Lender
Book value at the end of "m" years of
using, Cr:
Cm
7
Zr
Zr
Zr
Depreciation is the reduction of fall in the
value of an asset or physical property
during the course of its working life and
due to the passage of time.
Co
-Dm
where: D, = total depreciation after "m"
years
D, = d(m)
B. SINKING
Zr
:
FUND METHOD
ln this method of computing depreciation,
it is assumed that a sinking fund is
established in which funds will accumulate
for replacement purposes.
Annual depreciation charge, d
Pz
I
,
320
Question Bank
-
where: Co = first cost
Cn = cost after "n" years
(salvage/scrap value)
n = life of the property
Book value at the end of "m" years of
using, C.:
;9, ='%tD.,
= Co
ZYears
METHOD
!co
or
k=1-im
years
n(n+1)
=-n-
--',_
0.4507(10,000)
5,493.00
= 4,507.00
0:4507(5,493)
= 2,475.69
0.4507(3,017.30)
= 1.359.90
0.4507(1,657.40)
746.99
0.4507(910.41)
410.32
2
1,556
076[(1+o
1)3
-1]
Dg=
3
0.1
Ds = 5,'t 50.61 16
4
=
=
5
3,0'17.30
1,657.40
910.41
500.09
Note; S/rght difference is a result of
rounding off of values.
n
10,000_500
D. Using SYD method
I,Year
Tabulation of book value:
Period
I
Depreciation
Book
value
2
3
zyears
4
5
B
1,556.076[(1+0.05
Ds=
1,900.00
'1,900 00
1.900.00
1,900.00
1.900.00
Using sinking fund method:
zyears
8,100.00
6.200.00
4,300.00
2.400.00
500.00
2
1.556 08
3.267 76
3
5,150.6',l
Book
value
10,000 00
8.443.92
6.732.24
4.849.39
4
7.221.75
9,500.00
500.00
0
1
Assume 10% interest rate
o-(Co-C")i
('l + i)n
-'l
dl
Tabulation of book value:
Depreciation
5
year =
q =(co
Ds = 9'500 00
Period
- 1Dt;+tl
(5X1+
1)
= '' u
-1]
0.'1
'10,000 00
0
d1=(co-c.)---n
and so on. ..
'10,000.00
1
d = 1,900
1
zyears
Book
value
5
Respective depreciation charges:
Third year:
Depreciatibn
Period
0
METHOD
tr:2
d3 = (co - c.)-
Tabulation of book value:
6-'9u-C"
!co
a4
k = 1- 0.5493
k = 0.4507
Using straight line method:
*='t-,8
d2=(co-c.)#
o[(r*i)'-r]
Dz=3,267.7596
First cost, Co
= P 10,000
Salvage value, Cn = P 500
Lifeof property, n = 5 years
D. SUM-OF-YEARS' DlclT (SYD)
Second year:
oz=
ln order to establish the comparison
between the depreciation methods
mentioned above, let us consider the
following data:
The value k is the constant percentage.
Hence k must be decimal and a value less
than 1. ln this method, the salvage or
scrap value must not be zero.
First year:
\lUo
Dz=
A
Matheson Formula:
Uring decllning balance method
o, +'-'+ dr)
DIFFERENT METHODS OF
DEPREGIATION:
ln this method of computing depreciation,
it is assumed that the annual cost of
depreciation is a fixed percentage of the
book value at the beginning ofthe year.
This method is sometimes known as
constant percentage method or lhe
Matheson Formula.
k=1-"8
)
321
6k=1-d+
COMPARISON BETWEEN THE
I
C. DECLINING BALANCE
(a,r +
s
,
o[tr*if
-1]
L
^
-
Sum of year's digit,
where: D. = total depreciation after "m"
years
n_-
C
Book value at the end of "m" years of
using, C,.
C.
l'\lrmulus
'l'hooriott rtnd
Engineering Economics by Jaime R. Tiong
'2.778.29
-"")tk
=(1o,ooo-5oo)*
dr = 3'166'67
d2=(co-.^)#
d2
=(1o,ooo-5oo)+
dz = 2'533'33
z
322
Question llank
d3
-
Dngineeriug Economics by Jaime R. ,I.iong
=(co_c")+
depreciation charge while the sinkingJunr)
method has the smallest annual
depreciation charges.
!year
d3
'l'huorior and l'lrrntulrrn 323
=(1o,ooo-5oo)*
ds = 1,900.00
What is Deoletion Cost?
d4=(Co-c")a
Depletion cosl is the reduction of th€ value
of certain natural resources such as
mines, oil, timber, quarries, etc. due to the
gradual extraction of its contents
Lyeat
d4
=(1o,oo\-5oo)#
Methods of Computing Depletion
Charge for a year:
d4 = 1,266.61
-c"[=ii Lyeat
l,
0,000 L soo)
*
ds =
(co
ds =
(1
A.
Tabulation of book value:
Period
Depreciation
2
3.166.67
2,533 33
3
'1,900 00
4
1.266.67
633.33
1
5
6.833.33
4,300 00
2,400.00
1.133.33
500.00
Book Values obtained by Various
Depreciation Method
10,000
U.S. = units sold
The depletion cost during any year is:
lC'
T_U
ts I
\v.e.,
rt
Method
The depletion charge (or simply depletion)
under this method, is computed as follows:
Depletibn = Fixed % of gross income
SYD Method
6,000
or
Declining
4,000
Balance
Method
Depletion = 90 :%iof 1et ta;able incci4e
2,000
Note: Use the smaller depletion charge
500
Fixed Percentages Allowed for Ce(ain
Natural Resources:
1
2
3
4
The diagram shows that ltrc declining
balance method has the largest annual
5Yrs
llrcttk-even analysis is used to determine
the break-even cost, which is the cost at
which the total income is exactly equal
to the total expenses incurred in the
business.
1o
5
The diagram below is the most
convenient chart used in break-even
analysis. This is known as the breakeven chart. lt represents the graphs of
fixed costs, variable costs, the expected
income, etc.
or
ure
and the
cement,
long time
Ke!
CQ=Frntcrd+Cdd
structure or
the annual
rest ofthe first cost
A
p
d
a
Y
Ltr
ing and maintenance
costs, or
Straight Line Method
8,000
15
or forever, or
B. Percentage or Depletion Allowance
Sinking Fund Method
22
Chapter I
l.C. = initial cost of property
T.U. = total units in property
Depletion __-.
cost _
=
22
* Based on Gross lncome
Unit or Factor Method
Let:
'10.000 00
0
Sulfur, Cobalt, Lead,
Nickel, Zinc. etc
Oil and Gas wells
Gold, Silver, CoPPer,
lron ore
Sodium Chloride
-Coal.
Gravel, Sand, CleY-
following formula:
Book
value
AC
9
Percentage'
This method is dependent on the initial
cost ofthe prope(y and the number of
units in the property. The depletion charge
during any year is calculated using the
ds = 633.33
Chapter
Maximttttt
Natural Resource
Production
p = profit
L = loss
324
Question Bank
-
(lloannry 325
Engineering Economics by Jaime R. Tiong
amalgamation (syn merger) lhe creation of a
new company by a combination of business
units
asset a property or item owned by an indlvldual
or organization which has a monetary valuo
auction the selling of goods or producls and
amortization as applied to capitalized cost, the
distributi
charges
the redu
inegular
A
I cost by
periodic
in depreciation or
y either periodic or
gram
annual general meeting (AGM) the required by
law yearly meeting of shareholders of
corporation or organization
accounls the financial slatements
of
annual ratb of return is the ratio of the annual
an
individual or organization prepared from a
syslem of recorded financial transactions
acid-test hatio (syn. current ratio) the accounting
measure of individuals, or company's ability
to pay its short{erm liabilities
actuary the chief executive of an assurance
company whose job is to calculate
premtums
administered price the price for a product whish
is set by an individual producer or group of
producers
salaries, lease of space and offices
occupied, electricity and water utilities,
mailing costs and equipment costs
agent an individual or organization who acts iir
of a
net profit to the initial capital investment
leg
compan
trading
unOerlyi
ndividual or
audit the
sheet and
ccount and
and records
auditor an accountant duly licensed by PRC
tasked to check the accuracy of a joint
stock company's ledger account and annual
report and accounts, and to present an
independent report to shareh0lders on
annuity a series ofpqual payments occurring at
wheiher the accounts present
equal interval of time. Types of annuity are:
fair vies of the comPanY's affair
Ordlnary annuity, Annuity due, Deiened
annuity and Perpetuity
annuity due
a
type
of
annuity where the
payment is made al the beginning of each
period starting from the first period
anticompetitve prac{ice the practice which
effect the restricting, distorting and
auth
client and buys
products/properties
in
or
return
a corporation can
by the SecuritY and
Exchange Commission (SEC)
automatic saving the unspent income of a
person if his net income is too large to
spent
products and capital goods
average cost price the selling price for a good
oi service which is equated to the average
COSt
product between two or more markets in
order to take profitable advantage of any
differences in the prices quoted in these
aggregate supply the total amount of domestic
goods and services supplied by
.
both
increase in the value of a currency against
other curencies under a floating exchinge_
rate system
a
for
consumei
in
which human
operations are substituted by mechanical or
electronic oPeration or both
average cost is the ratio of the total cost to the
total number of units
arbitrage the buying or selling of goods or
markets
articles of incorporation the legal constitution
of a corporation which is submitted to the
Security and Exchange Commission (SEC)
receives
deposits of money from the public and fulfils
its obligation to return that money to the
depositor when withdrawn
bank deposit an amount of money deposited
bank loan credit made available to individuals or
corporations bY commercial bank
bank note the paper currency issued by central
bank which forms part of the country's
moneY suPPlY
bank rate the rate of interest charged by central
bank
to its bonowers, mainlY the
commercial banks
banke/s yeat ayeat. which is equivalent to-12
mont-hs of 30 days each and is equal to 360
daYs
bankruptcy (syn. insolvency) the
average revenue is the ratio of the revenue to
the unit of commoditY sold
B
BIR Bureau of lnternal Revenues
bad debt money owed which is unlikely to be
paid due to the customer being insolvent
condition
individual or company is
insolvent, i.e. unable to meet its debts on
wnere
appreciation (ant, depreciation) the increase in
the value of an assel. lt also refers to lhe
sells
commission
businesses, including
analysis to real world economic situations
a true and
maximum amount
eliminating competition in a market
applied economics the application of economic
bank a commercial institution which
with the bank
examined bY a qualified auditor
automation the process
administration costs the cost of maintaining a
firm within which goods and services are
produced. Administration costs includes
behalf
i
services bY comPetitive bidding
statemenl of a
flrm's assel and llabilities on the last day of
a iradlng Period
brlrncr rhcct an accounting
an
due dates.
barGr exchange of goods or services without the
use of moneY
bearer bond a financial security which are not
registered under the name of a particular
holder but where possession serves as
proof of ownershiP
bilaterat monopoly
a
market situation where
there is only one seller with only one buyer
bilatenl oligopoly a market situation
where
there are few sellers and with few buyers
billion refers to one million million (1012) in the
United Kingdom and Germany - and one
thousand mlttion (10g) in the United States