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
124
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=yx
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(yx) 0.07x:0.58y0.58x
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 subassembly 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 * 109 100
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 manday 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 secondhand 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=1600010400 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=5000032400 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.kwhr for the first 50 kwhr, P 2.90 per kwhr for the second 50 kwhr, P 2.80 for the third kwhr and so on until a minimum of P 2.50 per kwhr 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 8meter wide concrete road pavement s00 meters long is desired to be constructed over a wellcompacted gravel road, together wiih 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 10inch 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 kwhr per month of electric energy from
National Power Cbrporation at P 2.OO per kwhr and sells all this to consumers per kwhr should afteiJeOucting distribution loss es of 2OYo. At what average rate expenses in monthly other are besold 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 kwhr 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 breakeven; 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 orchoosino sites on w.rtrr'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 minihydroeiectric 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 Hydroelectric 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 Tbills earning 12% interest 35% in money market placement earning 35% in blue chip stock earning 13%
14o/o
lf her annual earnings from the tbills, 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 Tbills, 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=2300000y 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
nrm0", 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=60005370 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 =108000012000 = 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
1d 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 noninterest 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 Octoberl12
= 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
June2130
=$
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 125 =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
Decemberl25=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 lfthe 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
Marchl31=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+ift
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
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 semiannually, 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.
$ermine the exact simpre interest on p 5,000 fnuary 15, 1e96 to october rz,1sigo, ;tiJI]t;invested P 664.89
",B. c. P 665.21 D. P 666.3e
P 1,302.67
F=P(1+in) l= 1,250 =p[r.
1
o.oa[i9)l (s60i.l
Pin
Solving for the total number of days money was invested, d
P = 1,233.55
January
.'. The cash price of the television set is p 1,233.55.
A.
B.
C. D.
.V;z*
15.47 o/o 15.69 % 15.80 o/o 15.96 %
to repay
1,500.00
"i
1S31
February March April
man.bollory"d Iamount rgney from a roan shark. He receives from the roan shark and of p 1,u2.00 and^promised p
What is the simple interest rate?
+ Eq. 1
tn" Lna
otl
quarters.
May June July August September October 1  12
= 16 =
29 (since 19g6 is a leap
=31
=30 = 31 = 30 = 3.1 = 31
=30 =12 271 days
Substitute values in Eq. r
=
s,ooo(o.18,{!1\ ' '\366'
t= 666.39
1
year)
38
(1)ucstrorr
llrrrrlr
Simplc lnt,ereel, & Discoupt 39
lr)rrgrrrrrr.rirrg l,lt:ortorrricx by ,lnime R. Tiong
The exact clmplo lnterest ls P 666.39.
'Aaa
F = P(1+
The exact simple interest of P 5,000 invested from June 21, 1995 to.December 25, 1995 is P 100. What is the rate of interest?
A. B. c. D.
)
April223O
3.95 % 3.98 %
+
Eq.
1
=$ = 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)
Decemberl25=25
1
247 days Solving for the total number of days, d
June2130
JulY August September October November December
125
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. lfthe 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
Marchl31=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",
semiannually, "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? ffilO,, 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 semiannually 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.9119)'r ER = 0.1225
Flnd the nominal rate, which if converted quarterly could be used instead oI 12oh compounded semiannually.
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=FP 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
lJ2z /../t//rn
18% compounded semiquarterly? Wnrt 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
ifrr,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 semiannually?
,'. 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
SoZ
r !tF'
50
Quesl,iurr
llntrk
I,)rrginooring l,lt:onuurit:x by Jairne Il,. Ti
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 semiannually?
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
''
Pn"""""'
[,nl)o t=(t+o06)21 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 semiannually 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,irrn
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=FP
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 semiannually. 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=FP
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
Lb
=
(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 semiannually. What was the interest
o/o
o/o
tala peid in this account?
A, B, C, D,
9"2*r;.
a.
= 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

.liuginocring
li)<:()norr(.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 semiannually. 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)rt 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+ urvvv4vrvvv[\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 semiannually
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'npl."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=FP 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 semiannually?
Oourllounrl
,'__
lrrlrrorl tn
A mrn expects to receive P20,000 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 semiannually 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
Itltllannually. 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
'3A+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=FP 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 27,78236.
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
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
,*,[emailprotected] 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 fiveyear 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.
::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 yearbyyear 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 {unlflllrlndneering
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
""""""""'
,,,1
000
t;t
rl t 
l']rlncipalat end of 2001
 .lr I Fr+FsF+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 laborsaving 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=AB 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)
60'000 (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 yeai 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+BC D=1,200+500600
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*ifr] 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 fiveyear 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 semiannuaily is to be amortized by semiannuar payment ore," tirn"rt 5 years. The first due in 6 months. Determine the semiannual 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)lt 1+
.'. The semiannual payment is p 1,295.05
a lending institution which willbe paid after 10 lZV3 compounded annually lf money is worth g% per .o! ld he deposit to a bank monthly in ordei to discharge his
OO0.t1o.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)10rl
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) 1o1] (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 deposited 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 year 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
rl1 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.
ur]
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+)'or] ,[emailprotected]
_
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 ofland. 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) '?40r]
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)
3r]
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 semiannual 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)11 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 semiannual amount that will be set aside is p 7,425.72.
Pt =16,528'92
40,000
^'
(1+ 0.10)"
Pz
=27'32054
^ '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 III 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)'z4r]
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 16. P4,0OO at the end of your 1 7. , "i year't in the account at the end ofthe 21"'
41]
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)121 = 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
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] L10
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=Fr2,000
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 10year 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)nrl
..
___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 after three " furthLr payment established now. For tax purposes' 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)ur] 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'zor] 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 2year 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*oorsyourl 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' rflr
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)20l"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 semiannual payments over the next three years, the first due in 6 months. How much is the semiannual 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 semiannual 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
'
 tll
11+
\
o'20
)4 4) 
1
= 0.1 025
Solving for semiannual 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 = FzFt 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 semiannual 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 payoff 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=AB x=700,00020,000
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 endoftheyear 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 29inch flatscreened 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 endofyear 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 endofyear 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
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 semiannual 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
l1i=
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=F1,000,000
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 PLJ
'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=350,000135,000 A = 215,000
.'. Buying in cash is economical by P 45.95.
Bon Fing Y Hermanos, lnc. has offered for sale its twostorey 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
=40,0007,000
33,000
Labor = 70,000  10,000== 60,000 Overhead = 10,0008,000 + 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=150,000130,000 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 semiannual 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 semiannually, 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 j35
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
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(1if] 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
^ "
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 =rli"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

,t.
ll,,
I
'.
I.500
I 1.1
1l
10.05
(1+ 0.05)"
 1.1
Ar = 719'53 Let A = equivalent uniform payment of the. amortization:
A:5,000A1 A=5,000719.53 A = 4,280.47
s,000 5,000 s,000
s,000 :
P
+
=PtPz
Eq.
1
P,r=
,
4280.474280.474,280.47 4,28.0.47 :
7
P;:,<"""""'
=17,729.75
n. i fl.'r]
o_c[t1t*i;" ''TL
D_soo[t1t*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
.9224,. ,%:;*,.". 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
=1Fz
+ Eq. 1
Substituting values to Eq.
1:
F =21,550.623,101.25 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
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=5,00041 A=5,000719.53
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 lA Lakers B. Chicago Bulls offer is exactly the same as lA Lakers, chicago Bulls offer is just few dollars more p6r year than that of lA 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 f1x") o_ '
1+i[ 1x
 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 1or818 ]
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'thegi'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 (tr"') D_ '  t+rIrr.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 1oss1s 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 (tr") H=t 1+i[ 1x
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{ lifeat ase +0. neVisnis to have upon his retirement the sum of P 5,000,000. On his 21"tbirthday, 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 10s64, ]
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 newlyagq.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_ '  1l[tr") ir.,J 1o,ooo
(t
t.o+s+a,o') ' _ 1*ols[1l.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.
2
4
6
8
1011
12
A P 1,000,000 issue of 3%, 1Syear 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
,52034.
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 [emailprotected]!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 i7.56%
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=1FE 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
. CoCn 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.
8.000500 10
d=750 Solving for book value after 8 years, C6
C. =CoDm c8 =8,000750(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 =
rr)
"[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)43
6=
ds =857'14
CoCn n
.50.00010.000 .'. 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=CoDs cs =50,0004,000(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 sumofyears' 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,:(coc")=+.years z
d1
226
Question Bank

Engineering Economics by Jaime R. Tiong
Deprocintion and Depletion 227
But
)
vears
r
vears
n(n +

Cr=CoD, cr =530,00096,000(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
=(9,0001,000)#
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 straightline 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
_ _ 530,00050,000
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 straightline 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=CoDs cs =50,0004,000(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
[emailprotected]
.'. The annual rate of depreciation is 32.25
o/o
o/o.
V;1itr.. Let d = annual depreciation
dcocn 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 sumofyears' 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 = (120,00012,000)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
=
[emailprotected]'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=CoDs 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
,..CoCn
The annual depreciation charge is P 76'500'00'
n
r_
530,000x(530,000) 10
d:53,00053000x 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 aightline depreciation' O" of six end machine at the Years?
=(53,00053,000x)5
Book value at the end of 5 years, Cs
D.
Cs=CoDs 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 straightline 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 =800,00076,500(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=CoCn 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 straightline 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=CoD16 cro = 41o,0oo  10(17,500) Cro'= 235' o0o
6=CoCn
...Theappraisalvalueofthemillattheendofl0yearsisP235,000.00.
n
,_45,0002,500 .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 bookvalue 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 sumofyears' 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 depreciationcharge?
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,0001,ooo)*
= 1,454'54
"")thk
=
(e,ooo1,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_(CoCn)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,00050,000  Cn =274,OOO
Since, there is a given interest rate, use sinking fund method in solving for annual depreciation charge. i
'[tr*irr] ^_(coc")
\ ' .{ 274.000(0.06) [t',.
o.oo)'  r]
d62,634.07 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)ur]
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 3ffordto 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."
, (coc")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,

107'393'60xyegl
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 =
jgqqq!= + 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
i2.2071
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,  50'900)(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
(640'000 0X0' 16)

(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
_ 300'00030'000
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)(912)
 +(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 kwhr. 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.twnr/
day /( year /(kwhr/)
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)(0'10)
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)fiT
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
fGff+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 breakeven
ftt*i)" t1
= =
(190
000
Material cost = 100x
0)(0'05)
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
BreakevenAnalysis 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 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?
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 vlll
Let: x = number of motors to be sold each month to breakeven
+ 20,000
)/ Sdavs\(Thours) lll
\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 breakeven
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 breakeven, 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
BreakevenAnalysis 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 breakeven.
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 breakeven,
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 multiplepunch 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) 10=
= 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=250,00080,000 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 krrvt'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'nkcvon Analvsis 279
'l'iorrg
'/nl,t;,, The following data for year 2000 are drqllable for Cagayan Automotive Company which manufactures and sells a single arltomotive 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. Tbills 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 lnplace 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 turnover profit margin ratio current ratio acidtest 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 Semimonopoly 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 coowners for profit.
Workin 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 [emailprotected]
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. Tbill C. Preferred stock D. Common stock 87.
Queslion
A form of fixedinterest security issued by central or local governments, companies, banks or other institutions. They are usually a form of longterm 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
Tbills 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 "shortterm,, 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 sumofyears' 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 semivariable cost
c. D.
128. Question What type Qf cost increases stepwise?
A. SuPervision cost B. Direct labor cost C. Semivaiiable 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. priceearning 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.
Tbills 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 socalled "Banker's Rule",
157. Queslion A government bond which have an indefinite life rather than a specific maturity.
A. B. C. D
Coupon
Tbi[ 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 (n1) 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 andinterest 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
Andinterest 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.
priceearning 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 measuringrod 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 wellknown 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
incomeelasticity 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 desired 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 incomeelasticity 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 welldefined 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 marketentry 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 yacting 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 pricesupply
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 pricesupplydemand 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 pricesupplydemand 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 oneyear 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'.' ri, [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)nl
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 ITI
12341
I
n1
AAAA
AA
Substituting the value of F:
A a. SUM OF ORDINARY ANNUITYT
01
where:
ltr*ifrl 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=+ (n2)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*ifr'] 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+(n2)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
nl
l,.,*,,
]
(n2)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*iIr_nl rL ,
U
+
rrrl l'irlrrt
n
(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)n1
234
2)G<.
(n'l)G('1
+
^G2G3G (1 + i)' (1 + i)' =f
b
Substitute values in Eq. 3
2G
i)n
..............Tf
' ',Pi G[trt*il'n I i
'..
(1
+ i)*
(n2)G (nl)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)
,rrTllr
01234...nln 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
1 2
3 4 ... nl
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
G1t*r1"2 : \'r'' G(1+r)z iclt .................i (1i)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)n2
G(l + r)n
I,", i
iA=c[]^ I (1+if_jl
,/
.
Li
'1.':l
Uniform geometrio gradient is a series consisting of endofperiod payments, where each payment increases or decreases by a fixed percentage.
(1
r)n2
+ifr 
G(1t)rt
rn)
,(tx") / \ 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
1r
(1 + i)n
.l:
x2
G(1+r)"2
a,(1
^ i
{..................
Substitute in Eq.
1:
+rf
p=9f tr)[tl"
1+r\1rilf tx
, I
]
,=erL{l 1+i( 1x
P=G
*1'. =1*' 1+i F = P(1+ i)n
o x(1x" , 1+r (1x)
G(1 + r)n1 (1
c(t.+
Sum of geometric progression, S:
: 1
II
l=X
!
(1 + i)n1
t?
t
Solving for the common ratio, r:
(1 + i)2
lrrtl l"orrurrlits 317
=9[L{],, (1+i)l 1x 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('!+0nl
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 longterm 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
,CoCn 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=1im
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(CoC")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
d1=(coc.)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:
6'9uC"
!co
a4
k = 1 0.5493 k = 0.4507
Using straight line method:
*='t,8
d2=(coc.)#
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. SUMOFYEARS' 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=1d+
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,ooo5oo)*
dr = 3'166'67
d2=(co.^)# d2
=(1o,ooo5oo)+
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,ooo5oo)*
ds = 1,900.00
What is Deoletion Cost?
d4=(Coc")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
llrcttkeven analysis is used to determine the breakeven 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 breakeven 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
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
acidtest 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 to12 months 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
FAQs
What are the four fundamentals of engineering economics? ›
Principle 1: A dollar earned today is worth more than a dollar earned in the future. Principle 2: The only thing that matters is the difference between alternatives. Principle 3: Marginal revenue must exceed marginal cost. Principle 4: Additional risk is not taken without the expected additional return.
What is engineering economics briefly explain? ›Engineering economics is a field that addresses the dynamic environment of economic calculations and principles through the prism of engineering. It is a fundamental skill that all successful engineering firms employ in order to retain competitive advantage and market share.
Is economics for engineers hard? ›Its difficulty depends on grasping the basics. If you got the basics down, everything will be chill but if your basics are not solid, you will end up getting confused on whats going on and why its going on !! My advice is to use this book Engineering Economy by Sullivan.
What are the 4 three basic questions of economics? ›Students will read and take notes on the three main questions of economics. These are what to produce, how to produce it, and who to produce it for.
What are six fundamentals of engineering? › Civil Engineering. ...
 Mechanical Engineering. ...
 Chemical Engineering. ...
 Electrical Engineering. ...
 Aerospace and Aeronautical Engineering. ...
 Computer Engineering.
The change in the amount of money over a given time period is called the time value of money; it is the most important concept in engineering economy.
What is the importance of engineering economics? ›One of the major importance of engineering economics is the proper and efficient use of limited and scarce resources. Engineering economics requires the application of technical and economic analysis with the goal of deciding best meets technical performance criteria and uses scarce capital in a prudent manner.
What are the objectives of engineering economics? ›It can analyze personal finances and investments in a fashion similar to corporate project finances. Engineering economy is a collection of techniques that simplify comparisons of alternatives on an economic basis. In defining what engineering economy is, it might also be helpful to define what it is not.
What is an example of engineering economics worked? ›Examples include the choice between a concrete and a steel structure, between various insulation thicknesses,between possible loans for a car or a robot, and between prices at which to sell a duplex, afirm, or a product. Engineering economy can be applied by an engineer to size a pump or to buy a home.
Is economics one of the hardest degrees? ›So why do people consider economics a difficult major? In this area, you will take a lot of math and statistics classes and do plenty of critical thinking and synthesizing of data – so it can definitely be hard!
What is cash flow in engineering economics? ›
Engineering Economics. Cash Flow. Cash flow is the sum of money recorded as receipts or disbursements in a project's financial records. A cash flow diagram presents the flow of cash as arrows on a time line scaled to the magnitude of the cash flow, where expenses are down arrows and receipts are up arrows.
Who is father of engineering economics? › Time Value of Money.
 Interest Rates and Project Returns.
 Inflation and Foreign Exchange.
 Depreciation and Tax.
 Replacement Analysis.
 Risk and Uncertainty.
 Capital Budgeting.
LCM: Compare the PW of alternatives over a period of time equal to the least common multiple (LCM) of their estimated lives. Study period: Compare the PW of alternatives using a specified study period of n years. This approach does not necessarily consider the useful life of an alternative.
What are the five 5 basic economics questions? › What to produce and what quantity to produce?
 How to produce?
 For whom to produce the goods?
 How efficient are the resources being utilised?
 Is the economy growing?
Economic systems answer three basic questions: what will be produced, how will it be produced, and how will the output society produces be distributed? There are two extremes of how these questions get answered.
What are the 3 major economic questions? ›Economists address these three questions: (1) What goods and services should be produced to meet consumer needs? (2) How should they be produced, and who should produce them? (3) Who should receive goods and services? The answers to these questions depend on a country's economic system.
What's the most paid engineer? › Systems Engineer. ...
 Electrical Engineer. ...
 Chemical Engineer. ...
 Big Data Engineer. ...
 Nuclear Engineer. ...
 Aerospace Engineer. ...
 Computer Hardware Engineer. ...
 Petroleum Engineer.
So in the spirit of organizing how to think about Engineering leadership I am in this article putting forward the five "essential elements" of an engineering leader: People, execution, strategy, technology and self awareness.
What are the 10 most important concepts of economics? ›
Some of the concepts are scarcity, supply & demand, incentives, tradeoff and opportunity cost, economic systems, factors of production, production possibilities, marginal analysis, circular flow, and international trade.
What is the single most important concept in economics? ›Supply and Demand
The relationship between supply and demand sits at the heart of most economic theory, for a simple reason: They are inextricably linked. The law of supply and demand can be explained as follows: When supply of a good or service exceeds its demand, prices will fall until an equilibrium is reached.
The engineering economics is concerned the systematic evaluation of the benefits and costs of projects involving engineering design and analysis. Engineering economics quantifies the benefits and costs associating with engineering projects to determine if they save enough money to warrant their capital investments.
What are the 4 economic objectives? ›Students study the principal economic objectives of stable prices, economic growth, full employment and the Balance of Payments.
What are five economic objectives? ›Government has adopted a macroeconomic policy framework which will deliver job creation, better export performance, more investment, greater efficiency and equity of government spending and enhanced human resource development.
How do you calculate present worth in engineering economics? ›In Engineering Economic Analysis, where: Present Worth = A(P/A, i%, n) P = present value A =...
What is the difference between economic and engineering economics? ›What is the difference between economics and engineering economy? Economics is the social science that describes the factors that determine the production, distribution and consumption of goods and services. Engineering economy is a subset of economics for application to engineering projects.
How does engineering affects the economy? ›Engineering plays a key role in supporting the growth and development of a country's economy as well as in improving the quality of life for citizens. As such, there is an important link between a country's engineering capacity and its economic development.
What's the hardest bachelor's degree? ›A number of engineering majors are known to be extremely difficult including mechanical engineering, petroleum engineering, bioengineering, biomedical engineering, aerospace engineering, and chemical engineering (but not limited to these, either!).
What are the three 3 major types of cash flow? ›There are three cash flow types that companies should track and analyze to determine the liquidity and solvency of the business: cash flow from operating activities, cash flow from investing activities and cash flow from financing activities. All three are included on a company's cash flow statement.
What are the 4 cash flows? ›
He introduces the Cashflow Quadrant, which, as the name indicates, has four sections: Employees, Selfemployees, Business Owners, and Investors. Each quadrant has advantages and disadvantages, and as the author explained, they're not created equally.
What is the 5th principle of engineering economy? ›PRINCIPLE 5  CONSIDER ALL RELEVANT CRITERIA: Selection of preferred alternative (decision making) requires the use of a criterion (or several criteria).
Which branch is daddy of engineering? ›Mechanical branch is father of engineering so we teach father of engineering..
Who are the three fathers of economics? ›The Big Three in Economics: Adam Smith, Karl Marx, and John Maynard Keynes: Adam Smith, Karl Marx, and John Maynard Keynes.
What are the five main types of engineering economic decisions? › The term engineering economic decision refers to all investment decisions relating to engineering projects.
 The five main types of engineering economic decisions are (1) service improvement, (2) equipment and process selection, (3) equipment replacement, (4) new product and product expansion, and (5) cost reduction.
Demand is a principle of economics that captures the consumer's desire to buy the product or service. The demand is calculated as the price the consumers are willing to pay for the product or service.
What is future worth in engineering economics? ›Conversely, the future value of today's equipment is given by the formula: F = P(1 + i)^{n} where F is the future worth of the equipment, P is its present value, i is the average rate of inflation or the interest rate, and n is the number of years.
What are the three types of engineering economy problems? ›Simple problems. Intermediate problems. Complex problems. Where is Engineering Economic Analysis most suitable?
What are the three 3 measures of worth used in the analysis of engineering economy? ›There are several different measures of worth used to select an alternative for a specific set of estimates, some of these are: Present worth (PW), Future worth (FW) and Rate or return (ROR).
Why is Marr important in engineering economy? ›The Minimum Attractive Rate of Return (MARR) is a reasonable rate of return established for the evaluation and selection of alternatives. A project is not economically viable unless it is expected to return at least the MARR.
Should an engineer study economics? ›
Engineers must understand the economic viability of their projects. Fundamentally, engineering economics involves formulating, estimating, and evaluating the economic outcomes when alternatives to accomplish a defined purpose are available.
Does economics go well with engineering? ›1. Engineers have a natural aptitude for economics. “Engineers are ideal economics students because of their ability to think logically and their strong mathematical skills,” says Carl.
Is economics a good minor for engineering? ›Adding an Economics Minor to your Engineering degree will allow you to better understand the business side of Engineering. This can help you to pursue management or logistics roles in your chosen field.
Is it hard to pass economics? ›A collegelevel economics class can be challenging because you need to grasp new concepts like supply and demand, scarcity, diminishing returns, and opportunity costs. It requires you learn new vocabulary and to use critical thinking skills.
Who is the father of engineering economics? ›Economics is the social science that study market and people interactions whereas Engineering economics is the subset of economics that involves economic decision making for engineering system.
Do engineers or economists make more money? ›Up until about the 57^{th} percentile, engineers make the most. But then the earnings curve for economics grads basically goes parabolic. At the 95^{th} percentile, they can expect to earn more than $3 million more during their lifetime than an engineering grad.
What all topics are in engineering economics? ›Some other topics that may be addressed in engineering economics are inflation, uncertainty, replacements, depreciation, resource depletion, taxes, tax credits, accounting, cost estimations, or capital financing. All these topics are primary skills and knowledge areas in the field of cost engineering.
Who hires economists? ›What kind of companies hire economists? Economists are hired by research institutes, universities, healthcare, and government institutions. Private companies like finance consultancies and banks also need economists. In addition, they can be hired by an Internet company like Netflix or Microsoft.
What minor goes best with economics? › Data science. Data science minors can help students develop the skills to collect and analyze data using a variety of tools. ...
 Mathematics. ...
 Political science. ...
 Behavioral science. ...
 Statistics. ...
 Internet technology. ...
 Analytics. ...
 Psychology.
What is the best minor with economics? ›
Suggested Minors for students who are majoring in Economics might include Mathematics, Statistics, Politics, Environmental Studies, Business, IDS, or Psychology to name a few. A Minor in Economics is also available for students majoring in other Departments or Programs.
Is economics a good major to get a job? ›A degree in economics can provide a variety of job opportunities in nearly every industry, from local government and the sciences to finance and insurance. While some of these jobs might require additional certifications or education, others are available to those with a bachelor's degree.
Can I major in economics if I bad at math? ›According to the United States Bureau of Labor Statistics (BLS), four important qualities economists should have are analytical skills, criticalthinking skills, speaking skills and writing skills. You cultivate strong skills in all of these areas even if you are “bad” at math.
Does economics have a lot of math? ›There are many diagrams in economics, but there is not a large amount of math. A proviso: The amount of math in the economics curriculum varies across colleges and universities. Some economics departments do not require their students to learn much math or statistics, but others do.
Is economics a STEM major? ›Nonetheless, recognizing economics as a STEM major welcomes a mathbased option for learners with broader interests. They can acquire the skills and lessons that STEM content promotes but implement them in different fields such as entrepreneurship and finance.